PHYSICS FOR SCIENTISTS AND ENGINEERS WITH MODERN PHYSICS (VOLUME 3) 5TH EDITION GLOBAL EDITION BY DOUGLAS C. GIANCOLI CHAPTER 36_44 SOLUTIONS MANUAL CHAPTER 36: The Special Theory of Relativity Responses to Questions 1.
No. Since the windowless car in an exceptionally smooth train moving at a constant velocity is an inertial reference frame and the basic laws of physics are the same in all inertial reference frames, there is no way for you to tell if you are moving or not. The first postulate of the special theory of relativity can be phrased as “no experiment can tell you if an inertial reference frame is at rest or moving uniformly at constant velocity.”
2.
The fact that you instinctively think you are moving is consistent with the relativity principle applied to mechanics. Even though you are at rest relative to the ground, when the car next to you creeps forward, you are moving backward relative to that car.
3.
Since the railroad car is traveling with a constant velocity, the ball will land back in his hand. Both the ball and the car are already moving forward (relative to the ground), so when the ball is thrown straight up into the air with respect to the car, it will continue to move forward at the same rate as the car and fall back down to land in his hand.
4.
Whether you say the Earth goes around the Sun or the Sun goes around the Earth depends on your reference frame. It is valid to say either one, depending on which frame you choose. The laws of physics, though, won’t be the same in each of these reference frames, since the Earth is accelerating as it goes around the Sun. The Sun is nearly an inertial reference frame, but the Earth is not.
5.
The starlight would pass at c, regardless of your spaceship’s speed. This is consistent with the second postulate of relativity, which states that the speed of light through empty space is independent of the speed of the source or the observer.
6.
The clocks are not at fault and they are functioning properly. Time itself is actually measured to pass more slowly in moving reference frames when compared to a rest frame. Any measurement of time (heartbeats or decay rates, for instance) would be measured as slower than normal when viewed by an observer outside the moving reference frame.
7.
Time actually passes more slowly in the moving reference frame, including aging and other life processes. It is not just that it seems this way–time has actually been measured to pass more slowly in the moving reference frame, as predicted by special relativity.
8.
This situation is an example of the “twin paradox” applied to parent–child instead of to twins. This situation would be possible if the woman was traveling at high enough speeds during her trip. Time would have passed more slowly for her and she would have aged less than her son, who stayed on Earth. (Note that the situations of the woman and son are not symmetric; she must undergo
acceleration during her journey.) 9.
You would not notice a change in your own heartbeat, mass, height, or waistline. No matter how fast you are moving relative to Earth, you are at rest in your own reference frame. Thus, you would not notice any changes in your own characteristics. To observers on Earth, you are moving away at 0.6c, which gives = 1.25. If we assume that you are standing up, so that your body is perpendicular to the direction of motion, then to the observers on Earth, it would appear that your heartbeat has slowed by a factor of 1/1.25 = 0.80 and that your waistline has decreased by a factor of 0.80 (due to time dilation and length contraction). Your height would be unchanged (since there is no relative motion between you and Earth in that direction). Also note the comments in Section 36–9 of the text on “Rest Mass
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and Relativistic Mass” for comments about mass change and relativity. Your actual mass has not changed. 10. Yes, they do occur. However, at a speed of only 90 km/hr, v c is extremely small, and therefore γ is very close to one, so the effects would not be noticeable. 11. Length contraction and time dilation would not occur. If the speed of light were infinite, v c would be 0 for all finite values of v, and therefore γ would always be 1, resulting in t t0 and l l 0 . 12. Both the length contraction and time dilation formulas include the term 1 v 2 c2 . If c were not the limiting speed in the universe, then it would be possible to have a situation with v c. However, this would result in a negative number under the square root, which gives an imaginary number as a result, indicating that c must be the limiting speed. Also, assuming the relativistic formulas were still correct, as v gets very close to c, an outside observer should be able to show that l l 0 1 v2 c2 is getting smaller and smaller and that the limit as v c is l 0. This would show that c is a limiting speed, since nothing can get smaller than having a length of 0. A similar to analysis for time dilation should show that t is getting longer and longer and that the 2 2 1 v c limit as v c is t . This would show that c is a limiting speed, since the slowest that time can pass is that it comes to a stop. 13. If the speed of light was 25 m/s, then we would see relativistic effects all the time, something like the Chapter opening figure or Figure 36–16 with Question 21. Everything moving relative to us would be length contracted and time dilation would have to be taken into account for many events. There would be no “absolute time” on which we would all agree, so it would be more difficult, for instance, to plan to meet friends for lunch at a certain time. Many “twin paradox” kind of events would occur, and the momentum of moving objects would become very large, making it very difficult to change their motion. One of the most unusual changes for today’s modern inhabitants of Earth would be that nothing would be able to move faster than 25 m/s, which is only about 56 mi/h. 14. No. The relativistic momentum of the electron is given by p mv
mv
. At low speeds 1 v2 c2 (compared to c) this reduces to the classical momentum, p mv. As v approaches c, γ approaches infinity so there is no upper limit to the electron’s momentum.
15. No. To accelerate a particle with nonzero rest mass up to the speed of light would require an infinite amount of kinetic energy, according to Eq. 36–10a, and so is not possible. 16. No, E = mc2 does not conflict with the conservation of energy, it actually completes it. Since this equation shows us that mass and energy are interconvertible, it says it is now necessary to include mass as a form of energy in the analysis of energy conservation in physical processes. 17. Every observer will measure the speed of a beam of light to be c. Check it with Eq. 36–7d. “Away” from the Earth is taken as the positive direction, so “towards” the Earth is the negative direction.
Chapter 36
The Special Theory of Relativity
v u c 0.70c c. vu 1 10.70 1 2 c The beam’s speed (magnitude of velocity), relative to Earth, is c. u
18. Yes. One way to describe the energy stored in the compressed spring is to say it is a mass increase (although it would be so small that it could not be measured). This “mass” will convert back to energy when the spring is uncompressed. 19. Matter and energy are interconvertible (matter can be converted into energy and energy can be converted into matter). Thus we should say, “Energy can neither be created nor destroyed.” 20. No, our intuitive notion that velocities simply add is not completely wrong. Our intuition is based on our everyday experiences, and at these everyday speeds our intuition is correct regarding how velocities add. Our intuition does break down, though, at very high speeds, where we have to take into account relativistic effects. Relativity does not contradict classical mechanics, but it is a more general theory whereas classical mechanics is a limiting case. 21. (a) From the girlfriend’s frame of reference, she and her Vespa are at rest while the observer and the streetscape are moving to the left at 70 km/h. As a result the observer and the streetscape will be narrower (in the horizontal direction), and she and her Vespa appear at their original width. The observer and streetscape will appear unchanged in the vertical direction.
Responses to MisConceptual Questions 1.
(e) Answer (e) is one of the postulates of special relativity: Light propagates through empty space with a definite speed c independent of the speed of the source or observer. The other answers contradict this postulate.
2.
(c, d) Page 1078 says: “Rotating or otherwise accelerating frames of reference are noninertial frames,” and “A reference frame that moves with constant velocity with respect to an inertial frame is itself also an inertial frame.” So answers (a) and (b) describe inertial frames (answer a with a relative velocity of 0), and answers (c) and (d) describe noninertial frames.
3.
(a) Proper length is the length measured by a person at rest with the object measured. The ship’s captain is at rest with the ship, so that measurement is the proper length.
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4.
(c) A common misconception is that the distance between the objects is important when measuring relativistic effects. The important parameter is the relative velocity. The ship’s captain is at rest with the flashlight and therefore the captain measures the proper time between flashes. The space-dock personnel measure the dilated time, which is always longer than the proper time. Since the dilated time is 1.00 s, the proper time must be shorter, which is answer (c).
5.
(b) The ship’s crew will have aged less. While the ship is moving at constant speed, both the crew and the space-dock personnel record that the other’s clock is running slow. However, the ship must accelerate (and therefore change frames of reference) when it turns around. The spacedock personnel do not accelerate and therefore they are correct in their measurement that they have aged more than the ship’s crew.
6.
(d) A common misconception is that the relativistic formulas are only valid for speeds close to the speed of light. The relativistic formulas are always valid. However, for speeds much smaller than the speed of light, time dilations, length contractions, and changes in momentum are insignificant.
7.
(d) Although different values are measured for energy and momentum in different frames, the values of the particle’s momentum and energy are conserved in each of the respective reference frames. The laws of conservation of energy and momentum pertain to events such as collisions and energy exchanges. If a collision or energy exchange is observed from a single inertial reference frame, the momentum and/or energy will be conserved in that reference frame, meaning the initial and final energy and/or momentum will be the same before and after the event. Observers from different frames will measure different values for the energy or momentum according to the equations of special relativity (Eqs. 36–7 for speeds, for example). But the observers in a different frame will still find that the final energy or momentum in their frame is equal to the initial energy or momentum in that frame, confirming that energy and momentum are conserved in that different frame. In other words, conservation of energy and conservation of momentum apply WITHIN a frame of reference, not between different frames of reference.
8.
(a, d, e) The speed of light in a vacuum is the same in all reference frames, so answer (a) is NOT “relative”–its value does not change from one reference frame to another. The time interval does change (i.e., time dilation), so answer (b) is relative. The length of an object in its direction of motion does change (i.e., length contraction), so answer (c) is relative. The length of an object perpendicular to its direction of motion does not change from one reference frame to another, and so answer (d) is NOT relative. The laws of physics have the same form in all inertial reference frames–that’s the first postulate of special relativity–and so answer (e) is NOT relative. Since time intervals are relative, so is simultaneity, so answer (f) is relative.
9.
(c) It might be assumed that the speed of the ship should be added to the speed of light to obtain a relative speed of 1.5c. However, the second postulate of relativity is that the speed of light is 1.0c, as measured by any observer.
10. (f)
A common misconception is that the person on the ship would observe the change in their measurement of time, or that the person on the spacecraft would see the Earth clocks running fast. Actually, the person on the spacecraft would observe their time running normally. Since the spacecraft and Earth are moving relative to each other, observers on the spacecraft and observers on the Earth would both see the other’s clocks running slow.
Chapter 36
The Special Theory of Relativity
11. (d, e) Due to relativistic effects, the observers will not necessarily agree on the time an event occurs, the distance between events, the time interval between events, or the simultaneity of two events. However, the postulates of relativity state that both observes agree on the validity of the laws of physics and on the speed of light. 12. (d) It is common to think that one frame of reference is preferable to another. Due to relativistic effects, observers in different frames of reference may make different time measurements. However, each measurement is correct in that frame of reference. 13. (d) A common misconception is that there is a stationary frame of reference and that motion relative to this frame can be measured. Special relativity demonstrates that a stationary frame does not exist, but all motion is relative. If the spaceship has no means to observe the outside (and make a reference to another object), then there is no way to measure your velocity. 14. (b) To an observer on Earth, the pendulum would appear to be moving slowly. Therefore the period would be longer than 2.0 seconds. 15. (a) A common misconception is to add the velocities together to obtain a speed of 1.5c. However, due to relativistic effects, the relative speed between two objects must be less than the speed of light. Since the objects are moving away from each other the speed must be greater than the speed of each object, so it must be greater than 0.75c. The only option that is greater than 0.75c and less than 1.0c is 0.96c.
Solutions to Problems 1.
We find the lifetime at rest from Eq. 36–1a. 2
2.70 108 m s 6 t0 t 1 v c 5.16 10 s 1 2.25 10 s 8 3.00 10 m s 2
6
2
2.
You measure the contracted length. Find the rest length from Eq. 36–3a. l 112 m 169 m l 0 2 2 2 1 v c 1 0.750
3.
The measured distance is the contracted length. Use Eq. 36–3a. 2
l l0
2.90 108 m s 1 v c 115 ly 1 29.44 ly 29ly 8 3.00 10 m s 2
2
4.
The speed is determined from the time dilation relationship, Eq. 36–1a. t0 t
2 2 2.60 108 s t 0 8 v c 1 0.841c 2.52 10 m s c 1 4.80 10 s t
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5.
The speed is determined from the length contraction relationship, Eq. 36–3a. 2
l 32ly v c 1 c 1 0.757c 0.76c 2.3108 m s l 0 49ly
l l0 6.
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2
The speed is determined from the length contraction relationship, Eq. 36–3a. 2
l 2 v c 1 c 1 0.900 0.436 c 1.31108 m s l
l l0 7.
(a) We use Eq. 36–3a for length contraction with the contracted length 99.0% of the rest length. 2
l 2 v c 1 c 1 0.990 0.141c l l0 l (b) We use Eq. 36–1a for time dilation with the time as measured from a relative moving frame 1.00% greater than the rest time. t 1 t0 t 1 v c v c 1 0 c 1 0.140c 1.0100 t We see that a speed of 0.14 c results in about a 1% relativistic effect. 2
2
8.
2
The speed is determined from the length contraction relationship, Eq. 36–3a. Then the time is found from the speed and the contracted distance. l l0 vc
l ; t l 1 v l 0 2
l
25ly
25 yc 26.9 y
27 y 2 c 0.930 l 2 c 1 25ly c 1 l 68ly 0 That is the time according to you, the traveler. The time according to an observer on Earth would be as follows. l 68 yrc t 0 73 yr v c 0.930 9.
(a) The measured length is the contracted length. We find the rest length from Eq. 36–3a. l 4.80 m 6.47 m l 0 2 2 2 1 v c 1 0.670 Distances perpendicular to the motion do not change, so the rest height is 1.35m . (b) The time in the spacecraft is the proper time, found from Eq. 36–1a. t0 t 1 v 2 c2 20.0s 1 0.670 14.8s 2
(c) To your friend, you moved at the same relative speed: 0.670 c . (d) She would measure the same time dilation: 14.8 s .
Chapter 36
The Special Theory of Relativity
10. (a) To an observer on Earth, 23.5 ly is the rest length, so the time will be the distance divided by the speed. l 23.5ly 24.74 yr t 0 24.7 yr Earth v 0.950c (b) We find the proper time (the time that passes on the spacecraft) from Eq. 36–1a.
t0 t 1 v 2 c2 22.74 yr 1 0.950 7.725 yr 7.73 yr (c) To the spacecraft observer, the distance to the star is contracted. Use Eq. 36–3a. 2
l l0
23.5ly 1 0.950 7.338ly 7.34ly 2
(d) To the spacecraft observer, the speed of the spacecraft is v
l t
7.338ly 0.95c , 7.725 yr
as expected. 11. (a) In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 36–1a. t0 t 1 v 2 c2 5.0 yr 1 0.80 3.0 yr (b) Now we assume the 5.0 years is the time as measured on the Enterprise. Again use Eq. 36–1a. t0 5.0 yr 8.3 yr t0 t 1 v2 c2 t 2 2 2 1 v c 1 0.80 2
12. The dimension along the direction of motion is contracted, and the other two dimensions are unchanged. Use Eq. 36–3a to find the contracted length. l l0
; V l l 0 l 0 2
2.4 m
3
3
1 0.80 8.29 m3 8.3m3 2
Note that the original volume was 2.4 m 13.8m3 . 3
13. The change in length is determined from the length contraction relationship, Eq. 36–3a. The speed is very small compared to the speed of light, so we use the binomial expansion. l l 0 1 v 2 c2
1/ 2
v2 11.2 103 m s v2 10 1 v 2 c2 1 2 1 12 2 1 1 6.97 10 c c 3.00 10 m s l0 l
2
So the percent decrease is 6.97 108 % . 14. We find the speed of the particle in the lab frame, and use that to find the rest frame lifetime and distance. x 1.00 m v lab 2.6316 108 m s 0.8772 c tlab 3.80 109 s (a) Find the rest frame lifetime from Eq. 36–1a. 9 t0 tlab 1 v2 c2 3.80109 s 1 0.8772 1.824109 s 1.8210 s (b) In its rest frame, the particle will travel the distance given by its speed and the rest lifetime. 2
x0 vt 0 2.6316 108 m s1.824 109 s 0.480 m
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15. In the Earth frame, the average lifetime of the pion will be dilated according to Eq. 36–1a. The speed of the pion will be the distance moved in the Earth frame divided by the dilated time. d v d 1 v2 c2 t t0 vc
1
1
0.9633c 2 3.00 108 m s2.6 108 s 1 Note that the significant figure addition rule give 4 sig. figs. for the value under the radical sign. ct 1 d
c
16. The length of all three sides of the triangle will change due to length contraction, but the vertical dimension will not change. We designate the rest length of the equal sides as l , and so the rest length of the base is 1.5 l . When moving, we designate the length of each side as s. Note that the altitude of the triangle, designated as h, does not change. See the diagram. Use the Pythagorean theorem to determine h2 in each triangle, and equate those two expressions to find the relationship between l and s. Then determine the speed from the length contraction formula, Eq. 36–3.
h2 l 2 0.75l
h2 s2 0.50s 7 2 3 2 s2 7 2 2 2 2 2 2 2 2 l 0.75l s 0.50s 1 0.75 l 1 0.50 s l s 16 4 l 2 12 s2 v2 v2 s2 7 7 20 20 1 v c 0.86c 1 1 1 2 2 2 2 2 c c 27 27 12 1.5 27 1.5 l 1.5 l 2
2
17. Since the number of particles passing per second is reduced from N to N / 2, a time T0 must have elapsed in the particles’ rest frame. The time T elapsed in the lab frame will be greater, according to Eq. 36–1a. The particles moved a distance of 2cT0 in the lab frame during that time. x 2cT0 T0 ; v T0 T 1 v 2 c2 T v 45 c 0.894 c 2 2 T 1 v c T0 2 2 1 v c 18. We take the positive direction as the direction of motion of the Enterprise. Consider the alien vessel as reference frame S, and the Earth as reference frame S. The velocity of the Earth relative to the alien vessel is v 0.70c. The velocity of the Enterprise relative to the Earth is u 0.90c. Solve for the velocity of the Enterprise relative to the alien vessel, u, using Eq. 36–7d. v u 0.90c 0.70c 0.54c u 0.70 0.90 vu 1 1 2 c We could also have made the Enterprise as reference frame S, with v 0.90c, and the velocity of the alien vessel relative to the Earth as u 0.70c. The same answer would result.
Chapter 36
The Special Theory of Relativity
Choosing the two spacecraft as the two reference frames would also work. Let the alien vessel be reference frame S, and the Enterprise be reference frame S. Then we have the velocity of the Earth relative to the alien vessel as u 0.70c, and the velocity of the Earth relative to the Enterprise as u 0.90c. We solve for v, the velocity of the Enterprise relative to the alien vessel. u v u u 0.70c 0.90c u vu v uu 0.90c0.70c 0.54c 1 1 2 1 2 2 c c c 19. The Galilean transformation is given in Eq. 36–4. (a)
x, y, z x vt, y, z 25m 30 m s 3.5s , 20 m,0 130 m, 20 m, 0
(b)
x, y, z x vt, y, z 25 m 30 m s10.0s, 20 m,0 325 m, 20 m, 0
20. (a) The person’s coordinates in S are found using Eq. 36–6, with x 25 m, y 20 m , z 0, and t 3.5106s. We set v 1.60 108 m/s. 25 m 1.60 108 m/s3.5 106 s x vt x 690 m 2 2 8 8 1 v 2 c2 1 1.60 10 m/s 3.00 10 m/s y y 20 m ; z z 0 (b) We repeat part (a) using the time t 10.0106 s. 25 m 1.60 108 m/s10.0 106 s x vt x 1920 m 2 2 2 2 1 v c 1 1.60 108 m/s 3.00 108 m/s y y 20 m ; z z 0 21. We determine the components of her velocity in the S frame using Eq. 36–7, where ux uy 1.30 108 m/s and v 1.60 108 m/s. Then using trigonometry we combine the components to determine the magnitude and8 direction. ux v 1.30 10 m/s 1.60 108 m/s u 2.356 10 8 m/s x 1 vux / c2 1 1.60 108 m/s1.30 108 m/s / 3.00 108m/s2 8 8 8 2 2 u u 1 v c 1.30 10 m/s 1 1.60 10 m/s 3.00 10 m/s 8.932 107 m/s y 2 8 8 8 1 vux / c2 1 1.60 10 m/s1.30 10 m/s / 3.00 10 m/s 2
2
u u2 u2 2.356 108 m/s 8.932 107 m/s 2.52 108 m/s u 1 y 1 8.932 107 m/s 20.8 tan tan 8 2.356 10 m/s ux 2
2
22. (a) Take the positive direction to be the direction of motion of spaceship 1. Consider spaceship 2 as reference frame S, and the Earth reference frame S. The velocity of the Earth relative to spaceship 2 is v 0.50c. The velocity of spaceship 1 relative to the Earth is u 0.50c. Solve for the velocity of spaceship 1 relative to spaceship 2, u, using Eq. 36–7d.
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v u 0.50c 0.50c u 0.800 c 1 vu 1 0.50 0.50 c2 (b) Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship 1 is v 0.60c. The velocity of spaceship 2 relative to the Earth is u 0.60c. Solve for the velocity of spaceship 2 relative to spaceship 1, u, using Eq. 36–7d. v u 0.50c 0.50c 0.800 c u 0.50 0.50 vu 1 1 2 c As expected, the two relative velocities are the opposite of each other. 23. (a) We take the positive direction in the direction of the first spaceship. We choose reference frame S as the Earth, and reference frame S as the first spaceship. So v 0.66c. The speed of the second spaceship relative to the first spaceship is u 0.87c. We use Eq. 36–7d to solve for the speed of the second spaceship relative to the Earth, u. v u 0.66c 0.87c u 0.972 c vu 1 0.66 0.87 1 c2 (b) The only difference now is that u 0.87 c. v u 0.66c 0.87c 0.49 c u 0.66 0.87 vu 1 1 2 c The problem asks for the speed, which would be 0.49c . 24. (a) The Galilean transformation is given in Eq. 36–4.
x x vt x vt 100 m 0.883.00 108 m s1.00 106 s 364 m
(b) The Lorentz transformation is given in Eq. 36–6. Note that we are given t, the clock reading in frame S. vx t vx t t t 2 c c2 t vx v ct vx x x vt x v 2 x c c c
1
1 0.88 3.00 10 m s1.00 10 s 0.88100 m
100 m 0.88 1 0.88 2
2
8
6
311m 25. Choose frame S as the frame at rest with the spaceship. In this frame the module has speed u uy 0.85c. Frame S is the frame that is stationary with respect to the Earth. The spaceship, and therefore frame S moves in the x-direction with speed 0.76c in this frame, or v 0.76c. We use Eqs. 36–7a and 36–7b to determine the components of the module velocity in frame S. Then using trigonometry we combine the components to determine the speed and direction of travel. ux v 0 0.76c u 1 v2 c2 0.85c 1 0.762 0.552c ux 0.76 c ; u y 1 vu / c2 1 0 1 vu / c2 1 0 x
x
Chapter 36
The Special Theory of Relativity
0.552 c 1 uy tan1 36 u u2 u2 0.76 c2 0.552 c2 0.94 c ; tan ux 0.76 c 26. We assume that the given speed of 0.85c is relative to the planet that you are approaching. We take the positive direction in the direction that you are traveling. Consider your spaceship as reference frame S, and the planet as reference frame S. The velocity of the planet relative to you is v 0.85c. The velocity of the probe relative to the planet is ux 0.95c. Solve for the velocity of the probe relative to your spaceship, ux , using Eq. 36–7a. ux v 0.95c 0.85c 0.5195c ux 0.52c vux 1 0.85 0.95 1 2 c 27. (a) In frame S the horizontal component of the stick length will be contracted, while the vertical component remains the same. We use the trigonometric relations to determine the x- and y-components of the length of the stick. Then using Eq. 36–3a we determine the contracted length of the x-component. Finally, we use the Pythagorean theorem to determine stick length in frame S. l l cos ; l l sin l ; l l 1 v2 c2 l cos 1 v2 c2 x
0
y
l l 2 l 2 l
0
y
x
cos 1 v c
x
l sin l
0
1 v cos c
2
(b) We calculate the angle from the length components in the moving frame. tan l y l 0 sin 1 1 1 tan tan1 tan tan tan 2 2 l cos 1 v c2 1 v c2 l x 0 28. We set up the two frames such that in frame S, the first event takes place at the origin and the second event takes place 180 meters from the origin, so xA 0 and xB 180 m. We set the time when event A occurred equal to zero, so tA 0 and tB 0.80 s. We then set the location of the two events in frame S equal, and using Eq. 36–6 we solve for the velocity. x xB x x x vt x vt ; v A 0 180 m 2.25108 m/s 0.75c A B A A B B tA tB 0 0.80s 29. (a) We choose the train as frame S and the Earth as frame S. Since the guns fire simultaneously in S, we set these times equal to zero, that is tA tB 0. To simplify the problem we also set the location of gunman A equal to zero in frame S when the guns were fired, xA 0. This places gunman B at xB 55.0 m. Use Eq. 36–6 to determine the time that each gunman fired his weapon in framevx S. v0 t t A 0 0 A A c2 c2 1 vx 0 35m/s55.0 m2 2.14 1014 s t B tB 2B 2 8 8 c 3.00 10 m/s 1 35.0 m/s 3.00 10 m/s Therefore, in Frame S, A fired first. (b) As found in part (a), the difference in time is 2.14 1014s .
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(c) In the Earth frame of reference, since A fired first, B was struck first. In the train frame, A is moving away from the bullet fired toward him, and B is moving toward the bullet fired toward him. Thus B will be struck first in this frame as well. 30. The velocity components of the particle in the S frame are ux u cos and uy u sin . We find the components of the particle in the S frame from the velocity transformations given in Eqs. 36–7a and 36–7b. Those transformations are for the S frame moving with speed v relative to the S frame. We can find the transformations from the S frame to the S frame by simply changing v to –v and primed to unprimed variables. ux v ux ux v ; u y uy 1 v2 c2 uy u y 1 v2 c2 ux 1 vux c2 1 vux c2 vu x vu x u tan
uy ux
1 v2 c2
u 1 v c u sin 1 v c sin 1 v c 2
u x v
1 vu c
2
ux v
2
2
2
u cos v
2
cos v u
2
x
31. We set frame S as the frame moving with the observer. Frame S is the frame in which the two light bulbs are at rest. Frame S is moving with velocity v with respect to frame S. We solve Eq. 36–6 for the time t in terms of t, x, and v. Using the resulting equation we determine the time in frame S that each bulb is turned on, given that in frame S the bulbs are turned on simultaneously at tA tB 0. Taking the difference in these times gives the time interval as measured by the observing moving with velocity v. x x x vt x vt v x v2 vx t vx t vx vx t c2 = t 2 vt t 1 2 2 c2 t t 2 c c c c vxA v0 vxB vl vl t t 0 0 ; t t A B A c2 B c2 0 c2 c2 c2 t tB tA
vl
According to the observer, bulb B turned on first. 32. From the boy’s frame of reference, the pole remains at rest with respect to him. As such, the pole will always remain 13.0 m long. As the boy runs toward the barn, relativity requires that the (relatively moving) barn contract in size, making the barn even shorter than its rest length of 10.0 m. Thus it is impossible, in the boy’s frame of reference, for the barn to be longer than the pole. So according to the boy, the pole will never completely fit within the barn. In the frame of reference at rest with respect to the barn, it is possible for the pole to be shorter than the barn. We use Eq. 36–3a to calculate the speed that the boy would have to run for the contracted length of the pole, l, to equal the length of the barn. l l 0 1 v2 c2
v c 1 l 2 l 2 c 1 10.0 m
2
13.0 m 0.6390 c 2
If persons standing at the front and back door of the barn were to close both doors exactly when the pole was completely inside the barn, we would have two simultaneous events in the barn’s rest frame
Chapter 36
The Special Theory of Relativity
S with the pole completely inside the barn. Let us set the time for these two events as tA tB 0. In frame S these two events occur at the front and far side of the barn, or at xA 0 and xB 10.0 m. Using Eq. 36–6, we calculate the times at which the barn doors close in the boy’s frame of reference. t t vxA 0 v 0 0 A A c2 c2 0.6390 10.0 m 1 2.769 108 s t t vxB 0 8 B 3.00 10 m/s B c2 1 0.63902 Therefore, in the boy’s frame of reference the far door of the barn closed 27.7 ns before the front door. If we multiply the speed of the boy by this time difference, we calculate the distance the boy traveled between the closing of the two doors. x vt 0.63903.00 108 m/s2.769 108 s 5.31 m. We use Eq. 36–3a to determine the length of the barn in the boy’s frame of reference.
l l0 10.0 m 1 0.63902 7.69 m Subtracting the distance traveled between closing the doors from the length of the pole, we find the length of the barn in the boy’s frame of reference. l 0,pole x 13.0 m 5.31 m 7.69 m l barn Therefore, in the boy’s frame of reference, when the front of the pole reached the far door it was closed. Then 27.7 ns later, when the back of the pole reached the front door, that door was closed. In the boy’s frame of reference these two events are not simultaneous. 33. The momentum of the proton is given by Eq. 36–8. p mv
mv 2
1.67 1027 kg0.723.00 108 m s
2
1 v c
1 0.72
2
5.2 1019 kgm s
34. (a) We compare the classical momentum to the relativistic momentum (Eq. 36–8). pclassical mv 2 1 0.10 0.995 prelativistic mv 1 v2 c2 The classical momentum is about 0.5% in error. The negative sign means the classical momentum is smaller than the relativistic momentum (b) We again compare the two momenta. pclassical mv 2 1 v2 c2 1 0.50 0.866 mv prelativistic 1 v 2 c2 The classical momentum is 13% in error. 35. The momentum at the higher speed is to be twice the initial momentum. Designate the initial state with a subscript “0,” and the final state with a subscript “f.” Use Eq. 36–8 for relativistic momentum.
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v2 mvf 2f 2 2 2 1 vf c 1 v c p 2 f 4 f p0 v20 mv 0 2 1 v 0 c2 2 2 1 v c 2 v2f 4 v20 4 0.32c 0.456c2 v2 0.456 c2 v 0.56c f 1.456 2 f 2 2 2 2 1 v f c 1 v0 c 1 0.32
36. The two momenta, as measured in the frame in which the particle was initially at rest, will be equal to each other in magnitude. The lighter particle is designated with a subscript “1,” and the heavier particle with a subscript “2.” Use Eq. 36–8 for relativistic momentum. m2v2 m1v1 p p 1 2 1 v2 c2 1 v2 c2 2 2 2 m 6.68 1027 kg 0.53c 6.25c2 v21 v22 2 27 2 2 2 1 v1 c2 m1 1 v 2 c2 1.67 10 kg 1 0.53 7.25 v2 6.25c2 v 6.25 c 0.93c 1 1 7.25
37. We find the proton’s momenta using Eq. 36–8. mpv1 mp 0.45c mp 0.80 c mpv2 p0.45 0.5039mp c ; p0.80 1.3333mpc 2 2 2 2 v v 1 0.45 1 0.80 1 12 1 22 c c mp 0.98c mpv3 p0.98 4.9247mp c 2 v2 1 0.98 1 23 c 1.3333mpc 0.5039mpc (a) p2 p1 100 164.6 160% 100 0.5039mpc p1 4.9247mpc 1.3333mpc (b) p2 p1 100 269.4 270% 100 1.3333mpc p1
38. The rest energy of the electron is given by Eq. 36–12. E mc2 9.111031 kg3.00 108 m s 8.199 1014 J 8.20 10 2
14
J
8.199 10 J 0.5124 MeV 0.512 MeV 1.60 10 J MeV 14
13
This does not exactly agree with the “endpaper” value because of significant figures. If more significant digits were used for the given values, a value of 0.511 MeV would be obtained. 39. We find the loss in mass from Eq. 36–12. E 13 m 200 MeV1.60 10 J MeV 3.56 1028 kg 2
c
3.00 10 m s 8
2
4 1028 kg
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Chapter 36
The Special Theory of Relativity
40. We find the mass conversion from Eq. 36–12. 11020 J E m 1111kg 1000 kg 2 2 8 c 3.00 10 m s 41. Each photon has momentum 0.75 MeV/c. Since the photon is massless, Eq. 36–14 says each photon has an energy of 0.75 MeV. Assuming the photons have opposite initial directions, then the total momentum is 0, and so the product mass will not be moving. Thus, all of the photon energy can be converted into the rest mass energy of the particle, and so the heaviest particle would have a mass of 1.50 MeV c2. Convert this to kg using values from the front of the book for the atomic mass unit. 27 kg 2 1.6605 10 1.50 MeV c 2.67 1030 kg 2 931.49 MeV c 42. (a) The work is the change in kinetic energy. Use Eq. 36–10b. The initial kinetic energy is 0. 2 1 W K K final 1mc 1 0.9952 1938 MeV 8.454 10 3 MeV 8.45GeV (b) The momentum of the proton is given by Eq. 36–8. 1 p mv 938MeV c2 0.995c 9.3447 103 MeV c 9.34GeV c 2 1 0.995 43. Use Eq. 36–10b to calculate the kinetic energy of the proton. Note that the classical answer would be that a doubling of speed would lead to a four-fold increase in kinetic energy. Subscript “1” represents the lower speed v 1 c and subscript “2” represents the higher speed v 2 c . 3
1 1 2
2
K 1mc2 K2 2 1 mc K1 1 1mc2
2
1/2
3
3
5 9
1/2
1
1 1 8 9 1 1 3
2
1/2
1/2
0.3416 5.6 0.0607
44. Since the electron was accelerated by a potential difference of 28 kV, its potential energy decreased by 28 keV, and so its kinetic energy increased from 0 to 28 keV, or 0.028 MeV. Use Eq. 36–10b to find the speed from the kinetic energy. 2 1 K 1 mc 2 2 2 1 mc 1 v c v c 1
1 K 1 mc
c 1
1 0.028MeV 1 0.511MeV
0.32 c
45. The work is the change2 in kinetic energy. Use Eq. 36–10b. The initial kinetic energy is 0. W 1mc ; W K K 1mc2 1mc2 1
W 2
W1
0.90
2
0.99c
0.99 1mc2 0.90 1mc2 0.90 1mc2
0.90c
0.99
0.99
0.90
0.90 1
0.90
1 1 0.99
2
1
1 1 0.902 3.7
1 0.902
1
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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46. We find the energy equivalent of the mass from Eq. 36–12. E mc2 1.0 103 kg3.00 108 m s 9.0 10 J 2
13
We assume that this energy is used to increase the gravitational potential energy. E 9.0 1013J 9.2 109 kg E mgh m hg 1.0 103 m9.80 m s2 47. The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 36–14 to find the momentum. E K mc2 ;
pc2 E 2 mc2 K mc2 mc2 K 2 2K mc2 2
2
2
1638 MeV K 1 2 mcK 950 MeV 1 2 938.3MeV 950 MeV 2
pc
p 1638 MeV c 1.6GeV c 48. The kinetic energy is given by Eq. 36–10b. K 1mc2 mc2
1
2
v
1 v c We see that the mass of the particle does not affect the result.
2
2
3
c 0.866c
4
49. We use Eq. 36–10b to find the speed from the kinetic energy. 2 1 K 1mc 2 2 2 1 mc 1 v c v c 1
1 K 1 mc
c 1
1 1.45 MeV 1 0.511MeV
0.9655c
50. We let M represent the rest mass of the new particle. The initial energy is due to both incoming particles, and the final energy is the rest energy of the new particle. Use Eq. 36–11b for the initial energies. 2m 2m E 2 mc2 Mc2 M 2 m 1 v 2 c2 We assumed that energy is conserved, and so there was no loss of energy in the collision. The final kinetic energy is 0, so all of the kinetic energy was lost. 2 Klost Kinitial 2 1mc
1
1 2mc
Chapter 36
The Special Theory of Relativity
51. We use Eqs. 36–11a and 36–14 in order to find the mass. E 2 p2c2 m2c4 K mc2 K 2 2Kmc2 m2c4 2
m
p2c2 K 2
28 121MeV c c 45 MeV kg 2 2.5 10 140 MeV c 2 2 45 MeVc 2
2Kc2
2
2
The particle is most likely a probably a 0 meson. 52. (a) Since the kinetic energy is half the total energy, and the total energy is the kinetic energy plus the rest energy, the kinetic energy must be equal to the rest energy. We also use Eq. 36–10b. K 12 E 12 K mc2 K mc2 K 1mc2 mc2 2
1
(b) In this case, the kinetic energy is half the rest energy. 1 K 1mc2 1 mc2 3 2
v
v
2
3 4
c 0.866c
5 9
c
0.745c
53. We use v Eq. 36–10b for the kinetic energy and Eq. 36–8 for the momentum. 7.85 107 m s 0.2617 c 3.00 108 m s 1 1 K 1 mc2 1mc2 1938.3MeV 2 2 1 v c 2 1 0.2617 33.88Mev 34 MeV mv 1 mc2 v c 1 938.3MeV0.2617 p mv 254 MeV c 2 2 2 2 2 c c 1 v c 1 v c 1 0.2617 Evaluate with the classical expressions. 2 2 v K c 21 mv2 12 mc2 21 938.3MeV0.2617 32.1MeV c 1 v p mv mc2 938.3MeV c0.2617 246 MeV c c c c Calculate the percent error. error Kc K 100 32.1MeV 33.9 MeV 100 5% K K 33.9 MeV error pc p 100 246 MeV c 254 MeV c 100 3% p p 254 MeV c
54. (a) The kinetic energy is found from Eq. 36–10b. 2 2 4 8 1 1 K 1mc 1 mc 1 1.7 10 kg 3.00 10 m s 1 v2 c2 1 0.222 3.8431019 J 3.8 1019 J
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(b) Use the classical expression and compare the two results. 2 K 1 mv 1 1.7 104 kg0.223.00 108 m s 3.7031019 J 2 2
3.70310 J 3.84310 J100 3.6% 3.84310 J 19
% error
19
19
The classical value is 3.6% too low. 55. By conservation of energy, the rest energy of the americium nucleus is equal to the rest energies of the other particles plus the kinetic energy of the alpha particle. mAm c2 mNp m c 2 K 5.5MeV 1u K m m m 241.05682 u 4.00260 u 237.0483u Np Am 2 c2 c2 931.49 MeV c 56. (a) For a particle of non-zero mass, we derive the following relationship between kinetic energy and momentum. E K mc2 ; pc E 2 mc2 K mc2 mc2 K 2 2K mc2 2
2
2
2
2mc2 4 mc2 4 pc 2
K 2K mc pc 0 K 2
2
2
2
2 For the kinetic energy to be positive, we take the positive root. 2mc2 4 mc2 4 pc 2
K
2
mc2 2 This is the relationship that is graphed in the first graph below. But notice these 2 “extreme” cases. First, if the momentum is large, we have the following relationship. K mc2
pc mc2
Thus, there should be a linear relationship between kinetic energy and momentum for large values of momentum, with a negative y-intercept on the graph. If the momentum is small, we use the binomial expansion to derive the classical relationship. pc mc2 mc2 1 2 mc 2 2 p pc mc2 mc2 1 1 2 2 mc 2m Thus, we expect a quadratic relationship for small values of momentum. The adjacent graph verifies these approximations. K mc2
(b) For a particle of zero mass, the relationship is simply K pc. See the second graph.
Chapter 36
The Special Theory of Relativity
57. All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use Eq. 36–11. Both masses are the same. 1 1 E E E mc2 mc2 mc2 105.7 MeV total 1 2 1 2 1 2 2 1 0.552 1 0.38 240.8 MeV 240 MeV 58. We use Eqs. 36–11 and 36–13. E K mc2 ; pc E 2 mc2 K mc2 mc2 K 2 2K mc2 2
2
p
2
2
K 2 2K mc2 c
59. Use Eq. 36–10b for kinetic energy, and Eq. 36–12 for rest energy. K 1 mEnterprise c 2 mconverted c 2 9 7 1 1 mconverted 1 m 1 4 10 kg 2.015 10 kg Enterprise 2 1 v 2 c2 2 107 kg 1 0.10 60. (a) We assume the mass of the particle is m, and we are given that the velocity only has an x-component, ux . We write the momentum in each frame using Eq. 36–8, and we use the velocity transformation given in Eq. 36–7. Note that there are three relevant velocities: ux , the velocity in reference frame S; ux , the velocity in reference frame S; and v, the velocity of one frame relative to the other frame. There is no velocity in the y or z-directions, in either frame. 1 , and also use Eq. 36–11b for energy. We reserve the symbol for
ux
1 u c u v 2
y
2
z
u
x
1 vux c2
p x
; p 0 ; p 0
mux
px
x
ux v
; u u 1 vu c 0 ; u u 1 vu c 0 y y z z 1 v2 c2 1 v 2 c2 2
1 vu x c2
2
; p 0 since u 0 ; p 0 since u 0
mux
y
1 u c 2
2
y
z
z
Substitute the expression for ux into the expression for px. ux v m ux v mux m px 2 2 1 u c 1 vux c2 u v 1 1 2 c 1 vu c2 2
1 c2
c m ux v 2
u v m 1 xvu c2 x
1 1
1 vu c 2
x
c
u v
u v
2
2
c
c2
u v
2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
m ux v
px
1 2
x
mux
m ux v
1 v c
1 u c
1 u c
2
v 1 u c c
px
1 v c 2
vu 1 2x c mv
2
mc2
v 1 u c c
1 v c 2
2
x
1 v c
mc2
1 u c
1 u c
mu
2
2ux v
m ux v
2
vu 2x c
vu x
Instructor Solutions Manual
2
p vE c2
1 v c 2
2
It is obvious from the first few equations of the problem that py py 0 and pz pz 0. mc2 mc2 mc2 E 2 2 2 1 1 ux 2 c2 ux v 1 1 vu c2 ux v 2
2
c 1 vux c2 2
1 vu c x
2
2
x
mvu
x
2
2
2
E px v
1 v c 2
x
1 u c mc mvu 1 u c 1 v c 1 u c 1 v c 2
u v
1 vu
c 2 1 vu c2 mc2
mc2 1 vu
2
(b) We summarize these results, and write the Lorentz transformation from Eq. 36–6, but solve in terms of the primed variables. That can be easily done by interchanging primed and unprimed quantities, and changing v to v. 2 p p vE c ; p p ; p p ; E E px v x
1 v c 2
x
x vt
1 v c 2
y
y
y
y
2
2
; y y ; z z ; t
1 v c 2
2
t vx c2
1 v c 2
2
These transformations are identical if we exchange px with x, py with y, p z with z, and E c2 with t (or E c with ct). 61. The galaxy is moving away from the Earth, and so we use Eq. 36–16b. f0 f 0.1015 f0 f 0.8985 f0 2 1 f f0 1 0.89852 c f f c v v c 0.1066c 2 0 cv 1+ f f0 2 1 0.8985
Chapter 36
The Special Theory of Relativity
62. For source and observer moving towards each other, use Eq. 36–15b. cv 1 v c 1 0.65 f f0 f0 206 MHz 210 MHz 95.0 MHz cv 1 0.65 1 v c 63. We use Eq. 36–16a, and assume that v
c. 1 c v 1 v c 1 v c1 v c 1 v c 0 0 0 0 cv 1 v c 1 v c1 v c 1 v2 c2 1 v c1 v2 0
c2
1/ 2
1 v c v c 0 v 0 0 0 0 0 c
64. (a) We apply Eq. 36–15b to determine the received/reflected frequency f. Then we apply this same equation a second time using the frequency f as the source frequency to determine the Dopplershifted frequency f . We subtract the initial frequency from this Doppler-shifted frequency to obtain the beat frequency. The beat frequency will be much smaller than the emitted frequency when the speed is much smaller than the speed of light. We then set c v c and solve for v. f f c v f f c v f c v c v f c v 0 0 0 cv cv cv cv c v cfbeat f f f f c v f c v f 0 2v f 0 2v beat 0 0 cv c v c v 0 c v 2 f0
3.00 10 m/s6670 Hz 27.8 m/s 236.0 10 Hz 8
v
9
(b) We find the change in velocity and solve for the resulting change in beat frequency. Setting the change in the velocity equal to 1 km/h we solve for the change in beat frequency. cfbeat cf v f 2 f0 v v 2beat beat c f 2f 0 0 236.0 109 Hz1km/h 1m/s fbeat 70 Hz 3.00 108 m/s 3.600 km/h 65. (a) We consider the difference between Doppler-shifted frequencies for atoms moving directly towards the observer and atoms moving directly away. Use Eqs. 36–15b and 36–16b. 2v f f0 c v f0 c v f0 c v c v f0 f0 2v c cv cv cv cv We take the speed to be the rms speed of thermal motion, given by Eq. 18–5. We also assume that the thermal energy is much less than the rest energy, and so 3kT mc2. 1/ 2 3kT f v 3kT 3kT 2 3kT2 1 2 2 3kT2 v vrms 2 c mc m mc mc mc f0 (b) We evaluate for a gas of H atoms (not H2 molecules) at 650 K. Use Appendix G to find the mass. f 31.38 1023 J K650 K 2 3kT2 2 2.7 105 2 27 8 mc f0 1.008u1.66 10 kg u3.00 10 m s
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66. We choose the North Pole location as a stationary frame of reference, so the clock at the North Pole is at rest, while the clock on the equator travels the circumference of the Earth each day. We divide the circumference of the Earth by the length of the day to determine the speed of the equatorial clock. We set the dilated time equal to 1.5 years and solve for the change in rest times for the two clocks. 6 2 R 2 6.38 10 m v 464 m/s
24 hr3600s/hr t0,eq v2 2 2 t t0,eq t 1 v / c t 1 2 1 v2 / c2 2c t t 0,pole t 0,pole t 1 0 v2 t t 0,eq 0,pole t 1 2 t 2c T
v 2 1.5 yr464 m/s 3.156 10 s/yr t 2 57 s 2 2c 23.00 108 m/s 2
7
67. We treat the Earth as the stationary frame, and the airplane as the moving frame. The elapsed time in the airplane will be dilated to the observers on the Earth. Use Eq. 36–1a. 2 rEarth 2 rEarth tEarth ; tplane tEarth 1 v2 c2 1 v2 c2 v v t tEarth tplane 2 rEarth 1 2 rEarth 1 1 1 v2 rEarth v 1 v2 c2 c2 v v 2 c2 1m s 6.38 6 10 m 1500 km h 3.6 km h 9.3108 s 2 8 3.00 10 m s
68. (a) To travelers on the spacecraft, the distance to the star is contracted, according to Eq. 36–3a. This contracted distance is to be traveled in 5.5 years. Use that time with the contracted distance to find the speed of the spacecraft. xspacecraft x 1 v2 c2 Earth v tspacecraft tspacecraft 1 1 vc c 2 0.6159c 0.62c 5.5ly ct 1 1 x 4.3ly Earth (b) Find the elapsed time according to observers on Earth, using Eq. 36–1a. tspaceship 5.5 y tEarth 6.981y 7.0 y 1 0.61592 Note that this agrees with the time found from distance and speed. x 4.3ly tEarth Earth 7.0 yr v 0.6159c
Chapter 36
The Special Theory of Relativity
69. The increase in kinetic energy comes from the decrease in potential energy: 2 1 K 1 mc 2 2 2 1 mc 1 v c 1 1 v c 1 2 0.8465c K c 1 2 2 1 1 mc 7.20 1014 J 31 8 2 9.1110 kg3.00 10 m s 1/ 2
1/ 2
70. We assume that some kind of a light signal is being transmitted from the astronaut to Earth, with a frequency of the heartbeat. That frequency will then be Doppler shifted, according to Eq. 36–16b. We express the frequencies in beats per minute. f f0
2 2 f02 f 2 60 52 cv vc c 0.14c cv f 2 f 02 602 522
71. (a) The velocity components of the light in the S frame are ux 0 and uy c. We transform those velocities to the S frame according to Eq. 36–7. 2 2 ux v 0v 2 2 u v ; u u 1 v c c 1 v c c 1 v2 c2 x y 1 vux c2 1 0 1 0 1 vu c2 2 2 u c2 tan1 y tan1 c 1 v c tan 1 2 v ux v
(b) u
v2 c2 1 v2 c2
c
(c) In a Galilean transformation, we would have the following. u u v v ; u u c ; u v2 c2 c ; tan1 c x x y y v 72. We take the positive direction as the direction of motion of rocket A. Consider rocket A as reference frame S, and the Earth as reference frame S. The velocity of the Earth relative to rocket A is v 0.65c. The velocity of rocket B relative to the Earth is ux 0.90 c. Solve for the velocity of rocket B relative to rocket A, ux , using Eq. 36–7a. u v 0.90c 0.65c ux x 0.60c vux 1 0.65 0.90 1 2 c Note that a Galilean analysis would have resulted in ux 0.25c. 73. (a) We find the speed fromEq. 36–10a. 1 2 2 2 K 1mc 2 2 1mc 14, 000mc 1 v c
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c 1 2 1 c 1 2 14, 001 14, 001 2 2 8 c v c 1 3.00 10 m s 1 2 14, 001 2 14, 001 0.77 m s (b) The tube will be contracted in the rest frame of the electron, according to Eq. 36–3a. 1 2 2 2 3 l 0 l 1 v c 3.0 10 m 1 1 0.21m 14, 001 v c
74. The electrostatic force provides the radial acceleration. We solve that relationship for the speed of the electron. 1 e2 m v2 F F electron electrostatic centripetal 2 40 r r 8.99 10 N m C 1.60 10 C 2.18 10 m s 0.0073c r 9.1110 kg 0.5310 m 9
v
1 40 melectron
2
2
19
2
6
31
10
Because this is much less than 0.1c, the electron is not relativistic . 75. The minimum energy required would be the energy to produce the pair with no kinetic energy, so the total energy is their rest energy. They both have the same mass. Use Eq. 36–12. E 2mc2 20.511MeV 1.022 MeV 1.637 1013 J
76. The wattage times the time is the energy required. We use Eq. 36–12 to calculate the mass. Pt 15W3.16 107 s 1000g E Pt mc2 m 5.3106 g 2 c2 3.00 108 m s 1kg
77. Use Eqs. 36–14, 36–8, and 36–11b. E2 p2c2 m2c4 E p2c2 m2c4 2 2 mvc dE 1 2 2 pc 1/ 2 2 p c m2 c4 2 pc 2 v 2 dp E mc 1/ 2
78. The kinetic energy available comes from the decrease in rest energy.
K mnc2 mpc2 m ec2 m vc2 939.57 MeV 938.27 MeV 0.511MeV 0 0.79 MeV
79. (a) We find the rate of mass loss from Eq. 36–13. E mc2 E m c 2 m 1 E 4 1026 J s 4.44 109 kg s 4 109 kg s t c2 t 3.00 108 m s2 (b) Find the time from the mass of the Sun and the rate determined in part (a). 5.98 1024 kg m t Earth 4.27 10 7 y 4 107 y m t 4.44 109 kg s3.156 107 s y
Chapter 36
The Special Theory of Relativity
(c) We find the time for the Sun to lose all of its mass at this same rate. 1.99 1030 kg m Sun t 1.42 1013 y 11013 y m t 80. Use Eq. 36–8 for the momentum to find the mass. mv p mv 2 2 1 v c 2
8 1 2.24 108 m s 3.07 10 kg m s 3.00 10 m s 2 2 9.12 1031 kg m p 1 v c 8 v 2.24 10 m s This particle has the mass of an electron, and a negative charge, so it must be an electron. 22
81. The total binding energy is the energy required to provide the increase in rest energy. E 2mp+e 2mn mHe c2
931.5MeV c2 2 1.00783u 2 1.00867 u 4.00260 u c 2 28.32 MeV u
82. The momentum is given by Eq. 36–8, and the energy is given by Eqs. 36–11b and 36–14. mc2v Ev pc2 pc p mv v 2 2 c c E
83. (a) The magnitudes of the momenta are2equal. We use Eq. 36–8. mv 1 mc v c 1 938.3MeV0.945 p mv 2711MeV c 2 2 2 2 2 c c 1 v c 1 v c 1 0.945 1c 1.602 1010 J GeV 2.71GeV c 2.711GeV c 1GeV 3.00 10 m s 1.45 1018 kg m s (b) Because the protons are moving in opposite directions, the vector sum of the momenta is 0 . (c) In the reference frame of one proton, the laboratory is moving at 0.945c. The other proton is moving at 0.945c relative to the laboratory. We find the speed of one proton relative to the other, and then find the momentum of the moving proton in the rest frame of the other proton by using that relative velocity. v ux 0.945c 0.945c 20.945c 0.9984 c ux 2 vux 1 0.9450.945 1 0.945 1 c2 1 mc2 u c 1 938.3MeV0.9984 mux 16567 MeV c p mux 1 u2 c2 c 1 u2 c2 c 1 0.99842 1.602 1010 J GeV 1c 1GeV 16.6GeV c 16.567 GeV c 3.00 108 m s 8.85 1018 kg m s
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84. The kinetic energy is given by Eq. 36–10b. 2 1 1 14,500 kg3.00 108 m s2 K 1mc2 1 mc 1 1 v2 c2 2 1 0.80 8.7 10 J The spaceship’s kinetic energy is almost 9 times as great as the annual U.S. energy use. 20
85. The pi meson decays at rest, and so the momentum of the muon and the neutrino must each have the same magnitude (and opposite directions). The neutrino has no rest mass, and the total energy must be conserved. We combine these relationships using Eq. 36–14. Ev pv 2c2 mv 2c4
1/ 2
pvc ; p pv p
E E Ev m c2 p 2c2 m 2c4
p vc p2c2 m 2c4
1/ 2
m c pc p c m c 2
2 2
1/ 2
2 4
1/ 2
m c pc p c m c 2
2
2 2
pc
2 4
Solve for the momentum. m 2c2 m 2c2 m c 2m c pc p c p c m c 2m Write the kinetic energy of the muon using Eqs. 36–11a and 36–14. 2 4
2
2 2
2 2
pc
2 4
K E mc2 ; E E E v m c2 pc
m c m c K m c pc m c m c m c 2m
2
2
2m m c2 mc2
2m
2 2
2
2 2
2
2 2
2 2
2m
2m 2m m m m c
2
m c m c
2
2
2
m 2m m m c 2
2m
2
2m
2
m m c 2m
86. We find the loss in mass from Eq. 36–13. E 484 103 J m 5.38 1012 kg c2 3.00 108 m s2 Two moles of water has a mass of 36.0 103 kg. Find the percentage of mass lost. 5.38 1012 kg 1.49 10 10 1.49 108 % 3 36.0 10 kg 87. (a) We use Eq. 36–16a. To get a longer wavelength than usual means that the object is moving away from the Earth.
0
1.0852 1 cv 1.0850 v c 0.0814c cv 1.0852 1
(b) We assume that f0 the quasar is moving and the Earth is stationary. Then we use Eq. 16–9b. c c 1 0 1 v c 1.0850 v f 1 v c 0.085c 1 v c 0
Chapter 36
The Special Theory of Relativity
88. (a) We set the kinetic energy of the spacecraft equal to the rest energy of an unknown mass, m. Use Eqs. 36–10b and 36–12. K 1 mship c2 mc2 1 1 m 1 mship 1 mship 11.6 10 5kg 2 2 2 6.4 104 kg 1 v c 1 0.70 This is 40% of the spacecraft’s entire mass. (b) From the Earth’s point of view, the distance is 32 ly and the speed is 0.70c. That data is used to calculate the time from the Earth frame, and then Eq. 36–1a is used to calculate the time in the spaceship frame. d 32 yrc t 45.7 yr ; t t 45.7 yr 1 0.702 32.6 yr 33 yr 0 v 0.70c 89. We consider the motion from the reference frame of the spaceship. The passengers will see the trip distance contracted, as given by Eq. 36–3a. They will measure their speed to be that contracted distance divided by the year of travel time (as measured on the ship). Use that speed to find the work done (the kinetic energy of the ship). l 2 2 l v 1 1 v 0 1 v c c 0.9887 c 2 2 ct t0 t0 1.0ly 0 1 l 0 1 6.6ly 1 W K 1mc 2 1 mc 2 1 v2 c2 2 4 8 1 1 4.2 10 kg 3.00 10 m s 2.11022 J 1 0.98872 Note that, according to Problem 84, this is about 200 the annual energy consumption of the U.S. 90. For the classical expression, use K 2 mv . For v 0.6c, the value is 1
2
K 12 mv 2 121.0 kg0.6 3.00 108 m s 1.6 1016 J. 2
2
For the relativistic expression, use Eq. 36–10a, K mc 2 2 K mc
1 2 2 1. Evaluate for v 0.6c. 1 v c 2 1 8 16 1 1 1.0 kg 3.00 10 m s 1 2.310 J 1 v2 c2 1 0.62
The requested plot is shown here.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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91. In O2’s frame of reference, O2 observes that both trains are the same length. But O2 is observing the contracted length of O1’s train. Thus, the proper length of O1’s train is longer than O2’s train. In O1’s frame of reference, O2 is moving and therefore O2’s train is length contracted, making it even shorter. When the end of O2’s train passes the end of O1’s train, the first lightning bolt strikes and both observers record the strike as occurring at the end of their train. A short while later O1 observes that O2’s train moves down the track until the front ends of the trains are at the same location. When this occurs, the second lightning bolt strikes and both observers note that it occurs at the front of their train. Since to observer O2 the trains are the same length, the strikes occurred at the same time. To observer O1, the trains are not the same length and therefore train O2 had to move between strikes so that the ends were aligned during each strike. 92. The change in length is the difference between the initial length and the contracted length. The binomial expansion is used to simplify the square root expression.
l l 0 1 v2 c2 l 0 1 v2 2 1/ 2 l 0 1 12 v 2 c 2 2 2c l l l l 1 1 1 v c 1 l v2 c2 0
1
0
2
500 m100 km/hr 2
m/s
2
0
2
3.6 km/hr
2
3.00 10 m/s 8
2
2.14 1012 m 9
This length contraction is less than the size of an atom, which is on the order of 10 m. 93. If the ship travels 0.90c at 35 above the horizontal, we can consider the motion as having two components: horizontal motion of 0.90c cos 35 0.737c and vertical motion of 0.90c sin 35 0.516c. The length of the painting will be contracted due to the horizontal motion, and the height will be contracted due to the vertical motion. 2 2 2 l l 0 1 v x c 1.50 m 1 0.737 1.01m 1.0 m 2 2 h h0 1 v y c2 1.00 m 1 0.516 0.857 m 0.86 m
94. Using non-relativistic mechanics, we would find that the muon could only travel a maximum distance of 3.00 108 m s2.20 106 s 660 m before decaying. But from an Earth-based reference frame, the muon’s “clock” would run slowly, so the time to use is the dilated time. The speed in the Earth’s reference frame is then the distance of 30 km divided by the dilated time. 2 2 l l 1 v c v 0 l 0 t0 t t0 3.0 104 m v l 0 2 2 2 c c t l 2 3.00 108 m s 2.20 106 s 3.0 104 m
0.9997581 v 0.99976 c
Chapter 36
The Special Theory of Relativity
Alternatively, we could find the speed for the muon to travel a contracted length during it’s “normal” l l 0 1 l 0 lifetime. The contracted length is l l 0 , so the speed would be v t t t , 0 0 0 which is the same as the earlier expression. The kinetic energy is found from Eq. 36–10a. 1 K 1 mc2 1 mc2 1 v 2 c2
1
1105.7 MeV
4700 MeV 95. (a) Earth observers see the ship as contracted, as given by Eq. 36–3a. l l 0 1 v2 c2 35m 1 0.75 23.15m 23m 2
(b) Earth observers see the duration of the lunch as lengthened in time, as given by Eq. 36–1a. 25min t0 t 37.8min 38min 2 1 v 2 c2 1 0.75 96. Use the intensity of sunlight reaching the Earth times the area of a sphere with the radius of the Earth’s orbit 1.51011 m to calculate the power (rate of energy) leaving the sun. Then use
Eq. 36–12 to relate energy radiated per2second to the mass loss per second. 1300the J s P IA 4 1.5 1011m 3.676 1026 J s 1m2 E mass energy second 3.676 1026 J s m 4.084 109 kg s 4.1109 kg s 2 2 8 c2 second c 3.0 10 m s
97. The “gamma” factor for the motion of the meter stick is calculated from its contracted length using Eq. 36–3,l and then its kinetic energy from Eq. 36–10. 2 1.000ismcalculated 0 ; K 1mc2 1 0.30 kg3.0 108 m s l 0.480 m 2.9 1016 J 98. (a) The relative speed can be calculated in either frame, and will be the same value in both frames. The time as measured on the Earth will be longer than the time measured on the spaceship, as given by Eq. 36–1a. t tspaceship x v Earth ; t Earth spaceship 2 2 2 tEarth 1 v c x 1 Earth ctEarth 2 tEarth 2 xEarth 2 t spaceship c 2 2 x 2 2 tEarth Earth tspaceship 6.00 y 3.25 y 6.82 y c
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(b) The distance as measured by the spaceship will be contracted. 3.25 y x t x v t Earth t spaceship xspaceship spaceship x Earth 6.82 y 6.00 ly 2.86 ly tEarth Earth spaceship
99. (a) The kinetic energy is 5.00 times as great as the rest energy. Use Eq. 36–10b. 1 1 6.00 v c 1 K 1mc2 5.00mc2 0.986c 2 2 2 6.00 1 v c (b) The kinetic energy is 999 times as great as the rest energy. We use the binomial expansion. 1 K 1 mc2 999mc2 1000 1 v2 c2 1 2 7 v c 1 1 c 1 1 c 1 5.0 10 2 2 1000 1000 100. (a) To observers on the ship, the period is non-relativistic. Use Eq. 14–7b. 2.42 kg m 1.0652s 1.07s T 2 2 k 84.2 N m (b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run slow. T 1.0652s TEarth 2.44s 2 2 2 1 v c 1 0.900 101. We use the Lorentz transformations relating two reference frames moving with respect to each other along the x-axis to derive the result. vx vx t t x x vt x x vt ; t t c2 c2 2 2 2 vx vx 2 2 2 ct x c t x vt x vt 2 ct c c 2 vx 2 2 2 vx 2 c2 t 2ct x 2xvt vt c c 2 v 2 2 2 1 c v t2 c 1x 2 1 2 2 2 2 2 2 c 1 v c t 1 v c x 1 v2 c2 1 v2 c2 2 2 2 2 ct x ct x 2 2 1 v c
Chapter 36
The Special Theory of Relativity
102. We assume that the left edge of the glass is even with point A when the flash of light is emitted. There is no loss of generality with that assumption. We do the calculations in the frame of reference in which points A and B are at rest, and the glass is then moving to the right with speed v. If the glass is not moving (the “v 0” case), we would have this “no motion” result. distance in glass distance in vacuum d l d t t t v0 glass vacuum speed in glass speed in vacuum vglass c d l d nd l d nd l d l n 1d c n c c c c c If the index of refraction is n 1, then the glass will have no effect on the light, and the time would simply be the distance divided by the speed of light. distance in glass distance in vacuum d l d d l d l t t t n1 glass vacuum speed in glass speed in vacuum c c c c This is the same result we obtain from the “v 0” if n 1.
Now, let us consider the problem from a relativistic point of view. The speed of light in the glass will be the relativistic sum of the speed of light in stationary glass, c n , and the speed of the glass, v, by Eq. 36–7a. We expressions. c define c to simplify further vn 1 vn v v 1 c c c c n v n n v light cv2 1 n 1 v 1 v in moving 1 nc nc nc nc glass The contracted width of the glass, from the Earth frame of reference, is given by Eq. 36–3a. d dmoving d glass Again, we assume the light enters the block when the left edge of the block is at point A, and write simple equations for the displacement of the leading edge of the light, and the leading edge of the block. Set them equal and solve for the time when the light exits the right edge of the block. c d x v t t ; x vt ; right light light n in glass edge c d d n t vt t x x light right glass glass c nv edge n glass
d n with either c nv expression–for the leading edge of the light, or the leading edge of the block. cd cd n x v tglass light light n c nv c nv in glass d c nv vdn cd d d d n xright vtglass v c nv c nv c nv edge cd The part of the path that is left, l , will be traveled at speed c by the light. We express c nv that time, and then find the total time. Where is the front edge of the block when the light emerges? Use t glass
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l tvacuum
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cd c nv c
cd l c nv l d n d n ttotal tglass tvacuum c c c nv c nv
l n 1 d c v cv c c We check this for the appropriate limiting cases. l 1d c v l n 1d c c l Case 1: ttotal c cc c c cv c c vc This result was expected, because the speed of the light would always be c. Case 2: t total l 1d c v l n 1d 1 l n 1d c cv c c c c v0 This result was obtained earlier in the solution. l cv l Case 3: t total c cv c n1 c This result was expected, because then there is no speed change in the glass. 103. The kinetic energy of 998 GeV is used to find the speed of the protons. Since the energy is over 1000 times the rest mass, we expect the speed to be very close to c. Use Eq. 36–10b. 2 1 K 1 mc 2 2 2 1 mc 1 v c 1
1
c to 7 sig. figs. 2 998GeV 1 K 1 mc 0.938GeV 998GeV K 27 8 2 0.938GeV 11.67310 kg3.00 10 m s mv mv 2 1 mc B mc 3.3T rqv rq rq 1.0 103 m1.60 1019 C v c 1
c 1
CHAPTER 37: Early Quantum Theory and Models of the Atom Responses to Questions 1.
Bluish stars are the hottest, whitish-yellow stars are hot, and reddish stars are the coolest. This follows from Wien’s law, which says that stars with the shortest wavelength peak in their spectrum have the highest temperatures. Blue is the shortest visible wavelength and red is the longest visible wavelength, so blue is the hottest, and red is the coolest.
2.
The difficulty with seeing objects in the dark is that although all objects emit radiation, only a small portion of the electromagnetic spectrum can be detected by our eyes. Usually objects are so cool that they only give off very long wavelengths of light (infrared), which our eyes are unable to detect.
3.
The lightbulb will not produce light as white as the Sun, since the peak of the lightbulb’s emitted light is in the infrared. The lightbulb will appear more yellowish than sunlight. The Sun has a spectrum that peaks in the visible range.
4.
According to the wave theory, light of any frequency can cause electrons to be ejected as long as the light is intense enough. A higher intensity corresponds to a greater electric field magnitude and more energy. Therefore, there should be no frequency below which the photoelectric effect does not occur. According to the particle theory, however, each photon carries an amount of energy which depends upon its frequency. Increasing the intensity of the light increases the number of photons but does not increase the energy of the individual photons. The cutoff frequency is that frequency at which the energy of the photon equals the work function. If the frequency of the incoming light is below the cutoff, then the electrons will not be ejected because no individual photon has enough energy to eject an electron.
5.
If the threshold wavelength increases for the second metal, then the second metal has a smaller work function than the first metal. Longer wavelength corresponds to lower energy. It will take less energy for the electron to escape the surface of the second metal.
6.
Cesium will give a higher maximum kinetic energy for the ejected electrons. Since the incident photons bring in a given amount of energy, and in cesium less of this energy goes to releasing the electron from the material (the work function), it will give off electrons with a higher kinetic energy.
7.
Individual photons of ultraviolet light are more energetic than photons of visible light and will deliver more energy to the skin, causing burns. UV photons also can penetrate farther into the skin and, once at the deeper level, can deposit a large amount of energy that can cause damage to cells.
8.
No, fewer electrons are emitted, but each one is emitted with higher kinetic energy, when the 400-nm light strikes the metal surface. The intensity (energy per unit time) of both light beams is the same, but the 400-nm photons each have more energy than the 450-nm photons. Thus there are fewer photons hitting the surface per unit time. This means that fewer electrons will be ejected per unit time from the surface with the 400-nm light. The maximum kinetic energy of the electrons leaving the metal surface will be greater, though, since the incoming photons have shorter wavelengths and more energy per photon, and it still takes the same amount of energy (the work function) to remove each electron. This “extra” energy goes into higher kinetic energy of the ejected electrons.
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(a) No. The energy of a beam of photons depends not only on the energy of each individual photon but also on the total number of photons in the beam. It is possible that there could be many more photons in the IR beam than in the UV beam. In this instance, even though each UV photon has more energy than each IR photon, the IR beam could have more total energy than the UV beam. (b) Yes. A photon’s energy depends on its frequency: E = hf. Since IR light has a lower frequency than UV light, a single IR photon will always have less energy than a single UV photon.
10. Yes, an X-ray photon that scatters from an electron does have its wavelength changed. The photon gives some of its energy to the electron during the collision and the electron recoils slightly. Thus, the photon has less energy and its wavelength is longer after the collision, since the energy and wavelength are inversely proportional to each other E hf hc . 11. In the photoelectric effect, the photons (typically visible frequencies) have only a few eV of energy, whereas in the Compton effect, the photons (typically X-ray frequencies) have more than 1000 times greater energy and a correspondingly smaller wavelength. In the photoelectric effect, the incident photons eject electrons completely out of the target material while the photons are completely absorbed (no scattered photons). In the Compton effect, the photons are not absorbed, but are scattered from the electrons. 12. According to both the wave theory and the particle theory, the intensity of a point source of light decreases as the inverse square of the distance from the source. In the wave theory, the intensity of the waves obeys the inverse square law. In the particle theory, the surface area of a sphere increases with the square of the radius, and therefore the density of particles decreases with distance, obeying the inverse square law. The variation of intensity with distance cannot be used to help distinguish between the two theories. 13. The proton will have the shorter wavelength, since it has a larger mass than the electron and therefore a larger momentum h p . 14. Light demonstrates characteristics of both waves and particles. Diffraction and interference are wave characteristics, and are demonstrated, for example, in Young’s double-slit experiment. The photoelectric effect and Compton scattering are examples of experiments in which light demonstrates particle characteristics. We can’t say that light IS a wave or a particle, but it has properties of each. 15. We say that electrons have wave properties since we see them act like waves when they are diffracted, or exhibit 2-slit interference. We say that electrons have particle properties since we see them act like particles when they are bent by magnetic fields or accelerated and fired into materials where they scatter other electrons. 16. Both a photon and an electron have properties of waves and properties of particles. They can both be associated with a wavelength and they can both undergo scattering. An electron has a negative charge and a rest mass, obeys the Pauli exclusion principle, and travels at less than the speed of light. A photon is not charged, has no rest mass, does not obey the Pauli exclusion principle, and travels at the speed of light. Property Photon Electron 31 9.11 10 kg Mass None Charge None 1.60 1019 C 8 8 3 10 m s 3 10 m s Speed There are other properties, such as spin, that will be discussed in later chapters.
Chapter 37
Early Quantum Theory and Models of the Atom
17. In Rutherford’s planetary model of the atom, the Coulomb force (electrostatic force) keeps the electrons from flying off into space. Since the protons in the center are positively charged, the negatively charged electrons are attracted to the center by the Coulomb force and orbit around the center just like the planets orbiting a sun in a solar system due to the attractive gravitational force. 18. To tell if there is oxygen near the surface of the Sun, you need to collect light coming from the Sun and spread it out using a diffraction grating or prism so you can see the spectrum of wavelengths. If there is oxygen near the surface, there will be dark (absorption) lines at the wavelengths corresponding to electron transitions in oxygen. 19. At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light passes through the gas, photons are absorbed, causing electrons to make transitions to higher states and creating absorption lines. These lines correspond to the Lyman series since that is the series of transitions involving the ground state or n = 1 level. Since there are virtually no atoms in higher energy states, photons corresponding to transitions from n > 2 to higher states will not be absorbed. 20. The closeness of the spacing between energy levels near the top of Fig. 37–26 indicates that the energy differences between these levels are small. Small energy differences correspond to small wavelength differences, leading to the closely spaced spectral lines in Fig. 37–21. 21. (a) The Bohr model successfully explains why atoms emit line spectra; it predicts the wavelengths of emitted light for hydrogen; it explains absorption spectra; it ensures the stability of atoms (by decree); and it predicts the ionization energy of hydrogen. (b) The Bohr model did not give a reason for orbit quantization; it was not successful for multielectron atoms; it could not explain why some emission lines are brighter than others; and it could not explain the “fine structure” of some very closely-spaced spectral lines. 22. It is possible for the de Broglie wavelength h p of a particle to be larger than the dimension of the particle. If the particle has a very small mass and a slow speed (like a low-energy electron or proton) then the wavelength may be larger than the dimension of the particle. It is also possible for the de Broglie wavelength of a particle to be smaller than the dimension of the particle if it has a large momentum and a moderate speed (like a baseball). There is no direct connection between the size of a particle and the size of the de Broglie wavelength of a particle. For example, you could also make the wavelength of a proton much smaller than the size of the proton by making it go very fast. 23. The lines in the spectrum of hydrogen correspond to all the possible transitions that the electron can make. The Balmer lines, for example, correspond to an electron moving from all higher energy levels to the n = 2 level. Although an individual hydrogen atom only contains one electron, a sample of hydrogen gas contains many atoms and all the different atoms could be undergoing different transitions, leading to many lines being visible simultaneously. 24. Both particles will have the same kinetic energy. But since the proton has a larger mass, it will have a larger momentum than the electron
K p 2m p 2mK . Wavelength is inversely 2
proportional to momentum, so the proton will have the shorter wavelength, and the electron will have the longer wavelength. 25. The Balmer series spectral lines are in the visible light range and could be seen by early experimenters without special detection equipment. It was only later that the UV (Lyman) and IR (Paschen) regions were explored thoroughly, using detectors other than human sight.
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26. When a photon is emitted by a hydrogen atom as the electron makes a transition from one energy state to a lower one, not only does the photon carry away energy and momentum, but to conserve momentum, the atom must also take away some momentum. If the atom carries away some momentum, then it must also carry away some of the available energy, which means that the photon takes away less energy than Eq. 37–9 predicts. 27. (a) (b) (c) (d) (e)
continuous line, emission continuous line, absorption continuous with absorption lines (like the Sun)
28. No, the two spectra will not contain identical lines. At room temperature, virtually all the atoms in a sample of hydrogen gas will be in the ground state. Thus, the absorption spectrum will contain primarily just the Lyman lines, as photons corresponding to transitions from the n = 1 level to higher levels are absorbed. Hydrogen at very high temperatures will have atoms in excited states. The electrons in the higher energy levels will fall to all lower energy levels, not just the n = 1 level. Therefore, emission lines corresponding to transitions to levels higher than n = 1 will be present as well as the Lyman lines. In general, you would expect to see only Lyman lines in the absorption spectrum of room-temperature hydrogen, but you would find Lyman, Balmer, Paschen, and other lines in the emission spectrum of high-temperature hydrogen. 29. On average, the electrons of helium are closer to the nucleus than the electrons of hydrogen. The nucleus of helium contains two protons (positive charges), and so attracts each electron more strongly than the single proton in the nucleus of hydrogen. (There is some shielding of the nuclear charge by the “other” electron, but each electron still feels an average attractive force of more than one proton’s worth of charge.) 30. (a) (b) (c) (d)
particle wave particle wave
Responses to MisConceptual Questions 1.
(b) A common misconception is that the maximum wavelength increases as the temperature increases. However, the temperature and maximum wavelength are inversely proportional. As the temperature increases the intensity increases and the wavelength decreases.
2.
(b) The blue light has a shorter wavelength and therefore more energy per photon. Since both beams have the same intensity (energy per unit time per unit area), the red light will have more photons.
3.
(d) A common misconception is that violet light has more energy than red light. However, the energy is the product of the energy per photon and the number of photons. A single photon of violet light has more energy than a single photon of red light, but if a beam of red light has more photons than the beam of violet light, the red light could have more energy.
Chapter 37
Early Quantum Theory and Models of the Atom
4.
(d) The energy of the photon E is equal to the sum of the work function of the metal and the kinetic energy of the released photons. If the energy of the photon is more than double the work function, then cutting the photon energy in half will still allow electrons to be emitted. However, if the work function is more than half of the photon energy, then no electrons would be emitted if the photon energy were cut in half.
5.
(c) An increase in intensity of the incoming light means that more photons are incident, so more electrons will be ejected, but since the photon energy is unchanged, the maximum energy of the ejected electrons will also be unchanged.
6.
(a) Since no electrons are being ejected, the incoming photons do not have enough energy to overcome the work function. Higher energy photons are needed, which means a higher frequency of photon, which is a shorter wavelength photon.
7.
(b) The momentum of a photon is inversely proportional to its wavelength. Therefore doubling the momentum would cut the wavelength in half.
8.
(d) A common misconception is to think that only light behaves as both a particle and as a wave. De Broglie postulated, and experiments have confirmed that in addition to light, electrons and protons (and many other particles) have both wave and particle properties that can be observed or measured.
9.
(d) This is the central idea of the “wave-particle duality” discussed in Section 37–6.
10. (d) A thrown baseball has a momentum on the order of 1 kg m s. The wavelength of the baseball is Planck’s constant divided by the momentum. Due to the very small size of Planck’s constant, the wavelength of the baseball would be much smaller than the size of a nucleus. 11. (d) This chapter demonstrates the wave–particle duality of light and matter. Both electrons and photons have momentum that is related to their wavelength by p h . Young’s double-slit experiment demonstrated diffraction with light and later experiments have demonstrated electron diffraction. Therefore, all three statements are correct. 12. (a, c, d) Alpha particles are positive. To produce large angle scattering, the nucleus of the atom would have to repel the alpha particle, thus the nucleus must be positive. If the nuclear charge was spread out over a large area there would be more small angle scattering, but insufficient force to create large angle scattering. With a small nuclei, large scattering forces are possible. Since most of the alpha particles pass through the foil undeflected, most of the atom must be empty space. The scattering does not require quantized charge, but would be possible with a continuous charge distribution. Therefore answer (b) is incorrect. 13. (d) The current model of the atom is the quantum mechanical model. The plum-pudding model was rejected as it did not explain Rutherford scattering. The Rutherford atom was rejected as it did not explain the spectral lines emitted from atoms. The Bohr atom did not explain fine structure. Each of these phenomena are explained by the quantum mechanical model. 14. (d) A photon is emitted when the electron transitions from a higher state to a lower state. Photons are not emitted when an electron transitions from 2 5 or from 5 8. The other two transitions are to states that are three states below the original state. The energy levels change more rapidly for lower values of n, so the 5 2 transition will have a higher energy, and therefore a shorter wavelength than the 8 5 transition.
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15. (b) The lowest energy cannot be zero, if zero energy has been defined when the electron and proton are infinitely far away. As the electron and proton approach each other, their potential energy decreases. The energy levels are quantized and therefore cannot be any value. As shown after Eq. 37–14b, the lowest energy level of the hydrogen atom is –13.6 eV.
Solutions to Problems In several problems, the value of hc is needed. We often use the result of Problem 24, hc 1240eV 1.
We use Wien’s law, Eq. 37–1. 2.90 103 m K 2.90 103 m K 1.06 105 m (a) 10.6 m P T 273K This wavelength is in the far infrared . 3 3 (b) 2.90 10 m K 2.90 10 m K 8.29 107 m 829 nm P T 3500 K This wavelength is in the infrared . 3 3 (c) 2.90 10 m K 2.90 10 m K 6.90 104 m 0.69 mm P T 4.2 K This wavelength is in the microwave region. 3 3 (d) 2.90 10 m K 2.90 10 m K 1.06 103 m 1.06 mm P T 2.725 K This wavelength is in the microwave region.
2.
Use Wien’s law, Eq. 37–1, to find the temperature for a peak wavelength of 490 nm. T
3.
490 10 m 9
5900 K
Because the energy is quantized according to Eq. 37–2, the difference in energy between adjacent levels is simply E hf . E hf 6.63 1034 J s8.11013Hz 5.37 1020 J 5.4 1020 J 5.37 1020 J 1eV 1.60 1019 J 0.34 eV
4.
Use Wein’s law, Eq. 37–1. 2.90 103 m 2.90 103 m K (a) T 1.81105 K 9 16.0 10 m 2.90 103 m K 2.90 103 m K (b) 1.318 106 m P
T
2200 K
1.3m
Chapter 37
5.
Early Quantum Theory and Models of the Atom
The potential energy is “quantized” in units of mgh.
(a) U1 mgh 68.0 kg9.80 m s2 0.200 m 133.28J 133J (b) U2 mg 2h 2U1 2133.28J 267 J (c) U 3 mg 3h 3U1 3133.28J 4.00 10 J 2
(d) U n mg nh nU 1 n 133.28J 133nJ (e)
6.
E U2 U6 2 6133.28J 533J
Use Eq. 37–1 with a temperature of 98.6F 37.0C 273 37 K 310 K 3 sig. figs. 2.90 103 m K 2.90 103 m K 9.35 106 m 9.35 m P T 310 K
7.
(a) Wien’s displacement law says that PT constant. We must find the wavelength at which I ,T is a maximum for a given temperature. This can be found by setting I 0. 5 I 2 hc2 5 2 hc 2 hc/kT e 1 ehc/kT 1 hc hc/kT 6 5 hc/kT 15 e e kT 2 2 2 hc 2 hc/kT e 1 2 hc2 5 e hc/kT hc 5 0 5 e hc/kT 5 hc 2 6 hc/kT kT kT e 1
ex 5 x 5 ; x
hc
PkT
This transcendental equation will have some solution x = constant, and so
hc constant, and PkT
so PT constant . The constant could be evaluated from solving the transcendental equation. (b) To find the value of the constant, we solve ex 5 x 5, or 5 x 5ex. This can be done graphically, by graphing both y 5 x and y 5e x on the same set of axes and finding the intersection point. Or, the quantity 5 x 5ex could be calculated, and find for what value of x that expression is 0. The answer is x = 4.966. We use this value to solve for h. hc 4.966 PkT Tk 2.90 103 m K1.38 1023 J K 6.62 1034 J s h 4.966 P 4.966 3.00 108 m s c (c) We integrate Planck’s formula over all wavelengths. radiation 2hc25 hc hc hc ; I ,T d d ; d 2 dx 0 0 ehc/kT 1 ; let kT x x kT xkT
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hc 5 2 hc x 2 5 2 hc 2 I ,T d 2 k 4T 4 x 3 dx xkT hc d 0 x 2 dx x kT ehc / kT 1 e 1 e 1 h3c2 0 0 0 2k 4 x3 4 4 dxT T 3 2 x h c 0 e 1
Thus the total radiated power per unit area is proportional to T 4. expression is constant with respect to temperature. 8.
Everything else in the
We use Eq. 37–3 to find the energy of the photons. 26 E hf 6.626 1034 J s88.5 106 Hz 5.86 10 J
9.
We use Eq. 37–3 along with the fact that f c for light. The longest wavelength will have the lowest energy. hc E hf 4.85 1019 J 1eV 3.03eV 1.60 1019 J 1 1 1 410 109 m hc 1eV E hf 1.66eV 2.65 1019 J 19 2 2 2 1.60 10 J 750 109 m Thus, the range of energies is 2.71019 J E 4.91019 J or 1.7eV E 3.0eV .
10. At the minimum frequency, the kinetic energy of the ejected electrons is 0. Use Eq. 37–4a. W 5.2 1019 J 0 7.8 1014 Hz K h fmin W0 0 fmin 34 h 6.626 10 J s 11. The longest wavelength corresponds to the minimum frequency. That occurs when the kinetic energy of the ejected electrons is 0. Use Eq. 37–4a. We check our result with a result from Problem 24. c W 0 K hf W 0 f min 0 min max h 3.00 108 m s6.626 1034 J s ch 3.50 107 m 350 nm or max W0 3.55eV 1.60 10 19 J eV
hc 1240eV nm 349 nm 3.55eV W0
max
12. We use Eq. 37–3 with the fact that f c for light. We utilize a result from Problem 24. c hc 1240eV nm 2.952 103 nm 3.0 103 nm f E 420 103 eV Significant diffraction occurs when the opening is on the order of the wavelength. Thus, there would be negligible diffraction through the doorway.
Chapter 37
Early Quantum Theory and Models of the Atom
13. We use Eq. 37–3 with the fact that f c for light. E 2.411013 Hz 2 1013 Hz E hf f min min min min h 6.63 1034 J s c 1.24 105 m 1105 m max f min 2.411013 Hz 14. The energy of the photon will equal the kinetic energy of the baseball. We use Eq. 37–3. c 2hc 2 3.76 1027 m K hf 1 mv h 2 2 2 mv 0.145 kg27.0 m s
15. We divide the minimum energy by the photon energy at 550 nm to find the number of photons. 1018 J550 109 m E E min min 2.77 3 photons E nhf Emin n hf hc 16. The photon of visible light with the maximum energy has the least wavelength. We use 400 nm as the lowest wavelength of visible light, and calculate the energy for that wavelength. hc 6.631034 J s3.00 108 m/s hfmax 3.11eV min Electrons will not be emitted if this energy is less than the work function. The metals with work functions greater than 3.11 eV are copper and iron . 17. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so the work function is equal to the energy of the photon. We use a result from Problem 24. hc 1240eV nm 2.296eV W hf K hf 2.3eV 0 max 540 nm (b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy. We use Eq. 37–4b to calculate the maximum kinetic energy, along with a result from Problem 24. 1240eV nm hc 2.296eV 0.287eV only 1 significant figure K hf W W max 0 0 480 nm Kmax 0.287eV V 0.287 V 0.3V 0 e e
18. The photon of visible light with the maximum energy has the minimum wavelength. We use Eq. 37–4b to calculate the maximum kinetic energy, with a result from Problem 24. hc 1240eV nm K hf W W 2.48eV 0.54 eV max 0 0 410 nm 19. We use Eq. 37–4b to calculate the work function, along with a result from Problem 24. hc 1240eV nm W hf K K 1.55eV 2.80eV 0 max max 285nm
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20. Electrons emitted from photons at the threshold wavelength have no kinetic energy. We use Eq. 37–4b with the threshold wavelength to determine the work function. hc W K hc 1240eV nm 3.758eV max 0 max 330 nm (a) We now use Eq. 36–4b with the work function determined above to calculate the kinetic energy of the photoelectrons emitted by 290 nm light. hc 1240eV nm K W 3.758eV 0.5179eV 0.5eV max 0 290 nm The subtraction rule for significant figures is used here. (b) Because the wavelength is greater than the threshold wavelength, the photon energy is less than the work function, so there will be no ejected electrons .
21. The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy of the photoelectrons. We use Eq. 37–4b to calculate the work function where the maximum kinetic energy is the product of the stopping voltage and electron charge. We also use a result from Problem 24. 1240eV nm hc hc 1.74 V e 3.651eV W K eV 3.7eV 0 0 max 230 nm The subtraction rule for significant figures is used here.
22. The energy required for the chemical reaction is provided by the photon. We use Eq. 37–3 for the energy of the photon, where f c / . We also use a result from Problem 24. hc 1240eV nm 1.851eV E hf 1.9eV 670 nm Each reaction takes place in a molecule, so we use the appropriate conversions to convert eV/molecule to kcal/mol. 19 23 1.60 10 J 6.02 10 molecules kcal E 1.851eV molecule mol 42.59 kcal mol eV 4186 J
43kcal/mol
23. (a) Since f c , the photon energy given by Eq. 37–3 can be written in terms of the wavelength as E hc . This shows that the photon with the largest wavelength has the smallest energy. The 700-nm photon then delivers the minimum energy that will excite the retina. –34 8 hc 6.6310 J s3.00 10 m s 1 eV 1.776eV 1.8eV E –19 1.60 10 J 700 10–9 m (b) The eye cannot see light with wavelengths less than 410 nm. Obviously, these wavelength photons have more energy than the minimum required to initiate vision, so they must not arrive at the retina. So we speculate that photons with a wavelength less than 410 nm are absorbed by the eye before the photon reaches the retina. The threshold photon energy is that of a 410-nm photon. –34 8 hc 6.63 10 J s3.00 10 m s 1 eV E –19 3.032 eV 3.0eV 1.60 10 J 410 10–9 m We assumed both wavelengths were measured to the nearest 10 nm.
Chapter 37
Early Quantum Theory and Models of the Atom
24. (a) Since f c , the energy of each emitted photon is E hc . We insert the values for h and c and convert the resulting units to eV nm. 6.626 10–34 J s2.998 108 m s 1eV 1.602 10–19J E hc 1239.997 eV nm in nm 10–9m 1nm
1.240 10 eV nm in nm 3
(b) Insert 610 nm into the above equation. 1240 eV nm E 2.03eV 2.0eV 610 nm 25. (a) Since f c , the photon energy is E hc and the largest wavelength has the smallest energy. In order to eject electrons for all possible incident visible light, the metal’s work function must be less than or equal to the energy of a 750-nm photon. Thus the maximum value for the metal’s work function Wo is found by setting the work function equal to the energy of the 750-nm photon. –34 8 hc 6.63 10 J s3.00 10 m s 1 eV Wo 1.60 10–19J 1.7 eV 750 10–9 m (b) If the photomultiplier is to function only for incident wavelengths less than 410 nm, then we set the work function equal to the energy of the 410-nm photon. –34 8 hc 6.6310 J s3.00 10 m s 1 eV 3.0 eV Wo –19 1.60 10 J 410 10–9 m
26. Since f c , the energy of each emitted photon is E hc . We multiply the energy of each photon by n 1.3106 s to determine the average power output of each atom. At a distance of r 25 cm, the light sensor measures an intensity of I 1.6 nW 1.0 cm2. Since light energy emitted from atoms radiates equally in all directions, the intensity varies with distance as a spherical wave. Thus, from Section 15–3 in the text, the average power emitted is P 4r2I. Dividing the total average power by the power from each atom gives the number of trapped atoms. 2 4 25cm 1.6 109 W/cm2 P 4r2 I N Patom nhc 1.3 106 /s6.63 1034 J s3.00 108 m/s / 780 109 m 3.8 10 atoms 7
27. We set the kinetic energy in Eq. 37–4b equal to the stopping voltage, eV0 , and write the frequency of the incident light in terms of the wavelength, f c . We differentiate the resulting equation and solve for the fractional change in wavelength, and we take the absolute value of the final expression. hc hc eV W edV d d edV0 e V 0 0 0 2 hc hc
1.60 10 C550 10 m 0.01V 0.004 6.63 10 J s3.00 10 m s
19
34
9 8
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28. We use Eq. 37–6b. h 1 cos mec m c 9.111031 kg3.00 108 m s1.2 1013 m 1 cos1 1 e cos 18 1 34 6.63 10 J s h 29. The Compton wavelength for a particle of mass m is h mc. h 6.631034 J s 2.431012 m (a) mec (b)
h
6.6310 J s 34
mpc
1.32 1015 m
(c) The energy of the photon is given by Eq. 37–3. hc hc Ephoton hf mc2 rest energy h mc 30. We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 37–6b. The Compton wavelength of a free electron is given in the text right after Eq. 37–6b, and was calculated in Problem 29 above. The angles are taken as exact, and so do not limit the number of significant figures h in the answers. 1 cos 1 cos 2.43103 nm1 cos mc C e (a)
a 2.43103 nm1 cos 60 1.22 10 nm 3
3 (b) b 2.43103 nm1 cos 90 2.4310 nm
(c)
3 c 2.43 103 nm1 cos180 4.86 10 nm
31. (a) In the Compton effect, the maximum change in the photon’s wavelength occurs when the scattering angle 180. We use Eq. 37–6b to determine the maximum change in wavelength. Dividing the maximum change by the initial wavelength gives the maximum fractional change. h 1 – cos mec
6.6310–34 J s1 cos180 h 1 – cos 8.8 106 –31 8 9 mec 9.1110 kg3.00 10 m s550 10 m
(b) We replace the initial wavelength with 0.10 nm. h 1 – cos mec
6.63 10 J s1 cos180 –34
0.049
Chapter 37
Early Quantum Theory and Models of the Atom
32. We find the change in wavelength for each scattering event using Eq. 37–6b, with a scattering angle of 0.50. To calculate the total change in wavelength, we subtract the initial wavelength, obtained from the initial energy, from the final wavelength. We divide the change in wavelength by the wavelength change from each event to determine the number of scattering events. h 6.6310–34 J s1 cos 0.5 9.24 10 m 9.24 10 nm 1 – cos 0.5 o 9.1110–31 kg3.00 108 m s –17 8 mec
0 n
hc E0
6.6310 J s3.00 10 m s 1.24 10 m 0.00124 nm. 1.0 10 eV1.60 10 J eV –34
6
8
–12
–19
0 555 nm – 0.00124 nm 9 6 10 events –8 9.24 10 nm
33. (a) We use conservation of momentum to set the initial momentum of the photon equal to the sum of the final momentum of the photon and electron, where the momentum of the photon is given by Eq. 37–5 and the momentum of the electron is written in terms of the total energy (Eq. 36–14). Wehmultiply this equation by the speed of light to simplify. h hc 0 p hc e Using conservation of energy, we set the initial energy of the photon and rest energy of the electron equal to the sum of the final energy of the photon and the total energy of the electron. hc E hc E 0 By summing these two equations, we eliminate the final wavelength of the photon. We then solve the resulting equation for the kinetic energy of the electron, which is the total energy less the rest energy. 2 hc 2 E 2 +E 2 hc E E E 2 E 2 2 E E 0 0 0 2 hc 2 2 E 0 E0 hc 2 hc 2 2 E 2E 2 E0 E E 2 E 02 E 0 hc 2 2 E 0 2 2 hc hc hc 2 2 2 2 E0 E0 2 E0 E0 K E E0 hc hc hc 2 2 E0 2 2 E0 2 E0 2 1240eV nm 2 0.140 nm nm 297 eV 1240eV 5 2 5.1110 eV 0.140 nm
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(b) We solve the energyhc equation for the final wavelength. E hc E 0 1 1 297eV 1 hc 1 K 0.145nm hc E E hc 0.140 nm 1240eV nm 0 34. First we use conservation of energy, where the energy of the photon is written in terms of the wavelength, to relate the initial and final energies. Solve this equation for the electron’s final energy. hc hc hc hc E E E E mc2 E E mc2 e e initial initial final final photon electron photon electron Next, we define the x-direction as the direction of the initial motion of the photon. We write equations for the conservation of momentum in the horizontal and vertical directions, where is the angle the photon makes with the initial direction of the photon and is the angle the electron makes. h h h p : p cos cos p : 0 p sin sin x e y e To eliminate the variable we solve the momentum equations for the electron’s momentum components, square the resulting equations, and add the two equations together using the identity cos2 sin2 1. 2 2 2 2 h h h p cos p sin cos sin e e 2 2 2 2 h h h pe cos pe sin cos sin 2 2 h h 2 2 2 2 2 sin h 2h2 cos h p h 2h cos cos e 2 We now apply the relativistic equation relating energy and momentum, E p2c2 m2c4 (Eq. 36–14), to write the electron momentum in terms of the electron energy. Then using the electron energy obtained from the conservation of energy equation (the first line of equations in this solution), we eliminate the electron energy and solve for the change in wavelength. Ee2 m2c4 2 2 2 2 4 2 E p c m c p e e e 2 c 2 2 2 E2e m2c4 h h h 2h2 h cos mc m2c2 2 c 2 2 2 h h 1 1 = m2c2 2hmc 2h m2c2 2 2h2 1 1 2h cos 2hmc multiply by 2h
h cos mc h
h 1 cos mc
Chapter 37
Early Quantum Theory and Models of the Atom
35. The photon energy must be equal to the kinetic energy of the products plus the mass energy of the products. The mass of the positron is equal to the mass of the electron. Ephoton K products mproducts c 2 K E m c2 E 2m c2 2.85MeV 20.511MeV 1.83MeV products
photon
products
photon
electron
36. The photon with the longest wavelength has the minimum energy in order to create the masses with no additional kinetic energy. Use Eq. 37–5. hc hc h 6.63 1034 J s 6.62 1016 m max Emin 2mc2 2mc 2 1.67 1027 kg3.00 108 m s This must take place in the presence of some other object in order for momentum to be conserved. 37. The minimum energy necessary is equal to the rest energy of the two muons. Emin 2mc2 22070.511MeV 212 MeV The wavelength is given by Eq. 37–5. hc 6.63 1034 J s3.00 108 m s 5.86 1015 m E 1.60 1019 J eV212 106 eV 38. Since v 0.001c, the total energy of the particles is essentially equal to their rest energy. Both particles have the same rest energy of 0.511 MeV. Since the total momentum is 0, each photon must have half the available energy and equal momenta. E E m c2 0.511MeV ; p photon 0.511MeV c photon electron photon c
39. The energy of the photon is equal to the total energy of the two particles produced. Both particles have the same kinetic energy and the same mass. Ephoton 2 K mc2 20.345MeV 0.511MeV 1.712 MeV The wavelength is found from Eq. 37–5. 34 8 hc 6.63 10 J s3.00 10 m s 7.26 1013 m E 1.60 1019 J eV1.712 106eV hc Or, using Problem 24, 1240 eV nm 7.24 104 nm 7.24 1013 m 6 E 1.712 10 eV 40. We find the wavelength from Eq. 37–7. h h 6.631034 J s 3.2 10 p mv 0.21kg 0.10 m s
m
41. The neutron is not relativistic, so we can use p mv. We also use Eq. 37–7.
6.631034 J s h h 2.51012 m p mv 1.67 1027 kg1.6 105 m s
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42. We use the relativistic expression for momentum, Eq. 36–8. h mv mv p 2 2 2 2 1 v c 1 v c
6.631034 J s 1 0.95 h 1 v2 c2 8 1013 m 31 8 mv 9.1110 kg0.953.00 10 m s The subtraction rule for significant figures is used here. 2
43. Since the particles are not relativistic, we may use K p2 2m. We then form the ratio of the kinetic energies, using Eq. 37–7. h2 2 mp 1.67 1027 kg K e 2me 2 2 K p h 2 ; h2 m 2m 2m K 9.111031 kg 1840
p
2mp 2
e
44. For diffraction, the wavelength must be on the order of the opening. Find the speed from Eq. 37–7. 6.631034 J s 3.9 1038 m s h h h v m 1400 kg12 m p mv This is on the order of 1039 times ordinary highway speeds. 45. We assume the neutron is not relativistic. If the resulting velocity is small, our assumption will be valid. We use Eq. 37–7. h h h 6.631034 J s 1323m s 1000 m s v m p mv This is not relativistic, so our assumption was valid. 46. (a) We find the momentum from Eq. 37–7. 34 h p 6.6310 J s 9.51025 kg m s 7.0 1010 m (b) We assume h thehspeed is non-relativistic. h 6.631034 J s v 1.0 106 m s 31 10 p mv m 9.1110 kg7.0 10 m Since v c 3.47 103 , our assumption is valid. (c) We calculate the kinetic energy classically. K 21 mv2 12 mc2 v c 21 0.511MeV3.47 103 3.08 106 MeV 3.08eV 2
2
This is the energy gained by an electron if accelerated through a potential difference of 3.1 V. 47. Because all of the energies to be considered are much less than the rest energy of an electron, we can use non-relativistic relationships. We use Eq. 37–7 to calculate the wavelength. p2 h h K p 2mK ; 2m p 2mK h 34 6.63 10 J s 2.7 10 10 m 31010 m (a) 31 19 2mK 2 9.1110 kg20eV1.60 10 J eV
Chapter 37
(b)
(c)
Early Quantum Theory and Models of the Atom
h
2mK h
6.63 1034 J s
2mK
2 9.1110 kg200eV1.60 10 31
19
J eV
8.7 10 11 m 9 1011 m
6.63 1034 J s
2 9.1110 kg2.0 10 eV1.60 10 J eV 31
19
3
2.7 1011 m
48. Since the particles are not relativistic, we may use K p2 2m. We then form the ratio of the wavelengths, using Eq. 37–7. h 2m pK h h ; p me 1 h mp p 2mK e 2me K Thus the proton has the shorter wavelength, since me mp. 49. We assume the electrons are non-relativistic, and then check the result in light of that assumption. The electron’s wavelength is found from Eq. 34–2a. The speed can then be found from Eq. 37–7. h h d sin m d sin ; order m p mv
order
6.63 10 J s2
e
34
hm
690 m s me d sin This is far from being relativistic, so our original assumption was fine. v
order
50. We relate the kinetic energy to the momentum with a classical relationship, since the electrons are non-relativistic. We also use Eq. 37–7. We then assume that the kinetic energy was acquired by electrostatic potential energy. p2 h2 K eV 2m 2m 2
6.63 10 J s V 13.85V 14 V 2me 2 9.11 10 kg1.60 10 C0.33 10 m 34
h2
2
31
19
2
9
2
51. The kinetic energy is 6450 eV. That is small enough compared to the rest energy of the electron for the electron to be non-relativistic. We use Eq. 37–7. h h hc 1.240 103 eV nm 1/2 1/2 1/2 p
2mK
2mc K 2
6 2 0.511 10 eV6450eV
1.527 102 nm 15.3pm 52. The energy of a level is En
13.6 eV
. n2 (a) The transition from n = 1 to n' = 3 is an absorption , because the final state , n' = 3, has a higher energy. The photon energy is the difference between the energies of the two states. 1 1 hf En En 13.6 eV 2 2 12.1 eV 3 1
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(b) The transition from n = 6 to n' = 2 is an emission , because the initial state , n' = 2, has a higher energy. The photon energy is the difference between the energies of the two states. 1 1 hf En En 13.6 eV 2 2 3.0 eV 2 6 (c) The transition from n = 4 to n' = 5 is an absorption , because the final state , n' = 5, has a higher energy. The photon energy is the difference between the energies of the two states. 1 1 hf En En 13.6 eV 2 2 0.31 eV 5 4 The photon for the transition from n = 1 to n' = 3 has the largest energy. 53. To ionize the atom means removing the electron, or raising it to zero energy. E 0 E 0 13.6eV 13.6eV 1.51eV ionization n n2 32 54. We use the equation that appears above Eq. 37–15 in the text. (a) The second Balmer line is the transition from n = 4 to n = 2. hc 1240eV nm 490 nm E4 E2 0.85eV 3.4 eV (b) The third Lyman line is the transition from n = 4 to n = 1. hc 1240eV nm 97.3nm E E 4 1 0.85 eV 13.6 eV 55. Doubly ionized lithium is similar to hydrogen, except that there are three positive charges (Z = 3) in the nucleus. Use Eq. 37–14b with Z = 3. Z 2 13.6eV 32 13.6eV 122eV En n2 n2 n2 122 eV E 0 E 0 122 eV ionization 1 2 1 56. We evaluate the Rydberg constant using Eqs. 37–8 and 37–15. We use hydrogen so Z = 1. 1 R 1 1 Z 2e4m 1 1 2 2 2 3 2 2 8 n 0 h c n n n 2 4 Z em R 2 3 8 0h c
19 12 1.602176634 10 C 9.1093837015 1031 kg 2 3 4
88.8541878128 1012 C2 N m2 6.62607015 1034 J s 2.99792458 108 m s 1.0973731568 107
C4 kg J3s3 m s
7 1 1.0974 10 m
Chapter 37
Early Quantum Theory and Models of the Atom
57. The longest wavelength corresponds to the minimum energy, which is the ionization energy. hc 1240eV nm 91.2 nm Eion 13.6eV 58. Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. Use Eq. 37–14b. Z 2 13.6 eV 22 13.6 eV 54.4 eV En n2 n2 n2 We find the energy of the photon from the n = 6 to n = 2 transition in singly-ionized helium. 1 1 E E E 54.4 eV 12.1 eV 6 2 62 2 2 Because this is the energy difference for the n = 1 to n = 3 transition in hydrogen, the photon can be absorbed by a hydrogen atom which will jump from n = 1 to n = 3 . 59. The energy of the photon is the sum of the ionization energy of 13.6 eV and the kinetic energy of 17.5 eV. The wavelength is found from Eq. 37–3. hc hc hf E 1240eV nm 39.9 nm total E 1 3.6 17.5eV total 60. A collision is elastic if the kinetic energy before the collision is equal to the kinetic energy after the collision. If the hydrogen atom is in the ground state, then the smallest amount of energy it can absorb is the difference in the n = 1 and n = 2 levels. So as long as the kinetic energy of the incoming electron is less than that 13.6eV difference, the collision must be elastic. 13.6eV KE E 2 1 10.2eV 4 61. Singly-ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. Use Eq. 37–14b. Z 2 13.6eV 22 13.6eV 54.4 eV En n2 n2 n2 E1 54.5eV, E2 13.6eV, E3 6.0eV, E4 3.4 eV
62. The potential energy for the ground state is given by the charge of the electron times the electric potential caused by the proton. U eV
e proton
9.00 109 N m2 C2 1.60 1019 C 1eV 1.60 1019 J r 0.529 1010 m 40 1 1
e
27.2eV The kinetic energy is the total energy minus the potential energy. K E1 U 13.6eV 27.2eV 13.6eV
2
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63. The value of n is found from rn n2 r1, and then find the energy from Eq. 37–14b. 1 0.10 103 m rn 2 2 972 rn n r1 n 0.529 1010 m r1 13.6 eV 13.6 eV 13.6 eV E 1.4 105 eV 2 2 n 972 13752 64. The velocity is found from Eq. 37–10 evaluated for n = 1. nh mvrn 2 h 6.631034 J s v 2.190 106 m s 7.30 103 c 2 r1me 2 0.529 1010 m9.11 1031 kg Since v
we say yes, non-relativistic formulas are justified. The relativistic factor is as follows.
2 2 1 1 2.190 10 m s 5 0.99997 1 v c 1 2 v c 1 2 1 2.66 10 8 3.00 10 m s 6
2
We see that
2
2
is essentially 1. The answer is yes, non-relativistic formulas are justified.
65. Hydrogen atoms start in the n 1 orbit (“ground state”). Using Eqs. 37–9 and 37–14b, we determine the state to which the atom is excited when it absorbs a photon of 12.75 eV via collision with an electron. Then, using Eq. 37–15, we calculate all possible wavelengths that can be emitted as the electron cascades back to the ground state. E E E E 13.6 eV E E L U L U n2 13.6 eV n 13.6 eV 4 13.6 eV + 12.75 eV EL E Starting with the electron in the n 4 orbit, the following transitions are possible: n 4 to n 3 ; n 4 to n 2 ; n 4 to n 1 ; n 3 to n 2 ; n 3 to n 1 ; n 2 to n 1. 1 1 1 1.097 107 m–1 – 5.333105 m–1 2 2 1875 nm 3 4 1 1 1 1.097 107 m–1 – 2.057 106 m–1 486.2 nm 22 42 1 1 1 1.097 107 m–1 – 1.028107 m–1 97.23 nm 12 42 1 1 1 1.097 107 m–1 – 1.524 106 m–1 656.3 nm 22 32 1 1 1 1.097 107 m–1 – 9.751106 m–1 102.6 nm 12 32 1 1 1 1.097 107 m–1 – 8.228106 m–1 121.5 nm 12 22
Chapter 37
Early Quantum Theory and Models of the Atom
66. When we compare the gravitational and electric forces we see2 that we can use the same expression for the Bohr orbits, Eqs. 37–11 and 37–14a, if we replace Ze 4 with Gm m . h2 h2 4 0 r1 m Ze2 4 2m Ze2 0 e
0
p
e
6.626 10 J s N m / kg 9.1110 kg 1.67 10 34
h2 r1 2 2 4 Gm e m p 4
6.67 10
1.20 1029 m 2 Ze2 2 2m e Z 2e4m e E 1 2 8 02h2 4 h 0
E
2
h2
2 2 6.67 1011 N m2 kg2 9.111031 kg 1.67 1027 kg 3
2
6.626 10 J s
2
34
kg
2 2G2m e3m p2
1
2
e
97 4.22 10 J
67. We know that the radii of the orbits are given by rn n2 r1. Find the difference in radius for adjacent orbits. 2 r rn rn1 n2r1 n 1 r1 n2r1 n2 2n 1r1 2n 1r 1 r If n 1, we have r 2nr 2n n 2rn . 1 n2 n In the classical limit, the separation of radii (and energies) should be very small. We see that letting n accomplishes this. If we substitute the expression for r1 from Eq. 37–11, we have this. 2nh2 0 r 2nr1 me2 We see that r h2 , and so letting h 0 is equivalent to considering n . 68. (a) Calculate the energy from the light bulb that enters the eye by calculating the intensity of the light at a distance of l 1.0 m, by dividing the power in the visible spectrum by the surface area of a sphere of radius l 1.0 m . Multiply the intensity of the light by the area of the pupil (diameter = D) to determine the energy entering the eye per second. Finally, divide this energy by the energy of a photon (Eq. 37–3) to calculate the number of photons entering the eye per second (n). 2 P D I P2 P I D2 / 4 16 l 4 l
e
n
Pe
2 D P
4.5W550 109 m
hc / 16hc l 34 8 166.626 10 J s3.00 10 m s 1.2 1013 photons/sec
2
4.0 103 m
1.0 m
(b) Repeat the above calculation for a distance of2 l 1.0 km instead of 1.0 m. P P D I 2 2 P I D / 4 16 l 4 l e
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
n
Pe
P
4.5 W550 109 m
2
D
hc / 16hc l
Instructor Solutions Manual 2
4.0 103 m
166.626 1034 J s3.00 108 m s 1.0 103 m
1.2 107 photons/sec 69. To produce a photoelectron, the hydrogen atom must be ionized, so the minimum energy of the photon is 13.6 eV. We find the minimum frequency of the photon from Eq. 37–3. 13.6eV1.60 1019 J eV E hf f E f Emin 3.28 1015 Hz min 34 h h 6.6310 J s 70. From Section 35–11, the spacing between planes, d, for the first-order peaks is given by Eq. 35–20, 2d sin. The wavelength of the electrons can be found from their kinetic energy. The electrons are not relativistic at the energy given. p2 h2 h K 2d sin 2m 2m 2 2mK
d
h 2sin 2mK
8.11011 m
2 sin 43 2 9.1110 kg125eV1.60 10 J/eV 31
19
71. The power rating is the amount of energy produced per second. If this is divided by the energy per photon, then the result is the number of photons produced per second. 830 W12.2 102 m hc P P 5.11026 photons s Ephoton hf ; Ephoton hc 72. The intensity is the amount of energy per second per unit area reaching the Earth. If that intensity is divided by the energy per photon, the result will be the photons per second per unit area reaching the Earth. We use Eq. 37–3. hc Ephoton hf Isunlight I sunlight 1350 W m2 550 109 m I photons 3.7 1021 photons s m2 34 8 Ephoton hc 6.6310 J s3.00 10 m/s 73. First find the area of a sphere whose radius is the Earth–Sun distance.
2.83 1023 m2 2
A 4r2 4 150 109 m
Multiply this by the given intensity to find the Sun’s total power output.
2.8310 m 1350 W m 3.82 10 W 23
2
2
26
Multiplying by the number of seconds in a year gives the annual energy output.
E 3.82 1026 W3600s h24 h d365.25d yr 1.20 1034 J yr Finally we divide by the energy of one photon to find the number of photons per year. 34 9 E E 1.20 10 J yr550 10 m 3.31052 photons yr hf hc
Chapter 37
Early Quantum Theory and Models of the Atom
74. We find the peak wavelength from Wien’s law, Eq. 37–1. 2.90 103 m K 2.90 103 m K 1.1103 m 1.1mm P T 2.7 K 75. If we ignore the recoil motion, at the closest approach the kinetic energy of both particles is zero. The potential energy of the two charges must equal the initial kinetic energy of the particle: 1 Z e Z Ag e K U rmin 4 0 1 Z e Z e 9.00 109 N m2 C2 279 1.60 1019 C 2 4.74 1014 m Ag rmin 13 40 KE 4.8 MeV1.60 10 J MeV The distance to the “surface” of the gold nucleus is then 4.74 1014 m 7.0 1015 m 4.0 1014 m . 76. The electrostatic potential energy is given by Eq. 23–5. The kinetic energy is given by the total energy, Eq. 37–14a, minus2the potential 2energy.2 The Bohr radius is given by Eq. 37–11. 1 Ze mZe 1 Ze Z 2e4m U eV 2 2 2 4n h 4 r 4 n2h2
0
n
0
0
0
Z e m Z e m Z e m K E U 8 2h2n2 4n2h2 2 8n2h2 2 0 0 0 2 4
2 4
2 4
Z 22e42m 2 4n h Z 2e4m 8n2h2 2 2 4 0 ; U K Z e m 4n2h2 2 Z 2e4m 0 2 8n2h2 02
0
77. We calculate the ratio of the forces. Gmemp r2 Gm m F 6.67 1011 N m2 kg2 9.111031 kg1.67 1027 kg gravitational e p 2 ke2 Felectric ke2 9.00 109 N m2 C2 1.60 1019 C 2 r 4.40 1040 Yes , the gravitational force may be safely ignored. 78. (a) The electron has a charge e, so the potential difference produces a kinetic energy of eV. The shortest wavelength photon is produced when all the kinetic energy is lost and a photon is emitted. hc eV hc hfmax 0 eV 0 hc (b) 0 eV
79. The wavelength is found from Eq. 35–13. The velocity of electrons with the same wavelength (and thus the same diffraction pattern) is found from their momentum, assuming they are not relativistic. We use Eq. 37–7 to relate the wavelength and momentum.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
d sin 0.010 103 m sin 3.5 610.5 nm d sin n d sin h h n n p mv hn 6.631034 J s 1 v 1192 m s 1200 m s md sin
Instructor Solutions Manual
80. (a) See the adjacent figure. (b) Absorption of a 5.1 eV photon represents a transition from the ground state to the state 5.1 eV above that, the third excited state. Possible photon emission energies are found by considering all the possible downward transitions that might occur as the electron makes its way back to the ground state. 6.4eV 6.8eV 0.4eV 6.4eV 9.0eV 2.6eV 6.4eV 11.5eV 5.1eV 6.8eV 9.0eV 2.2eV 6.8eV 11.5eV 4.7eV 9.0eV 11.5eV 2.5eV 81. (a) We use Eq. 37–4b to calculate the maximum kinetic energy of the electron and set this equal to the product of the stopping voltage and the electron charge. K hf W eV V hf W0 hc / W0 max 0 0 0 e e 1240eV nm 408 nm 2.28eV 0.7592 V V0 0.76 V e (b) We calculate the speed from the non-relativistic kinetic energy equation and the maximum kinetic energy found in part (a). K 1 mv2 max
vmax
2
max
20.7592eV1.60 1019 J eV 31
5.164 105 m s 5.2 105 m/s
m 9.1110 kg This is only about 0.002c, so we are justified in using the classical definition of kinetic energy. (c) We use Eq. wavelength. h 37–7 h to calculate the de Broglie 9 6.631034 J s 1.409 10 m 1.4 nm p mv 9.111031 kg5.164 105 m s 82. (a) Use Bohr’s analysis of the hydrogen atom, replacing the proton mass with Earth’s mass, the ke2 electron mass with the Moon’s mass, and the electrostatic force Fe 2 with the gravitational r GmE mM 2 . To account for the change in force, replace ke with Gm Em M . With force, Fg r2 these replacements, write expressions similar to Eqs. 37–11 and 37–14a for the Bohr radius and energy.
Chapter 37
Early Quantum Theory and Models of the Atom
rn
h2n2 4 2mke2
rn
h2n2 6.626 1034 J s n2 2 2 2 4 GmM m E 4 2 6.67 1011 N m2 / kg2 7.35 1022 kg 5.98 1024 kg
2
n2 5.16 10129 m 2 2e4mk 2 En 2 2 nh 2 2 3 2 2G2m2 m3 2 2 6.67 1011 N m2 / kg2 5.98 1024 kg 7.35 1022 kg E M En 2 n2h2 n2 6.626 1034 J s
2.84 10165 J
(b) Insert the known masses and Earth–Moon distance into the Bohr radius equation to determine the Bohr state. n
4 Gm m r 4 2 6.67 1011 Nm2 / kg2 7.35 1022 kg 5.98 1024 kg3.84 108 m 2
6.626 10
Js
2
34
2.731068 Since n 1068 , a value of n 1 is negligible compared to n. Hence the quantization of energy and radius is not apparent . 83. We use Eqs. 36–14, 36–11a, and 37–7 to derive the expression. p2c2 m2c4 E 2 ; E K mc2 p2c2 m2c4 K mc2 K 2 2mc2 K m2c4 h2c2 2 hc h2c2 2 2 2 2 K 2mc K p c 2 K 2 2mc2 K K 2 2mc2 K 2
84. The theoretical resolution limit is the wavelength of the electron. We find the wavelength from the momentum, and find the momentum from the kinetic energy and rest energy. We use the result from Problem 83. The kinetic energy of the electron is 78 keV. 1240eV nm hc 2 2 K 2mc K 4.233 103 nm 4.2 103 nm
85. As light leaves the flashlight it gains momentum. This change in momentum is given by Eq. 31–20a. Dividing the change in momentum by the elapsed time gives the force the flashlight must apply to the light to produce this momentum. This is equal to the reaction force that light applies to the flashlight. p U P 2.8W 9.3109 N 8 t ct c 3.00 10 m s
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
86. (a) We write the Planck time as tP G h c , and the units of tP must be T . L3 ML2 L t G h c T L 2 M T 2 P MT 2 T T
5
T 3 .
T
2
There are no mass units in T , and so , and T L length units in T , and so 5 and T T
3 5
There are no
1 . Thus, and
2
. 5
2
Gh c5
t
6.67 10 N m kg 6.631034 J s 11
Gh (b) tP c5
2
2
3.00 10 m s
5
8
1.351043 s
(c) We write the Planck length as P G h c , and the units of P must be L .
P G h c
L3 ML2 L L L 2 M T 2 2 MT T T
There are no mass units in L , and so , and L L units in L , and so 3 and L L
(d) P
Gh c3
L
2
T 3. There are no time
1 3 . Thus and .
2
2
Gh c3
t
5 3
5
6.67 10 N m kg 6.631034 J s 4.051035 m 11
2
2
3.00 10 m s 8
5
87. For standing matter waves, there are nodes at the two walls. For the ground state (first harmonic), the wavelength is twice the distance between the walls, or l 12 (see Fig. 15–26b). We use Eq. 37–7 to find the velocity and then the kinetic energy. 2 1 h h h p2 ; K l 12 2l ; p 2 2l 2m 2m 2l 8ml For the second harmonic, the distance between the walls is a full wavelength, and so l . 2 h h p2 1 h l p ; K l 2m 2m l 2ml 2 88. Find the energy of 37–3 and Eq. 37–7. c the photon from Eq. 1eV 28 8 0.662 eV E hf h pc 3.5310 kg m s3.00 10 m s 19 1.60 10 J
Chapter 37
Early Quantum Theory and Models of the Atom
Looking at Fig. 37–26, the Lyman series photons have 13.6eV E 10.2eV. The Balmer series photons have 3.4eV E 1.9eV . The Paschen series photons have 1.5eV E 0.65eV. It appears that this photon belongs to the Paschen series, ejected from energy level 4. 89. (a) We solve Eq. 34–2a for the slit separation with the wavelength given in terms of the electron momentum from Eq. 37–5. The electron momentum is written in terms of the kinetic energy. m d sin 1240 eV nm hc d h 2 sin p sin 2mc eV sin sin10 2 0.511 106 eV18 eV 1.665 109 m 1.7 nm (b) No. The slit separation distance is only about 10 atomic diameters. That would probably be impossible to create. 90. The impulse on the wall is due to the change in momentum of the photons. Each photon is absorbed, and so its entire momentum is transferred to the wall. 0 np F t p p np photon nh on wall wall photons photon
9 9 n F 7.5 10 N63310 m 7.2 1018 photons s 34 t h 6.6310 J s
91. The kinetic energy of the hydrogen gas would have to be the difference between the n = 1 and n = 2 states of the hydrogen atom, 10.2 eV. Use Eq. 18–4. 2K 2 10.2 eV1.60 1019 J eV 3 T 7.88 104 K K 2 kT 23 3k 31.38 10 J K 92. (a) The minimum energy necessary to initiate the chemical process on the retina is the energy of a single photon of red light. Red light has a wavelength of approximately 750 nm. c 3.00 108 m/s 1eV 34 h 19 9 Emin hf 1.60 10 J 1.7 eV 6.626 10 J s 750 10 m
(b) The maximum energy is the energy of a photon of violet light of wavelength 400 nm. 3.00 108 m/s 1eV Emax 6.626 1034 J s 400 109 m 1.60 1019 J 3.1eV
93. (a) Here are standing wave diagrams for the first three modes of vibration.
(b) From the diagram we see that the wavelengths are given by n
2L
,
n = 1, 2, 3, … .
n
2 2 2 The momentum is p h nh , and so the kinetic energy is K pn n h . n n 2L 2m
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
(c) Because the potential energy is zero inside the box, the total energy is the kinetic energy. The ground state energy has n = 1. 2 2 1 6.631034 J s 1eV 19 E1 n h 2 2 150 eV 31 10 8mL 8 9.1110 kg 0.50 10 m 1.60 10 J 2
2
(d) Do the same calculation for the baseball, and then find the speed from the kinetic energy. 2 2 1 6.631034 J s 2 2 E nh 1.297 1066 J 1.31066 J 2 1 2 8mL 80.14 kg0.55m 66 E 1 mv2 v 2K 21.297 10 J 4.31033 m s 1 2 m 0.14 kg
(e) Find the width of the box from E1
n2 h2 . 8mL2
n2h2 8mL2 16.63 1034 J s nh L 1.447 1010 m 0.14 nm 8mE1 89.111031 kg18eV1.60 1019 J eV
E1
nh
L
8mE1
nhc 8mc2 E
11240eV nm 1.446 101 nm 0.14 nm 6 80.51110 eV18eV
94. The photon will be moving at the speed of light, c, regardless of its wavelength. The speed of the electron can be found from Eq. 37–7. We assume the electron is non-relativistic. We also use a result fromhProblem h 24. h hc2 v e e p mv m m c2 e e e e e e e ve 1 1 hc 1240eV 4.3 106 2 c mc e 0.51110 eV 560 nm
95. The work function is the amount of energy needed to ionize a hydrogen atom, which is the ionization enegy of 13.6eV . 96. (a) Apply conservation of momentum before and after the emission of the photon to determine the recoil speed of the atom, where the momentum of the photon is given by Eq. 37–7. The initial momentum is 0. h h 6.6310–34 J s v 6.0 10–3 m s 0 mv m (b) We solve Eq. 18–5 for the lowest achievable temperature, where the recoil speed is the rms speed of the rubidium gas. 2 –27 –3 3kT mv 2 85 1.66 10 kg 6.0 10 m s v T 1.2 10–7 K 0.12 K –23 m 3k 31.38 10 J K
Chapter 37
Early Quantum Theory and Models of the Atom
97. Each time the rubidium atom absorbs a photon its momentum decreases by the momentum of the photon. Dividing the initial momentum of the rubidium atom by the momentum of the photon, Eq. 37–7, gives the number of collisions necessary to stop the atom. Multiplying the number of collisions by the absorption time, 25 ns per absorption, provides the time to completely stop the atom. 27 9 mv mv 85u1.66 10 kg/u260 m s780 10 m n 43,160 6.631034 J s h h 3 T 43,16025 109 s 1.110 s
98. The decrease in mass occurs because a photon has been emitted. We calculate the fractional change. Since we are told to find the amount of decrease, we use the opposite of the change. The mass of an H atom is its atomic weight (1.008 u) times the value of “u” in MeV. 1 1 13.6 eV E 2 m E 12 32 c 1.29 108 2 6 m0 m0 m0 c 1.008 u931.5 10 eV u
99. The intensity is the amount of energy hitting the surface area per second. That is found from the number of photons per second hitting the area, and the energy per photon, from Eq. 37–3. 12 12 34 8 I 1.0 10 photons hc 1.0 10 6.626 10 J s3.00 10 m s 3.815 107 W m2
1m 1s 2
1s
3.8 107 W m2 The magnitude of the electric field is found from Eq. 31–19a. I 1 cE2 E
2
0
0
0
2I 0c
23.815 107 W m2
8.85 10
12
1.7 102 V m
C N m 3.00 10 m s 2
2
8
CHAPTER 38: Quantum Mechanics Responses to Questions 1.
(a) A matter wave ψ does not need a medium, but a wave on a string does. The square of the wave function for a matter wave ψ describes the probability of finding a particle within a certain spatial range, whereas the expressions equation for a wave on a string describes the displacement of a piece of string from its equilibrium position. (b) Neither an EM nor a matter wave needs a medium in which to exist. The expression for the EM wave describes the way in which the amplitudes of the electric and magnetic fields change as the wave passes a point in space. An EM wave represents a vector field and can be polarized. A matter wave is a scalar and cannot be polarized.
2.
According to Bohr’s theory, each electron in an atom travels in a circular orbit and has a precise position and momentum at any point in time. This view is inconsistent with the postulates of quantum mechanics and the uncertainty principle, which does not allow both the position and momentum to be known precisely. According to quantum mechanics, the “orbitals” of electrons do not have precise radii but describe the probability of finding an electron in a given spatial range.
3.
4.
As a particle becomes more massive, the uncertainty of its momentum p mv becomes larger, and so the Heisenberg uncertainty principle predicts a reduction in the uncertainty of its position (due to x p ). Thus, with a small uncertainty in its position, we can do a better job of predicting its future position. Because of the very small value of , the uncertainties in both a baseball’s momentum and position can be very small compared to typical macroscopic positions and momenta without being close to the limit imposed by the uncertainty principle. For instance, the uncertainty in the baseball’s position 10
10
could be 10 m, and the uncertainty in the baseball’s momentum 10 kg m s , and still easily satisfy the uncertainty principle. Yet we never try to measure the position or momentum of a baseball to that level of precision. For an electron, however, typical values of uncertainty in position and momentum can be close to the uncertainty principle limit, so the relative uncertainty can be much higher for that small object. 5.
No. According to the uncertainty principle, if the needle were balanced, then the position of the center of mass would be known exactly, and there would have to be some uncertainty in its momentum. The center of mass of the needle could not have a zero momentum, and therefore would fall over. If the initial momentum of the center of mass of the needle were exactly zero, then there would be uncertainty in its position, and the needle could not be perfectly balanced (with the center of mass exactly over the tip).
6.
Yes, some of the air escapes the tire (and goes into the tire gauge) in the act of measuring the pressure. It is impossible to avoid this escape. The act of measuring the air pressure in a tire therefore actually changes the pressure, although not by much since very little air escapes compared to the total amount of air in the tire. This is similar to the uncertainty principle, in which one of the two factors limiting the precision of measurement is the interaction between the object being observed, or measured, and the observing instrument. By making the measurement, you have affected the state of the system.
Chapter 38
7.
Quantum Mechanics
Yes, this is consistent with the uncertainty principle for energy Et . The ground-state energy can be precisely determined because the electrons remain in that state for a very long time. Thus, as t , E 0 . However, the electrons do not remain in the excited states for very long, thus the t in this case is relatively small, making the E , the energy width, relatively large.
8.
If Planck’s constant were much larger than it is, then the consequences of the uncertainty principle would be noticeable with macroscopic objects. For instance, attempts to determine a baseball’s speed would mean that you could not find its position very accurately. Using a radar gun to find the speed of a pitcher’s fastball could significantly change the actual course of the ball.
9.
According to Newtonian mechanics, all objects have an exact position and momentum at a point in time. This information can be used to predict the future motion of an object, and so Newtonian mechanics is said to be “deterministic”. According to quantum mechanics, there is unavoidable uncertainty in the position and momentum of all objects. It is impossible to exactly determine both position and momentum at the same time, which introduces uncertainty into the prediction of the future motion of the object. Newtonian mechanics is still relevant for macroscopic objects because there even small macroscopic uncertainties means that the consequences of the uncertainty principle are not noticed. Thus we don’t notice quantum effects in macroscopic situations like the motion of a car. This is an example of the correspondence principle.
10. If you knew the position precisely (meaning a very, very small uncertainty), then you would know essentially nothing about the momentum (because it would have a very large uncertainty). 11. No. Some of the energy of the soup would be used to heat up the thermometer, so the temperature registered on the thermometer (the equilibrium temperature of the soup and thermometer) would be slightly less than the original temperature of the soup. This is a situation where making a measurement affects the physical situation and causes uncertainty in the measurement. 12. No. However, the greater the precision of the measurement of position, the greater the uncertainty in the measurement of the object’s momentum will be. 13. A particle in a box is confined to a region of space. Since the uncertainty in position is limited by the box, there must be some uncertainty in the particle’s momentum, and the momentum cannot be zero. The zero-point energy reflects the uncertainty in momentum. 14. Yes, the probability of finding the particle at these points is zero. It is possible for the particle to pass by these points. Since the particle is acting like a wave, these points correspond to the nodes in a standing wave pattern in the box. 15. For large values of n, the probability density varies rapidly between zero and the maximum value. It can be averaged easily to the classical result (a uniform probability density for all points in the well) as n becomes large.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
16. As n increases, the energy of the corresponding state increases, and the separation between states increases, but the ratio ΔE/E is approximately 2/n, which approaches zero. 2 h2 n 21 2 h2 n22 E n 1 n2 2n 1 2 2 8ml 8ml hn 2 ; E En1 En ; En 2 E h2 n2 n2 n 8ml 2 8ml E 2n 1 2 lim lim lim 0 2 n E n n n n 17. As the potential decreases, the wave function extends into the forbidden region as an exponential decay function. When the potential drops below the particle energy, the wave function outside the well changes from an exponential decay function to an oscillating function with a longer wavelength than the function within the well. When the potential is zero, the wavelengths of the wave function will be the same everywhere. The ground state energy of the particle in a well becomes the energy of the free particle.
Responses to MisConceptual Questions 1.
(a) In considering electrons, if only the particle property of electrons is considered, then the electron is viewed as a point particle passing through the slit. The electron also has wave properties and therefore undergoes diffraction as it passes through the slit.
2.
(c) The uncertainty principle allows for a precise measurement of the position, provided that there is a large uncertainty in the momentum. It also allows for a precise measurement of the momentum, provided that the simultaneous measurement of the position has a large uncertainty. It does not allow for the precise measurement of position and momentum simultaneously.
3.
(c) The Heisenberg uncertainty principle deals with our inability to know the position and momentum of a particle precisely. The more precisely you can determine the position, the less precisely you can determine the momentum, and vice versa.
4.
(b) If you knew the position precisely, then you would know nothing about the momentum.
5.
(e) According to the uncertainty principle, it is impossible to measure a system without disturbing, or affecting, the system. The more precisely you measure the position, the greater the disturbance in the momentum, and hence the greater uncertainty in the momentum. The more precisely you measure the momentum, the greater the disturbance in the position, and hence the greater the uncertainty in the position. When measuring position and momentum, there is a minimum value for the product of the uncertainties in momentum and position. Similar statements could be made for energy and time (instead of position and momentum).
6.
(a) The hydrogen atom will have a greater probability of tunneling through the barrier because it has a smaller mass and therefore a larger transmission coefficient. (See Eqs. 38-17a and 3817b.)
7.
(b) The de Broglie wavelength of the baseball is given by h p. For a moving baseball, the momentum is on the order of 1kg m s , and so the wavelength is on the order of 1034 m. This is far too small to be detected.
Chapter 38
Quantum Mechanics
8.
(a) Energy and momentum are conserved in both Newtonian and quantum mechanics, so answers (b) and (d) are not true. And quantum mechanics can be used for macroscopic objects – it gives the same results as Newtonian mechanics. Answer (a) highlights a main difference between the two formulations of mechanics.
9.
(d) The “waving” of , or it’s variation in space and time, is a variation in the probability of finding a particle (like an electron).
10. (b), (d) The electron cannot be at rest. The ground state energy is greater than 0, and so the electron is always moving, and thus answer (a) is false. The energy values are quantized (take on specific discrete values), and so answer (c) is also false. Answer (b) is true (for the same reason that answer (c) is false). And, the electron has a wave function that describes the relative probability of the electron being at a certain position. 11. (b) Classically, the answer would be (a), because the particle could not escape from the potential well. But, due to quantum mechanical “tunneling”, the particle may be found occasionally outside of the well (answer (b)). However, it cannot completely escape from the well. 12. (c) For a free particle, the wave function of the particle is a simple sinusoidal function. Because there are no boundaries on the particle, the energy is not quantized. The energy of a free particle can have any value. 13. (c) The normalization condition calculates the probability of finding the particle somewhere in all of space. That probability is “1” since the particle must be found somewhere.
Solutions to Problems 1.
We find the wavelength of the neutron from Eq. 37–7, using the classical relationship between momentum and kinetic energy for the low-energy neutrons. The peaks of the interference pattern are given by Eq. 34–2a and Fig. 34–10. For small angles, we have sin tan . h h ; d sin n, n 1, 2, ... ; x l tan p 2mK n l , n 1, 2, ... sin tan n x x d d l 6.631034 J s1.0 m l hl x d d 2mK 3.5 104 m 21.675 1027 kg0.025eV1.60 1019 J eV 5.17 107 m 5.2 107 m
x The ratio 2.
l
is very small, so our use of sin tan is justified.
We find the wavelength of a pellet from Eq. 37–7. The half-angle for the central circle of the 1.22 , where D is the diameter of the diffraction pattern is given in Section 35–5 as sin D opening. Assuming the angle is small, the diameter of the spread of the bullet beam is
d 2l tan 2l sin.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
h
h
; d 2l tan 2l sin 2l
1.22
2l
1.22h
p mv 2.55 Dmv 3.0 103 m2.0 103 kg150 m s0.010 m l Dmvd 5.6 1027 m 34 2.44h 2.446.6310 J s
Instructor Solutions Manual
This is almost 1012 light years. The small angle approximation is definitely valid. 3.
We find the wavelength of the electrons from their kinetic energy, using nonrelativistic mechanics. Classical formulas are justified because the kinetic energy of 45 keV = 0.045 MeV is much less than the rest energy of 0.511 MeV. Eq. 37–7 is used to find the wavelength, and Eq. 34–2a is used for the two-slit interference, with a small-angle approximation. h h ; d sin m, m 0,1, 2, ... ; x l tan p 2me K ml , m 0,1, 2, ... sin tan m x x d d l 6.631034 J s0.450 m l hl x d d 2me K 1.6 106 m 29.111031kg45 103 eV1.60 1019 J eV 1.63106 m 1.6m
4.
The uncertainty in the velocity is given. Use Eq. 38–1 to find the uncertainty in the position. x
5.
p mv
1.67 10
27
3.9 1011 m kg 1600 m s
The uncertainty in position is given. Use Eq. 38–1 to find the uncertainty in the momentum. 1.055 1034 J s p mv v 5515m s 5.5 103 m s 31 8 x mx 9.1110 kg 2.110 m
6.
The minimum uncertainty in the energy is found from Eq. 38–2. 1.055 1034 J s 1eV 8 E 1.055 1026 J 6.59 10 eV 107 eV 19 t 1108s 1.60 10 J
7.
We find the uncertainty in the muon’s energy from Eq. 38–2, and then find the uncertainty in the mass. E ; E m c2 t 1.055 1034 J s 4.7955 1eV m 1029 J 2 2 19 2 6 c t c 1.60 10 J c 2.20 10 s
8.
We find the uncertainty in the energy of the free neutron from Eq. 38–2, and then the mass uncertainty from Eq. 36–13. We assume that the lifetime of the neutron is good to 2 significant figures. The current experimental lifetime of the neutron is between 881 and 882 seconds.
Chapter 38
Quantum Mechanics
E 9.
t
m
54
2
8
The uncertainty in the position is found from the uncertainty in the velocity and Eq. 38–1. The uncertainty in velocity is found by multiplying the velocity by its precision. xelectron
xbaseball
1.055 10 J s 1.310 kg 3.00 10 m s 880s 34
; E mc2
xelectron xbaseball
1.287 103 m 1.3103 m
9.1110 kg120 m s7.5 10 8.37310 m 8.4 10 m p mv 0.14 kg120 m s7.5 10 p mv
31
4
33
33
4
melectron v
mbaseball v
mbaseball melectron
0.14 kg
9.1110
31
1.5 1029
kg
The uncertainty for the electron is greater by a factor of 1.5 1029. 10. The uncertainty in the energy is found from the lifetime and the uncertainty principle. 34 5.49 1011eV E 10.55 10 J 1eV t 12 106 s 1.60 1019 J E 5.49 1011eV 1.056 1017 1.11017 1.11015% E 5.2 106 eV
11. Use Eq. 38–1 (the uncertainty principle) and Eq. 37–7 (the de Broglie wavelength). 34 p 11.55 10 J s 4.0581027 kg m s 4.11027 kg m s 8 x 2.6 10 m 6.631034 J s h 1.634 107 m 1.6 107 m 27 p 4.058 10 kg m s
12. We use the radius as the uncertainty in position for the neutron. We find the uncertainty in the momentum from Eq. 38–1. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the lowest possible kinetic energy (non-relativistic) by the following calculation. 2 2 2 p x 1.055 1034 kg m s 1MeV E 2.314 1012 J 2 13 27 15 2m 2m 1.60 10 J 2 1.67 10 kg1.2 10 m
14.46 MeV 14 MeV 13. We assume the electron is non-relativistic. The momentum is calculated from the kinetic energy, and the position uncertainty from the momentum uncertainty, Eq. 38–1. Since the kinetic energy is known to 1.00%, we have K K 1.00 102.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
p
x
1 dp 2m 2mK 2K 2 K 2mK ; dK
Instructor Solutions Manual
p 2mK K 2K
2mK
K K
1.055 10 J s kg 4.75 keV 1.60 10 J keV 1.00 10 34
p
K 2mK K
2 9.11 10
2
5.67 1010 m 14. Use the radius as the uncertainty in position for the electron. We find the uncertainty in the momentum from Eq. 38–1, and then find the energy associated with that momentum from Eq. 36–14. 34 p 14.55 10 J s 1.055 1019 kg m s. x 1.0 1015 m
If we assume that the lowest value for the momentum is the least uncertainty, we can estimate the lowest possible energy. 1/ 2 1/ 2 2 E p2c2 m 2c4 p c2 m 2c4 0 0 2 2 2 4 1/ 2 19 8 1.055 10 kg m s 3.00 10 m s 9.11 1031 kg 3.00 108 m s 3.175 1011 J 1MeV 200 MeV 13 1.60 10 J 15. (a) The minimum uncertainty in the energy is found from Eq. 38–2. 1.055 1034 J s 1eV 6.59 108eV E 1.055 1026 J 107 eV 19 8 t 110 s 1.60 10 J
(b) The transition energy can be found from Eq. 37–14b. Z 2= 1 for hydrogen. 2 Z2 1 1 En 13.6eV 2 E2 E1 13.6eV 2 13.6eV 2 10.2 eV n 2 1 8 E 6.59 10 eV 6.46 109 108 E2 E1 10.2 eV (c) The wavelength is given by Eq. 37–3. hc E hf hc 1.22 107 m 122 nm 100 nm E 1.60 1019 J 10.2eV eV Take the derivative of the above relationship to find . hc d hc dE hc E E 15.55 E2 E2 E E 6 122 nm6.46 109 7.88 107 nm 10 nm E
Chapter 38
Quantum Mechanics
16. Assume the electron has an initial x momentum px , and thus a wavelength of h px . The maxima of the double-slit interference pattern occur at locations satisfying Eq. 34–2a, d sin m, m 0,1,2, . If the angles are small, then we replace sin by , and so the maxima are given by m d . The angular separation of the maxima is then d , and the angular separation between a maximum and the adjacent minimum is 2d . The separation of a maximum and the adjacent minimum on the screen is then yscreen l 2d , where l is the distance from the slits to the detection screen. This means that many electrons hit the screen at a maximum position, and very few electrons hit the screen a distance l 2d to either side of that maximum position. If the particular slit that an electron passes through is known, then y for the electrons at the location of the slits is d 2. The uncertainty principle says py 1 d d We assume that y slits
2
py for the electron must be at least that big. Because of this uncertainty in y momentum, the
h d l . yscreen yscreen l electron has an uncertainty in its location on the screen, as h d l px Since this is about the same size as the separation between the maxima and minima, the interference pattern will be “destroyed.” The electrons will not be grouped near the maxima locations. They will instead be “spread out” on the screen, and no interference pattern will be visible. py
17. We are given that 1 x,t and 2 x,t are solutions to the Schrödinger equation. Substitute the function A1 x,t B2 x,t into the Schrödinger equation. 2 2 2 2 A 1 A B U x A B 2m x2
1
2
1
2
2m
x2
2 21 22 B AU x 1 BU x 2 2 2m x2 2m 2 x2 2 2 2 1 U x 2 2 U x B 1 2 A x x 2m 2m 1 A1 B 2 B i 2 A i t i t t
A
2
2m
2 B 2 U x A U x B x2
1
2
2 A1 B2 U xA1 B2 i A1 B2 , the combination 2m x t A1 x,t B2 x,t is also a solution to the time-dependent Schrödinger equation. So, since
2
2
18. (a) Substitute x,t Aeikxt into both sides of the time-dependent Schrödinger equation, Eq. 38–7, and compare the functional form of the results. 2 2 Aeikxt ikx t i kx t U0 Ae U Ae U 0 2 2 0 2m x 2m x 2m ikx t Ae i Ae ikxt i t t
2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
Both sides of the equation give a result of constant Aeikxt, and so t x,t Aeikxt is a valid solution if the constants are equal. Now repeat the process for x,t Acoskx t .
2 2 Acoskx t U U0 Acoskx t 2m x2 x2 2m 2k 2 2m U 0 A coskx t Acoskx t i Asinkx t i i t t
Because coskx t sin kx t for arbitrary values of x and t, x,t Acoskx t is NOT a valid solution. Now repeat the process for x,t Asin kx t .
2 2 Asin kx t U U0 Asin kx t 2m x2 x2 2m 2k 2 2m U 0 A sin kx t Asin kx t i Acoskx t i i t t
Because coskx t sin kx t for arbitrary values of x and t, x,t Asin kx t is NOT a valid solution. (b) Conservation of energy gives the following result. 2 2 h hk p2 k k U E K U U0 ; p 0 2m 2m 2 We equate the two results from the valid solution. 2 2 2k 2 ikxt k ikxt U Ae Ae U 0 0 2m 2m The expressions are the same. 19. The wave function is given in the form x Asin kx. 2 2 2.618 1010 m (a) 2.6 1010 m 10 1 k 2.4 10 m h 6.631034 J s 24 2.532 10 kg m s 2.51024 kg m s (b) p 2.6181010 m 24 (c) v p 2.532 10 kg m s 6 2.779 10 m s
m
9.111031 kg
Note that the electron is not relativistic – its speed is less than 1% of the speed of light. 24 p2 2.532 10 kg m s2 1 21.99eV 22eV (d) K 2m 29.111031 kg 1.60 1019 J eV
Chapter 38
Quantum Mechanics
20. The general expression for the wave function of a free particle is given by Eq. 38–3a. The particles are not relativistic. 9.111031kg 2.7 105 m s 2.3109 m1 (a) k 2 2 p mv
(b)
1.055 10 J s 34
h
Asin 2.3109 m1 x B cos 2.3109 m1 x 1.67 1027 kg 2.7 105 m s 2 2 p mv k 4.31012 m1
1.055 10 J s 34
h
Asin 4.31012 m1 x B cos 4.31012 m1 x
21. This is similar to the analysis performed in Chapter 16 Section 16–6 for beats. Referring to Fig. 16– 17, we see the distance from one node to the next can be considered a wave packet. We add the two wave functions, employ the trigonometric identity for the sine of a sum of two angles, and then find 2 . Since the distance between the nodes. The wave numbers are related to the wavelengths by k
1 2 , it is also true that k1 k2 and so 2k1 k2 kavg. We define k k1 k2 . 1 2 Asin k1x Asin k2 x Asin k1x sin k2x 2 Asin 1 k1 k2 xcos 1 k1 k2 x 2 2 2 Asin kavg x cos 1 k x 2 The sum function will take on a value of 0 if 12 k x n 12 , n 0,1,2 . The distance between these nodal locations is found as follows. 2 n 12 2n 1 2 n 1 1 2n 1 2 1
x
, n 0,1, 2 . x k k k k Now use the de Broglie relationship between the wavelength and momentum. h 2 2 p k k p ; x xp h k p
k
22. The minimum speed corresponds to the lowest energy state. The energy is given by Eq. 38–13. 2 h 2 6.631034 J s Emin h 1 mvmin vmin 1.5106 m s 8ml 2 2 2ml 29.111031 kg0.25 109 m Note that this is less than 1% of the speed of light, so the classical formula for kinetic energy is justified.
23. We assume the particle is not relativistic. The energy levels are given by Eq. 38–13, and the wave functions are given by Eq. 38–14. n 2 hn n x En h2n22 p2 ; n 2 sin 8ml 2m pn kn 2l l l l n 2l h , which is the de Broglie wavelength n n pn
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
24. We assume the particle is not relativistic. The energy levels provide the kinetic energy of the particles in the box. 2 2 h2 h h 2 1 h p E1 p1 p1 p 2 p1 8ml 2 2m 4l 2 2l l h xp l h l This is consistent with the uncertainty principle.
25. The energy levels for a particle in an infinite potential well are given by Eq. 38–13. The wave 2 functions are given by Eq. 38–14 with A . l 2 6.631034 J s E h 6.544 102 eV 0.065eV 1 2 2 8ml 89.111031 kg2.4 109 m 1.60 1019 J eV 2 E 22 E 46.544 10 eV 0.26 eV 2
2
1
E3 32 E1 96.544 102 eV 0.59 eV E4 42 E1 166.544 102 eV 1.0 eV 2 n sin sin x 2 2.4 nm n l x ; 1 l 0.91nm1/2 sin 1.3nm1 x 2.4 nm
2 0.91nm1/2 sin 2.6 nm1 x ; 3 0.91nm1/2 sin 3.9 nm1 x ;
4 0.91nm1/2 sin 5.2 nm1 x 26. The wave functions for an infinite square nwell are given by Eq. 38–14. 2 n Asin x ; A2 sin2 x n l n l (a) The maxima occur at locations where n 2 A2. n n sin2 x 1 x m 12 , m 0,1, 2, l l
n 1
The values of m are limited because x l . (b)
The minima occur at locations where n 2 0. n n sin2 x 0 x m , m 0,1, 2, l l
n xmin
m l , m 0,1, 2, n
n
Chapter 38
Quantum Mechanics
27. The energy levels for a particle in a rigid box are given by Eq. 38–13. We substitute the appropriate mass in each part of the problem. (a) For an electron, we have the following: 2 6.631034 J s h2 n2 E 1163MeV 8ml 2 8 9.111031kg 1.8 1014 m 2 1.60 1013 J MeV
1200 MeV (b) For a neutron, we have the following:
6.631034 J s E 0.63MeV 27 14 13 8ml 81.675 10 kg1.8 10 m 1.60 10 J MeV 2
h2 n2
2
2
(c) For a proton, we have the following:
6.631034 J s E 0.63MeV 27 14 13 8ml 81.67310 kg1.8 10 m 1.60 10 J MeV 2
h2 n2
2
2
28. (a) We use the ground state energy and Eq. 38–13 to find the width of the well. h2 E1 8ml 2 6.631034 J s h l 2.11010 m 0.21nm 31 19 8mE1 89.1110 kg8.4eV1.60 10 J eV (b) The longest wavelength photon will be the photon with the lowest frequency, and thus the lowest energy. The difference between energy levels increases with high states, so the lowest energy transition is from n = 2 to n = 1. The energy levels are given by Eq. 38–13. 2 2 h 2 n E1 En n 2 8ml c hc hc hc E E E 4E E 38.4 eV1.60 1019 J eV 2 1 1 1 4.9 108 m 49 nm 29. The energy levels for a particle in a rigid box are given by Eq. 38–13. Use that equation, evaluated for n = 4 and n = 1, to calculate the width of the box. We also use Eq. 37–3. hc h2 2 2 E hf E4 E1 4 1 8ml 2 l
370 109 m 1.3109 m 1.3nm 89.111031 kg3.00 108 m s
15 6.631034 J s
30. The longest wavelength photon will be the photon with the lowest frequency, and thus the lowest energy. The difference between energy levels increases with high states, so the lowest energy transition is from n = 2 to n = 1. The energy levels are given by Eq. 38–13.
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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
h2
hc E hv
l
E2 E1
2
Instructor Solutions Manual
2
2 1 8ml 2
580 109 m 10 7.310 m 0.73nm 31 8 89.1110 kg3.00 10 m s 3 6.631034 J s
31. The energy released is calculated by Eq. 38–13, with n = 2 for the initial state and n = 1 for the final state. We switch the order of the initial and final states in order to get a positive answer, since the problem asks for the “energy released”. 2 h2 3 6.631034 J s 2 1 E E2 E1 2 1 8ml 2 8 1.67 1027 kg 1.2 1014 m 2 1.60 1013 J MeV
4.3MeV 32. (a) The ground state energy is given by Eq. 38–13 with n = 1. An oxygen molecule has a mass of 32 amu. 6.631034 J s 1 832 u1.66 1027 kg u3.0 103 m 1.60 1019 J eV 2
E1
8ml 2 n1
2
19 7.1831019 eV 7.2 10 eV (b) We equate the thermal energy expression to Eq. 38–13 in order to find the quantum number. 2 2 1 kT h n
2
8ml 2
n 2 kTm
l h
2 1 .38 1023 J K 300 K 32 u 1.66 1027 kg u
6.6310 J s 34
1.342 108 1108 (c) Use Eq. 38–13 with a large-n approximation. 2 h h2 2 2 h2 E En1 En n 1 n 8ml 2 2n 1 2n 8ml 2 2nE1 8ml 2
2 1.342 108
7.1831019 eV 1.9 1010 eV
33. Because the wavenfunction is lnormalized, the probability 38–8. Change the x n l in Example is found as variable to dx or dx d . The lower limit is or x , and then d l n l n x1 0.60 nm 0.15nm 0.45nm, and the upper limit is x2 0.60 nm 0.15nm 0.75nm. The limits x 0.45nm 0.75nm on the variable are n 1 n 0.375n and n 0.625n . 1 2 l 1.20 nm 1.20 nm x2 x2 n x2 l 0.625n 2 l sin2 d 2 sin2 d P n 2 dx 2 sin2 n x dx l n 0.375n x x l l n n x l
1
2
1
0.625n 12 14 sin 2 0.375n
n
1
Chapter 38
Quantum Mechanics
(a) For the n = 12 state, we have the following: 0.625 2 P 1 1 sin 2 1 0.25 1 sin1.25 sin 0.75 0.47508 0.48 n1 4 0.375 2 4 2
(b) For the n = 52state, we have the following: 3.125 2 P 1 1 sin 2 1 1.25 1 sin 6.25 sin 3.75 0.20498 0.20 n5 4 4 1.875 5 2 5 2 (c) For the n = 20 2state, we have the12.5following: 1 P 1 1 sin 2 1 5 1 sin 25 sin15 n20 4 0.25 4 7.5 10 2 20 2 (d) The classical prediction would be that the particle has an equal probability of being at any 0.75 nm 0.45 nm location, so the probability of being in the given range is P 0.25 . We 1.20 nm see that the probabilities approach the classical value for large n. 34. Consider Fig. 38–9, partially copied here. To consider the problem with the boundaries shifted, we would not expect any kind of physics to change. So, we expect the same wave functions in terms of their actual shape, and we expect the same energies if all that is done is to change the labeling of the walls to x 12 l and x 12 l . The mathematical descriptions of the wave functions would change because of the change of coordinates. All we should have to do is shift the origin of coordinates to the right by 12 l . Thus, we might expect the following functions and energies. 2 nwave sin x n l l 2 n 2 n sin x 1 l sin x 1 n n 2 l 2 l l l 2 n 1 1: 1 2 sin x 2 2 cos x ; E1 h l l 8ml 2 l l 2 2 n 2 : 2 2 sin x 2 sin 2 x ; E2 4h l l 8ml 2 l l 2 3 2 3 2 3 n 3: 3 sin x 3 sin x 1 sin x 1 l l 2 2 2 l l l l 2 3 cos x ; E3 9h 2 8ml l l 4 2 n 4 : 4 2 sin x 2 2 sin 4 ; E4 16h l l 8ml 2 l l
2
For any higher orders, we simply add another 2 of phase to the arguments of the above functions. They can be summarized as follows. n x n1/ 2 n2 h2 n odd: 1 cos , En 2
l
8ml
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
n x n2 h2 sin , En 8ml 2 l
n/ 2 n even: 1
Of course, this is not a “solution” in the sense that we have not derived these solutions from the Schrödinger equation. We now show a solution that arises from solving the Schrödinger equation. We follow the development as given in Section 38–8. As suggested, let x Asin kx . For a region where U x 0, k
(Eq. 38–11a). The
1 1 boundary conditions are l Asin kl 0 and 1 l Asin 1 kl 0. To 2
2
2
2
guarantee the boundary conditions, we must have the following:
Asin 1 kl 0 1 kl m ; Asin 1 kl 0 2
2
1
2
kl n
2
Both n and m are integers. Add these two results, and subtract the two results, to get two new expressions. 1 1 2 kl n 12 n m ; k n m 1 2 kl m l So again, we have an energy quantization, with E
2 n m
2m
2ml 2
h2 n m
2
2
2
m n is not allowed, because this leads to k 0, n , and x 0.
. Note that
8ml 2
Next, we normalize the wave functions. We use an indefinite integral from Appendix B–4. 1 x Asinkx Asin n m x 1 n m 1
l
2
2 dx A2 sin2 l n m x 1 n m dx 1 1l 2
2
let
1
21 l
n m x n m 1
2
l
l
dx
n m
x l n ; x l m 2 2 n sin2 d A2l 1 dx A l 1 2
1 2
2
n m m
d
1 1 sin 2
n m 2
n
4
m
A2l
n m
n m
2
A2l 2
2 l l This is the same as in Section 38–8. Finally, let us examine a few allowed cases. 2 n 1, m 0 : k , 21 1,0 2 sin x 1 2 cos x ; E1,0 h 2 l l 8ml 2 l l l 2 2 2 ; E2,0 4h n 2, m 0 : k l , 2,0 2 sin x 2 sin 2 x l 8ml 2 l l l 3 2 3 2 3 n 3, m 0 : k , 3 sin x 3 sin x 1 l l 2 l 2 3,0 2 l l 3 2 2 1 sin x 2 2 cos 3 x ; E3,0 9h 8ml 2 l l l l A2
A
Chapter 38
Quantum Mechanics
4 4 2 ; E4,0 16h n 4, m 0 : k l , 2 4,0 2 sin x 2 2 sin 4 l 8ml 2 l l l These are the same results as those obtained in the less formal method. Other combinations of m and n would give essentially these same results for the lowest four energies and the associated wave functions. For example, consider n 4, m 1. 3 2 3 2 3 , 5 sin x 5 sin x 1 n 4, m 1: k l l 2 2 2 4,1 l l l
2 9h2 4 cos ; E 4,1 8ml 2 l l
We see that 4,1 3,0 and that both states have the same energy. Since the only difference in the wave functions is the algebraic sign, any physical measurement predictions, which depend on the absolute square of the wave function, would be the same. 35. We choose the zero of potential energy to be at the bottom of the well. Thus in free space, outside the well, the potential is U0 48eV. Thus, the total energy of the electron is E K U0 180eV 48eV 228 eV. We assume both energies are to the nearest 1 eV. (a) In free space (regions I and III in the diagram), the kinetic energy of the particle is 180 eV. Use that to find the momentum and then the wavelength. 2 K p p 2mK h 2m 6.631034 J s h 9.15 1011m 1/ 2 31 19 2 9.1110 kg180eV1.60 10 J eV (b) Over the well (region II in the diagram), the kinetic energy is 228 eV. h 8.131011m 1/ 2 2mK 2 9.1110 31 kg 228eV 1.60 10 19 J eV
(c) The diagram is qualitatively the same as Fig. 38–14, reproduced here. Notice that the wavelength is longer when the particle is not over the well, and shorter when the particle is over the well.
36. We pattern our answer after Fig. 38–13.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
37. (a) We assume that the lowest three states are bound in the well, so that E U0 . See the diagrams for the proposed wave functions. Note that, in the well, the wave functions are similar to those for the infinite well. Outside the well, for x l , the wave functions are drawn with an exponential decay, similar to the right side of Fig. 38–13. (b) In the region x 0, 0. In the well, with 0 x l , the wave function is similar to that of a free particle or a particle in an infinite potential well, since U = 0. 2mE Thus Asin k x B cos k x , where k .
Instructor Solutions Manual
3
2 1 x0
xl
2m U0 E .
In the region x l , DeGx , where G
38. We will consider the “left” wall of the square well, using Fig. 38–12, and assume that our answer is applicable at either wall due to the symmetry of the potential well. As in Section 38–9, let CeGx for x 0, with G given in Eq. 38–16. Since the wave function must be continuous, x 0 C. The energy of the electron is to be its ground state energy, approximated by Eq. 38–13 for the infinite well. If that energy is much less than the depth of the well, our approximation will be reasonable. We want to find the distance x for which x 0.010 0. E
6.631034 J s 14.73eV 8ml 2 89.111031 kg0.16 109 m 1.60 1019 J eV
x 0
2
h2
2
CeGx C
e Gx 0.010 x
ln0.010 ln 0.010 G 2m U0 E
1.055 10 J s ln 0.010 x 2.11011 m 31 19 29.1110 kg1800eV 14.73eV1.60 10 J eV 34
The wave function will be 1.0% of its value at the walls at a distance of 2.11011 m 0.021nm from the walls. 39. We use Eqs. 38–17a and 38–17b.
Chapter 38
Quantum Mechanics
; G 2mU0 E ln T 2 2l 2l 2 34 2 ln 0.00080 1.055 10 J s ln T 14eV 9 E U0 31 2m 2l 2 9.1110 kg 2 0.85 10 m 14eV 0.672eV 13.328eV 13eV G
T e2Gl
ln T
2
119 1.60 10 J eV
40. We use Eqs. 38–17a and 38–17b to solve for the particle’s energy. ln T G ; G 2m U0 E ln T T e2Gl 2 2l 2l 2 34 2 ln 0.010 1.055 10 J s E U0 2 ln T 18eV 9 31 2m 2l 2 9.1110 kg 2 0.48 10 m
2
119 1.60 10 J eV
18eV 0.879eV 17.12eV 17 eV 41. We use Eqs. 38–17a and 38–17b to solve for the transmission coefficient, which can be interpreted in terms of probability. The energy difference U0 E 25.0 MeV 7.5MeV 17.5MeV. For the mass of the helium nucleus, we take the mass of 2 protons and 2 neutrons, ignoring the (small) binding energy. Proton: G
2m U0 E
21.67 1027 kg17.5 MeV1.60 1013 J MeV
1.055 10 J s e 2Gl 29.166 10 m 3.6 10 m 6.600 ; T
9.166 1014 m1
34
14
1
15
2Gl
proton
e6.600 1.4 103
Helium: Mass = 2m proton 2m neutron 21.6731027 kg 2 1.675 1027 kg 6.70 1027 kg. G
2m U0 E
26.70 1027 kg17.5 MeV1.60 1013 J MeV
1.055 10 J s 2Gl 21.836 10 m 3.6 10 m 13.219 ; T e
1.836 1015 m1
34
15
1
15
He
2Gl
e13.219 1.8 106
It is much less likely that the heavier particle will tunnel through the barrier. 42. (a) The probability of the electron passing through the barrier is given by Eqs. 38–17a and 38–17b. The energy difference is U0 E 9.2eV 7.2eV 2.0eV . T e 2l
2l
2mU0 E
1019 J eV 2m U0 E 29.111031 kg2.0eV1.60 9 3.619 20.25 10 m 1.055 1034 J s
T e3.619 2.681102 2.7% (b) The probability of reflecting is the probability of NOT tunneling, and so it is 97.3% .
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
43. The transmitted current is caused by protons that tunnel through the barrier. Since current is directly proportional to the number of charges moving, the transmitted current is the incident current times the transmission coefficient. We use Eqs. 38–17a and 38–17b. T e 2l
2l
2mU0 E
2mU0 E 2 2.8 1013 m
1.055 1034 J s
I I 0T 1.5mA e122.71 7.7 10
T e122.71
122.71
13 J eV 2 1.67 1027 kg 1.0 MeV 1.60 10 54
mA
This is so incredibly small that it could be considered 0 – no protons will get through the barrier. Note that using more digits on the constants will produce a slightly different answer.
44. The transmission coefficient is given by Eqs. 38–17a and 38–17b. (a) The barrier height is now 1.0370eV 72.1eV, so the energy difference is now 22.1 eV/ 2l
2m U0 E
20.10 109 m
1.055 10
34
J s
4.812 2l
2mU E
T e
e4.812 8.132 103 ;
T
8.132 103
0.81 19% decrease 0.010 (b) The barrier width is now 1.030.10 nm, 0.103nm, and the energy difference is again 20 eV. T0
2l
2m U0 E
20.10310 m 9
29.111031 kg20eV1.60 1019 J eV
1.055 10
34
J s
4.715 2 l
2mU0 E
e4.715 8.960 103 T e 3 T 8.960 10 89.6 10% decrease 2 sig fig T0 0.010 45. We assume that the wave function inside the barrier is given by a decaying exponential, so x AeGx .
T
x l
2
x 0
2
Ae e Gl
2
2Gl
A2
46. (a) We assume that the transmission coefficient for a single barrier is small; Eq. 38–17a applies, and so T e2Gl . With two separate barriers, there can be multiple reflections at each barrier,
Chapter 38
Quantum Mechanics
assuming the possibility that the electron may pass through the second barrier on the first time it encounters that barrier, or the second time, or the third, etc. (assuming it doesn’t pass through the first barrier having reflected from the second barrier initially). Note that the separation distance between the barriers does not enter the calculation. Consider the diagram. We take the probability of the electron encountering the first barrier as “1”. Then the probability that the electron passes through the first barrier is T, and thus the probability that the electron encounters the second barrier is also T. Then, the probability that the electron passes T 2 R2 through the second barrier on this encounter is T 2 , and the probability that the electron reflects from the second barrier is TR. If the electron has reflected from the second barrier, then the probability that it reflects upon reaching the first barrier again is TR R TR2 . The probability that this “twice reflected” electron will reflect again (from the second barrier) is
T 2 R4
T 2 R6
TR2 R TR3 , and the probability that this “twice reflected” electron will pass through the second barrier is TR2 T T 2R2 .
So, the probability of the electron passing through the second barrier on either the first or second encounter is T 2 T 2R2 . From the diagram, we can see that with each pair of reflections, the “additional” probability of passing through is R2 times the previous probability of passing through. Thus, the total probability of passing through the two separated barriers is T 2 times an infinite 2 geometric sum, with a first term of 1 and a common ratio of R . Note that R < 1, T < 1, and R = 1 – T. T2 2 2 2 2 4 2 6 2 2 n 2 1 R 2 1 0 2 P T T R T R T R T R T 2 T 1 R 1 R 2T 1 12 T n0
Note that if T << 1 (a small transmission possibility), then the probability of passing through both barriers is 12 T , half the probability of going through just the first barrier. (b)
Now there is only one transmission, with twice the thickness, so we use Eq. 38–17a again. e2G2l e4Gl e2Gl e2Gl T 2 T double
If T is small, 12 T T 2 , and the transmission probability is greater through two thinner barriers than one thick barrier. 47. Please note – because the final answer is very sensitive to the actual numbers used, we have used more significant figures in the constants than we typically do in these solutions. (a) We assume that the alpha particle is at the outer edge of the nucleus. The potential energy is electrostatic potential energy and is found from Eq. 23–10. 2 2901.602 1019 C 8.988 109 N m2 C2 1 Q1Q2 8 1015 m U surface 1MeV13 4 rsurface of nucleus 1.602 10 J 32.40 MeV 30 MeV
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(b) The kinetic energy of the free alpha particle is also its total energy. Since the free alpha has 4 MeV, by conservation of energy the alpha particle had 4 MeV of potential energy at the exit from the barrier. See the diagram, a copy of Fig. 38–17, modified to show U = 0 inside the barrier, and stated in part (c). 1 QQ 1 Q Q rsurface of nucleus 1 2 1 2 U exit r r r 4 exit from 4 surface of exit from barrier
nucleus
barrier
rsurface of U surface nucleus rexit from
barrier
rexit from
Usurface
barrier
Uexit
rsurface of
32.40 MeV
8fm 64.80fm
4 MeV
nucleus
r rexit from rsurface of 64.80fm 8fm 56.80fm 57 fm barrier
nucleus
(c) We now model the barrier as being rectangular, with a
width of rbarrier 0.3356.80fm 18.74fm. The barrier exists at both “edges” of the nucleus, if we imagine the alpha particle as moving along a diameter inside the nucleus, from wall to wall. See the diagram (not to scale). We calculate the speed of the alpha particle and use that to find the frequency of collision with the barrier.
“out”
“in” 18.7 fm
“out” 16 fm
18.7 fm
E K 21 mv2 2E v m
2 4 MeV1.602 1013 J MeV 41.67310 kg 27
1.3839 107 m s 1.4 107 m s
Note that the speed of the alpha is less than 5% of the speed of light, so we can treat the alpha without using relativistic concepts. The time between collisions is the diameter of the nucleus (16 fm) divided by the speed of the alpha particles. The frequency of collision is the reciprocal of the time between collisions. v 1.3839 107 m s 8.6494 1020 collisions s 8.6 1020 collisions s f d 16 1015 m If we multiply this collision frequency times the probability of tunneling, T, then we will have an estimate of “effective” collisions/s, or in other words, the decays/s. The reciprocal of this effective frequency is an estimate of the time the alpha spends inside the nucleus – the life of the uranium nucleus. 2m U0 E 241.6731027 kg28.40 MeV1.602 1013 J MeV G 1.0551034 J s 2.3390 1015 m1 2.34 1015 m1
2 2.3391015 m1
T e2Gl e
18.7410 m 8.4579 1039 15
Chapter 38
Quantum Mechanics
Lifetime
1
1
fT 8.6494 10 collisions s8.4579 1039 1yr 1.3669 1017 s 4.33109 yr 4.3109 yr 3.156 107 s While the agreement with experiment is quite good for this model, the calculated answer is highly dependent on the width of the barrier used. For instance, if we used a factor of 0.32 instead of 0.33, the lifetime calculation gives about 3.0 108 yr , and if a factor of 0.34 is used, the calculation gives about 6.2 1010 yr . The textbook points this out in Example 38–11. 20
48. We find the lifetime of the particle from Eq. 38–2.
t
E
2.5GeV1.60 10
10
2.6 1025 s.
J GeV
49. We find the wavelength of the protons from their kinetic energy, and then use the two-slit interference formulas from Chapter 34, with a small angle approximation. If the protons were accelerated by a 650-volt potential difference, then they will have 650 eV of kinetic energy, and thus be non-relativistic. h h ; d sin m, m 1, 2, ... ; y l tan p 2m0 K ml , m 1, 2, ... sin tan m y y d d l 6.631034 J s15m l hl y d d 2m0 K 8.0 104 m 21.67 1027 kg650eV1.60 1019 J eV 2.1108 m 21nm 50. We assume that the particles are not relativistic. Conservation of energy is used to find the speed of each particle. That speed can then be used to find the momentum and finally the de Broglie wavelength. We let the magnitude of the accelerating potential difference be V. h h h h U K eV 1 mv2 v 2eV ; x initial final 2 p mv m 2eV 2meV m m h 2h xp p 2 x h 2h 27 pproton xproton xelectron 2melectron eV 1.67 10 kg 42.8 2h h p x 9.111031 kg electron
xelectron
proton
2mproton eV
51. The energy levels for a particle in an infinite potential well are given by Eq. 38–13. The wave 2 functions are given by Eq. 38–14. with A . l
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
6.631034 J s E1 32.90 MeV 33MeV 8ml 81.67 1027 kg2.5 1015 m 1.60 1013 J MeV 2
h2
(a)
Instructor Solutions Manual
2
2
E 22 E 4 32.90 MeV 130 MeV ; E 32 E 9 32.90 MeV 300 MeV 2
1
3
2 sig. fig.
1
E4 42 E1 1632.90 MeV 530 MeV (b) 2 sin n x n
l
l
sin
x 2.5 10 m 2.8 107 m1/ 2 sin 1.3 1015 m1 x 2.5 1015 m 2 sin x 2 2.5 1015 m 2.8 107 m1/ 2 sin 2.5 1015 m1 x 2.5 1015 m 3 sin x 3 15 2.5 10 m 2.8 107 m1/ 2 sin 3.8 1015 m1 x 2.5 1015 m 4 sin x 4 15 2.5 10 m 2.8 107 m1/ 2 sin 5.0 1015 m1 x 2.5 1015 m
1
15
(c)
E E2 E1 4E1 E1 3E1 332.90 MeV 98.7 MeV 99 MeV
E
hc
hc E
1.31014 m 13fm
This is in the gamma-ray region of the EM spectrum, as seen in Fig. 31–12. 52. The uncertainty in the energy is found from the lifetime and the uncertainty principle. h ; E hv hc E t 2t h E 2t 400 109 m 2.12 10 8 2 108 hc E 2ct hc hc hc E E dE d E E 2 2 E 2 108 The wavelength uncertainty is the absolute value of this expression, and thus
53. We assume the alpha particle is in the ground state. The energy is given by Eq. 38–13. 2 6.631034 J s h2 E1 0.3343MeV 8ml 2 84 1.67 1027 kg 1.24 1014 m 2 1.60 1013 J MeV
0.334 MeV The speed can be found from the kinetic energy. The alpha is non-relativistic.
Chapter 38
Quantum Mechanics
E1
h2
1 2
2
2 mv
v
h
6.6310
34
Js
2ml
8ml The result justifies the assumption that the alpha is non-relativistic.
4.00 106 m s
54. We use Eq. 37–10, Bohr’s quantum condition, with n = 1 for the ground state. h mvrn n mvr1 mv p 2 r1 xp
x
The uncertainty in position is comparable to the Bohr radius. 2 d 2 E. 55. The time-independent Schrödinger equation with U = 0 is 2m dx2 2
2
2 2 2 2 d2 k d 2 2 2 2m dx 2m dx 2m 2m We see that the function solves the Schrödinger equation.
2
56. The wave functions for the particle in the infinite well are n
2
2mE E 2m
sin
l
n as derived in Section x , l
38–8. A table of integrals was consulted to find x2 sin2 ax dx. l n l n 2 dx 2 x2 sin2 x dx x x2 x2 n 2 dx x2 sin l 0 l l 0 x cos 2ax x3 x2 1 . 3 sin 2ax Note that x 2 sin2 ax dx 4a2 6 4a 8a l 2n x l 1 2n 2 x x x cos 3 2 2 2 n sin x x dx 2 l 2 x2 x sin l 6 3 l l n n l n 0 4 8 4 l l l 0
See the adjacent graph.
0.34 0.33 0.32 0.31 0.30 0.29 0.28 0
2
4
6
8
10
n
12
14
16
18
20
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57. The de Broglie h hwavelength is h determined by34Eq. 37–3. 6.6310 J s 35 v 2.07 10 m s 2.11035 m s m 64.0 kg 0.50 m p mv
Since is comparable to the width of a typical doorway, yes , you would notice diffraction effects. However, assuming that walking through the doorway requires travel through a distance of 0.2 m, the time t required is huge. d vt d t 0.20 m 9.7 1034 s 31026 yr v 2.07 1035 s
58. (a) See the diagram. 2 (b) We use the solution x AeBx in the Schrödinger equation. 2 d AeBx ; 2 ABxeBx2 dx 2 d 2 ABe Bx 2 ABx 2Bxe Bx 2 dx
2
2
d 222Bx2 1 ABe2Bx
U (x )
This is many orders of magnitude larger than the approximate age of the universe 11010 yr .
2
x
2
2
Bx
2
1
Bx
2
2
Bx2
U EAe 22Bx 1 ABe 2 Cx Ae 2m dx2 2m 2B 2 2B2 2 1 E 2C x 0 m m 2 2 2 This is a solution if B E and 1 C 2 B . Solve these two equations for E in terms of C, 2 m m and let C m. 1
C
2
2 2B2
B
m
mC 2
2
; E
B
m
2
m
mC 2
1
C m 1
2
2
B
59. From energy conservation, the speed of a ball after falling a height H, or the speed needed to rise to a height H, is v 2gH . We say that the starting height is H 0 , and so the speed just before the ball hits the ground before the first bounce is v0 2gH0 . After that bounce, the ball rebounds to H1 0.65H0 , and so the speed right after the first bounce, and right before the second bounce, is v1
2g 0.65 H0 . Repeated application of this idea gives the maximum height after n
bounces as H n 0.65 H 0, and the maximum speed after n bounces as v n n
2g 0.65 H 0 . The n
uncertainty principle will come into play in the problem when the maximum speed after a bounce is of the same order as the uncertainty in the speed. We take the maximum height as the uncertainty in the position.
Chapter 38
Quantum Mechanics
mvy py ; ypy
mvy
m 2g 0.65 H 0.65 n H
2m2 g 0.65n H 0 0.65 2n H 2 2
n
0
0
2 2 1.055 1034 J s ln 2 3 ln 2 3 6 2 3n 2 3.0 10 kg 9.80 m s 2.0 m 2m gH 0.65 0 n 105 2 3 2m gH0 3ln0.65 3ln0.65 After about 105 bounces, the uncertainty principle will be important to consider.
60. The difference in measured energies is found from Eq. 37–3. E E E hc hc 2 1 2 1 1 1 6.631034 J s3.00 108 m s 1.670 1021J 9 9 487 10 m 489 10 m The lifetime of the excited state is determined from Eq. 38–2. 1.0551034 J s 6.317 10 14 s 6.32 1014 s Et t 21 E 1.670 10 J This solution assumes that the entire spread in energy is the uncertainty. 61. Find the uncertainty in the position from Eq. 38–1.
x
mv
3.2 10 37 m
62. We model the electrons as being restricted from leaving the surface of the sodium by an energy barrier, similar to Fig. 38–15a. The difference between the barrier’s height and the energy of the electrons is the work function, and so U0 E W0 2.28eV. But quantum mechanically, some electrons will “tunnel” through that barrier without ever being given the work function energy, and thus get outside the barrier, as shown in Fig. 38–15b. This is the tunneling current as indicated in Fig. 38–18. The distance from the sodium surface to the tip of the microscope is the width of the barrier, l . We calculate the transmission probability as a function of barrier width by Eqs. 38–17a and 38–17b. The barrier is then increased to l l , which will lower the transmission probability. 2Gl 2l
2m U0 E
9
2 0.02 10 m
0.3091
2Gl l
2Gl l
2 9.111031kg 2.28eV 1.60 1019 J eV
1.055 1034 J s
T e 2G l 0.3091 e e 0.734 2Gl T0 e The tunneling current is caused by electrons that tunnel through the barrier. Since current is directly proportional to the number of electrons making it through the barrier, any change in the transmission probability is reflected as a proportional change in current. So we see that the change in the transmission probability, which will be reflected as a change in current, is a decrease of 27%. Note that this change is only a fraction of the size of an atom.
T0 e 2Gl ; T e
;
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Let x . l l 1 1 1 1 1 so the region of interest extends from to xcenter 2 l , xmin 2 l 2 x xmax 2 l 2 x . We are to 2
63. The ground state wave function for the particle in the infinite well is
sin
find the largest value of x so that the approximate probability of x 1 l x 2
2x
(from
2
l 2 Example 38–7) is no more than 10% different than 2 sin x dx , the exact probability. We l x l have calculated the integral result two ways. xmax
min
First, we calculated the value of the integral using numeric integration (as described in Appendix C), first finding the number of steps needed between xmin and xmax that gives a stable value. Then we compared the integral to the approximation. (Note that the integral could be evaluated exactly.) To aid in the evaluation of the integral, we make the substitution that u x l . Then the integral u
max
becomes as follows: I
x 1 1 2 l
2sin2 u du . In doing the numeric integrations, we found that
umin 1 1 x 2 l
for any value of x up to l , breaking the numeric integration into 50 steps gave the same answer to 3 significant digits as breaking it into 100 steps. So we did all numeric integrations with 50 steps. We then numerically calculated the integral for values of x l , starting at 0.01, and increasing by steps of 0.01, until we found a 10% difference between the approximation and the numeric integration. This happens at x l 0.34, and so the approximation is good within 10% up to x 0.34l 0.340.10nm 0.034 nm . This is much broader than we might have guessed initially, indicating that the wave function is varying rather slowly over the central region of the potential well. Secondly, we calculated the integral exactly. Here is that calculation.
Chapter 38
Quantum Mechanics 2
1 l 1 x 2 2 2 2 2 dx sin x sin x dx 1 l 1 x l x l x l l l 2 2 2 2 1 2 1 Note that sin ax dx x sin 2ax 2 4a 2 2 1 l x l 2 1 1 l x sin 1 l x 22 4 l 2 l 1 2 l I x sin 2 x 1 2 2 2 l x 4 a l l x l l 12 1 1 l x sin 2 l 22 4 l 2 1 l l x sin l x l x l sin l x 4 4 l 14 4 l l 1 1 x 1 x x x 1 2 1 l sin 1 l sin 1 l 1 x l x x sin 1 sin 1 l 2 l l 2x 2x for various values of . We see that the 10% The following table compares I to I approx l l discrepancy shows up at about the same place as with the numeric integration.
xmax 12 l 12 x
max
1
1
64. (a) See the graph. (b) From the graph (and the spreadsheet used to create the graph), these results are found.
1.0 0.8
Probability
Transmission Reflection
0.6
0.4 0.2 0.0 0
2
4
E /U 0
6
8
10
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T 10% at E U 0 0.146 T 20% at E U0 0.294 T 50% at E U0 0.787 T 80% at E U0 1.56
65. Note that
has SI units of J . And J s
E t has “obvious” SI units of J s .
kg m2 kg m2 s . s2 s 2
xpx has SI units of m kgs m kg sm . kg m2 , and is unitless, so Lz has SI units of kg m2 . Lz has SI units of s s 66. The duration of shortest pulse that can be produced is found from the uncertainty principle for energy and time, converted to relate wavelength to time. The wavelength is found from Eq. 37–3, and then the derivative of that relationship is used to relate to t . We also take an absolute value to eliminate the negative sign from the differentiation. hc hc dE hc hc E hf 2 E E t 2 t d 2 2 9 2 704.110 m t 2.63 1012 s 3 1012 s 2c 2 3.00 108 m s 0.110 9 m The length of the pulse is the distance that it would travel, at the speed of light, times its time duration. x ct 3.00 108 m s2.63 1012 s 7.89 104 m 7.89 105 nm 8 10 nm 5
67. The minimum speed corresponds to the lowest energy state (the ground state). The energy is given by Eq. 38–13. 2 h 2 6.631034 J s Emin h 1 mvmin vmin 2.9 1031 m s 2 2 3 2 8ml 2ml 215.0 10 kg7.5 10 m We can check this result to see if it satisfies the uncertainty principle, Eq. 38–1. 1.055 1034 J s x p x m v v 9.4 10 x x x mx
The speed has to be at least as big as its uncertainty, so the uncertainty principle is satisfied. 68. The uncertainty in the electron position is x r1 . The minimum uncertainty in the velocity, v , can be found by solving Eq. 38–1, where the uncertainty in the momentum is the product of the electron mass and the uncertainty in the velocity. xp min r1 mv v
mr1
1.055 1034 J s
9.11 10 kg0.529 10 m 31
10
2.19 10 m/s 6
Chapter 38
Quantum Mechanics
69. (a) To check that the wave function is normalized, we calculate
x dx. 2
x dx 1 x e bx dx 2 xe bx dx 2 b2 e bx 0 1 1 bb b2 2 b2
2
2
22
2 2
2
0
0
We see that the function is normalized, independent of the value of b. 2 (b) The most probable position is that for which x is maximized. That point can be found by solving
d x
2
0 for x. Since we are only considering
x 0, we need not use the absolute
dx
value signs in the function. 2 x bx2 x2 d 2 e 2x 2x2 2 x2 b 2 2 1 x d x dx b2 e b b2 b 2 e b 0 1 2 dx b b 1.0 nm 0.71nm x 2 2 This value for x maximizes the function because the function must be positive, and the function is 0 at x 0 and x . Thus, this single local extreme point must be a maximum. (c) To find the probability, we integrate the probability density function between the given limits. 2 0.25 nm 0.25 nm 0.25 nm x 1 e bx 1 e0.25 1 P x dx e bx dx 0.030 b2 2 2 2 0 0 0
2
22
2
70. (a) We assume the pencil is a uniform rod, and that it makes an angle of with the vertical. If the bottom point is fixed, then the torque due to its weight about the bottom point will cause an angular acceleration. See the diagram. d 2 l 1 2 1 I mg l sin ml 2 3 dt2 From the equation, if the pencil is exactly upright, so that 0, then the mg angular acceleration will be exactly 0, and the pencil will remain stationary. But, according to the uncertainty principle (as expressed near the end of Section 38–4), the angle cannot be known with 0 uncertainty. Let the z axis be coming out of the page. Lz
Thus, the pencil cannot have exactly 0, and so there will be a torque and hence rotation. (b) For the initial part of the motion, the angle will be very small, and so the differential equation 3g d 2 can be expressed as . The solutions to this differential equation are of the form dt2 2l 39.80 m s2 3g Aekt Bekt , where k 9.037s1 . Since the angle will be 2l 2 0.18m increasing in time, we ignore the second term, which decreases in time. Thus Aekt , with d kt t 0 0 A. The angular velocity of the pencil is approximated as kAe , and dt
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the initial angular velocity is 0 kA. We take the initial position and the initial angular velocity as their smallest possible values, which are their uncertainties – the magnitude of a quantity must be at least as big as its uncertainties. Apply the uncertainty principle in angular form. Lz I 13 ml 2 kA A 31 ml 2kA2 A
3
3 1.055 1034 J s
4.650 1016 rad
5.0 103 kg0.18m 9.037 s1 kA 9.037 s1 4.650 1016 rad 4.202 1015 rad s ml 2k
2
0
0
With this initial position and initial angular velocity, we can then perform a numeric integration to find the time when the angle is 2 rad. For a step size of 0.01 s, the time of fall is about 4.05 s. For a step size of 0.001 s, the time of fall is about 3.97 s. This is only about a 2% change in the final result, so the time is quite stable around 4 s. Even changing the starting angle to a value 100 times bigger than that above (so 0 3.930 1014 rad ) still gives a time of fall of 3.46 s. So, within a factor of 2, we estimate the time of fall as 4 seconds. Note that if the solution of the approximate differential equation is used, Aekt , we get the following time of fall. 1 1 t ln max 16 3.96s. 1 ln 4.650 10 max 9.037s k A
CHAPTER 39: Quantum Mechanics of Atoms Responses to Questions 1.
(a) The quantum-mechanical theory retained the aspect that electrons only exist in discrete states of definite energy levels. It also retained the feature that an electron absorbs or emits a photon when it changes from one energy state to another. (b) In the Bohr model, the electrons moved in circular orbits around the nucleus. In the QM model, the electrons occupy orbitals about the nucleus and do not have exact, well-defined orbits.
2.
The quantity is maximum at r = 0 because of its dependence on the factor e r r . In the ground 2
0
state, the electron is expected to be found near the nucleus. The radial probability density 4r gives the probability of finding the electron in a thin spherical shell located at r. Since r = 0 is at the center of the nucleus, the radial probability density is zero at the center. 2
2
3.
The quantum-mechanical model predicts that the electron spends more time near the nucleus. In the Bohr model, the electron in the ground state is in a fixed orbit of definite radius. The electron cannot come any closer to the nucleus than that distance. In the quantum-mechanical model, the most probable location of the electron is the Bohr radius, but it can also be found closer to the nucleus (and farther away). See Fig. 39–1 for a visual representation of this question.
4.
As the number of electrons goes up, the number of protons in the nucleus increases, which increases the attraction of the electrons to the center of the atom. Even though the outer electrons are partially screened from the increased nuclear charge by the inner electrons, they are all pulled closer to the more positive nucleus. Also, more states are available in the upper shells to accommodate many more electrons at approximately the same radius.
5.
Because the nuclei of hydrogen and helium are different, the energy levels of the atoms are different. The presence of the second electron in helium will also affect its energy levels. If the energy levels are different, then the energy difference between the levels will be different and the spectra will be different.
6.
The two levels have different orbital quantum numbers. The orbital quantum number of the upper level is 𝓁 = 2. This results in five different possible values of 𝑚ℓ (–2, –1, 0, 1, and 2), so the energy level is split into five separate levels in the presence of a magnetic field. The lower level shown has an orbital quantum number of 𝓁 = 1, so only three different values of 𝑚ℓ (–1, 0, and 1) are possible. Therefore, the energy level is split into only three separate levels.
7.
The Zeeman effect is the splitting of an energy level in the presence of a magnetic field. In the reference frame of the electron, the nucleus orbits the electron. The “internal” Zeeman effect, as seen in sodium, is caused by the magnetic field produced by the “orbiting” nucleus.
8.
Configuration (a) is not allowed. Only six electrons are allowed in the 2p state. Configurations (b) and (c) are allowed for atoms in an excited state.
9.
The complete electron configuration for a uranium atom is as follows. 1s2 2s2 2p6 3s2 3p6 3d 10 4s2 4p6 4d 10 4f 14 5s2 5p6 5d 10 5f 3 6s2 6p6 6d 1 7s2
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10. For noble gases, the electrons fill all available states in the outermost shell. Helium has two electrons, which fill the 1s shell. For the other noble gases, there are six electrons in the p subshells, filling each of those shells. For each of these gases, the next electron would have to be added to an s subshell at much higher energy. Since the noble gases have a filled shell, they do not react chemically with other elements. Lithium, which has three electrons and is the first alkali metal, has its third electron in the 2s shell. This shell has significantly higher energy than the 1s shell. All of the alkali metals have one electron in their outer s subshell and therefore react well with halogens. Halogens all have five electrons in their outer p subshell. They react well with atoms that have a donor electron in their outer shell, like the alkali metals. 11. (a) (b) (c) (d)
Group II. Outer electrons are 3s2: magnesium. Group VII. Outer electrons are 2s2 2p5: fluorine. Group VIII. Outer electrons are 3s2 3p6: argon. Group I. Outer electron is 4s1: potassium.
12. The periodicity of the periodic table depends on the number and arrangement of the electrons in the atom. The Pauli exclusion principle requires that each electron have distinct values for the principal quantum number, angular orbital quantum number, magnetic quantum number, and spin quantum number. As electrons are added to the atom, the electrons fill the lowest available energy states. Atoms with the same number of electrons in the outer shells have similar properties, which accounts for the periodicity of the periodic table. The quantization of the angular momentum determines the number of angular momentum states allowed in each shell. These are represented by the letters s, p, d, f, etc. The magnitude of the angular momentum also limits the number of magnetic quantum states available for each orbital angular state. Finally, the electron spin allows for two electrons (one spin up and one spin down) to exist in each magnetic quantum state. Thus, the Pauli exclusion principle, quantization of angular momentum, direction of angular momentum, and spin all play a role in determining the periodicity of the periodic table. 13. If there were no electron spin, then according to the Pauli exclusion principle, s-subshells would be filled with one electron, p-subshells with three electrons, and d-subshells with five electrons. The first 20 elements of the periodic table would look like the following: H 1 1s1 He 2 2s1 C 6 3s1 Ne 10 4s1 K 19 5s1
Na 11 3d1 Ca 20 4d1
Mg 12 3d2
Al 13 3d3
Si 14 3d4
P 15 3d5
Li 3 2p1 N 7 3p1 S 16 4p1
Be 4 2p2 O 8 3p2 Cl 17 4p2
B 5 2p3 F 9 3p3 Ar 18 4p3
14. Neon is a “closed shell” atom—a noble gas. All of its shells are completely full and spherically symmetric, so its electrons are not readily ionized. Sodium is a Group I atom (an alkali metal) that has one lone electron in the outermost shell, and this electron spends a considerable amount of time far away and shielded from the nucleus. The result is that sodium can be ionized much easier than neon, as demonstrated by their two ionization energies.
Chapter 39
Quantum Mechanics of Atoms
15. Both chlorine and iodine are in the same column of the periodic table (group VII). Thus, both of their outer electron configurations are p5 (5 electrons in their outer p shell), which makes them both halogens. Halogens are one electron short of a filled shell, so both of their outer electron orbit shapes are similar and can readily accept an additional electron from another atom. This makes their chemical reaction properties extremely similar. 16. Both sodium and potassium are in the same column of the periodic table (group I). Thus, both of their outer electron configurations are s1 (1 electron in their outer s shell), which makes them both alkali metals. Alkali metals have one lone electron in their outermost shell, so both of their outer electron orbit shapes are similar, and they can readily share this electron with another atom (especially since this outermost electron spends a considerable amount of time very far away from the nucleus). This makes their chemical reaction properties extremely similar. 17. Noble gases are nonreactive because they have completely filled shells or subshells. In that configuration, other electrons are not attracted nor are electrons readily lost. Thus, they do not react (share or exchange electrons) with other elements. Alkali metals all have a single outer s electron. This outer electron is easily removed and can spend much of its time around another atom, forming a bond. Thus, alkali metals are highly reactive. 18. Rare earth elements have similar chemical properties because the electrons in the filled 6s or 7ssubshells serve as the valence electrons for all these elements. They all have partially filled inner fsubshells, which are very close together in energy. The different numbers of electrons in the fsubshells have little effect on the chemical properties of these elements. 19. When we use the Bohr theory to calculate the X-ray line wavelengths, we estimate the nuclear charge seen by the transitioning electron as Z – 1, assuming that the second electron in the ground state is partially shielding the nuclear charge. This is only an estimate, so we do not expect the calculated wavelengths to agree exactly with the measured values. 20. In helium and other complex atoms, electrons interact with other electrons in addition to their interactions with the nucleus. The Bohr theory only works well for atoms that have a single outer electron in an s state. X-ray emissions generally involve transitions to the 1s or 2s states. In these cases, the Bohr theory can be modified to correct for screening from a second electron by using the factor Z – 1 for the nuclear charge and can yield good estimates of the transition energies. Transitions involving outer electrons in more complex atoms will be affected by additional complex screening effects and cannot be adequately described by the Bohr theory. 21. The continuous portion of the X-ray spectrum is due to the “bremsstrahlung” radiation. An incoming electron gives up energy in the collision and emits light. Electrons can give up all or part of their kinetic energy. The maximum amount of energy an electron can give up is its total amount of kinetic energy. In the photon description of light, the maximum electron kinetic energy will correspond to the energy of the shortest wavelength (highest energy) photons that can be produced in the collisions. The result is the existence of a “cut-off” wavelength in the X-ray spectrum. An increase in the number of electrons will not change the cut-off wavelength. According to wave theory, an increase in the number of electrons could result in the production of shorter wavelength photons, which is not observed experimentally. 22. To figure out which lines in an X-ray spectrum correspond to which transitions, you would use the Bohr model to estimate the differences in the sizes of the energy level jumps that the falling electrons will make. Then, you just need to match the n = 2 n = 1 energy to the K line, the n = 3 n = 1 energy to the K line, the n = 3 n = 2 jump energy to the L line, and so on.
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23. The characteristic X-ray spectra occur when inner electrons are knocked out of their shells. X-rays are high-energy photons emitted when other electrons fall to replace the knocked-out electrons. Because the shells involved are close to the nucleus, Z will have a direct influence on the energies. The visible spectral lines due to transitions between upper levels have energies less influenced by Z because the inner electrons shield the outer electrons from the nuclear charge. 24. The difference in energy between n = 1 and n = 2 levels is much bigger than that between any other combination of energy levels. Thus, electron transitions between the n = 1 and n = 2 levels produce photons of higher energy and frequency, which means these photons have shorter wavelengths (since frequency and wavelength are inversely proportional to each other). 25. The electron has a negative charge. 26. Consider a silver atom in its ground state for which the entire magnetic moment is due to the spin of only one of its electrons. In a uniform magnetic field, the dipole will experience a torque that would tend to align it with the field. In a non-uniform field, each pole of the dipole will experience a force of different magnitude. Consequently, the dipole will experience a net force that varies with the spatial orientation of the dipole. The Stern-Gerlach experiment provided the first evidence of space quantization since it clearly indicated that there are two opposite spin orientations for the outermost electron in the silver atom. 27. Spontaneous emission occurs when an electron is in an excited state of an atom and it spontaneously (with no external stimulus) drops back down to a lower energy level. To do this, it emits a photon to carry away the excess energy. Stimulated emission occurs when an electron is in an excited state of an atom, but a photon strikes the atom and causes or stimulates the electron to make its transition to a lower energy level sooner than it would have done so spontaneously. The stimulating photon has to have the same energy as the difference in energy levels of the transition. The result is two photons of the same frequency—the original one and the one due to the emission. 28. No, the intensity of light from a laser beam does not drop off as the inverse square of the distance. Laser light is much closer to being a plane wave rather than a spherically symmetric wave, and so its intensity is nearly constant along the entire beam. It does spread slightly over long distances, so it is not a perfect plane wave. 29. Different: Laser light is coherent (all of the photons are in the same phase), and ordinary light is not coherent (the photons have random phases relative to each other). Laser light is nearly a perfect plane wave, while ordinary light is spherically symmetric (which means that the intensity of laser light remains nearly constant as it moves away from the source, while the intensity of ordinary light drops off as 1/r2). Laser light is always monochromatic, while ordinary light can be monochromatic, but it usually is not. Similar: Both travel at c, and both display the wave–particle duality of photons. Both are created when electrons fall to lower energy levels and emit photons. Both are electromagnetic waves. 30. Since laser light is a plane wave, its intensity remains approximately constant with distance. The light produced by a street lamp spreads out with an intensity that decreases as 1/r2. Thus, at a sufficient distance, the laser light will be more intense than the light from a street lamp.
Chapter 39
Quantum Mechanics of Atoms
Responses to MisConceptual Questions 1.
(b) A common error is to add up the coefficients (principal quantum numbers), which would give an answer of 15. However, the exponents represent the number of electrons in each subshell. In this case, there are 19 electrons.
2.
(b) The orbital quantum numbers are represented by the subshell letters: s = 0 and p =1. Since only s and p subshells are present, only the orbital quantum numbers 0 and 1 are in the electron configuration.
3.
(c) A common misconception is that all of the electrons have their lowest quantum numbers when in the ground state. However, due to the Pauli exclusion principle, no two electrons can have the same quantum numbers. The ground state of an atom is when each electron occupies the lowest energy state available to it without violating the exclusion principle.
4.
(e) The Pauli exclusion principle applies to all electrons within the same atom. No two electrons in an atom can have the same set of four quantum numbers. Electrons from different atoms can have the same set of quantum numbers (for example, two separate atoms in the same container, or two atoms from different parts of the periodic table).
5.
(a) In an atom, multiple electrons will be in the same shell and can therefore have the same value of n. Electrons in different shells can have the same value of 𝑚𝓁. Electrons in different atoms can also have the same set of four quantum numbers. The Pauli exclusion principle does not allow electrons within the same atom to have the same identical set of four quantum numbers.
6.
(e) The Heisenberg uncertainty principle prohibits the exact measurement of position and velocity at the same time; the product of their uncertainties will always be greater than a constant value.
7.
(d) Laser light is monochromatic (all photons have nearly identical frequencies), is coherent (all photons have the same phase), moves as a beam, and is created by a population inversion. A laser beam usually is intense, but it does not have to be brighter than other sources of light.
8.
(d) As stated in Table 39–3, the letters s, p, d, f, g, … designate the orbital quantum number, 𝓁. They do not summarize all of the quantum numbers for electrons, and they do not designate the principal quantum number. Also, they apply to electrons in any element, not just hydrogen.
9.
(c) The “forbidden transitions” are discussed on page 1207. On that page, it is mentioned that such transitions can occur but with low probability as compared to the “allowed transitions”.
10. (c) From Eq. 39–2, we see that En states as a function of n.
E1 n2
. Consider the energy difference between two adjacent
n2 n 12 2n 1 E1 E1 1 2 2 2 E1 2 2 E En1 En E 2 n n 1 n n 1 n 1 n 2n 1 As n increases, the value of 2 n 12 decreases since the denominator grows faster than the n numerator, and so E decreases.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
Solutions to Problems 1.
The value of 𝓁 ranges from 0 to n 1. Thus, for n = 6, ℓ = 0,1,2,3,4,5 .
2.
The value of ml can range from to . Thus, for ℓ = 3, 𝑚ℓ = −3, −2, −1,0,1,2,3 . The possible values of ms are
3.
.
The value of 𝑚ℓ can range from −ℓ to +ℓ, so we have 𝑚ℓ = −4, −3, −2, −1,0,1,2,3,4 . The value of ℓ can range from 0 to n 1. Thus, we have n 5. .
There are two values of ms : 4.
The value of 𝑚ℓ can range from −ℓ to +ℓ, so ℓ ≥ 5 . The value of ℓ can range from 0 to n 1. Thus, 𝑛 ≥ ℓ + 1(min𝑖𝑚𝑢𝑚6) . There are two values of ms :
5.
s
2
2 .
The value of ℓ ranges from 0 to n 1. Thus for n 4, ℓ = 0,1,2,3. For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ, or 2ℓ + 1 values. For each 𝑚ℓ, there are 2 values of ms . Thus, the number of states for each ℓ is 2(2ℓ + 1). The number of states is therefore given by the following. N 2 0 1 2 2 1 2 4 1 2 6 1 32states .
We start with ℓ = 0, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ, 𝑚𝑠). (4, 0, 0, 12 ), (4, 0, 0, 1 ), (4, 1, –1, 12 ), (4, 1, –1, 1 ), (4, 1, 0, 12), (4, 1, 0, 1 ), (4, 1, 1, 12 ), (4, 1, 1, 1 ), (4, 2, – 2, 12 ), (4, 2, – 2, 1 ), (4, 2, –1, 12), (4, 2, –1, 1 ), (4, 2, 0, 12 ), (4, 2, 0, 1 ), (4, 2, 1, 12 ), (4, 2, 1, 1 ), (4, 2, 2, 12), (4, 2, 2, 1 ), (4, 3, – 3, 12 ), (4, 3, – 3, 1 ), (4, 3, – 2, 12 ), (4, 3, – 2, 1 ), (4, 3, 1, 12), (4, 3, – 1, 1 ), (4, 3, 0, 12 ), (4, 3, 0, 1 ), (4, 3, 1, 12 ), (4, 3, 1, 1 ), (4, 3, 2, 12), (4, 3, 2, 1 ), (4, 3, 3, 12 ), (4, 3, 3, 1 ). 6.
The magnitude of the angular momentum depends only on l , which is 4. 𝐿 = √ℓ(ℓ + 1)ℏ = √20ℏ = √20(1.055 × 10−34𝐽𝑠) = 4.718 × 10−34𝐽𝑠
7.
8.
(a) The principal quantum number is n 8. (b) The energy of the state is found from Eq. 39–2. 13.6eV 13.6eV E7 0.213eV n2 82 (c) The “g” subshell has ℓ = 4 . The magnitude of the angular momentum depends on l only: 𝐿 = ℏ√ℓ(ℓ + 1) = √20ℏ = √20(1.055 × 10−34𝐽𝑠) = 4.718 × 10−34𝐽𝑠 (d) For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ: 𝑚ℓ = −4, −3, −2, −1,0,1,2,3,4 . Photon emission means a jump to a lower state, which has a lower principal quantum number, so for the final state, n = 1, 2, 3, or 4. For a d subshell, ℓ = 2, and because 𝛥ℓ = ±1, the new value of ℓ must be 1 or 3. (a) ℓ = 1 corresponds to a p subshell, and ℓ = 3 corresponds to an f subshell. Keeping in mind that
Chapter 39
Quantum Mechanics of Atoms
0 ≤ ℓ ≤ 𝑛 − 1, we find the following possible destination states: 2 p,3p,4 p,4 f . Using the notation requested in the problem, the destination states are 2,1, 3,1, 4,1, 4,3. (b) In a hydrogen atom, ℓ has no appreciable effect on energy (ignoring fine structure), and so for energy purposes, there are four possible destination states, corresponding to n = 2, 3, and 4. Thus, there are three different photon wavelengths corresponding to three possible changes in energy. 9.
(a) For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ, or 2ℓ + 1 values. For each of these, there are two values of ms. Thus, the total number of states in a subshell is 𝑁 = 2(2ℓ + 1) . 0, 1, 2, 3, 4, 5, and 6, N 2, 6, 10, 14, 18, 22, and 26 , respectively.
(b) For
10. For a given n, 0 ≤ ℓ ≤ 𝑛 − 1. Since for each ℓ the number of possible states is 2(2ℓ + 1), as shown in Problem 9, the number of possible states for a given n is as follows. 𝑛(𝑛−1)
∑𝑛−1 2(2ℓ + 1) = 4 ∑𝑛−1 ℓ + ∑𝑛−1 2 = 4 ( ℓ=0
ℓ=0
ℓ=0
) + 2𝑛 =
2
2𝑛2
The summation formulas can be found (and proven) in a pre-calculus mathematics textbook. 11. We use Eq. 39–3 to find l and Eq. 39–4 to find ml . 𝐿2
2
(6.84×10−34𝐽𝑠)
𝐿 = √ℓ(ℓ + 1)ℏ → ℓ(ℓ + 1) = ℏ2 = (1.055×10−34𝐽𝑠)2 = 42 ℓ2 + ℓ − 42 = 0 → ℓ =
−1 ± √1 − 4(1)(−42) −1 ± 13 = = 6, −7 → ℓ = 6 2 2 𝐿𝑧
2.11 × 10−34𝐽𝑠
𝐿𝑧 = 𝑚ℓℏ → 𝑚ℓ = ℏ = 1.055 × 10−34𝐽𝑠 = 2 Since ℓ = 6, we must have n 7. 12. The state 𝑛 = 2, ℓ = 0 must have 𝑚ℓ = 0, and so the wave function is 200 11
r r 2r 2 e 0 . 3 r 32 r0 0 1
To obtain numerical values, we use r0 0.0529 nm 5.29 10 m . 1 4r 4r0 (a) 2 0 e 2r0 1 7.02 1013 m1.5 200 r 4r0 r0 8 r 32r03 2 1 4r 2 4r0 (b) 200 2 r 4r 32r3 2 r 0 e 2r0 1 3 e4 4.92 1027 m3 8r 0 0 0 0 2 1 2 2 4 4r e4 8 4 2.77 109 m1 (c) P 4 r e r 200 0 3 r 4r0 8 r0 r0
13. The ground state wave function is 100 7
r0 0.0529 nm 5.29 10 m .
1
r
e r / r0 . To obtain numerical values, we use
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(a)
1
100 r 1.5r
3.27 1014 m1.5
r
0
Instructor Solutions Manual
1r e 1.07 10 m (c) P 4 r 4 1.5r 1 e 9 e 8.47 10 m 2 100
(b)
3
29
100
3
2
2
2
r
3
3 0
r 1.5r0
0
r 1.5r0
r0
3
9
3
1
r0
14. To show that the ground-state wave function is normalized, we integrate 100 over all space. Use substitution of variables and 2ran integral from Appendix B–5. 1 2r 2 3 e r0 4r 2 dr ; dV let x r 21 r0 x , dr 12 r0 dx r 100 r 0 0 all 0 2
space
Note that if r 0, x 0 and if r , x . 2 x 2 x 2 2 dV 1 e 2rr0 4r2dr 4 e 1 r x 1 r dx 1 e x dx 1 2! 1 r3 4 0 2 0 2 2 100 r 3 all space
0
0
0 0
0
And so, we see that the ground-state wave function is normalized. 15. The factor is found from the ratio of the radial probability densities for 100 . Use Eq. 39–7. r22 2rr 0 r2 2rr 0 4 e 0 0 P r r r2 4 r e 2 0 0 r r0 e 1.85 P rr 2r0 2 2r rr0 22r 2 2 4 e 4e 2r0 r0 r0 4 4 2 e r 0 4 rr2 e r 0 0 r2 r0 r2r 0 0
16. (a) To find the probability, integrate the radial probability distribution for the ground state. We follow Example 39–4 and use the last integral in Appendix B–4. r0 2r 2 r r P 4 3 e r0 dr ; let x 2 r 12 r0 x , dr 12 r0 dx r r0 0 0 2 2r 2 1 r0 x r 2 1 r dx 1 2 x2e xdx 1 e x x 2 2x 2 P 0 4 r 2 e r0 dr 2 21 r0 x 2r0 2 2 0 2 0 r03 0 4 r03 e 0 0 1 5e2 0.32 32% (b) We follow the same process here. 2r 2r0 r2 r0 r e dr ; let x 2 P 4 r3 r0 0 r0 P 1 4 x2e xdx 1 e x x 2 2x 24 5e2 13e4 0.44 2
2
2
44%
2
17. (a) To find the probability for the electron to be within a sphere of radius r, we must integrate the radial probability density for the ground state from 0 to r. The density is given in Eq. 39–7.
Chapter 39
Since r
Quantum Mechanics of Atoms
r , we approximate e2r r0 1.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
P
Instructor Solutions Manual
15 1.1fm 4 2 2 2r 3 r0 4 1.110 m r 4 r r 3 4 e dr 3 3 1.2 1014 dr 3 10 r0 r0 3 r0 3 0.529 10 m 0 0 3
1.1 fm
(b) The Bohr radius, r0 , is inversely proportional to the mass of the particle. So now, the Bohr radius is smaller by a factor of 207. 3 2r 1.1fm 1.1fm 2 2 1.11015 m 3 4 r33 4 P 4 r e r0 dr 4 r 3 1.1107 dr 3 10 r r 3 r 3 0.529 10 m 207 0 0 0 0 0 18. To show that 200 is normalized, we integrate 200 2 over all space. Use substitution of variables and an integral from Appendix B–5. r 1 r r 200 2 dV 32 r3 2 r e r0 4r2dr ; let x r r r0 x , dr r0dx 0 0 0 all 0 space
Note that if r 0, x 0 and 1 if r , x2 . 2 x 2 2 2 2 r e r0 4r2dr 2 x e r x r dx dV 4 0 0 32r3 r 32r3 200 0 0 0 0 all 0 space
1 1 1 1 1 x e dx 2 x 2 2 3 4 x 4x 4x x e dx x e dx x e dx 2 x e x dx 8 0 2 0 2 x 2 0 3 x 8 0 4 x 8 0 1 1 1 2! 3! 4! 1 3 3 1 2 2 8 And so, we see that 200 is normalized.
19. We follow the directions as given in the problem. We use the first integral listed in Appendix B–5. r 3 r 2 rr0 1 2 r0 r 2 2 2 r r 100 4r dr r 3 e 4r dr 4 3 e dr ; let x 2 r 12 r0 x , dr 12 r0 dx r0 r0 r 0 0 0 0 3 21 0 3 x x r 4 r3 e2 rr0 dr 4 r x e 1 r dx 1 r x e dx 1 r 3! 3 r 2 0 4 0 4 0 r3 r3 0 0 0 0 0 20. The probability is found by integrating the radial probability density over the range of radii given. We use integrals from Appendix B–4. 2r 1.01 r0 r 2 r0 r 0.99r0 1.98 P 4 3 e dr ; let x 2 r 12 r0 x , dr 12 r0 dx ; r 0.99r0 x 2 r r0 r0 0 0.99 r0
1.01 r0
P
0.99 r0
x2.02 r2 r0 3 4 r e dr
1 r x
2
2r
0
4
x1.98 2.02
3 2 0
r
2 x2.02
ex 12 r0 dx 1
x2exdx
x1.98
0
ex x 2 2x 2 e1.98 4.9402 e2.02 5.0602 0.0108 1.1% 2 1.98 1
Because the range of radii is small and the radial probability density is relatively constant over that range (see Fig. 39–7), we could have approximated the probability as follows.
Chapter 39
Quantum Mechanics of Atoms r r 2 P 1.01r0 4 r2 e 2r0 dr P r r r 4 r02 e 20r0 0.02r 0.08e 0.0108
r03
0.99r0
0
0
r03
1.1%
21. We follow the directions as given in the problem. The three wave functions are given in Eq. 39–9. We explicitly 2show the2 expressions involving the complex conjugate. P 4 r 1 2 1 2 1 3 211 210 3 211 r 3 r r r 2 r 0 1 z e x iy x iy e r0 1 x iy x iy r0 2 1 4 r e 3 5 5 3 3 5 64 r 64 r 32 r 0 0 0 2 x y2 r x 2 y2 r r r z2 2 4 r2 1 e r0 1 e r0 1 e r0 r z 2 x2 y2 e r0 3 3 3 5 5 64r05 64r05 24r0 320r
r
r4 r0 e 24r50 r
r4 r0 e for the 2p state. We find the most probable distance 22. From Problem 21, we know that Pr 24r05 dPr 0 and solving for r. This is very similar to Example 39–3. by setting dr 4
r Pr 24r5 e
r r0
dP 4r ; drr 24r5 e 3
0
r r0
0
r
r
r3e r0 1 r e r0 4r0 r r6 24 0 r 4r 0 r 24r5
4
0
0
0
23. To find the probability for the electron to be within a sphere of radius r, we must integrate the radial probability density the ground state from 0 to r. The density is given in Eq. 39–7. 2for r rsphere r r2 r r P 4 e 0 dr ; let x 2 ; let 2 sphere x r03 x0 r0 xr0 2 P 1 x e xdx 1 e x x2 2x 2 1 e x x 2 2x 2 2 1 e x 1 x2 x 1 2
2
0
2
2
0
We solve this equation numerically for values of x that give P = 0.50, 0.90, and 0.99. (a) The equation for P = 0.50 is solved by x = 2.674, and so rsphere 1 22.674r0 1.3r0 . (b) The equation for P = 0.90 is solved by x = 5.322, and so rsphere 1 25.322r0 2.7 r0 . (c) The equation for P = 0.99 is solved by x = 8.406, and so rsphere 1 2 8.406r0 4.2 r0 . 24. The probability is found by integrating the radial probability density over the range of radii given. The radial probability density is given after Eq. 39–8. 2 2 r r 5.00r0 r P e r0 dr ; let x r 3 2 r0 r0 4.00r0 8r0 2 x 1 4 1 5.00 2 x 2 x e dx 5.00 x 4x3 4x2 exdx P 8
4.00
8
4.00
There are some difficult integrals to evaluate. We use integration by parts.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
x e dx : u x ; dv e ; du 2xdx; v e x e dx x e 2xe dx x e 2e x 1 e x 2x 2 x e dx : u x ; d v e ; du 3x dx; v e x e dx x e 3 x e dx e x 3x 6x 6 x e dx : u x ; dv e ; du 4x dx; v e x e dx x e 4 x e dx e x 4x 12x 24x 24 2 x
2
2 x
x
2 x
3 x
3
3 x
x
x
3 x
4 x
4
4 x
4 x
x
2 x
x
x
x
3
2
x
3
3 x
2
x
2
2 x
x
x
4
3
2
We substitute the integrals above into the expression for the probability. We are not showing the algebra. 5.00 P 1 ex x 4 4x2 8x 8 1 773e5 360e4 0.173 17.3% 8 8 4.00 25. The wave function is given in Eq. 39–5a. Note that r x 2 y2 z2 . We will need the derivative 1/ 2
relationship derived in the first line below. r r r 1 1 x r x 1 2 2 2 r0 1 r 2 x y z 2 x ; 100 e ; e 0 x r x r x r r0 r r03 2 r 1 1 rr0 x r 1 1 rr0 x 1 1 r0 1 e 2 3 e 3 e x r r r r r r3 r x x r r r 0 0 0 0 0 0 1 1 1 1 1 1 1 rr0 rr r3 e 1 x2 rr r2 rr 1 x2 rr r2 0 0 0 0 0 2 1 1 r z r y ; 2 1 y2 rr 1 y rr 2 ; z r ; and Similarly, we would have y r 0 0 r 1 2 1 2 1 rr . Substitute into the time-independent Schrödinger equation. 1 z z2 2 rr0 0 r 2 2 2 2 1 e2 E 2m x2 x2 x2 4 0 r 2 e2 1 1 1 2 1 1 2 1 2 1 2 4 r 1 x rr 2 1 y rr 2 1 z rr r rr0 2m2 0 0 r 2 0 r 0 1 1 e 2 2 2 1 3 x y z 2 rr 2m rr r 4 r 0 0 2 1 0 2 1 2 r2 e2 2 1 1 e2 2 4 r 2m rr rr 4 r 2m rr 3 r rr r 0 0 0 0 0 0 e r mrr 2mr2 4 0 0 0 Since the factor in square brackets must be a constant, the terms with the r dependence must cancel. 4 h mrr 4 r 0 r0 me2 me2
0
0
Chapter 39
Quantum Mechanics of Atoms
Note from Eq. 37–11 that this expression for r0 is the same as the Bohr radius. Since those two terms cancel, we are left with the following. 2 2 2 me4 E 2 E 2 2 8h2 02 2mr0 2mr02 0h 2m 2 me 26. (a) The probability is found by integrating the radial probability density over the range of radii given. The radial probability density is given after Eq. 39–8. 2 r0 r r 2 P r 2 e dr ; let x r
3
r0
8r0 r0 r0 0 2 x 4 1 1 1 2 P x 2 x e dx 1 x 4x3 4x2 e xdx 8
8
0
0
The following integrals are derived in the solution to Problem 24.
x e dx e x 2x 2 ; x e dx e x 3x 6x 6 4 x x 4 3 2 x e dx e x 4x 12x 24x 24 2 x
x
3 x
2
1
x
3
2
P 18 x 4 4x 3 4x 2 e xdx
e x 4x 12x 24x 24 4 x 3x 6x 6 4 x 2x 2
1
0
x
4
3
2
3
2
0 e x x 4 4x 2 8x 8 1 8 21e1 0.0343 3.4% 8 1 8 8
1
1
2
0
r
r4 (b) From Problem 21, we know that the radial probability density for this state is Pr 5 e r0 . We 24r0 proceed as in part (a). r0 r4 rr0 P e dr ; let x r
5
24r0 r0 x4e x dx 1 e x x 4 4x3 12x2 24x 24 1 1 24 65e1 3.66 103 P 24 24 0 24 0 1 1
0
0.37% 27. (a) The radial probability distribution is given by Eq. 39–6. Use the wave function given. 2 2 2 2r 2r 2r 1 2r 2 4r2 1 2r 2r2 e 3r0 2 Pr 4r2 300 4r2 27r3 1 3r 27r 0 3 2 e 27r 3r 27r 3r0 0 0 0 0 0
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Pr (nm-1)
(b) See the graph. (c) The most probable distance is the radius for which the radial probability distribution has a global maximum. We dPr 0 and find that location by setting dr solving for r. We see from the graph that the global maximum is approximately at r 13r0.
Instructor Solutions Manual
2
4r 2 r dP 8r 2r 2r2 2 r 8r2 2r 2r2 2 2 27r e 3r0 27r3 1 27r2 27r2 e 3r0 drr 27r3 1 3r0 3r0 3r0 0 0 0 0 0 2 2 2 2 r 4r 2r 2r 2 27r3 1 27r2 3r e 3r0 3r0 0 0 0 8r 2r 2r2 5r 12r2 2r3 3r2 r 27r3 1 27r2 1 27r2 81r3 e 0 0 3r 3r 0 0 0 0 0 0 The above system has 6 non-infinite solutions. One solution is r = 0, which leads to Pr 0, which is not a2 maximum for the radial distribution. The second-order polynomial, 2r 2r 1 2 , is a factor of the radial probability distribution, and so its zeros also give 3r0 27r0 locations where Pr 0. Thus, the maxima must be found from the roots of the third-order polynomial. A spreadsheet was used to find the roots of 1 53 x 1227 x2 281x3 0. Those roots
are x 0.74, 4.19, and 13.07. Therefore, the most probable distance is r 13.1r0 . 28. The number of electrons in the subshell is determined by the value of ℓ. For each 𝓁, the value of 𝑚ℓ can range from −ℓ to +ℓ which is 2ℓ + 1 values. For each 𝑚ℓ value, there are two values of ms . Thus, the total number of states for a given 𝓁 is 𝑁 = 2(2ℓ + 1). 𝑁 = 2(2ℓ + 1) = 2[2(4) + 1] = 18 electrons 29. For carbon, Z = 6. We start with the n 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ, 𝑚𝑠). 1,0,0, , 1,0,0, , 2,0,0, , 2,0,0, , 2,1, 1, , 2,1, 1, Note that without additional information, there are other possibilities for the last two electrons. The value of 𝑚ℓ for the last two electrons could be –1 (as shown), 0, or 1. For example, those last two electrons might be 2,1,0, 1 2 and 2,1,1, 12. 30. (a) For oxygen, Z = 8. We start with the n 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ, 𝑚𝑠). 1,0,0,
, 1,0,0,
2,1, 1,
, 2,1, 1,
, 2,0,0, , 2,1,0,
, 2,0,0, , 2,1,0,
,
Chapter 39
Quantum Mechanics of Atoms
Note that without additional information, there are two other possibilities that could substitute for any of the last four electrons. The third quantum number could be changed to 1. (b) For aluminum, Z = 13. We start with the n 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ, 𝑚𝑠). 1 ,0,0,
, 1,0,0,
, 2,0,0,
, 2,0,0,
, 2,1, 1,
2,1,0,
, 2,1,0,
, 2,1,1,
, 2,1,1,
, 3,0,0,
, 2,1, 1, , 3,0,0,
,
, 3,1, 1,
Note that without additional information, there are other possibilities for the last electron. The third quantum number could be –2, –1, 0, 1, or 2, and the fourth quantum number could be either +1/2 or –1/2. 31. Since the electron is in its lowest energy state, we must have the lowest possible value of n. Since 𝑚ℓ = 3, the smallest possible value of 𝓁 is ℓ = 3 , and the smallest possible value of n is n 4. 32. The third electron in lithium is in the 2s subshell, which is outside the more tightly bound filled 1s shell. This makes it appear as if there is a “nucleus” with a net charge of 1e. Use the energy of the hydrogen atom, Eq. 39–2. E2
13.6eV 13.6eV 3.4eV
n2 22 We predict the binding energy to be 3.4 eV. Our assumption of complete shielding of the nucleus by the 2s electrons is evidently not correct. The partial shielding means the net charge of the “nucleus” is higher than +1e, and so it holds the outer electron more tightly, requiring more energy to remove it. 33. The configurations can be written with the subshells in order by energy (4s before 3d, for example), or with the subshells in principal number order (3d before 4s, for example, as in Table 39–4). Both are shown here. (a) Cobalt has Z = 27. 1s2 2s2 2 p6 3s2 3 p6 3d 7 4s2 or 1s2 2s2 2 p6 3s2 3 p6 4s2 3d 7
(b) Gold has Z 79 . 1s2 2s2 2 p6 3s2 3 p6 3d 10 4s2 4 p6 4d 10 4 f 14 5s2 5 p65d 10 6s1 or 1s2 2s2 2 p6 3s2 3 p6 4s2 3d 10 4 p65s2 4d 105 p6 4 f 14 6s15d 10
(c) Uranium has Z = 92. 1s2 2s2 2 p6 3s2 3 p6 3d10 4s2 4 p6 4d10 4 f 145s2 5 p6 5d10 6s2 6 p6 5 f 3 6d17s2
or
1s2 2s2 2 p6 3s2 3 p6 4s2 3d10 4 p6 5s2 4d10 5 p6 4 f 146s2 5d10 6 p6 7s2 5 f 36d1 34. Limiting the number of electron shells to six would mean that the periodic table stops with radon (Rn) since the next element, francium (Fr), begins filling the seventh shell. Including all elements up through radon means 86 elements. 35. We use Eq. 37–13, which says that the radius of a Bohr orbit is inversely proportional to the atomic number. We also use Eq. 37–14b, which says that the energy of a Bohr orbit is proportional to the square of the atomic number. The energy to remove the electron is the opposite of the total energy. A simple approximation is used in which the electron is attracted by the full effect of all 92 protons in the nucleus.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
n2 rn
0.529 10 Z
10
Z2
En 13.6eV
m
1
0.529 10 m 5.75 10 m 92
13.6eV
n2
Instructor Solutions Manual
10
922 12
13
1.15105 eV 115keV
36. The energy levels of the infinite square well are given in Eq. 38–13. Each energy level can have a maximum of two electrons since the only quantum numbers are n and ms. Thus, the lowest energy level will have two electrons in the n = 1 state, two electrons in the n = 2 state, and 1 electron in the n = 3 state. 2 ( )2] ℎ [ ( )2 ( 2) = 19 𝐸 = 2𝐸1 + 2𝐸2 + 𝐸3 = 2 1 + 2 2 + 1 3 2 8𝑚ℓ
37. In a filled subshell, there are an even number of electrons. All of the possible quantum number combinations for electrons in that subshell represent an electron that is present. Thus for every m value, both values of ms are filled, representing a spin “up” state and a spin “down” state. The total angular momentum of that pair is zero, and since all of the electrons are paired, the total angular momentum is zero. 38. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the electron in the tube: 6.631034 J s3.00 108 m/s E hc 5.405 104 eV 54 keV
1.60 1019 J/eV0.023109 m Thus, the operating voltage of the tube is 54 kV . 39. The maximum energy is emitted when the shortest wavelength is emitted. That appears to be about 0.025 nm. The minimum energy is emitted when the longest wavelength is emitted. That appears to be about 0.2 nm. Use Eq. 37–3. hc 1eV Emax hfmax 49725eV min 0.025 109 m
5.0 104 eV hc E hf min min max
0.2 10 m 9
1eV
6215eV
6 103 eV
40. With the shielding provided by the remaining n 1 electron, we use the energies of the hydrogen atom with Z replaced by Z 1. The energy of the photon is found, and then the wavelength. 2 1 1 hf E 13.6 eV26 1 6.40 103 eV. 22 12 34 8 6.63 10 J s3.00 10 m/s hc 1.94 1010 m 0.194 nm 19 3 E 1.60 10 J/eV6.40 10 eV
Chapter 39
Quantum Mechanics of Atoms
41. The energy of the photon with the shortest wavelength must equal the maximum kinetic energy of an electron. We assume V is in volts. hc eV E hf0 0
0
34 8 9 hc 6.63 10 J s3.00 10 m s10 nm m 1243nm 1240 nm V eV V 1.60 1019 CV V
42. We follow the procedure of Example 39–6, of using the Bohr formula, Eq. 37–15, with Z replaced by Z – 1. 1 1 2 1 1 e4m Z 1 2 1 7 1 9 1 2 3 8 h c 2 2 1.097 10 m 27 1 2 2 5.562 10 m n n 1 2 0 1 1.798 1010 m 5.562 109 m1
43. The wavelength of the K line is calculated for molybdenum in Example 39–6. We use that same procedure. Note that the wavelength is inversely proportional to Z 1 . Z 12 Z 1 Fe unknown 26 1 Z unknown 1 2 Fe Zunknown 1 The unknown material has Z = 24, and so it is chromium. 2
1 24
44. We assume “shielding” is provided by the 1s electron that is already at that level. Thus, the effective charge “seen” by the transitioning electron 1 is 2 1 42 – 1 = 41. We use Eqs. 37–9 and 37–14b. hf E 13.6eVZ 1 n2 n2 34 hc hc J s3.00 108 m/s 6.63 10 E 1 2 1 2 1 13.6eVZ 1 13.6eV41 1 1.60 1019 J eV n2 n2 12 32 6.12 1011 m 0.0612 nm We do not expect perfect agreement because there is some partial shielding provided by the n 2 shell, which was ignored when we replaced Z by Z 1. That would make the effective atomic number a little smaller, which would lead to a larger wavelength. The amount of shielding could be estimated by using the actual wavelength and solving for the effective atomic number. 45. Momentum and energy will be conserved in any inertial reference frame. Consider the frame of reference that is moving with the same velocity as the electron’s initial velocity. In that frame of reference, the initial momentum of the electron is 0, and its initial total energy is mc2 . Let the emitted photon have frequency f, and let the direction of motion of that photon be considered the positive direction. The momentum of the photon is then p h hf , and so the momentum of the c hf electron must be p . The final energy of the photon is E hf p c, and the final energy of e c
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
the electron is, from Eq. 36–14, E
Instructor Solutions Manual
p2c2 m2c4 . We write the conservation conditions, and
electron final
e
then solve for the frequency of the emitted photon. hf hf p p Momentum: 0 e e c c mc2 hf p2e c2 m2c4
Energy:
2
hf mc hf c2 m2c2 h2 f 2 m2c4 c 2
m2c4 2mc2hf h2 f 2 h2 f 2 m2c4 2mc2hf 0 f 0 Since the photon must have f = 0, no photon can be emitted and still satisfy the conservation laws. Another way to consider this situation is that if an electron at rest emits a photon, the energy of the electron must decrease for energy to be conserved. But the energy of a stationary electron cannot decrease unless its mass were to change. Then it would no longer be an electron. Thus, we conclude that a third object (with mass) must be present in order for both energy and momentum to be conserved. 46. The Bohr magneton is given by Eq. 39–12. 1.602 1019 C1.054 1034 J s 9.268 1024 J T e 9.27 1024 J T B 31 2m 29.109 10 kg 47. We use Eq. 39–14 for the magnetic moment since the question concerns spin angular momentum. The energy difference is the difference in the potential energies of the two spin states. 24 9.27 10 J T spin down U B g Bm 2.0023 3.5T 1 1 4.1104 eV z B s 2 2 spin up 1.60 1019 J eV 48. (a) The energy difference is the difference in the potential energies of the two spin states. Use Eq. 39–14 for the magnetic moment. 24 spin down U B g Bm 2.0023 9.27 10 J T 2.0T 1 1 z B s 2 2 spin up 1.60 1019 J 2.320 104 eV 2.3104 eV (b) Calculate the wavelength associated with this energy change. c U E h 6.63 1034 J s3.00 108 m/s hc 5.358 103 m U
2.320 10 eV1.60 10 J eV 4
19
5.4 mm
(c) The answer would be no different for hydrogen. The splitting for both atoms is due to an sstate electron: 1s for hydrogen and 5s for silver. See the discussion on page 1216 concerning the Stern-Gerlach experiment.
Chapter 39
Quantum Mechanics of Atoms
49. (a) Refer to Fig. 39–14 and the equation following it. A constant magnetic field gradient will produce a constant force on the silver atoms. Atoms with the valence electron in one of the spin states will experience an upward force, and atoms with the valence electron in the opposite spin state will experience a downward force. That constant force will produce a constant acceleration, leading to the deflection from the original direction of the atoms as they leave the oven. We assume the initial direction of the atoms is the x direction, and the magnetic field gradient is in the z direction. If undeflected, the atoms would hit the screen at z = 0. 2 2 gBms dBz x 2 F x z dBz x 2 1 1 dz dz 1 z at 12 v 2 v 2 2 m m mAg v Ag Ag
One beam is deflected up, and the other down. Their separation is the difference in the two deflections due to the two spin states. dB g B 21 12 z x 2 dz z z 1 z 1 1 2 ms 2 ms 2 mAg v 12
2.00239.27 1024 J T1800 T m 0.040 m 2 4 2.45310 m 0.25 mm 27 107.87 u1.66 10 kg u 780 m s
(b) The separation is seen in the above equation to be proportional to the g-factor. So, to find the new deflection, divide the answer to part (a) by the original g-factor. 2.453104 m 1.225 104 m 0.12 mm zg 1 2.0023 1
7 9 50. For the 5g state, ℓ = 4 and s 12. Thus, the possible values of j are 𝑗 = ℓ ± 𝑠 = 4 ± = , . 2
Let j 72 . Then we have the following. m 7 , 5 , 3 , 1 , 1 , 3 , 5 , 7
J
1
3 7 2
J z mj 7 , 5 , 3 , 1 Let j 92 . Then we have the following. m 9 , 7 , 5 , 3 , 1 , 1 , 3 , 5 , 7 , 9
J
1
3 11 2
J z mj 9 , 7 , 5 , 3 , 1 51. (a) For the 4p state, ℓ = 1. Since s 12, the possible values for j are 𝑗 = ℓ + 𝑠 = 𝑗 = ℓ−𝑠 =
1 . 2
(b) For the 4f state, ℓ = 3. Since s 12, the possible values for j are 𝑗 = ℓ + 𝑠 = 𝑗 = ℓ−𝑠 =
and
5 . 2
(c) For the 3d state, ℓ = 2. Since s 12, the possible values for j are 𝑗 = ℓ + 𝑠 = 𝑗 = ℓ−𝑠 =
and
3 . 2
and
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
(d) The values of J are found from Eq. 39–15.
4 p : J j j 1
15 2
and
3 2
4 f : J j j 1
63 2
and
35 2
3d : J j j 1
35 2
and
15 2
52. (a) Gallium has Z = 31. We list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ, 𝑚𝑠). 1, 0, 0,
, 1, 0, 0,
, 2, 0, 0,
, 2, 0, 0,
, 2,1, 1,
, 2,1, 1,
,
2,1, 0, , 2,1, 0, , 2,1,1, , 2,1,1, , 3, 0, 0, , 3, 0, 0, , 3,1, 1, , 3,1, 1, , 3,1, 0, , 3,1, 0, , 3,1,1, , 3,1,1, , 3, 2, 2, , 3, 2, 2, , 3, 2, 1, , 3, 2, 1, , 3, 2, 0, , 3, 2, 0, , 3, 2,1, , 3, 2,1, , 3, 2, 2, , 3, 2, 2, , 4, 0, 0, , 4, 0, 0, , 4,1, 0, The last electron listed could have other quantum numbers for 𝑚ℓ and ms . (b) The 1s, 2s, 2p, 3s, 3p, 3d, and 4s subshells are filled. (c) For a 4p state, ℓ = 1. Since s 12, the possible values for j are 𝑗 = ℓ + 𝑠 =
and
1 . 2 (d) The 4p electron is the only electron not in a filled subshell. The angular momentum of a filled subshell is zero, so the total angular momentum of the atom is the angular momentum of the 4p electron. (e) When the beam passes through the magnetic field gradient, the deflecting force will be proportional to mj . If j 12 , the values of mj are 12, and there will be two lines. If j 32 , the values of mj are 12 , 32 , and there will be four lines. The number of lines indicates the value of 𝑗 = ℓ−𝑠 =
j. 53. (a) The additional term for the spin–orbit interaction is given in the text as Uspin μ Bn z Bn . orbit
The separation of the energy levels due to the two different electron spins is twice this. U B spin down g B m z
spin orbit
n spin up
B
n
s
5 10 eV1.60 10 J eV 0.431T 0.4T B g m 2.0023 9.27 10 J T Uspin
n
5
19
orbit
B
s
(b) If we consider the nucleus to be a loop of current with radius r, then the magnetic field due to
Chapter 39
Quantum Mechanics of Atoms
the nucleus at the center of the loop (the location of the electron) is given in Example 28–11 as I B 0 . Model the current as the charge of the nucleus moving in a circle, with a period as n 2r given by circular motion. q e e ev emer v I t T 2r v 2r 2r mer Note that classically, merv Le , the angular momentum of the electron, and so 𝑚𝑒𝑟𝑣 = √ℓ(ℓ + 1)ℏ with ℓ = 2. Thus, we have the following: 𝜇 𝐼 2
𝜇 𝑒√ℓ(ℓ+1)ℏ 𝑒ℏ 𝜇0√ℓ(ℓ+1) 𝜇 𝜇 √ℓ(ℓ+1) = = 𝐵 0 3 2𝑟 2𝜋𝑚𝑒𝑟2 2𝑚𝑒 2𝜋𝑟3 2𝜋𝑟
𝐵𝑛 = 0 = 0
From Fig. 39–9(b), we see that the most probable radius for the 𝑛 = 2, ℓ = 1 state is approximately r 4r0. We can now calculate the approximate magnetic field. 𝐵𝑛 =
𝜇 𝜇 √ℓ(ℓ+1) 𝐵 0
2𝜋(4𝑟0)3
The two values are about
1 2
=
𝐽
𝑇 𝑚
𝑇
𝐴
(9.27×10 −24 )(4𝜋×10 −7
2𝜋[4(0.529×10−10𝑚)]3
)√6
= 0.479𝑇 ≈ 0.5𝑇
0.479 T 0.431T 100 11% different, and so they are consistent. 0.479 T 0.431T
54. The energy of a pulse is the power of the pulse times the duration in time.
E P t 0.68 W 23103s 0.01564J 0.016 J The number of photons in a pulse is the energy of a pulse divided by the energy of a photon, as given in Eq. 37–3. 0.01564 J640 109 m E E N 5.0 1016 photons 34 8 hf hc 6.6310 J s3.00 10 m/s 55. The angular half-width of the beam can be found in r 1.22 Section 35–5 and is given by 1 / 2 , where d d is the diameter of the diffracting circle. The angular width of the beam is twice this. The linear diameter of the beam is then the angular width times the distance from the source of the light to the observation point, D r . See the diagram. 2.44 658 10 9 m 2.44 3 170 m (a) D r r 380 10 m 3.6 103 m d 9 2.44 658 10 m 2.44 1.7 105 m (b) D r r 384 106 m 3 3.6 10 m d
56. Intensity equals power per unit area. The area of the light from the laser is assumed to be in a circular area, while the area intercepted by the light from a light bulb is the surface area of a sphere. P P 0.50 103 W 2 70.74 W m 2 (a) I 71W m2 2 S r 1.5 103 m
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
(b) I
Instructor Solutions Manual
P
P 2.2 W 4.377 102 W m2 0.044 W m2 2 2 S 4 r 4 2.0 m
The laser beam is more intense by a factor of
70.74 W m2 1616 1600 . 4.377 102 W m2
57. Transition from the E3 state to the E2 state releases photons with energy 1.96 eV, as shown in Fig. 39–21. The wavelength is determined from the energy. hc 6.63 1034 J s3.00 108 m/s 6.34 107 m 634 nm 19 E 1.60 10 J/eV1.96eV 58. We use Eq. 39–16b. N 2
e
E E 2 0 kT
e
E E 2 0 kT
e
N0 N1
19 2.2 eV 1.6010 J eV 1.381023 J K 300 K
e
1.8 eV 1.601019 J eV 1.3810 23 J K 300 K
N0
e69.6 6.11031
59. The relative numbers of atoms in two energy states at a given temperature is given by Eq. 39–16b. N2 0.5. From Fig. 39–20, the energy difference between the two states is 2.2 eV. We set N0 19 E2 E0 2.2eV1.60 10 J eV E E N2 e kT T 23 3.7 104 K N N0 k ln 2 1.38 10 J Kln 0.5 N 0 2
e85.0 1.2 1037
0
60. Consider Eq. 39–16b, with En En. To have a population inversion means that Nn Nn. E E Nn E En Nn n 0 En En e kT 1 ln 0 Nn Nn kT kT Since En En, to satisfy the last condition, we must have T < 0, a negative temperature. n
n
This negative “temperature” is not a contradiction. The Boltzmann distribution assumes that a system is in thermal equilibrium, and the inverted system is not in thermal equilibrium. The inversion cannot be maintained without adding energy to the system. If left to itself, the excited states will decay, and the inversion will not be maintained. 61. An h subshell has ℓ = 5. For a given ℓ value, ml ranges from −ℓ to +ℓ, taking on 2ℓ + 1 different values. For each 𝑚ℓ, there are 2 values of ms . Thus, the number of states for a given ℓ value is 2(2ℓ + 1). Thus, there are 2(2ℓ + 1) = 2(11) = 22 possible electron states. 62. (a) Z = 24 is chromium. 1s2 2s2 2 p6 3s2 3 p6 3d 5 4s1
(b) Z = 35 is bromine. 1s2 2s2 2 p6 3s2 3 p6 3d10 4s2 4 p5
Chapter 39
Quantum Mechanics of Atoms
(c) Z = 40 is zirconium. 1s2 2s2 2 p6 3s2 3 p6 3d10 4s2 4 p6 4d 2 5s2
63. (a) Boron has Z 5, so the outermost electron has n 2. We use the Bohr result with an effective Z. We might naively expect to get Zeff 1, indicating that the other four electrons shield the outer electron from the nucleus, or Zeff 3, indicating that only the inner two electrons accomplish the shielding.
13.6eVZ eff
2
13.6eVZ eff
2
E2
8.26eV
Zeff 1.56 n2 22 This indicates that the other two electrons in the n 2 shell partially shield the electron that is to be removed. (b) We find the average radius from the expression below. 2 10 n2r 2 0.529 10 m r 1 1.36 1010 m Zeff 1.56
64. The value of ℓ can range from 0 to n 1. Thus, for n 6, we have 0 ≤ ℓ ≤ 5. The magnitude of L is given by Eq. 39–3, 𝐿 = √ℓ(ℓ + 1)ℏ.
65. (a) We treat the Earth as a particle in rotation about the Sun. The angular momentum of a particle is given in Eq. 37–10 as 𝐿 = 𝑚𝑣𝑟𝑛, where r is the orbit radius. We equate this to the quantum mechanical expression in Eq. 39–3. We anticipate that the quantum number will be very large, so we approximate √ℓ(ℓ + 1) as l . 1 𝐿 = 𝑀𝐸𝑎𝑟𝑡ℎ𝑣𝑟Sun− = 𝑀 2𝜋𝑟 𝑟 = ℏ[ℓ(ℓ + 1)]2 = ℏℓ → l
Earth
M Earth
𝑇
24 11 2r2 5.98 10 kg 2 1.496 10 m 2.52551074 2
1.055 10
34
T
J s3.156 10 s 7
2.531074
(b) There are 2ℓ + 1values of 𝑚ℓ for a value of ℓ, so the number of orientations is as follows. 𝑁 = 2ℓ + 1 = 2(2.5255 × 1074) + 1 = 5.051 × 1074 ≈ 5.05 × 1074 . 66. Eq. 37–15 gives the Bohr-theory result for the wavelength of a spectral line. For the Mosley plot, the wavelengths are for the K line, which has n 2 and n 1. We assume that the shielding of the other n = 1 electron present reduces the effective atomic number to Z – 1. We use the value of the Rydberg constant from Section 1 37–11. 1 Z 2e2 4m 1 1 e42m3 Z 2 1 1 1 3e42m3 Z 2 8 h c 32 h c 8 h3c n2 n2 1 4 0 0 0 4 4 1/ 2 1 3e m 3e m ,a 2 1 32 2h3c , b 1 a Z b 32 2h3c Z 1 40 0 3e m 1/ 2 1/ 2 1/ 2 7 1 3 a 3 32 2h3c 4 R 4 1.0974 10 m 2868.9 m1/ 2 0
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
67. The magnitude of the angular momentum is given by Eq. 39–3, and Lz is given by Eq. 39–4. The cosine of the angle between L and the z axis is found from L and Lz . 𝐿 𝑚ℓ 𝐿 = √ℓ(ℓ + 1)ℏ ; 𝐿 = 𝑚 ℏ ; 𝜃 = 𝑐𝑜𝑠−1 𝑧 = 𝑐𝑜𝑠−1 ℓ
𝑧
𝐿
√ℓ(ℓ+1)
(a) For ℓ = 1, 𝑚ℓ = −1, 0, 1. 1 0 cos1 45 ; cos1 90 ; 1,1 1,0 2 2 1 1,1 cos1 135 x 2 (b) For ℓ = 2, 𝑚ℓ = −2, −1, 0, 1, 2. 2 1 0 cos1 35.3 ; cos1 65.9 ; cos1 90 2,2 2,1 2,0 6 6 2 1 1 1 2 cos ; cos 114.1 144.7 2,1 2,2 6 6 (c) For ℓ = 3, 𝑚ℓ = −3, −2, −1, 0, 1, 2, 3. 3 2 1 cos1 30 ; cos1 54.7 ; cos1 73.2 3,3 3,2 3,1 12 12 12 2 0 1 1 ; cos1 ; cos1 ; cos 90 106.8 125.3 3,0 3,1 3,2 12 12 12 1 3 150 3,3 cos 12 (d) We see from the previous parts that the smallest angle occurs for 𝑚ℓ = 𝑙. 100 5.71 100,100 cos1 100101
106
10 ,10 cos 1 6
106 106 1
6
y
0.0573
This is consistent with the correspondence principle, which would say that the angle between L and the z axis could be any value classically, which is represented by letting ℓ → ∞ (which also means n ). 68. (a) Since Lz 0, , and so is unknown . We can say nothing about the value of . (b) Since is completely unknown, we have no knowledge of the component of the angular momentum perpendicular to the z-axis. Thus Lx and Ly are unknown . (c) The square of the total angular momentum is given by L Lx2 Ly2 Lz2. Use this with the quantization conditions for L and Lz given in Eqs. 39–3 and 39–4. 𝐿2 = 𝐿2 + 𝐿2 + 𝐿2 → ℓ(ℓ + 1)ℏ2 = 𝐿2 + 𝐿2 + 𝑚2ℏ2 → 𝐿2 + 𝐿2 = 𝑥
𝑦
𝑥
𝑧
[ℓ(ℓ + 1) − 𝑚2]ℏ2 ℓ
√𝐿2 + 𝐿2 = [ℓ(ℓ + 1) − 𝑚 𝑥
𝑦
1/2
𝑦
ℓ
𝑥
𝑦
Chapter 39
Quantum Mechanics of Atoms
Pr (nm-1)
69. This is very similar to Example 39–3. The radial probability distribution for the n = 3, ℓ = 0 wave function is derived in Problem 27(a), and then we need to find the position at which that distribution has a maximum. We see from the figure in the solution to Problem 27(b) that there are three local maxima in the probability distribution function, and the global maximum is at approximately 13r0. The wave function is given in Problem 27. r 2r2 1 2r 300 27 r3 1 3r 27r2 e 3r0 0 0 0 2 2 1 2r 2r2 2r 2r2 2r 4r2 2r 2 2 2 Pr 4r2 300 4r 2 27r3 1 27r e 3r0 27r3 1 27r e 3r0 3r 3r 0 0 0 0 0 0 We see that Pr is a sixth-order polynomial, multiplied by a decaying exponential. Finding the local dPr 0 , is at best very difficult (or perhaps even impossible). So, extrema analytically, by solving dr instead we simply use the Excel graph from Problem 27(b), calculating with a small step size near the global maximum. Doing that, we find that the maximum occurs at 13.074r0 .
13.074r0 13.0740.0529 nm 0.692 nm 70. The “location” of the beam is uncertain in the transverse (y) direction by an amount equal to the aperture opening, D. This gives a value for the uncertainty in the transverse momentum. The momentum must be at least as big as its uncertainty, so we obtain a value for the transverse momentum. py y
py
The momentum in the forward direction is related to the h wavelength of the light by p . See the diagram to x relate the momentum to the angle. p y D ; “spread” 2 D D px h 2 D
D
py px
71. (a) The mean value can be found as described in Problem 19. We use the first definite integral given in Appendix B–5, with n = 1 and a = 1. 1 2 1 1 1 2 rr 4 r 2 rr 2 2 0 100 4 r dr e 4r dr e 0 dr ; let x 2 r 2 3 r0 0 r0 r0 r 0 r 0 r r0 1 1 x 1 xe dx 1 r0 0 r0 r0 2 e2 1 2 1 e 2 2 U U 100 4r dr 1 e 40 r0 40 r 40 0 r
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(b) For the ground state of hydrogen, Eq. 37–14a gives the energy, and Eq. 37–11 gives the Bohr radius. Substitute those expressions into E U K. e4m 1 e2 E 8 2h2 ; U 4 r 0
0
0
e 4m h2 0 2 e4m 1 e2 e2 e4m e2 me K E U 2 2 r 4 r 8 2h2 4 r 8 2h2r 80 h 4 0 0 0 0 0 0 0 0 0 2 2 2 e e e 1 U 4 r 8 r 8 r 0 0
0 0
72. In the Bohr model, LBohr n
0 0
h 2
In quantum mechanics, 𝐿QM = √ℓ(ℓ + 1)ℏ. For n 2, ℓ = 0
or ℓ = 1, so that LQM 00 1 0 or LQM
2 .
73. (a) The 4 p 3p transition is forbidden, because 𝛥ℓ = 0 ≠ ±1. (b) The 3p 1s transition is allowed, be cause 𝛥ℓ = −1. (c) The 4d 3d transition is forbidden, because 𝛥ℓ = 0 ≠ ±1. (d) The 4d 3s transition is forbidden, because 𝛥ℓ = −2 ≠ ±1. (e) The 4s 2 p transition is allowed, because 𝛥ℓ = +1. 74. The binding energy is given by the opposite of Eq. 37–14b. 2 Z 13.6eV 3 6.72 10 eV En 13.6eV 2 452 n The radius is given by Eq. 37–13. n2 7 rn 0.529 1010 m 452 0.529 1010 m 1.07 10 m Z The effective cross-sectional area is as follows.
r2 1.07 107 m 3.60 10 2
14
m2
75. The K line is from the n 2 to n 1 transition. We use the energies of the hydrogen atom with Z replaced by Z 21, as in Example 39–6, using Eq. 37–15. 1 2 1 1 2 e4mk 2 2 h3c Z 1 n 2 n 1 1 1 1 1 1 1 2 2e4mk 2 1 1 Z 1 1 1 .097 10 m 12 22 3 n h c n 0.154 109 m 1 28.09 29 The element is copper.
Chapter 39
Quantum Mechanics of Atoms
1 76. From Eq. 37–15, 1 , and so we seek Z and n so that for every fourth value of n , 2 12 Z X X n n2 there is an nH such that the Lyman formula equals the formula for the unknown element. 1 Z2 Z2 1 1 Z 2 12 2 2 n . 1 n n2 n nX2 H X X X Consider atoms (with Z > 1) for which Z 2 n2 1 n Z , and for which n2 Z 2n2 X
X
X
H
nX ZnH for every fourth line. This is met when Z 4 and nx 4. The resulting spectral lines for nx 4, 8, 12, ... match the hydrogen Lyman series lines for n 1, 2,3, ...... The element is beryllium.
77. (a) We use Eq. 39–16b, with the “note” as explained in the problem, multiplying the initial expression times 82 . N2 8 2 e N1
E E 2 1 kT
28 e
13.6 eV 13.6 eV 19 1.6010 J eV 4 1 1.381023 J K 300 K
4e394.2
Many calculators will not directly evaluate e394.2 , so we do the following. x e394.2 ; log x 394.2log e 394.2 0.4343 171.2 ; x 10171.2 100.210171 0.63110171 N2 4e394.2 4 0.63110171 2.52 10171 310171 N1 There are 18 states with n = 3, so we multiply by 182 . E E 3kT 1
N 3
182 e
13.6 eV 13.6 eV 19 1.6010 J eV 9 1 1.381023 J K 300 K
182 e
N1 (b) We repeat the evaluations for the higher temperature. 13.6 eV 13.6 eV 19 1.6010 J eV 4 1 E E 2 1
N 2 8 kT 8 e 2 2 e N1 E E 3kT 1
N 3
182 e
182 e
1.3810 J K6000 K 23
1.1310 202 110202
4e19.71 1.10 108 1108
13.6 eV 13.6 eV 19 1.6010 J eV 9 1 1.381023 J K 6000 K
467.2
9e
23.36
9e
6.44 10 10 6 1010
N1 (c) Since the fraction of atoms in each excited state is very small, we assume that N1 is the number
of hydrogen atoms given. 1.0 g of H atoms contains 6.02 1023 atoms. N 2 N 1 1.10 108 6.02 1023 1.10 108 6.62 1015 7 10
15
N 3 N 16.44 1010 6.02 1023 6.44 1010 3.881014 4 10
14
(d) We assume the lifetime of an excited state atom is 108 s. Each atom would emit one photon as its electron goes to the ground state. The number of photons emitted per second can be estimated by the number of atoms, divided by the lifetime. 14 15 N N 4 10 ; n 2 7 10 n3 3 2 108 s 108 s 78. (a) The additional energy due to the presence of a magnetic field is derived in Section 39–7, as
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𝑈 = 𝜇𝐵𝑚ℓ𝐵𝑧. We use this to calculate the energy spacing between adjacent 𝑚ℓ values. 𝑈 = 𝜇𝐵𝑚ℓ𝐵𝑧 → 𝐽 (9.27 × 10−24 ) 𝑇 (2.6𝑇) 𝑈 = 𝜇𝐵𝐵𝑧𝛥𝑚ℓ = = 1.506 × 10−4eV ≈ 1.5 × 10−4eV 𝐽 −19 (1.60 × 10 eV) (b) As seen in Fig. 39–4, the 𝑛 = 3, ℓ = 2 level will split into 5 levels, and the 𝑛 = 2, ℓ = 1 level will split into 3 levels. With no restrictions, there would be 15 n 3, 2 different transitions possible. All transitions would have the same n 1. Thus, there are only three unique wavelengths possible: the one corresponding to a transition with 𝛥𝑚ℓ = +1 (a slightly larger energy change than in the B = 0 case); the one corresponding to a transition with 𝛥𝑚ℓ = 0 (the same n 2, 1 energy change as in the B = 0 case); and the one corresponding to a transition with 𝛥𝑚ℓ = −1 (a slightly smaller energy change than in the B = 0 case). ml See the diagram showing 9 possible transitions grouped into 3 actual energy changes. The value along the right side is the change in energy level due to the magnetic field interaction. (c) Eq. 37–15 gives the wavelength for hydrogen, considering only a change in principal quantum number. The energies for those transitions are on the order of eV. The energy change due to the magnetic field interaction is much smaller than that, so we can use an approximation, knowing E from part (a). We obtain 1E from Eq. 37–14b. 1 7 1 1 7 1 1 1 n32 R 2 2 1.0974 10 m 4 9 6.56096 10 m 656.10 nm 2 3 c hc E 1 1 h 2 E ; E 13.6eV 1.889eV 2 2 n32 E E 2 3 E 1.506 104 eV E 656.10 nm 0.052 nm E 1.889eV
m 1 n32 656.10 nm 0.052nm 656.05 nm l
m 1 n32 0 656.10 nm l
m 1 n32 656.10 nm 0.052nm 656.15 nm l
79. The uncertainty in the electron position is x r1 . The minimum uncertainty in the velocity, v , can be found by solving Eq. 38–1, where the uncertainty in the momentum is the product of the electron mass and the uncertainty in the velocity. xpmin r1 mv 1.055 1034 J s v 2.19 106 m/s mr1 9.1110 31 kg 0.529 10 10 m
80. (a) An electron in a 4s state has ℓ = 0, and so due to the selection rule of 𝛥ℓ = ±1, it must transition
Chapter 39
Quantum Mechanics of Atoms
to an ℓ = 1 state (a p state). Thus, it could transition to a 2p or a 3p state (there is no 1p state). 4s 3 p , 4s 2 p (b) To get to the ground state, it must get to a 1s state. Since again 𝛥ℓ = ±1, it must transition from either 3p or 2p to the 1s (ground) state. Thus, it must make 2 transitions to get to the ground state. 4s 3 p 1s , 4s 2 p 1s 81. The wave functions are given in Eq. 39–9. To show that the desired quantity is spherically symmetric, there must not be any dependence on x, y, or z except through x2 y2 z2 r 2 . r x iy r x iy r z 2r0 2r0 2r0 210 ; 211 ; 211 5 e 5 e 5 e 32 r 64 r 64 r 0
0
0
r z2 2 x iy x5 iy e rr x 2 y52 e rr 2 210 32 r5 e r ; 211 64 r 64 r 0
0
0
2
2
210
2
0
211
0
211
211
0
x iy x iy e 64r
r r0
5 0
x y e 2
2
64r50
0
r r0 z2 x 2 y2 e r0 x 2 y5 2 e r0 32 r5 e 64 r5 64 r r
2
r
r
0
r
0
0
r r 0
r 2 e , which is spherically symmetric 32r05
82. From Fig. 39–20, we see that the energy difference from E1 to E0 is 1.8 eV. Use Eq. 37–3 to calculate the wavelength. hc hc 6.63 1034 J s3.00 108 m/s E hf 6.906 107 m 690 nm 19 E 1.8eV1.60 10 J/eV The textbook quotes an actual wavelength of 694.3 nm.
83. The classical angular momentum of a particle moving in a circular path is given in Section 39–7 as 𝐿 = 𝑚𝑣𝑟. We want the magnetic moment to be equal to 1 Bohr magneton. And, note from Eq. 37– 11 that the radius of an electron in orbit around a hydrogen atom is given by rn n r2 . 𝜇=
1 𝑒 2𝑚
𝐿=
1𝑒
𝑚𝑣𝑟 → 𝑣 =
2 𝑚
2𝜇
𝐽
𝐵 =
𝑒𝑟
2(9.27×10−24 ) 𝑇
(1.60×10−19𝐶)4(5.29×10−11𝑚)
1
= 5.48 × 105 𝑚 𝑠
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CHAPTER 40: Molecules and Solids Responses to Questions 1.
(a) The bond in an N2 molecule is expected to be covalent. (b) The bond in the HCl molecule is expected to be ionic. (c) The bond between Fe atoms in a solid is expected to be metallic.
2.
A neutral calcium atom has 20 electrons. Its outermost electrons are in the 4s2 state. The inner 18 electrons form a spherically symmetric distribution and partially shield the outer two electrons from the nuclear charge. A neutral chlorine atom has 17 electrons; it lacks just one electron to have its outer shell filled. A CaCl2 molecule could be formed when the outer two electrons of the calcium atom are “shared” with two chlorine atoms. These electrons will be attracted by both the Ca and the Cl nuclei and will spend part of their time between the Ca and Cl nuclei. The nuclei will be attracted to this negatively charged area, forming a covalent bond. As is the case with other asymmetric covalent bonds, this bond will have a partial ionic character as well. The two electrons will partly orbit the Ca nucleus and partly orbit each of the two Cl nuclei. Since each Cl nucleus will now have an extra electron part of the time, it will have a net negative charge. The Ca nucleus will “lose” two electrons for part of the time, giving it a net positive charge.
3.
No, neither the H2 nor the O2 molecule has a permanent dipole moment. The outer electrons are shared equally between the two atoms in each molecule, so there are no polar ends that are more positively or negatively charged. The H2O molecule does have a permanent dipole moment. The electrons associated with the hydrogen atoms are pulled toward the oxygen atom, leaving each hydrogen with a small net positive charge and the oxygen with a small net negative charge. Because of the shape of the H2O molecule (see Fig. 40–6), one end of the molecule will be positive and the other end will be negative, resulting in a permanent dipole moment.
4.
The molecule H3 has three electrons. According to the Pauli exclusion principle, no two of these electrons can be in the same quantum state. Two of them will be 1s2 electrons and will form a “closed shell” and a spherically symmetric distribution, and the third one will be outside this distribution and unpaired. This third electron will be partially shielded from the nucleus and will thus be easily “lost,” resulting in an unstable molecule. The ion H3+ only has two electrons. These 1s2 electrons will form a closed shell and a spherically symmetric distribution, resulting in a stable configuration.
5.
Yes, H2 should be stable. The two positive nuclei share the one negative electron. The electron spends most of its time between the two positive nuclei (basically holding them together).
6.
The two diagrams are copied here for easy reference. The H2 is on the left, and the ATP on the right.
Note that both diagrams show a “dip” in potential energy indicating a stable bond is formed at bond length r0, but because the “dip” in the ATP diagram is above the U = 0 line, stored potential energy can be released from that bond when it is broken. That energy is used in other chemical processes in living cells. The ATP diagram also has an “activation energy,” which means that an energy input (perhaps from kinetic energy of the components) is necessary to make the bond. The H2 molecule does not have an activation energy. Thus, the two hydrogen atoms do not need an initial energy input (from kinetic energy, for example) for the atoms to bond. If the separation of the two H atoms is greater than r0, the atoms can attract each other and bond. 7.
The carbon atom (Z = 6) usually forms four bonds because carbon requires four additional electrons to form a closed 2p shell, and each hydrogen-like atom contributes one electron.
8.
The four categories of molecular energy are: translational kinetic energy, electrostatic potential energy, rotational kinetic energy, and vibrational kinetic energy.
9.
Metallic bond theory states that the free electrons in metallic elements can vibrate at any frequency, so when light of a range of frequencies falls on a metal, the electrons can vibrate in response and reemit light of those same frequencies. Hence, the reflected light will largely consist of the same frequencies as the incident light, and so it will not have a distinct color but will be “shiny” instead. Nonmetallic materials only absorb and emit light at specific frequencies, and so they have distinct colors.
10. The conduction electrons are not strongly bound to particular nuclei, so a metal can be viewed as a collection of positive ions and a negative electron “gas.” (The positive ions are just the metal atoms without their outermost electrons since these “free” electrons make up the gas.) The electrostatic attraction between the freely roaming electrons and the positive ions keeps the electrons from leaving the metal. 11. As temperature increases, the thermal motion of ions in a metal lattice will increase. More electrons collide with the ions, increasing the resistivity of the metal. When the temperature of a semiconductor increases, more electrons are able to move into the conduction band, making more charge carriers available and thus decreasing the resistivity. The thermal motion also increases the resistance in semiconductors, but the increase in the number of charge carriers is a larger effect. 12. For an ideal pn junction diode connected in reverse bias, the holes and electrons that would normally be near the junction are pulled apart by the reverse voltage, preventing current flow across the junction. The resistance is essentially infinite. A real diode does allow a small amount of reverse current to flow if the voltage is high enough, so the resistance in this case is very high but not infinite. A pn junction diode connected in forward bias has a low resistance (the holes and electrons are close together at the junction) and current flows easily. 13. The last valence electron of a sodium atom is shielded from most of the sodium nuclear charge and experiences a net nuclear charge of +1e. The outer shell of a chlorine atom is the 3p shell, which contains five electrons. Due to shielding effects, the 3p electrons of chlorine experience a net nuclear charge of +5e. In NaCl, the last valence electron of sodium is strongly bound to a chlorine nucleus. This strong ionic bonding produces a large energy gap between the valence band and the conduction band in NaCl, characteristic of a good insulator. 14. The general shape of Fig. 40–28 is the same for most metals. The scale of the graph (especially the x-axis scale and the Fermi energy) is peculiar to copper and will change from metal to metal.
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15. The main difference between n-type and p-type semiconductors is the type of atom used for the doping impurity. When a semiconductor such as Si or Ge (each atom of which has four electrons to share) is doped with an element that has five electrons to share (such as As or P), then it is an n-type semiconductor since an extra electron has been inserted into the lattice. When a semiconductor is doped with an element that has three electrons to share (such as Ga or In), then it is a p-type semiconductor since an extra hole (the lack of an electron) has been inserted into the lattice. 16. The phosphorus atoms will be donor atoms. Phosphorus has five valence electrons. It will form four covalent bonds with the silicon atoms around it and will have one “extra” electron that is weakly bound to the atom and can be easily excited up to the conduction band. This process results in extra electrons in the conduction band. Silicon doped with phosphorus is therefore an n-type semiconductor. 17. When the top branch of the input circuit is at the high voltage (current is flowing in this direction for half the cycle), then the bottom branch of the output is at the high voltage. The current follows the path through the bridge in the diagram on the left. When the bottom branch of the input circuit is at the high voltage (current is flowing in this direction during the other half of the cycle), then the bottom branch of the output is still at high voltage. The current follows the path through the bridge in the diagram on the right.
18. The energy comes from the power supplied by the collector/emitter voltage source ℰC. The input signal to the base just regulates how much current, and therefore power, can be drawn from the collector’s voltage source. 19. In the circuit shown in Fig. 40–49, the base-emitter is forward biased (the current will easily flow from the base to the emitter), and the base-collector is reverse biased (the current will not flow easily from the base to the collector). 20. No. Ohmic devices (those that obey Ohm’s law) have a constant resistance and therefore a linear relationship between voltage and current. The voltage–current relationship for diodes is not linear (see Fig. 40–38). The resistance of a diode operated in reverse bias is very large. The same diode operated in a forward-bias mode has a much smaller resistance. Since a transistor can be thought of as made up of diodes, it is also non-ohmic. 21. A diode cannot be used to amplify a signal. A diode does let current flow through it in one direction easily (forward biased), and it does not let current flow through it in the other direction (reversed bias), but there is no way to connect a source of power to use it to amplify a signal (which is how a transistor amplifies a signal). Combinations of diodes with additional power sources, as in a transistor, are able to amplify a signal. 22. The base current (between the base and the emitter) controls the collector current (between the collector and the emitter). If there is no base current, then no collector current flows. Thus,
Chapter 40
Molecules and Solids
controlling the relatively small base current allows the transistor to act as a switch, turning the larger collector current on and off.
Responses to MisConceptual Questions 1.
(c) In the H2 molecule, the two electrons orbit both atoms. In order for those two electrons to not violate the Pauli exclusion principle, they must have different quantum numbers. Each atom initially had one electron and the molecule has two, so no electrons are lost. When the atoms are separated by one bond length, the energy is a minimum (not a maximum), and the molecule has less total energy than the two atoms separately. This decrease in energy is the binding energy. Thus, energy has to be added to the molecule in order to separate it into individual atoms.
2.
(c) The shared electrons cannot have the same spin state. One electron must be spin up and the other spin down.
3.
(b) Covalent bonding is the sharing of atoms between molecules. When one atom has excess electrons in its outer shell and another atom lacks electrons in the outer shell, the atom with excess transfers the electrons to the other atom, making a positive ion and a negative ion, and thus creating an ionic “bond”—an attraction between the two ions.
4.
(a, d) Because the ADP molecule has a positive activation energy, as the phosphate group approaches the ADP molecule, it is first repelled and then attracted. The phosphate group must initially have kinetic energy to overcome the repulsion. Some of this kinetic energy is stored as potential energy in the ATP molecule. The binding energy is negative as some of the initial kinetic energy is stored as positive potential energy. When the ATP molecule is broken apart, this energy is released and is available to instigate other reactions.
5.
(d) For the DNA to replicate properly, the bond holding the two strands together must be a very weak bond. Ionic and covalent bonds are strong bonds. The Van der Waals bond is a weak bond that holds the DNA together.
6.
(a) As stated in Section 40–5, the outer electrons in a metallic solid “roam” rather freely among all the metal atoms, acting like an electron “gas.”
7.
(a) A common misconception is that metals have free electrons because they have more electrons than protons. Actually, metals are electrically neutral in that they have the same number of electrons as protons. In metals, the outer electrons are not tightly bound to a single atom but can move between atoms in the metal lattice. Since the electrons can move easily between atoms, metals make good conductors of electricity.
8.
(b) As discussed in Example 40–10, the speed of the electrons in a metal is determined by the Fermi energy. That speed is significantly lower than the speed of light and is much higher than the speed determined from the absolute temperature. And in an earlier chapter it was shown the drift speed of electrons in current is much smaller than the speed as determined by the Fermi energy.
9.
(a) In an insulator, the electrons fill the upper valence bands and are tightly bound to their atoms. A large energy gap exists between the filled valence band and the next higher (conduction) band,
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so a large energy would need to be added to the atom to move an electron to the conduction band. 10. (d) A hole is a positive region in the semiconductor that is formed when an electron is missing from the periodic molecular structure. 11. (a) To be used for doping silicon, the element should have one more or one less electron in its outer shell than silicon. Silicon has two electrons in its outer p shell. Boron and gallium each have one electron in their outer p shell, one fewer than silicon. Phosphorus and arsenic each have three electrons in their outer p shells, one more than silicon. Germanium has two electrons in its outer p shell, the same as silicon, and so would not be a good choice as a doping impurity. 12. (a) A diode allows current to flow through it in one direction, but not in the other (until the breakdown voltage is reached). The diode is not linear, so the current flowing is not proportional to the voltage applied. Thus, answer (b) is not correct. A diode is essentially a junction between n and p type semiconductors, and so it is labeled as a pn junction. Thus, answer (c) is not correct. And, from Fig. 40–38, the current in a forward-biased diode is highly dependent on the applied voltage, and so answer (d) is incorrect.
Solutions to Problems Note:
1.
A factor that appears in the analysis of electron energies is e2 2 9.00 109 N m2 C2 1.60 1019 C 2.30 1028 J m. 40
We calculate the binding energy as the opposite of the electrostatic potential energy. We use Eq. 23– 10 for the potential energy. 1 Q1Q2 e2 1 2.30 1028 J m Binding energy U 9 4 r 4 0.28 10 m 0.28 10 9 m
0
0
19 8.214 1019 J 8.2 10 J 1eV 8.214 1019 J 5.134 eV 5.1eV 1.60 1019 J
2.
From Problem 1, the “point electron” binding energy is 5.134 eV. With the repulsion of the electron clouds included, the actual binding energy is 4.43 eV. Use these values to calculate the contribution of the electron clouds. 5.134eV 4.43eV 0.704eV 0.70eV
3.
We follow the procedure outlined in the statement of the problem.
d d 74 pm 145pm 110 pm CN: d d 154 pm 145pm 150 pm NO: d d 145pm 121pm 133pm HN:
4.
12
H2
N2
1 2
1
1 2
C2
1 2
N
N
2
2
1
O
2
2
2
We convert the units from kcal/mole to eV/molecule.
Chapter 40
Molecules and Solids
kcal 4186 J 1eV 1mole eV 4.339 102 19 23 molecule mole 1kcal 1.602 10 J 6.022 10 molecules Now convert 4.43 eV per molecule into kcal per mole. eV 4.43 molecule 1
5.
We calculate the binding energy as the difference between the energy 0.66 e of two isolated hydrogen atoms and the energy of the bonded p combination of particles. We estimate the energy of the bonded combination as the negative potential energy of the two electron– d proton combinations, plus the positive potential energy of the proton– proton combination. We approximate the electrons as a single object with a charge of 0.33 of the normal charge of two electrons, since the electrons only spend that fraction of time between the nuclei. A simple picture illustrating our bonded model is shown. When the electrons are midway between the protons, each electron will have a potential energy Uep due to the two protons. Uep
20.33ke2
21 d
40.332.30 1028 J m
p
25.6eV
0.074 10 m1.60 10 J eV 9
19
The protons themselves have this potential energy:
2.30 1028 J m ke2 Upp 19.4eV. r 0.074 109 m1.60 1019 J eV When the bond breaks, each hydrogen atom will be in the ground state with an energy E1 13.6eV. Thus the binding energy is as follows.
Binding energy 2E1 2 PEep PEpp 213.6eV 225.6eV 19.4eV 4.6eV This is close to the actual value of 4.5 eV quoted in the text. 6.
According to the problem statement, 5.39 eV of energy is required to make an Li ion from neutral Li, and 3.41 eV of energy is released when an F atom becomes an F ion. That means that a net energy input of 5.39 eV – 3.41 eV = 1.98 eV is needed to form the ions. We calculate the negative potential energy of the attraction between the two ions. 1 e2 2.30 1028 J m U 9.21eV 40 r 0.156 109 m1.60 1019 J eV The binding energy should therefore be 9.21 eV – 1.98 eV = 7.23 eV. But, the actual binding energy is only 5.95 eV. Thus, the energy associated with the repulsion of the electron clouds is 7.23 eV – 5.95 eV = 1.28 eV.
7.
(a) The neutral He atom has two electrons in the ground state, 𝑛 = 1, 𝓁 = 0, 𝑚𝓁 = 0. Thus, the two electrons have opposite spins, ms 12 . If we try to form a covalent bond, we see that an electron from one of the atoms will have the same quantum numbers as one of the electrons on the other atom. From the exclusion principle, this is not allowed, so the electrons cannot be shared. (b) We consider the He2 molecular ion to be formed from a neutral He atom and an He ion. It will have three electrons. If the electron on the ion has a certain spin value, it can have the opposite spin as one of the electrons on the neutral atom. Thus, those two electrons can be in the
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same spatial region (because their quantum numbers are not identical), and so a bond can be formed.
J s
2
8.
The MKS units of
are I
J2
J2
kg m kg m s2 m N m
J2 J. The final unit is joules, which
2
J
is an energy unit. 9.
The reduced mass is given in Section 40–5 as
m1m2 . We calculate in atomic mass units. m1 m2
39.10 u35.45u 18.59 u m1 m2 39.10 u 35.45u 16.00 u16.00 u mm 1 2 8.00 u m1 m2 16.00 u 16.00 u 1.008 u35.45u mm 1 2 0.9801u m1 m2 1.008 u 35.45u
(a) KCl: (b) O2: (c) HCl:
m1m2
2
r 10. The moment of inertia of H2 about its CM is I 2m H 2 2 m r H . The bond length is given in the text as 74 pm. 2 2 2 27 11 r m r 2 1.008 1.66 10 kg 7.4 10 m H 4.6 1048 kg m2 I 2mH 2 2 2
11. (a) Use the quantization of angular momentum to determine the angular velocity. The moment of 2 r mr 2 . inertia of the dumbbell about its CM is I 2m 2 2 𝑚𝑟2 𝐼𝜔 = √𝓁(𝓁 + 1)ℏ = 𝜔 → 2
𝜔=
2√𝓁(𝓁 + 1)ℏ 2√1(2)(1.055 × 10−34𝐽 · 𝑠) rad = = 3.3 × 10−23 2 𝑚𝑟2 −6 −3 𝑠 (1.0 × 10 kg)(3.0 × 10 𝑚)
(b) We must convert the angular velocity to rad/s for the calculation. We anticipate that 𝓁 will be quite large, so we approximate 𝓁(𝓁2 + 1) ≈ 𝓁2. 𝑚𝑟 𝐼𝜔 = √𝓁(𝓁 + 1)ℏ ≈ 𝓁ℏ = 𝜔 → 2
𝓁=
(1.0 × 10−6kg)(3.0 × 10−3𝑚)2 𝑚𝑟2 rev 2𝜋rad 1min × × ) ≈ 2.2 × 1023 𝜔= (50 2(1.055 × 10−34𝐽 · 𝑠) min 2ℏ rev 60𝑠
Our assumption about 𝓁 is justified.
Chapter 40
Molecules and Solids
mO
12. (a) The moment of inertia of O2 about its CM is given by 2 I 2m r mO r 2 . O 2 2
2I
mOr
16 1.66 10
r
1.055 10 J s kg 0.12110 m 1.60 10 34
mO
CM
2
J eV
1.789 104 eV
1.79 10 eV (b) From Fig. 40–17, we see that the energy involved in the 𝓁 = 3 to 𝓁 = 2 transition is 3 4
2
E 3 E h
2
6 I c
2I
4
3
2
I.
61.789 10 eV 1.0734 10 eV 1.07 10 eV
hc E
6.63 10 J s3.00 10 m s 1.16 10 m 34
3
8
3
13. Use the rotational energy and the moment of inertia of N2 about its CM to find the bond length. See the adjacent diagram. 2 2 I 2m 1 r 1 m r ; E N 2 2 N rot 2I mN r2
mN
mN
CM
r
1.10 1010 m
14. The energies involved in the transitions are given in Fig. 40–17. We find the rotational inertial from Eq. 40–4. The basic amount of rotational energy is 2 1.055 1034 J s 2 2 I m m 2 m r mH r2 kg u0.074 10 m m r 1 2 2.429 1021 J (a) For 𝓁 = 1 to 𝓁 = 0: 2 1 21 2 E 2.429 10 J 1.5 10 eV 19 1.60 10 J eV I hc 8.2 105 m 21 2.429 10 J E (b) For 𝓁 = 2 to 𝓁 = 1: 2 1 3.0 102 eV 19 E 2 2 2.429 10 21 J I 1.60 10 J eV 34 6.6310 J s3.00 108 m s hc 4.1105 m 21 E 2 2.429 10 J 2
(c) For 𝓁 = 3 to 𝓁 = 2:
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
1 4.6 102 eV 32.429 10 21 J 19 I 1.60 10 J eV 6.631034 J s3.00 108 m s hc 2.7 105 m E 32.429 1021 J 2
E 3
15. Use the value of the rotational inertia as calculated in Example 40–5. We also use Eq. 40–3. ℏ
2
(6.63×10−34𝐽·𝑠)
2
𝛥𝐸 = 𝐼 𝓁 = 4𝜋2(1.46×10−46kg·𝑚2) (5) = 3.813 × 10−22𝐽
𝑚 (6.63 × 10−34𝐽 · 𝑠) (3.00 × 108 ) ℎ𝑐 ℎ𝑐 𝑠 = 5.22 × 10−4𝑚 𝜆= = = = 𝑓 𝑓 ℎ 𝛥𝐸 (3.813 × 10−22𝐽) 𝑐
16. We first find the energies of the transitions represented by the wavelengths. 6.63 1034 J s3.00 108 m/s 5.38105 eV E hc
1.60 10 J/eV23.110 m 6.63 10 J s3.00 10 m/s E hc 10.72 10 eV
1
19
34
3
8
5
1.60 10 J/eV11.6 10 m 6.63 10 J s3.00 10 m/s 16.12 10 eV E 1.60 10 J/eV7.7110 m hc
2
19
34
3
19
3
8
5
3
E3 16.12 3, from the energy levels indicated in Fig. 40–17, and Since E2 10.72 2 and E1 5.38 E1 5.38 from the selection rule that 𝛥𝓁 = ±1, we see that these three transitions must represent the 𝓁 = 1 to 𝓁 = 0 transition, the 𝓁 = 2 to 𝓁 = 1 transition, and the 𝓁 = 3 to 𝓁 = 2 transition. Thus, E1 We use that relationship along with Eq. 40–4 to find the bond length. E1
r
I
r2
E1
1.055 1034 J 22.990 u35.453u 27 1.66 10 kg u 22.990 u 35.453u
10
2.36 10
5.38 105eV1.60 1019 J eV
m
17. The longest wavelength emitted will be due to the smallest energy change. From Fig. 40–17, the smallest rotational energy change is E We find the rotational inertia from Eq. 40–4. E
hc I 2 hc 4 c mm 2 I h m 1 m2 r 2 1 2 4 2 3.00 108 m s 6.941u1.008u 2 1.66 10 kg u 0.16 10 m 27 9 6.631034 J s 6.941u 1.008u
4 6.7 10 m
Chapter 40
Molecules and Solids
18. The energy change for transitions between combined rotational and vibrational states is given above Eq. 40–8a. E Evib Erot hf Erot If E hf is to be in the spectrum, then Erot 0. But, the selection rules state that 𝛥𝓁 = ±1 for a transition. It is not possible to have 𝛥𝓁 = 0 for a transition. The only way to have Erot 0 is for 𝛥𝓁 = 0, which is forbidden. Thus, E hf is not possible. Here is a mathematical statement as 2 2 well. )ℏ ( )ℏ ( ( )( ℏ2 −𝓁 𝓁+1 = 𝛥𝓁 2𝓁 + 𝛥𝓁 + 1 ) 𝛥𝐸rot = 𝐸(𝓁+𝛥𝓁) − 𝐸𝓁 = 𝓁 + 𝛥𝓁 𝓁 + 𝛥𝓁 + 1 2𝐼
2𝐼
2𝐼
For Erot 0, mathematically we must have𝛥𝓁 = 0, which is forbidden. 19. (a) The reduced mass is defined in Eq. 40–4. 12.01u16.00 u 6.86 u mm C O mC mO 12.01u 16.00 u (b) We find the effective spring constant from Eq. 40–5. 1 k f 2 k 4 f 2 4 2 6.42 1013 Hz 6.86 u1.66 1027 kg u 1850 N m 2
The spring constant for H2 is estimated in Example 40–6 as 550 N/m. kCO 850 N m 3.4 kH 550 N m 2 20. The effective spring constant can be found from Eq. 40–5 using the vibrational frequency and the reduced mass. 1 k m1m2 f k 4 f 2 4 2 f 2 m1 m2 2 4 2 1.7 1013 Hz
2
6.941u79.904 u 1.66 1027 kg u 120 N m 6.941u 79.904 u
21. (a) The curve for the function 2 U r 1 k r r 4.5eV is shown in Fig. 40– 2
0
18 as a dotted line. Measuring on the graph in the textbook with a ruler gives the distance from the origin to the 0.1 nm mark as 38 mm. The measured distance from the origin to the largest r-intercept of the parabola is 45 mm. Taking a ratio gives the distance from the origin to the largest r-intercept as 0.118 nm. We fit a parabolic curve to data.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
U 1 k r r 4.5eV ; U 2
0
2
U r 0.118 nm
k
Instructor Solutions Manual
1 k 0 4.5eV 4.5eV check 2
r r0 2
2
1 k 0.118 nm r 4.5eV 0 0
2
24.5eV 4649 eV2 1.60 1019 J 109 nm 2 nm eV 0.118 nm 0.074 nm 1m 743.8 N m 2
740 N m (b) The frequency of vibration is given by Eq. 14–7a using the reduced mass. Use this relationship to find the wavelength. c f 1 k 2
2c
0.51.00794 u1.66 1027 kg u 6 2 3.00 108 m s 2.0 10 m k 743.8 N m
22. Consider the system in equilibrium to find the center of mass. See the first diagram. The dashed line represents the location of the center of mass.
m2
m1
𝓁 = 𝓁1 + 𝓁2 ; 𝑚1𝓁1 = 𝑚2𝓁2 = 𝑚2(𝓁 − 𝓁1) → 𝑚2
𝓁 𝓁 ; 𝓁2 = 𝑚1 + 𝑚2 𝑚1 + 𝑚2 Now let the spring be stretched to the left and right, but let the center of mass be unmoved. 𝓁1 =
2
1
𝑚1
m1
m2 x1
2
1
x2
𝑥 = 𝑥1 + 𝑥2 ; 𝑚1(𝓁1 + 𝑥1) = 𝑚2(𝓁2 + 𝑥2) → 𝑚1𝓁1 + 𝑚1𝑥1 = 𝑚2𝓁2 + 𝑚2𝑥2 → 𝑚1𝑥1 = 𝑚2𝑥2 This is the2 second relationship requested in the2 problem. Now use the differential relationships. d x1 d 2 x2 d x k d 2 x2 k m1 2 kx ; m2 2 kx x ; x 21 2 dt dt dt m dt m 1 2 d 2 x1 x2 1 1 m1 m2 k d 2x d 2 x1 d 2 x2 kx kx x kx 2 2 dt2 dt2 dt dt m1m2 m1 m2 This last equation is the differential equation for simple harmonic motion, as in Eq. 14–3, with m
replaced by . The frequency is given by Eq. 14–7a, f
2
, which is the same as Eq. 40–5.
23. The ionic cohesive energy is given right after Eq. 40–9 and derived in the solution to Problem 28. The text states that the Madelung constant is 1.75 for NaCl. e 2 1 1 1.75 2.30 1028 J m 1 1 7.9 eV U0 4 r0 m 0.28 109 m1.60 1019 J eV 8
Chapter 40
Molecules and Solids
24. Each ion is at the corner of a cube of side length s, the distance between ions. From Fig.40–24, a cube of side length s would have 4 NaCl pairs. But, each ion is part of 8 cubes that share a common corner. Thus, any one “cube” has only the equivalent of one-half of an NaCl molecule. Use this to find the density, which is mass per unit volume. 1 m 2 NaCl s3 1/3 1 1/3 1 molecule cube58.44g mole 2 mNaCl 1mole 23 3 s 2 6.022 10 molecule 2.165g cm
2.819 108 m Note that Problem 26 quotes this value as 2.4 108 m . 25. Each ion is at the corner of a cube of side length s, the distance between ions. From Fig. 40–24, a cube of side length s would have 4 KCl pairs. But, each ion is part of 8 cubes that share a common corner. So, any one “cube” has only the equivalent of one-half of a KCl molecule. Use this to find the density, which is mass per unit volume. From the periodic table, the molecule weight of KCl is 74.55. 1 m 2 3KCl s 1/3 1 1 1/3 m mole 1mole s 2 KCl 2 molecule cube74.55g 23 3 3.15 108 m 6.02 10 molecule 1.99 g cm
26. According to Section 40–6, the NaCl crystal is face-centered cubic. It is illustrated in Fig. 40–24. We consider four of the labeled ions from Fig. 40–24. See the adjacent diagram. The distance from an Na ion to a Cl ion is labeled as d, and the distance from an Na ion to the nearest neighbor Na ion is called D. Dd 0.24 nm 0.34 nm 27. See the diagram. Select a charge in the middle of the chain. There will be two charges of opposite sign a distance r away, two charges of the same sign a distance 2r away, etc. Calculate the potential 2energy of the chosen charge. 2 2 1 2e2 1 2e 1 2e 1 2e 4 3r 4 4r U 4 r 4 2r 0 0 0 0 2e2 1 21 31 14 40r From Appendix A, ln1 x x 12 x2 13 x2 14 x4 ln1 1 1 21 13 14
ln 2 U
r
. Evaluate this expansion at x = 1. 2
2e 1 4 r
2e2 ln 2 4 r
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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2 From Section 40–6, the potential energy is given as U e . Equate the two expressions for the 40r potential energy to evaluate the Madelung constant. 2 e2 U 2e ln 2 2 ln 2 40r 40r
28. (a) Start with Eq. 40–9 and find the equilibrium distance, which minimizes the potential energy. Call that equilibrium distance r0. e2 e2 B B e2 B dU U ; m m 0 2 m1 m rm1 rr dr rr0 40 r2 r r 4 40 r r 0 0 0 0 B
e2rm1 0 40 m
e2 rm1 e2 B e2 40 m0 e2 1 U 0 U r r m m 4 r 1 m r 4 r r 4 r 0 0 0 0 0 0 0 0 0 (b) For NaI, we evaluate U0 with m = 10, 1.75, and r0 280.33nm. U e2 1 2.30 10 J m 0 1 1.75 1 101 6.861eV 4 r m 9 19 0.3310 m 1.60 10 J eV 0 0 6.9 eV 0.21nm. (c) For MgO, we evaluate U 0 with m = 10, 1.75, and r0 28 U e2 1 2.30 10 J m 0 1 101 10.78eV 4 r 1 m 1.75 0.21109 m1.60 1019 J eV 0 0 11eV (d) Calculate the % difference using m = 8 instead of m = 10. e 2 e 2 1 1 4 r 1 8 4 r 1 10 1 1 1 1 1 1 U0 U0 m8 m10 0 0 0 0 8 10 10 8 0.0278 e2 1 U 1 10 1 101 1 0m10 10 1 40 r0 2.8% 29. We follow Example 40–9. The density of states (number of states per unit volume in an infinitesimal energy range) is given by Eq. 40–10. Because we are using a small energy range, we estimate the calculation with a difference expression. We let N represent the number of states and V represent the volume under consideration. 3 / 2 8 2m 1/ 2 E V E N g E V E h3
8 2 9.1110
31
kg
3/ 2
6.63 10 J s 34
9.0 10 states 20
3
7.025eV 1.0 106 m3 0.05eV1.60 1019 J eV3 / 2
Chapter 40
Molecules and Solids
30. We follow Example 40–9. The density of occupied states (number of states per unit volume in an infinitesimal energy range) is given by Eq. 40–10. Because we are using a small energy range, we estimate the calculation with a difference expression. We let N represent the number of states and V represent the volume under consideration. 8 2m3/ 2 1/ 2 N g E V E E V E h3 E 12 EF 0.975EF 0.9875E ; E EF 0.975EF 0.025EF 0.137 eV F 3/ 2 8 2 9.111031 kg N
0.98755.48eV 1.0 106 m3 0.137 eV1.60 1019 J eV
3/ 2
6.63 10 J s
3
34
2.2 10 states 21
31. The density of molecules in an ideal gas can be found 5from the ideal gas law. 1.01310 Pa P N 25 3 2.576 10 m PV NkT V gas kT 1.38 1023 J K285K We assume that each copper atom contributes one free electron and use the density of copper as given in Table 13–1. 28 3 N 1e 6.02 1023 Cu atoms 8.9 103 3 kg Cu atom 63.546 103 kg V m 8.431 10 m e's Compare the two densities. N V 2.576 1025 m3 gas 3.1104 28 3 N 8.43110 m V e's 32. We use Eq. 40–14 for the occupancy probability, and solve for the energy. The Fermi energy is 7.0 eV. (a) Evaluate for T = 295 K. 1 f E E E kT e 1 1 1 1.38 1023 J K E kT ln 1 EF 1 7.0eV 295Kln 0.150 1.60 1019 J f E F
7.04 eV (b) Evaluate for T = 950 K. 1 f E E E kT e 1 1 1 1.38 1023 J K E kT ln 1 EF 1 7.0eV 950 K ln 0.150 1.60 1019 J f E F
7.14 eV
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
33. We use Eq. 40–14 for the occupancy probability, and solve for the energy. The Fermi energy is 7.0 eV. (a) Evaluate for T = 295 K. 1 f E E E kT e 1 1 1 1.38 1023 J K E kT ln 1 EF 1 7.0eV 295Kln 0.850 1.60 1019 J f E F
6.96eV (b) Evaluate for T = 750 K. 1 f E E E kT e 1 1 1.38 1023 J K 750 K ln 1 1 7.0eV E kT ln 1 EF 0.850 1.60 1019 J f E F
6.89 eV 34. The occupancy probability is given by Eq. 40–14. The Fermi level for copper is 7.0 eV. 1 1 1 0.0159 f E E / kT 1.015E E / kT 19 23 K 1 e 1 e 1 F
F
F
1.6% 35. We follow Example 40–10. We need the number of conduction electrons per unit volume of sodium. 23 N 3 6.02 10 atoms mol 28 3 970 kg m 22.99 103 kg mol 1free electrons atom 2.540 10 m V The Fermi energy is given by Eq. 40–12. 2 1 2/3 6.631034 J s 3 2.540 1028 m3 2 / 3 3.159 eV h2 3 N EF 19 8m V 89.111031 kg 1.60 10 J eV 3.2eV The Fermi speed is the speed of electrons with the Fermi energy. vF
23.159eV1.60 1019 J eV 2EF 1.1106 m s 31 m 9.1110 kg
36. We follow Example 40–10. (a) Because each zinc atom contributes two free electrons, the density of free electrons is twice the density of atoms. 23 N 7100 kg m 3 6.02 10 atoms mol 2 free electrons atom 1.307 1029 m3 3 V 65.409 10 kg mol 29 3 1.3 10 m
(b) The Fermi energy is given by Eq. 40–12.
Chapter 40
Molecules and Solids
h2 3 N EF 8m V
2/3
6.6310 J s 3 1.307 10 m 1 89.1110 kg 1.60 10 J eV 34
2
29
3
2/3
19
31
9.414eV 9.4eV (c) The Fermi speed is the speed of electrons with the Fermi energy.
vF
29.414eV1.60 1019 J eV 2EF 1.8 106 m s 31 m 9.1110 kg
37. (a) Find the density of free electrons from Eq. 40–12. 2/3 h2 3 N EF 8m V 3/ 2 31 19 N 8mE 3 / 2 89.1110 kg11.63eV1.60 10 J eV F V 3 h2 3 6.63 1034 J s2
1.7945 1029 m3 1.79 1029 m3 (b) Let n represent the valence number, so there are n free electrons per atom. N 6.02 10 atoms mol 270 kg m n free electrons atom 27.0 10 kg mol V 3 27.0 10 kg mol 1 29 3 n 23 3 1.7945 10 m 2.981 3 6.02 10 atoms mol 270 kg m This agrees nicely with aluminum’s position in the periodic table, and its electron configuration of 1s22s22p63s23p1. The level 3 electrons are the valence electrons. 38. We calculate the given expression with T = 0 so that the maximum energy is EF. The value of f (E) at T = 0 is given below Eq. 40–14. EF EF EF 8 2m3 / 2 E1/ 2 1dE EF E dE Eg E f E dE En E 0 E 3 / 2dE 0 h3 0 0 0 E E E E E F F F F 1/ 2 8 3/ 2 E 1 dE 2 E dE g E f E dE 3m n 1/ 2 0 h 0 E dE 0 0 0 5/ 2 2 5 EF 3 EF 5 3/ 2 2 3 EF 39. We start with Eq. 38–13 for the energy level as a function of n. If we solve for n, we have the number of levels with energies between 0 and E. Taking the differential of that expression will give the number of levels with energies between E and dE. Finally, we multiply by 2 since there can be 2 electrons (with opposite spins) in each energy level. ℎ2
8𝑚𝓁2
8𝑚𝓁2 1 𝑑𝐸
𝐸 = 𝑛 2 8𝑚𝓁2 → 𝑛 = √ ℎ2 √𝐸 → 𝑑𝑛 = √ ℎ2
2 √𝐸
√8𝑚𝓁 1 𝑑𝐸 ℎ2 2 √𝐸
→
2
𝑑𝑛 𝑔𝓁 = 2
𝑑𝐸
=2
𝑑𝐸
=
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
40. We first find the density of neutrons, and then use Eq. 40–12. N 1 neutron 1 2.51.99 1030 kg 4.1031044 m3 4 12,000 m3 V 1.675 1027 kg 3 2 3 34 44 3 2/3 6.6310 J s 4.10310 m 2 / 3 h2 3 N 1 EF 1.60 1013 J MeV 8m V 81.675 1027 kg 109.8 MeV 110 MeV
41. We use Eq. 40–14, with E EF. 1 1 1 1 f 0 E E / kT e 1 e 1 11 2 The result is independent of the value of T. F
42. (a) We use Eq. 40–14 with the data as given in the problem. 0.12 eV1.60 1019 J eV EE F 4.74848 kT k 1.38 10 23 J K 293K 1 1 1 f 8.590 103 8.6 103 4.74848 E E / kT 1 116.409 e 1 e This is reasonable. Very few states this far above the Fermi energy are occupied at this relatively low temperature. (b) Use a similar calculation to part (a). 0.12 eV1.60 1019 J eV EE F 4.74848 23 kT k 1.38 10 J K 293K 1 1 1 f 0.991409 0.99 4.74848 E E / kT 1 1.008665 e 1 e (c) Since the probability that the state is occupied is 0.991409, the probability that the state is F
F
unoccupied is 1 0.991409 8.591103 8.6 103 . This is the same as part (a). 43. (a) Eq. 38–13 gives the energy levels as a function of n, the number of levels. Since there are 2 electrons in every energy level, n = N / 2. The Fermi energy will be the highest energy level occupied with N electrons. 𝐸𝑛 = 𝑛
2
2 ℎ2 1 ℎ2𝑁2 ℎ2 = ( 𝑁) = 8𝑚𝓁2 2 8𝑚𝓁2 32𝑚𝓁2
ℎ2𝑁2
→ 𝐸𝐹 = 32𝑚𝓁
(b) The smallest amount of energy that this metal can absorb is the spacing between energy levels. ℎ2
ℎ2
𝛥𝐸 = 𝐸𝑛+1 − 𝐸𝑛 = 8𝑚𝓁2 [(𝑛 + 1)2 − 𝑛2] = 8𝑚𝓁2 (2𝑛 + 1) =
(𝑁 + 1)
(c) We calculateℎ2the limit requested. ( ) 𝑁+1 𝛥𝐸 8𝑚𝓁2 = ℎ2𝑁2 = 𝐸𝐹 𝑁 32𝑚𝓁2
For large N, this is a very small change in energy. Thus, a very small change in energy will allow an electron to change energy levels, and so the metal conducts very easily.
Chapter 40
Molecules and Solids
44. We consider the cube to be a three-dimensional infinite well, with a width of 𝓁 in each dimension. We apply the boundary conditions as in Section 38–8 separately to each dimension. Each dimension gives a quantum number which we label as n1, n2 , and n3. We then have a contribution to the energy of the bound particle from each quantum number, as in Eq. 38–13. ℎ2 + 𝑛2 ℎ2 + 𝑛2 ℎ2 = ℎ2 (𝑛2 + 𝑛2 + 𝑛2), 𝑛1, 𝑛2, 𝑛3 = 1,2,3, ⋯ 2 3 1 8𝑚𝓁2 2 8𝑚𝓁2 3 8𝑚𝓁2 8𝑚𝓁2 1
𝐸 = 𝐸1 + 𝐸2 + 𝐸3 = 𝑛2
Specifying the three quantum numbers gives a state and the corresponding energy. Choosing axes as specified in the problem, the equation of a sphere of radius R in that coordinate system is R2 n21 n22 n23. Each state “contained” in that sphere could be indicated by a cube of side length 𝓁, and each state can have two electrons (two spin states). The “volume” of that sphere is 1 8 of a full sphere. From that we calculate the number of states in one octant, and then g(E). 1 4
3
𝜋
2
2
2 3/2
𝜋 8𝑚𝓁2
3/2
𝑁 = 2 ( ) 𝜋𝑅 = (𝑛1 + 𝑛2 + 𝑛3) = 3 ( ℎ2 𝐸) 8 3 3 1 𝑑𝑁 1 𝜋 8𝑚𝓁2 3/2 3 𝜋 8𝑚 3/2 8√2𝜋𝑚3/2 1/2 1/2 𝐸 = ( 2 ) 𝐸1/2 = 𝐸 𝑔(𝐸) = = 3 ( 2 ) 2 2 ℎ 𝓁 3 ℎ ℎ3 𝑉 𝑑𝐸 45. The photon with the minimum frequency for conduction must have an energy equal to the energy gap. hc 6.631034 J s3.00 108 m/s 2.0eV Eg hf 1.60 1019 J/eV620 109m 46. The photon with the longest wavelength or minimum frequency for conduction must have an energy equal to the energy gap. c hc hc 6.631034 J s3.00 108 m/s 1.11106 m 1.11m f hf Eg 1.60 1019 J/eV1.12 eV 47. The energy of the photon must be greater than or equal to the energy gap. The lowest energy corresponds to the longest wavelength that will excite an electron. c hc hc 6.63 1034 J s3.00 108 m/s 1.7 106 m 1.7 m f hf E 1.60 1019 J/eV 0.72 eV g
Thus, the wavelength range is 1.7m. 48. The minimum energy provided to an electron must be equal to the energy gap. Divide the total available energy by the energy gap to estimate the maximum number of electrons that can be made to jump. 710 103 eV N hf 9.9 105 Eg 0.72eV
49. (a) In the 2s shell of an atom, 𝓁 = 0, so there are two states: ms 1 2. When N atoms form bands, each atom provides two states, so the total number of states in the band is 2N. (b) In the 2p shell of an atom, 𝓁 = 1, so there are three states from the 𝑚𝓁 values: m 0, 1; each
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
of which has two states from the ms values: ms 12, for a total of six states. When N atoms form bands, each atom provides six states, so the total number of states in the band is 6N. (c) In the 3p shell of an atom, 𝓁 = 1, so there are three states from the ml values: 𝑚𝓁 = 0, ±1; each of which has two states from the ms values: ms 12, for a total of six states. When N atoms form bands, each atom provides six states, so the total number of states in the band is 6N. (d) In general, for a value of 𝓁, there are 2𝓁 + 1 states from the 𝑚𝓁 values: 𝑚𝓁 = 0, ±1, . . . , ±𝑙. For each of these there are two states from the ms values: ms 12 , for a total of 2(2𝓁 + 1) states. When N atoms form bands, each atom provides 2(2𝓁 + 1) states, so the total number of states in the band is 2𝑁(2𝓁 + 1). 50. Calculate the number of conduction electrons in a mole of pure silicon. Also calculate the additional conduction electrons provided by the doping, and then take the ratio of those two numbers of conduction electrons. 3 28.09 10 kg mol 16 3 11 N 10 electrons m 1.206 10 electrons mole Si
Ndoping
2330 kg m3
6.02 10 atoms 23
4.6311017 added conduction electrons per mole.
6
1.310 17 N doping 4.63110 3.84 106 4 106 NSi 1.206 1011
51. The wavelength is found from the energy gap. c hc hc 6.631034 J s3.00 108 m/s 6.91 107 m 690 nm f hf Eg 52. The photon will have an energy equal to the energy gap: hc 6.631034 J s3.00 108 m/s Eg hf 2.0eV 53. From the current–voltage characteristic graph in Fig. 40–38, we see that a current of 16 mA means a voltage of about 0.69 V across the diode. The battery voltage is the sum of the voltages across the diode and the resistor. Vbattery Vdiode VR 0.69V 0.016A860 14.45V 14V 54. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We approximate the average current as half of the full rms current. V 120 V I av 12 rms 12 1.4 mA R 42 k (b) For a full-wave rectifier without a capacitor, the current is positive all the time. We approximate the average current as equal to the full rms current. V 120 V I av rms 2.9 mA R 42 k
Chapter 40
Molecules and Solids
55. The battery voltage is the sum of the voltages across the diode and the resistor. Vbattery Vdiode VR ; Vdiode
2.0 V 120 120 This is the equation for a straight line which passes through the points (0 V, 16.7 mA) and (0.8 V, 10 mA). The line has a y-intercept of 16.7 mA and a slope of 8.33 mA/V. If we assume the operating voltage of the diode is about 0.7 V, then the current is given from the equation above. 0.7 V 2.0 V I 1.08102 A 11mA 120 120 There is some approximation involved in the answer. 2.0 V Vdiode I 120Ω I
56. The band gap is the energy corresponding to the emitted wavelength. hc 6.631034 J s3.00 108 m s E 1.13eV 1.1eV 6 19 1.110 m1.60 10 J eV 57. We have copied the graph for V > 0 and rotated it so that it shows V as a function of I. This is the first diagram below. The resistance is the slope of that first graph. The slope, and thus the resistance, is very high for low currents and decreases for larger currents, approaching 0. As an approximate value, we see that the voltage changes from about 0.55 V to 0.65 V as the current goes from 0 to 10 mA. That makes the resistance about 10 ohms when the current is about 5 mA. The second diagram is a sketch of the resistance. 0.8 V volts
0.6
0.4 0.2
I mA 10
20
30
58. There will be a current in the resistor while the ac voltage varies from 0.6 V to 9.0 V rms. Because the 0.6 V is small, the voltage across the resistor will be almost sinusoidal, so the rms voltage across the resistor will be close to 9.0V 0.6V 8.4 V. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We ignore the short time it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in the resistor for about half the time. We approximate the average current as half of the full rms current. V I av 12 rms 1 8.4 V 28mA 2 R 0.150 kΩ (b) For a full-wave rectifier without a capacitor, the current is positive all the time. We ignore the short times it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in the resistor all the time. We approximate the average current as the full rms current.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Iav
Instructor Solutions Manual
Vrms 8.4 V 56 mA 0.150 kΩ R
59. (a) The time constant for the circuit is 1 RC1 24 103
3510 F 0.84s. As seen in 6
Fig. 40–40(c), there are two peaks per cycle. The period of the rectified voltage is 1 T , the voltage across the capacitor will be essentially T 120 s 0.0083s. Because 1 constant during a cycle, so the average voltage is the same as the peak voltage. The average current is basically constant. Vavg Vpeak 2 120 V I 7.07 mA 7.1mA avg R R R 24 103 (b) With a different capacitor, the time constant for the circuit changes. 2 RC2 24 10 3 0.10 10 6F 0.0024s
Now the period of the rectified voltage is about 3.5 time constants, and so the voltage will decrease to about 3% e3 of the peak value during each half-cycle. We approximate the voltage as dropping linearly from its peak value to 0 over each half-cycle, and so take the average voltage as half the peak voltage. Vavg 1 2Vrms 1 2 120 V 3.54 mA I 3.5 mA avg 2 2 R R 3 24 10 60. For a pnp transistor, both the collector and the base voltages are negative, and holes move from the emitter to the collector. The diagram for a pnp amplifier looks just like Fig. 40–49, with the polarity of eB and eC reversed, I B and IC flowing in opposite directions, and the emitter arrow pointing toward the base.
61. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage. 0.40 V V i R R Vout Vout 4211 4200 out C C C I iB iC 6 951.0 10 A 62. (a) The voltage gain is the collector ac voltage divided by the base ac voltage. V i R R 7.8k V C C C I C 65 133.4 V i R R 3.8k 130 B B B B (b) The power amplification is the output power divided by the input power. i V C C 65133.4 8672 8700 P I V iBVB 63. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage. V 750.080 V V V i R i out V input 2.4 104 A 0.24 mA out C C R R 25 103
Chapter 40
Molecules and Solids
64. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage.
Vout iCRC Ii BR C 85 2.0 106 A 3900 0.663V 0.66 V
65. The arrow at the emitter terminal, E, indicates the direction of current IE . The current into the transistor must equal to the current out of the transistor. IB IC IE 66. (a) The potential energy for the point charges is found from Eq. 23–10. 1 e2 1.60 1019 C1.60 1019 C U 9.0 109 N
0.27 109 m 1.60 1019 J eV
40 r
5.33eV 5.3eV (b) Because the potential energy of the ions is negative, 5.33 eV is released when the ions are brought together. The other energies quoted involve the transfer of the electron from the K atom to the F atom. 3.41 eV is released, and 4.34 eV is absorbed in the individual electron transfer processes. Thus, the total binding energy is as follows. Binding energy 5.33eV 3.41eV 4.34eV 4.4eV
67. We find the temperature from the given relationship. 2 4.0eV1.60 1019 J eV (a) K 3 kT T 2K 3.1104 K 2 23 3k 31.38 10 J K
T
(b) K kT 3 2
2K 3k
2 0.12 eV1.60 1019 J eV 930 K 31.38 1023 J K
68. For an electron confined in one dimension, we find the uncertainty in the momentum from Eq. 38–1, p
x
. The momentum of the electron must be at least as big as the uncertainty in the momentum,
so we approximate p
. Finally, we calculate the kinetic energy by K
x between the two kinetic energies based on the two position uncertainties. 2 K p 2m 2m x 2 1 2 K Kin atoms Kmolecule 2 1 x molecule x 2 34 1 1 1.055 10 J s 2 9 2 9.1110 kg 0.053 10 m 0.074 10 m
p2
. Find the difference
2m
119 molecule 1.60 10 J eV
6.62 eV There are two electrons, and each one has this kinetic energy difference, so the total kinetic energy difference is 26.62eV 13.2eV 13eV .
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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69. The diagram here is similar to Fig. 40–9 and Fig. 40–11. The activation energy is the energy needed to get the (initially) stable system over the barrier in the potential energy. The activation energy is 1.4 eV for this molecule. The dissociation energy is the energy that is released when the bond is broken. The dissociation energy is 1.6 eV for this molecule. 70. Vibrational states have a constant energy difference of Evib 0.54eV, as found in Example 40–7. Rotational states have a varying energy difference, depending on the 𝓁 value, of 𝛥𝐸rot = represents the upper energy state, as given in Eq. 40–3. Erot
2
ℏ2𝓁 𝐼
, where l
1.055 10 J s r 0.51.008u1.66 10 kg u0.074 10 m 1.60 10 J eV
34
2
2
2
I
0
2
27
9
19
0.0152eV Each gap, as represented in Fig. 40–17, is larger. We add those gaps until we reach 0.54eV. 𝓁 = 1 → 𝓁 = 0: 𝛥𝐸rot = 0.0152eV
∑ 𝛥𝐸rot = 0.0152eV
𝓁 = 2 → 𝓁 = 1: 𝛥𝐸rot = 2(0.0152eV) ∑ 𝛥𝐸rot = 3(0.0152eV) = 0.0456eV 𝓁 = 3 → 𝓁 = 2: 𝛥𝐸rot = 3(0.0152eV) ∑ 𝛥𝐸rot = 6(0.0152eV) = 0.0912eV 𝓁 = 4 → 𝓁 = 3: 𝛥𝐸rot = 4(0.0152eV) ∑ 𝛥𝐸rot = 10(0.0152eV) = 0.152eV 𝓁 = 5 → 𝓁 = 4: 𝛥𝐸rot = 5(0.0152eV) ∑ 𝛥𝐸rot = 15(0.0152eV) = 0.228eV 𝓁 = 6 → 𝓁 = 5: 𝛥𝐸rot = 6(0.0152eV) ∑ 𝛥𝐸rot = 21(0.0152eV) = 0.3192eV 𝓁 = 7 → 𝓁 = 8: 𝛥𝐸rot = 7(0.0152eV) ∑ 𝛥𝐸rot = 28(0.0152eV) = 0.4256eV 𝓁 = 8 → 𝓁 = 7: 𝛥𝐸rot = 8(0.0152eV) ∑ 𝛥𝐸rot = 36(0.0152eV) = 0.5472eV So, we see that rotational states from 𝓁 = 0 to 𝓁 = 7 can be “between” vibrational states, or a total of 8 rotational states .
71. The kinetic energy of the baton is 12 I 2 , and the quantum number can be found from Eq. 40–2. Let the length of the baton be d. We assume the quantum number will be very large. The rotational inertia about the center of mass is the sum of the inertias for a uniform rod 1 m d 2 and two point
1d
2
masses 2m end
2
12
bar
Chapter 40
Molecules and Solids
1 2
l I 2 2I 2I I 2 2 f 2m 1 d 1 m d 2 end bar 2 12 2 2 1 20.35 kg0.16 m 0.26 kg0.32 m
2 1.6s1
1.92 1033
1.055 1034 J s
12
The spacing between rotational energy levels is given by Eq. 40–3. We compare that value to the rotational kinetic energy. 𝓁ℏ2 2 𝛥𝐸 2 = 𝐼 = = = 1.0 × 10−33 𝓁2ℏ2 𝐸 2𝓁 (1.92 × 1033) 2𝐼 This is such a small difference that it would not be detectable, so no, we do not need to consider quantum effects. 72. (a) The equilibrium position is the location where the potential energy is a minimum. We find that location by setting the derivative of the potential energy equal to 0. 2 dU a rr a r r U U 0 1 e ; 2U 0 1 earr aearr 0 1 e 0 dr arr e 1 a r r0 0 r r0 0
0
0
0
0
The dissociation energy is the energy difference between the two states of equilibrium separation and infinite separation. 2 2 2 2 a r a r r U U r Urr0 U 0 1 e U 0 1 e U 0 1 0 U 0 11 (b) See the included graph. 0
0
0
8
U (eV)
6 4 2 0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
r/r 0
73. The Boltzmann factor indicates that the population of a state decreases as the energy of the state increases, according to N n eEn / kT . The rotation energy of states increases with higher 𝓁 values, according to Eq. 40–2. Thus states with higher values of 𝓁 have higher energies, and so there are fewer molecules in those states. Since the higher states are less likely to be populated, they are less likely to absorb a photon. As an example, the probability of absorption between 𝓁 = 1 and 𝓁 = 2 is more likely than that between 𝓁 = 2 and 𝓁 = 3, and so the peak representing the transition between 𝓁 = 1 and 𝓁 = 2 is higher than that between 𝓁 = 2 and 𝓁 = 3. The molecule is not rigid, and so the distance between the two ions is not constant. The moment of inertia depends on the bond length, and the energy levels depend on the moment of inertia. Thus, the energy levels are not exactly equally spaced.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
74. From Fig. 40–17, a rotational absorption spectrum would show peaks at energies of 2 2 We use the photon frequency at that 3 I , etc. Adjacent peaks are separated by an energy of energy to determine the rotational inertia. E
2
I
6.6310 h I 2 E hf 4 f 2
34
2
J s
1.81047 kg m2
75. (a) The reduced mass is defined in Eq. 40–4. 1.008 u35.453u 0.9801u m m H Cl 1.008 u 35.453u mH mCl (b) We find the effective spring constant from Eq. 40–5. 1 k f 2 k 4 f 2 4 2 8.66 1013 Hz 0.9801u1.6605 1027 kg u 482 N m 2
The spring constant for H2 is estimated in Example 40–6 as 550 N/m. kHCl 0.88 kH 2
76. An O 2 molecule can be treated as two point masses, 16 u each, and each a distance of 6.05 1011 m from the molecule’s center of mass. I mr2 2 16 u1.66 1027 kg u6.05 1011m 1.94 10 2
46
kg m2
77. From the diagram of the cubic lattice, we see that an atom inside the cube is bonded to the six nearest neighbors. Because each bond is shared by two atoms, the number of bonds per atom is 3 (as long as the sample is large enough that most atoms are in the interior, and not on the boundary surface). We find the heat of fusion from the energy required to break the bonds: number of bonds number of atoms E Lfusion atom bond mol 36.02 1023 atoms mol3.6 103 eV1.60 1019 J eV 1040 J mol 1.0 103 J mol 78. (a) We calculate the Fermi temperature for a Fermi energy of 7.0 eV, the Fermi energy of copper (as calculated in Example 40–10). 19 EF 7.0eV 1.60 10 J/eV TF 8.1104 K k 1.38 1023 J K (b) We are given that T TF , and we assume that eE / kT 1. 1 1 1 1 f E E e E / kT E kT E E E / kT kT kT kT kT e 1 e 1 e 1 e F
F
Chapter 40
Molecules and Solids
This is not useful for conductors like copper because the Fermi temperature is higher than the melting point, and we would no longer have a solid conductor. 79. From Eq. 25–13, the number of charge carriers per unit volume in a current is given by n
I evdrift A
,
where vdrift is the drift velocity of the charge carriers, and A is the cross-sectional area through which eH , where eH is the Hallthe carriers move. From Eq. 27–14, the drift velocity is given by v drift Bd effect voltage and d is the width of the strip carrying the current (see Fig. 27–32). The distance d is the shorter dimension on the “top” of Fig. 40–54. We combine these equations to find the density of charge carriers. We define the thickness of the current-carrying strip by t A d . 0.28 103 A1.3T n I IBd IB 1.264 1020 electrons m3 evdrift A
eeH A
1.60 10 C0.018V1.0 10 m 19
eeHt
3
The actual density of atoms per unit volume in the silicon is found from the density and the atomic weight. We let that density be represented by N. That density is used to find the number of charge carriers per atom. 6.02 1023 atoms 1mole 3 28 3 N 2330 kg m 4.994 10 atoms m 3 1mole 28.0855 10 kg n 1.264 1020 electrons m3 2.5 109 electrons atom 28 3 4.994 10 atoms m N 80. To use silicon to filter the wavelengths, wavelengths below the IR should cause the electron to be raised to the conduction band, so the photon is absorbed in the silicon. We find the shortest wavelength that will cause the electron to jump. c hc hc 1.11106 m 1.11 m f hf Eg Because this is in the IR region of the spectrum, the shorter wavelengths of visible light will excite the electron, and the photon would be absorbed. Thus, silicon can be used as a window. 81. The longest wavelength will be for the photon with the minimum energy, which corresponds to the gap energy. hc hc 6.63 1034 J s3.00 108 m/s 3.5 107 m Eg max 19 max Eg 3.6eV1.60 10 J/eV So, the photon must have 3.5107 m . 82. We use Eq. 40–11 with the limits given in order to determine the number of states. E2 E2 8 2m3 / 2 2 3 / 2 3/ 2 8 1/ 2 V E2 E13 / 2 N V g E dE 23m V E dE 3 h 3 h E E1
16 2 9.1110 kg 36.6310 J s 34
3.365 1025 31025
1
3/ 2
31
3
0.1m3 6.2 eV3 / 2 4.0eV3 / 2 1.60 1019 J eV
3/ 2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
83. The photon with the longest wavelength has the minimum energy, which should be equal to the gap energy. hc 6.63 1034 J s3.00 108 m/s 1.130eV 1.1eV Eg 1.60 1019 J/eV1100 109m If the energy gap is any larger than this, some solar photons will not have enough energy to cause an electron to jump levels. So, they will not be absorbed, making the solar cell less efficient. 84. The energy gap is related to photon wavelength by Eg hf hc . Use this for both colors of LED. 6.63 1034 J s3.00 108 m/s Green: Eg 2.37 eV 1.60 1019 J/eV525 109 m Blue:
6.63 1034 J s3.00 108 m/s E 2.67 eV 1.60 1019 J/eV465 109 m g
85. The photon with the maximum wavelength for conduction has an energy equal to the energy gap. hc 6.631034 J s3.00 108 m/s Eg hf 5.50eV 1.60 1019 J/eV226 109 m 86. (a) For the glass to be transparent to the photon, the photon’s energy must be < 1.12 eV, so the wavelength of the photon must be longer than the wavelength corresponding to 1.12 eV. hc Eband gap min
min
6.63 1034 J s3.00 108 m s 1.11106 m 1.11106 m E 19 1.12 eV1.60 10 J eV band gap hc
The minimum wavelength for transparency is in the infrared region of the spectrum. Since IR has longer wavelengths than visible light, the silicon would not be transparent for visible light. The silicon would be opaque, as in Example 40–14. (b) The minimum possible band gap energy for light to be transparent would mean that the band gap energy would have to be larger than the most energetic visible photon. The most energetic photon corresponds to the shortest wavelength, which is 400 nm in this problem. We treat the wavelength as being accurate to 2 significant figures. hc 6.63 1034 J s3.00 108 m s 3.1078eV 3.1eV Eband gap E min 400 109 m1.60 1019 J eV min
87. We assume that the value of 120 mA has 3 significant figures. (a) The current through the load resistor must be maintained at a constant value. 125V 5.952 mA V Iload output Rload 21.0 k At the minimum supply voltage, there will be no current through the diode, so the current through R is also 5.952 mA. The supply voltage is equal to the voltage across R plus the output voltage. VR Iload R 5.952 mA2.80 k 16.67 V Vsupply VR Voutput 16.67 V 125V 142 V min
Chapter 40
Molecules and Solids
At the maximum supply voltage, the current through the diode will be 120 mA, and so the current through R is Iload Idiode 5.952 mA 120 mA 125.952 mA . VR Iload R 125.952 mA2.80 k 353V Vsupply VR Voutput 353V 125V 478V max
(b) The supply voltage is fixed at 175 V, and the output voltage is still to be 125 V. The voltage across R is then fixed at 175V 125V 50V. We calculate the current through R. V 50 V 17.9 mA I R R R 2.80 k If there is no current through the diode, this 17.9-mA current will be in the load resistor. Note that the value of 17.9 mA only has 2 significant figures. V 125V Rload load 6.98k 7.0 k Iload 17.9 mA If Rload is less than this, there will be a greater current through R, meaning a greater voltage drop across R and a smaller voltage across the load. Thus, regulation would be lost, so 7.0k is the minimum load resistance for regulation. If Rload is greater than 7.0 k, the current through Rload will have to decrease in order for the voltage to be regulated, which means current must flow through the diode. The current through the diode is 17.9 mA when Rload is infinite, which is less than the diode maximum of 120 mA. Thus, the range for load resistance is 7.0 k Rload . 88. The voltage as graphed in Fig. 40–40c decays exponentially according to Eq. 26–9b. As suggested in the problem, we use a linear approximation for the decay using an expansion from Appendix A–3. From Fig. 40–40c, we see that the decay lasts for approximately one-half of a cycle before it increases back to the peak value. 1 1 2 60 s Vmin et / 1 t 1 t 1 0.97 t / V V e min peak Vpeak RC 7.8 103 36 106 F The voltage will decrease 3% from its maximum, or 1.5% above and below its average. 89. (a) The total potential energy is due to the electron–electron interaction, the proton–proton interaction, and 4 electron–proton interactions. e2 e2 1 e2 1 1 U Ue-e Up-p 4Up-e 4 40 d 40 r0 40 d r
8 e2 1 1 2 2 40 d r0 r d 0
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
(b) U has a minimum at d 0.043nm . U 0 for the
200 150
approximate range 0.011nm d 0.51nm.
100
U (eV)
(c) To find the point of greatest stability, set the derivative of U with respect to d (indicated by U ) equal to 0 and solve for d.
Instructor Solutions Manual
50 0 -50 -100 0
0.2
0.4
d (nm)
e2 1 1 8 2 r d 2 40 d r0 0 2 8 8d 1 e 1 1 2d e2 2 2 0 U 4 d 2 r 2 d 2 3 / 2 40 r 2 d 2 3 / 2 d 0 0 0 3 / 2 8d 1 2 8d 3 r02 d 2 2d r d 2 2 3/ 2 r0 d d
U
4d 2 r02 d 2 d
r0 0.074 0.0427 nm 3 3
0.6
0.8
Chapter 40
Molecules and Solids
CHAPTER 41: Nuclear Physics and Radioactivity Responses to Questions 1.
Different isotopes of a given element have the same number of protons and electrons. Because they have the same number of electrons, they have almost identical chemical properties. Each isotope has a different number of neutrons from other isotopes of the same element. Accordingly, they have different atomic masses and mass numbers. Since the number of neutrons is different, they may have different nuclear properties, such as whether they are radioactive or not.
2.
With 88 nucleons and 50 neutrons, there must be 38 protons. The number of protons is the atomic number, and so the element is strontium. The nuclear symbol is 3888 Sr .
3.
Identify the element based on the atomic number. (a) Uranium (Z = 92) (b) Nitrogen (Z = 7) (c) Hydrogen (Z = 1) (d) Strontium (Z = 38) (e) Berkelium (Z = 97)
4.
The number of protons is the same as the atomic number, and the number of neutrons is the mass number minus the number of protons. (a) Uranium: 92 protons, 232 – 9 = 140 neutrons (b) Nitrogen: 7 protons, 18 – 7 = 11 neutrons (c) Hydrogen: 1 proton, 1 – 1 = 0 neutrons (d) Strontium: 38 protons, 86 – 38 = 48 neutrons (e) Berkelium: 97 protons, 252 – 97 = 155 neutrons
5.
The atomic mass of an element as shown in the periodic table is the average atomic mass of all naturally occurring isotopes. For example, chlorine occurs as roughly 75% 35 Cl and 25% 37 Cl , and 17
17
so its atomic mass is about 35.5 (= 0.75 35 + 0.25 37). Other smaller effects would include the fact that the masses of the nucleons are not exactly 1 atomic mass unit, and that some small fraction of the mass energy of the total set of nucleons is in the form of binding energy. 6.
The nucleus of an atom consists of protons (which carry a positive electric charge) and neutrons (which are electrically neutral). The electric force between protons is repulsive and much larger than the force of gravity. If the electric and gravitational forces are the only two forces present in the nucleus, the nucleus would be unstable as the electric force would push the protons away from each other. Nuclei are stable, and therefore, another force must be present in the nucleus to overcome the electric force. This force is the strong nuclear force.
7.
The strong force and the electromagnetic (EM) force are two of the four fundamental forces in nature. They are both involved in holding atoms together: the strong force binds quarks into nucleons and binds nucleons together in the nucleus; the EM force is responsible for binding negatively charged electrons to positively charged nuclei and for binding atoms into molecules. The strong force is the strongest fundamental force; the EM force is about 100 times weaker at distances on the order of 1017 m. The strong force operates at short range and is negligible for distances greater than about the size of the nucleus. The EM force is a long-range force that decreases as the inverse square of the distance between the two interacting charged particles. The EM force operates
only between charged particles. The strong force is always attractive; the EM force can be attractive or repulsive. Both these forces have mediating field particles associated with them – the gluon for the strong force and the photon for the EM force. 8.
Quoting from Section 41–3, “… radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong chemicals.” Chemical reactions are a result of electron interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to electron interactions. Another piece of evidence is the fact that the -particle emitted in many radioactive decays is much heavier than an electron and has a different charge than the electron, so it can’t be an electron. Therefore, it must be from the nucleus. Finally, the energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron orbital transitions. All of these observations support radioactivity being a nuclear process.
9.
The resulting nuclide for gamma decay is the same isotope in a lower energy state: 64 Cu* 64 Cu γ . 29
29
The resulting nuclide for beta-minus decay is an isotope of zinc, 6430Zn : 64 Cu 64 Zn e . 29
30
The resulting nuclide for beta-plus decay is an isotope of nickel, 6428Ni : 64 Cu 64 Ni e . 29
10.
238
28
U decays by alpha emission into 234 Th, which has 144 neutrons.
92
90
11. The alpha particle is a very stable nucleus. It has less energy when bound together than when split apart into separate nucleons. In most cases, more energy is required to emit separate nucleons than an alpha particle. It is usually true that the emission of a single nucleon is energetically not possible. See Example 41–5. 12. Alpha (α) particles are helium nuclei. Each α particle consists of 2 protons and 2 neutrons, and therefore, it has a charge of +2e and an atomic mass value of 4 u. They are the most massive of the
or positrons . Electrons have a charge of –e and
three. Beta (β) particles are electrons
positrons have a charge of +e. In terms of mass, beta particles are much lighter than protons or neutrons, by a factor of about 2000, so are lighter than alpha particles by a factor of about 8000. Their emission is always accompanied by either an anti-neutrino (in decay) or a neutrino (in decay). Gamma (γ) particles are photons. They have no rest mass and no charge. 24 13. (a) Magnesium is formed: 24 11 Na 12 Mg e . 22
82
2
(b) Neon is formed: 22 11 Na 10 Ne e . 210 (c) Lead is formed: Po 206Pb 4 He . 84
32 14. (a) Sulfur is formed: 32 15 Pb 16 S e . 35 (b) Chlorine is formed: 16 S 35 17Cl e . (c) Thallium is formed: 211Bi 207Tl 4 He . 83
81
2
Chapter 41
Nuclear Physics and Radioactivity
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Chapter 41
Nuclear Physics and Radioactivity
15. (a)
45 20
Ca
(b)
58 29
Cu*
(c)
46 24
45 21
Sc e v
Scandium-45 is the missing nucleus.
Cu
Copper-58 is the missing nucleus.
58 29
Cr 46 V e v
U
(d)
234 94
Pu
(e)
239
Np 239 Pu
93
The positron and the neutrino are the missing particles.
23
230 92
Uranium-230 is the missing nucleus. The electron and the anti-neutrino are the missing particles.
94
16. The two extra electrons held by the newly formed thorium will be very loosely held, as the number of protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will be easy for these extra two electrons to escape from the thorium atom through a variety of mechanisms. They are in essence “free” electrons. They do not gain kinetic energy from the decay. They might get captured by the alpha nucleus, for example. 17. When a nucleus undergoes either or decay it becomes a different element, since it has either converted a neutron to a proton or a proton to a neutron. Thus, its atomic number (Z) has changed. The energy levels of the electrons are dependent on Z, and so all of those energy levels change to become the energy levels of the new element. Photons (with energies on the order of a few eV) are likely to be emitted from the atom as electrons change energies to occupy the new levels. 18. Alpha particles from an alpha-emitting nuclide are part of a two-body decay. The energy carried off by the decay fragments is determined by the principles of conservation of energy and of momentum. With only two decay fragments, these two constraints require the alpha particles to be monoenergetic. Beta particles from a beta-emitting nucleus are part of a three-body decay. Again, the energy carried off by all of the decay fragments is determined by the principles of conservation of energy and of momentum. However, with three decay fragments, the energy distribution between the fragments is not determined by these two constraints. The beta particles will therefore have a range of energies. 19. In electron capture, the nucleus will effectively have a proton change to a neutron. This isotope will then lie to the left and above the original isotope. Since the process would only occur if it made the nucleus more stable, it must lie BELOW the line of stability in Fig. 41–2. 20. Neither hydrogen nor deuterium can emit an particle. Hydrogen has only one nucleon (a proton) in its nucleus, and deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither one has the necessary four nucleons (two protons and two neutrons) to emit an particle. 21. Many artificially produced radioactive isotopes are rare in nature because they have decayed away over time. If the half-lives of these isotopes are relatively short in comparison with the age of Earth (which is typical for these isotopes), there won’t be any significant amount of these isotopes left to be found in nature. Also, many of these isotopes have a very high energy of formation, which is generally not available under natural circumstances. 22. After two months, the sample will not have completely decayed. After one month, half of the sample will remain, and after two months, one-fourth of the sample will remain. Each month, half of the remaining atoms decay. 23. For Z > 92, the short range of the attractive strong nuclear force means that no number of neutrons is able to overcome the electrostatic repulsion of the large concentration of protons.
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24. There are a total of 4 protons and 3 neutrons in the reactants. The -particle has 2 protons and 2 3 neutrons, and so 2 protons and 1 neutron are in the other product particle. It must be 2 He . 6 1 4 3 Li p He 3
1
2
2
14
25. The technique of 6C could not be used to measure the age of stone walls and tablets. Carbon-14 dating is only useful for measuring the age of objects that were living at some earlier time. Stone walls and tablets were never alive. 26. The decay series of Fig. 41–12 begins with a nucleus that has many more neutrons than protons and lies far above the line of stability in Fig. 41–2. In a β+ decay, a proton is converted to a neutron, which would take the nuclei in this decay series farther from the line of stability and is not energetically preferred. 27. There are four alpha particles and four β– particles (electrons) emitted, no matter which decay path is chosen. The nucleon number drops by 16 as 222 Rn decays into 206 Pb , indicating the presence of four 86
82
alpha decays. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha decays would result in a decrease of eight protons. Four β– decays will convert four neutrons into protons, making the decrease in the number of protons only four, as required. (See Fig. 41–12.) 28. (i)
Since the momentum before the decay was 0, the total momentum after the decay will also be 0. Since there are only 2 decay products, they must move in opposite directions with equal magnitude of momentum. Thus, (c) is the correct choice: both the same. (ii) Since both products have the same momentum, the one with the smallest mass will have the greater velocity. Thus, (b) is the correct choice: the alpha particle. (iii) We assume that the products are moving slowly enough that classical mechanics can be used. In 𝑝2
that case, k = . Since both particles have the same momentum, the one with the smallest 2𝑚 mass will have the greater kinetic energy. Thus, (b) is the correct choice: the alpha particle. 29. Fig. 41–7 shows the potential energy curve for an alpha particle and daughter nucleus for the case of radioactive nuclei. The alpha particle tunnels through the barrier from point A to point B in the figure. In the case of stable nuclei, the probability of this happening must be essentially zero. The maximum height of the Coulomb potential energy curve must be larger and/or the Q-value of the reaction must be smaller so that the probability of tunneling is extremely low. 30. If the Earth had been bombarded with additional radiation several thousand years ago, there would have been a larger abundance of carbon-14 in the atmosphere at that time. Organisms that died in that time period would have had a greater percentage of carbon-14 in them than organisms that die today. Since we assumed the starting carbon-14 was the same, we would have previously underestimated the age of the organisms. With the new discovery, we would re-evaluate the organisms as older than previously calculated. 31. (a) An einsteinium nucleus has 99 protons and a fermium nucleus as 100 protons. If the fermium undergoes either electron capture or + decay, a proton would in effect be converted into a neutron. The nucleus would now have 99 protons and be an einsteinium nucleus. (b) If the einsteinium undergoes decay, a neutron would be converted into a proton. The nucleus would now have 100 protons and be a fermium nucleus.
Chapter 41
Nuclear Physics and Radioactivity
32. In decay, an electron is ejected from the nucleus of the atom, and a neutron is converted into a proton. The atomic number of the nucleus increases by one, and the element now has different chemical properties. In internal conversion, an orbital electron is ejected from the atom. This does not change the atomic number of the nucleus, nor its chemical properties. Also, in decay, both a neutrino and an electron will be emitted from the nucleus. Because there are three decay products (the neutrino, the particle, and the nucleus), the momentum of the particle can have a range of values. In internal conversion, since there are only two decay products (the electron and the nucleus), the electron will have a unique momentum and, therefore, a unique energy.
MisConceptual Questions 1.
(a) A common misconception is that the elements of the period table are distinguished by the number of electrons in the atom. This misconception arises because the number of electrons in a neutral atom is the same as the number of protons in its nucleus. Elements can be ionized by adding or removing electrons, but this does not change what type of element it is. When an element undergoes a nuclear reaction that changes the number of protons in the nucleus, the element does transform into a different element.
2.
(b) The role of energy in binding nuclei together is often misunderstood. As protons and neutrons are added to the nucleus, they release some mass energy, which usually appears as radiation or kinetic energy. This lack of energy is what binds the nucleus together. To break the nucleus apart, the energy must be added back in. As a result, a nucleus will have less energy than the protons and neutrons that constitute the nucleus.
3.
(c) Regarding (a), large nuclei (with many more neutrons than protons) are typically unstable, so increasing the number of nuclei does not necessarily make the nucleus more stable. Regarding (b), nuclei such as 14 O have more protons than neutrons, yet 14 O is not as stable as 16 O . 8
8
8
Therefore, having more protons than neutrons does not necessarily make a nucleus more stable. The large Coulomb repulsion between the protons helps lead to instability. Regarding (d), there are no electrons in the nucleus. Electrons do not affect the stability of nuclei. Regarding (e), large unstable nuclei, such as 238 94 Pu, have a much larger total binding energy than small stable nuclei such as 42 He, so the total binding energy is not a measure of stability. The correct answer is (c). Stable nuclei typically have large binding energies per nucleon, which means that each nucleon is more tightly bound to the others. 4.
(e) The Coulomb repulsive force does act inside the nucleus pushing the protons apart. Another larger attractive force is thus necessary to keep the nucleus together. The force of gravity is far too small to hold the nucleus together. Neutrons are not negatively charged. It is the attractive strong nuclear force that overcomes the Coulomb force to hold the nucleus together.
5.
(b) The exponential nature of radioactive decay is a concept that can be misunderstood. It is sometimes thought that the decay is linear, such that the time for a substance to completely decay is twice the time for half of the substance to decay. Radioactive decay is not linear but exponential. That is, during each half-life, 1/2 of the remaining substance decays. If half the original decays in the first half-life, then 1/2 remains. During the second half-life, 1/2 of what is left decays, which would be 1/4 of the initial substance. In each subsequent half-life, half of the remaining decays, so it takes many half-lives for a substance to effectively decay away. The decay constant is inversely related to the half-life.
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6.
(e) The half-life is the time it takes for half of the substance to decay away. The half-life is a constant determined by the composition of the substance and not on the quantity of the initial substance. As the substance decays, the number of nuclei decreases, the activity (number of decays per second) decreases, but the half-life remains constant.
7.
(d) A common misconception is that it would take twice the half-life, or 20 years, for the substance to completely decay. This is incorrect because radioactive decay is an exponential process. That is, during each half-life, 1/2 of the remaining substance decays. After the first 10 years, 1/2 remains. After the second ten-year period, 1/4 remains. After each succeeding half-life, another half of the remaining substance decays, leaving 1/8, then 1/16, then 1/32, and so forth. The decays stop when none of the substance remains. But after a time, there will be too few nuclei to reliably use statistics. Thus, the time cannot be exactly determined.
8.
(c) A common misconception is that after the second half-life, none of the substance remains. However, during each half-life, 1/2 of the remaining substance decays. After one half-life, 1/2 remains. After two half-lives, 1/4 remains. After three half-lives, 1/8 remains.
9.
(a) The decay constant is proportional to the probability of a particle decaying and is inversely proportional to the half-life. Therefore, the substance with the shorter half-life (Sr) has the larger decay constant and the larger probability of decaying. The activity is proportional to the amount of the substance (number of atoms) and the decay constant, so the activity of Tc will be smaller than the activity of Sr.
10. (d) The element with the largest decay constant will have the shortest half-life. Converting each of the choices to decays per second yields: (a) 100/s, (b) 1.610–7/s, (c) 2.510–9/s, (d) 1.2104/s. Answer (d) has the largest decay rate, so it will have the smallest half-life. 11. (b) The half-life of radium remains constant and is not affected by temperature. Radium that existed several billion years ago will have essentially completely decayed. Small isotopes, such as carbon-14, can be created from cosmic rays, but radon is a heavy element. Lightning is not energetic enough to affect the nucleus of the atoms. Heavy elements such as plutonium and uranium can decay into radium and thus are the source of present-day radium. 12. (a) The nature of mass and energy in nuclear physics is often misunderstood. When the neutron and proton are close together, they bind together by releasing mass energy that is equivalent to the binding energy. This energy comes from a reduction in their mass. Therefore, when the neutron and proton are far from each other, their net mass is greater than their net mass when they are bound together.
Solutions to Problems 1.
Convert the units from MeV c2 to atomic mass units. 1u m 775MeV c2 0.832 u 2 931.49 MeV c
Chapter 41
2.
Nuclear Physics and Radioactivity
The particle is a helium nucleus and has A = 4. Use Eq. 41–1.
15
r 1.2 10
3.
1
15
m A 1.2 10 3
1
m 4 3 1.9 10
15
m 1.9 fm
The radii of the two nuclei can be calculated with Eq. 41–1. Take the ratio of the two radii. 1/3 1/3 1.2 1015 m214 r 214 214 1.0128 1/3 15 r206 1.2 10 m206 206 206 So, the radius of 214 82 Pb is 1.28% 1% larger than the radius of 82 Pb.
4.
Use Eq. 41–1 for both parts.
15 5.9 10 m 5.9fm (a) r 1.2 10 m A 1.2 10 m 120 3 3 r 3.7 1015 m 15 1/ 3 29.3 29 A (b) r 1.2 10 m A 15 15 1.2 10 m 1.2 10 m
5.
15
1/3
15
1/3
To find the rest mass of an particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m mHe 2me 4.002603u931.5 MeV uc2 2 0.511MeV c2 3727 MeV c2
This is less than the sum of the masses of two protons and two neutrons because of the binding energy. 6.
Each particle would exert a force on the other through the Coulomb electrostatic force. The distance between the particles is twice the radius of one of the particles. The Coulomb force is given by Eq. 21–2. 2 9 2 2 19 8.988 10 N m C 21.60 10 C 1 F q q 63.41N 63 N 2 40 2r 2 24 1/3 1.2 10 15 m
The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha calculated in Problem 5. F 63.41N 9.5 1027 m s2 F ma a 27 kg m 2 1.6605 10 3727 MeV c 2 931.49 MeV c 7.
First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is approximately (A u) and its radius is r 1.2 1015 m A1/ 3. Calculate the density. m 27 27 A1.6605 10 kg A1.6605 10 kg 2.294 1017 kg m3
1.2 1015 m A We see that this is independent of A. The value has 2 significant figures. V
4 3
r3
4 3
3
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(a) We set the density of the Earth equal to the density of nuclear matter. Earth Earth nuclear 4 M R3 matter
3
Earth 1/3
1/3 5.98 1024 kg 183.9 m M Earth R 180 m 4 nuclear Earth 2.294 10 kg m 3 matter (b) Set the density of Earth equal to the density of uranium, and solve for the radius of the uranium. Then compare that to the actual radius of uranium, using Eq. 41–1, with A = 238. M M Earth mU 4 Earth 3 4 mU 3 R r 3 3 REarth rU 3 Earth 3 U 1/3 238u1.6605 1027 kg u 1/3 6 mU 6.38 10 m 2.58 1010 m r R 24 5.98 10 kg U Earth M Earth rU 2.58 1010 m 3.5 104 1/3 15 rU actual 1.2 10 238
8.
Use Eq. 41–1 to find the value for A. We use uranium-238 since it is the most common isotope. 1.2 1015 m A1/3 3 r unknown 0.5 A 2380.5 29.75 30 rU 1.2 1015 m2381/3 From Appendix G, a stable nucleus with A 30 is 3115P.
9.
The basic principle to use is that of conservation of energy. We assume that the centers of the two particles are located a distance from each other equal to the sum of their radii. That distance is used to calculate the initial electrical potential energy. Then, we also assume that since the Rf nucleus is much heavier than the alpha, the alpha has all of the final kinetic energy when the particles are far apart from each other (and so have no potential energy). 1 q qRf K 0 K U K U 0 i i f f 40 r rRf
21041.60 10 C 3.118 10 eV K 8.988 10 N m C 4 263 1.2 10 m1.60 10 J eV 19
9
2
2
7
2
1/3
1/3
15
19
31MeV 10. (a) The hydrogen atom is made of a proton and an electron. Use values from Appendix G. mp 1.007276 u 0.9994553 99.95% mH 1.007825u (b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by Eq. 41–1. For the atomic radius, we use the Bohr radius3 given in Eq. 37–12. 3 15 3 r 1.2 10 m Vnucleus 34 rnucleus nucleus 1.2 1014 3 4 Vatom 10 3 r atom ratom 0.5310 m
Chapter 41
Nuclear Physics and Radioactivity
11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.01 atomic mass unit. Therefore, in a 1.0-kg object, (1.0 kg) 6.022 1026 u kg N 5.96 1026 6 1026 nucleons 1.01u nucleon No , it does not matter what the element is because the mass of one nucleon is essentially the same for all elements. 12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha just touches the thorium nucleus. Assume the two nuclei are at rest when they “touch”. The distance between the two particles is the sum of their radii. 1 q qTh K U K U K 0 0 i i f f 40 r rTh K 8.988 10 N m C 9
2
2
2901.60 1019 C
2
2.790 107 eV
4 232 1.2 10 m1.60 10 J eV 1/3
1/3
15
19
28 MeV 13. From Fig. 41–1, we see that the average binding energy per nucleon at A = 63 is about 8.7 MeV. Multiply this by the number of nucleons in the nucleus. 638.7 MeV 548.1MeV 550 MeV
14. (a) From Fig. 41–1, we see that the average binding energy per nucleon at A 238 is 7.5MeV. Multiply this by the 238 nucleons. 2387.5MeV 1785MeV 1800 MeV (b) From Fig. 41–1, we see that the average binding energy per nucleon at A 84 is 8.7 MeV. Multiply this by the 84 nucleons. 848.7 MeV 730.8MeV 730MeV 15.
18 8
O consists of 8 protons and 10 neutrons. We find the binding energy from the masses of the components and the mass of the nucleus, from Appendix G. Note that the electron masses from the hydrogens cancel out the electron masses for the oxygen. 2 8 c Binding energy 8m 1 1H 10m 1 n0 m 18 O
81.007825u 101.008665u 17.999160 u c2 931.5MeV c2 0.150090 139.81MeV u
Binding energy per nucleon 139.81MeV 18 7.767 MeV
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16. Deuterium consists of 1 proton, 1 neutron, and 1 electron. Ordinary hydrogen consists of 1 proton and 1 electron. Use the atomic masses from Appendix G. The electron masses cancel. 1 2 2 1 m 0n m H 1 c Binding energy m 1 H
931.5 MeV c2 1.007825u 1.008665u 2.014102 u c 2 u 2.224 MeV
17. We find the binding energy of the last neutron from the masses of the isotopes. Binding energy m 3115P m 1 0n m 3215P c2
30.973762 u 1.008665u 31.973908 u c2 931.5 MeV / c2 7.935MeV
18. (a)
7 4
Be consists of 4 protons and 3 neutrons. We find the binding energy from the masses, using hydrogen atoms in place of protons so that we account for the mass of the electrons. 1 7 2 1 3m n 0 m Be 4 c Binding energy 4m 1 H
41.007825 u 31.008665 u 7.016929 u c2 931.5 MeV / c2 37.60 MeV
Binding energy 37.60 MeV 5.372 MeV nucleon nucleon 7 nucleons (b) 197 79 Au consists of 79 protons and 118 neutrons. We find the binding energy as in part (a). Binding energy 79m 11H 118m 1 n0 m 19779Au c2
791.007825 u 1181.008665 u 196.966570 u c2 931.5 MeV / c2 1559.4 MeV 1559 MeV
Binding energy 1559.4 MeV 7.916 MeV nucleon nucleon 197 nucleons 19.
31 15
P consists of 15 protons and 16 neutrons. We find the binding energy from the masses:
Binding energy 15 m 11H 16m 10n m 3115P c2
151.007825 u 161.008665 u 30.973762 u c2 931.5MeV uc2
262.9 MeV Binding energy 262.9 MeV 8.481MeV nucleon nucleon 31 We do a similar calculation for 3215P, consisting of 15 protons and 17 neutrons. 1 32 2 Binding energy 15 m 1H 1 17m n 0 m 15P c
151.007825u 17 1.008665u 31.973908 u c2 931.5 MeV uc2 270.85 MeV
Chapter 41
Nuclear Physics and Radioactivity
Binding energy 270.85 MeV 8.464 MeV nucleon nucleon 32 31 By this measure, the nucleons in P are more tightly bound than those in 32 P. Thus, we expect 31 P 15 15 15 to be more stable than 32 P. Appendix G bears that out – 32 P is radioactive, while 31 P is stable. 15
15
15
20. We find the required energy by calculating the difference in the masses. (a) Removal of a proton creates an isotope of nitrogen. To balance electrons, the proton is included as a hydrogen atom: 168 O 11H 157 N. 2 Energy needed m 157N m 11H m 16 O 8 c
15.000109 u 1.007825 u 15.994915 u 931.5 MeV uc2 12.13MeV
(b) Removal of a neutron creates another isotope of oxygen: 168O 10n 158O. 2 Energy needed m 158O m 10n m 16 O 8 c
15.003066 u 1.008665 u 15.994915 u 931.5 MeV uc2 15.66 MeV
The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the proton because there is also the repulsive electric force from the other protons “helping” to remove the proton. 21. (a) We find the binding energy from the masses. Binding Energy 2m
He m Be c2 4
2
8
4
2 4.002603u 8.005305u c2 931.5 MeV uc2 0.092 MeV Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy state as two alphas instead of a beryllium. (b) We find the binding energy from the masses. 2 Binding Energy 3m 42He m 12 C 6 c
34.002603u 12.000000 u c2 931.5 MeV uc2 7.3MeV Because the binding energy is positive, the nucleus is stable. 22. The wavelength is determined from the energy change between the states. c hc 2.6 1012 m E hf h E 0.48 MeV1.60 1013 J MeV
0 23. The nuclear decay is 31 H 32He 1 e v. When we add one electron to both sides in order to use
atomic masses, we see that the mass of the emitted particle is included in the atomic mass of 3 He. 2 The energy released is the difference in the masses. Energy released m 31H m 32He c2
3.016049 u 3.016029 u c2 931.5 MeV uc2 0.019 MeV
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24. For the decay 116C 105B 11p, we find the difference of the final and initial final masses. To balance the electrons (so that we can use atomic masses), we assume the proton is an atom of 11 H. m m 105B m 11 H m 116C
10.012937 u 1.007825 u 11.011433u 0.009329 u Since the final masses are more than the original mass, energy would not be conserved. 0 25. The decay is 01 n 11p 1 e v. The electron mass is accounted for if we use the atomic mass of 1 H as the combination of the proton and electron. If we ignore the recoil of the proton and the 1 neutrino, and any1 possible 1mass of the neutrino, we get the maximum kinetic energy. K m n m H c2 1.008665u 1.007825u c2 931.5 MeV uc2 max 1 0
0.782 MeV 26. For each decay, we find the difference of the initial and the final masses. If the final mass is more than the initial mass, the decay is not possible. 1 233 (a) m m 232 92 U m0 n m 92 U 232.037155u 1.008665u 233.039634 u 0.006816 u Because an increase in mass is required, the decay is not possible.
(b) m m 137 N m 01 n m 147 N 13.005739 u 1.008665u 14.003074 u 0.011330 u Because an increase in mass is required, the decay is not possible. 39 (c) m m 19 K m01 n m1940 K 38.963706 u 1.008665u 39.963998u 0.008373u
Because an increase in mass is required, the decay is not possible. 24 27. (a) From Appendix G, 11 Na is a emitter .
(b) The nuclear decay reaction is
. We add 11 electrons to both sides in
order to use atomic masses. Then, the mass of the beta is accounted for in the mass of the magnesium. The maximum kinetic energy of the corresponds to the neutrino having no kinetic energy (a limiting case) and no mass. We also ignore the recoil of the magnesium. K m 24Na m 24Mg c 2 11
12
23.990963u 23.985042 u c2 931.5 MeV uc2 5.515 MeV 28. (a) We find the final nucleus by balancing the mass and charge numbers. Z X Z U Z He 92 2 90 A X A U AHe 238 4 234 Thus, the final nucleus is 234 90 Th . (b) If we ignore the recoil of the thorium, the kinetic energy of the particle is equal to the Qvalue of the reaction. The electrons are balanced. K Q m
U m Th m He c2 238 92
234 90
4 2
Chapter 41
Nuclear Physics and Radioactivity
m 234 Th m 238 U m 4 He 90
92
K
2
c2 4.20 MeV 1u 238.050787 u 4.002603u 2 c2 931.5 MeV c 234.043675u
This answer assumes that the 4.20 MeV value does not limit the sig. fig. of the answer. 60 29. The nuclear reaction is 60 27 Co 28Ni . We add 27 electrons to both sides of the equation so that we can use atomic masses. The kinetic energy of the will be maximum if the (essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel.
K m 2760 Co m 6028Ni c2
931.5MeV c2 59.933816 u 59.930785u c 2 2.823MeV u
30. We add three electron masses to each side of the nuclear reaction 74 Be 10 e 73Li v. Then for the mass of the product side, we may use the atomic mass of 73 Li. For the reactant side, including the three electron masses and the mass of the emitted electron, we may use the atomic mass of 74Be. The energy released is the Q-value. Q m47 Be m 73Li c2 2 2 931.5 MeV c 0.863MeV 7.016929 u 7.016003u c u
31. For alpha decay, we have 21884 Po 21482Pb 42He. We find the Q value, which is the energy released. 214 4 2 Q m 218 84 Po m 82Pb m 2He c
218.008972 u 213.999804 u 4.002603uc 2 931.5MeV c u 2
6.115 MeV 218 0 For beta decay, the nuclear reaction is 218 84 Po 85 At 1e v. We add 84 electrons to both sides of the equations so that we can use the atomic masses. We assume the neutrino is massless, and find the Q value. 218 2 84 Po m 85At c Q m 218 2 218.008972 u 218.008694 uc 2 931.5 MeV c 0.259 MeV u
32. (a) We find the final nucleus by balancing the mass and charge numbers. Z X Z P Z e 15 1 16 A X A P Ae 32 0 32 Thus, the final nucleus is 3216S.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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(b) If we ignore the recoil of the sulfur and the energy of the neutrino, the maximum kinetic energy 32 of the electron is the Q-value of the reaction. The reaction is 15 P 32 16S v . We add 15 electrons to each side of the reaction, and then we may use atomic masses. The mass of the emitted beta is accounted for in the mass of the sulfur. 32 K Q m 1532 P m 16 S c2 K 1.71MeV 1u m 32 S m 32 P 31.973908 u 2 2 2 c 931.5 MeV c 16 15 31.97207 u c
33. We find the energy from the wavelength. 6.63 1034 J s3.00 108 m s E hc 11.8MeV 13 19 1.05 10 m1.602 10 J eV This has to be a ray from the nucleus rather than a photon from the atom. Electron transitions do not involve this much energy. Electron transitions involve energies on the order of a few eV.
34. The nuclear decay reaction is
. We add 10 electrons to both sides in order to
use atomic masses for the calculations. Then the mass of the beta is accounted for in the mass of the sodium. The maximum kinetic energy of the corresponds to the neutrino having no kinetic energy (a limiting case) and no mass, and it is the Q-value of the reaction. We also ignore the recoil of the sodium. K Q m 23 Ne m 23 Na c2 22.9945 u 22.9898 u c2 931.5 MeV uc2 max 11 10 4.4 MeV If the neutrino were to have all of the kinetic energy, the minimum kinetic energy of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be the Q-value, and so the neutrino energies are 0 and 4.4 MeV, respectively. 35. A emission can be modeled as a proton changing into a neutron and a positron. The nuclear reaction here is 137 N 136C 01 v . If we add 7 electrons to each side, the nitrogen atomic mass can be used, and the carbon atomic mass can be used, but there will the mass of an electron and the mass of a positron as products that must be included. Thus, the atomic reaction is 13 13 0 0 1 v . See Problem 36 for a general discussion. 7 N 6 C 1 1 e The kinetic energy of the particle will be maximum if the (almost massless) neutrino and the electron have no kinetic energy. We also ignore the recoil of the carbon. 0 2 K m 137 N m 136 C m 10 e m 10 c2 m 137 N m 136 C 2m 1e c
13.005739 u 13.003355 2 0.00054858 u c2 931.5 MeV uc2 1.199 MeV
If the and electron have no kinetic energy, the maximum kinetic energy of the neutrino is also 1.199 MeV . The minimum energy of each is 0 when the other has the maximum. 36. For the positron-emission process, A N Z
A Z 1
N e v. We must add Z electrons to the nuclear
mass of N to be able to use the atomic mass, and so we must also add Z electrons to the reactant side. On the reactant side, we use Z 1 electrons to be able to use the atomic mass of N. Thus, we
Chapter 41
Nuclear Physics and Radioactivity
have 1 “extra” electron mass and the -particle mass, which means that we must include 2 electron masses on the right-hand side. We find the Q-value given this constraint. Q M M 2m c2 M M 2m c2. P D e D e P 37. We assume that the energies are low enough that we may use classical kinematics. In particular, we will use p 2mK . The decay is 23892U 23490Th 42He. If the uranium nucleus is at rest when it decays, the magnitude of the momentum of the two daughter particles must be the same. p 2 p 2 2m K m K 4u 4.20 MeV p p ; K Th 0.0718MeV Th Th 2m m 234 u 2m 2m Th
Th
Th
Th
The Q-value is the total kinetic energy produced. Q K KTh 4.20MeV 0.0718MeV 4.27 MeV 38. Both energy and momentum are conserved. Therefore, the momenta of the product particles are equal in magnitude. We assume that the energies involved are low enough that we may use classical kinematics; in particular, p 2mK . m 2m K 4.002603 2 2 p p p pRn ; K Rn Rn 2m m K 222.017576 K 2m 2m Rn Rn Rn Rn The sum of the kinetic energies of the product particles must be equal to the Q-value for the reaction. 4.002603 226 222 4 2 K K K K m Ra m Rn m He c 88 86 2 Rn 222.017576 m 226 Ra m 222 Rn m 4 He c2 88 86 2 K 4.002603 1 222.017576
226.025409 u 222.017576 u 4.002603u c2 931.5 MeV uc2 4.79 MeV 4.002603 1 222.017576 2 The rest energy (mc ) of an alpha particle is about 4000 MeV, so our assumption of classical kinematics is valid.
39. (a) The decay constant can be found from the half-life, using Eq. 41–8. ln 2 ln 2 1.5 1010 yr1 4.9 1018 s1 T1/ 2 4.5 109 yr (b) The half-life can be found from the decay constant, 1h using Eq. 41–8. 18240s ln 2 ln 2 T 5.1h 1/ 2 3.8105 s1 3600s
40. We find the half-life from Eq. 41–7d and Eq. 41–8. ln 2 t R R et R e T1/2 T ln 2 t ln 2 4.2 h 1.4 h 0 0 1/ 2 R 140 ln ln 1120 R0 We can see this also from the fact that the rate dropped by a factor of 8, which takes 3 half-lives.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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41. We use Eq. 41–6 to find the fraction remaining. ln 22.2 yr12mo yr N t 9 mo t e e N N0e 0.1309 0.13 13% N0 42. The activity at a given time is given by Eq. 41–7b. The half-life is found in Appendix G. ln 2 dN N ln 2 N 5.6 1020 nuclei 2.1109 decays s 7 T1/ 2 dt 5730 yr3.16 10 s yr 43. Every half-life, the number of nuclei left in the sample is multiplied by one-half. N n 5 1 1 0.03125 1 32 3.125% 2 2 N0 44. We need the decay constant and the initial number of nuclei. The half-life is found in Appendix G and is about 8 days. Use Eq. 41–8 to find the decay constant. ln 2 ln 2 9.99668 107 s1 T1/2 8.0252 days24 h day3600s h 846 106 g 6.022 1023 atoms mol 3.8918 1018 nuclei. N 0 130.906126g mol (a) We use Eq. 41–7b to evaluate the initial activity.
Activity N N 0 et 9.99668 107 s1
3.8918 1018 e0 3.8905 1012 decays s
(b) We evaluate Eq. 41–7d at t 1.50 h. This is much less than one half life, so we expect that most of the initial nuclei remain, and the activity is essentially unchanged. 3600s R R e 3.8905 10 decays se 9.9966810 s 1.50 h 3.8696 10 decays s 7
t
1
12
h
12
0
3.87 1012 decays s (c) We evaluate Eq. 41–7d at t = 3.0 months. We use a time of 1/4 year for the 3.0 months – other approximations (like 1 month = 30 days) for that time period will give a slightly different final answer. In any event, this time period is more than 10 half-lives, so we expect the activity to be significantly lower than the initial value. 9.99668107 s1 0.25yr3.156107 s yr R R 0et 3.8905 1012 decays se 1.4607 109 decays s 1.46 109 decays s 45. The activity of a sample is given by Eq. 41–7a. There are two different decay constants involved. Note that Appendix G gives half-lives, not activities. t N N N e I N e Cot I e I Co t I I Co Co I 0 Co 0 Co
Chapter 41
Nuclear Physics and Radioactivity
t
I
5.2712 y365.25d y ln T I 8.0252d Co 1/2 1/2 63.715d ln 2 ln 2 1 1 ln 2 TI TCo 8.0252d 5.2712 y365.25d y 1/2 1/2
ln T
ln Co
I Co
46. We find the number of nuclei from the activity of the sample and the half-life. The half-life is found in Appendix G. dN N ln 2 N T1/ 2 dt T1/ 2 dN 4.468 10 yr3.156 10 s yr370decays s 7.5 1019 nuclei ln 2 dt ln 2 9
N
7
47. Each emission decreases the mass number by 4 and the atomic number by 2. The mass number changes from 235 to 207, which is a change of 28. Thus, 7 particles must be emitted. With the 7 emissions, the atomic number would have changed from 92 to 78. Each emission increases the atomic number by 1. Thus, to have a final atomic number of 82, 4 particles must be emitted. 48. We will use the decay constant frequently, so we calculate it here. ln 2 ln 2 0.022432s1 T1/ 2 30.9s (a) We find the initial number of nuclei from an estimate of the atomic mass. 9.6 106 g N0
124g mol
6.022 1023 atoms mol 4.662 1016 4.7 1016 nuclei.
(b) Evaluate Eq. 41–6 at t 2.6min. N N 0et 4.662 1016 e
0.022432s1
2.6 min60s min 1.409 1015 1.4 1015 nuclei
(c) The activity is found by Eq. 41–7a. N 0.022432s1 1.409 1015 3.1611013 (d) We find the time from Eq. 41–7a, relative to the time when we had the initial 9.6 g of 12455Cs .
N N 0et
N 1 1decay/s 16 ln 0.022432s 4.662 10 decay/s ln N0 1542s 26 min t 1 0.022432s This is about 50 half-lives.
49. We find the mass from the initial decay rate and Eq. 41–7b. 6.022 1023 nuclei mole dN N0 m dt 0 atomic weight g mole
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
m
dN
1 atomic weight
dt 0 6.022 1023
dN
Instructor Solutions Manual
T1/ 2 atomic weight
6.022 10 1.6 10 s 1.24810 yr3.156 10 s yr 39.963998g 0.6034g 0.60g ln 2 6.022 10 5 1
23
dt 0 ln 2
9
7
23
50. The number of nuclei is found from the mass and the atomic weight. The activity is then found from number of nuclei and the half-life, using Eq. 41–7b.6 7.8 10 g ln 2 ln 2 dN 6.022 1023 atoms mol N N 6 T1/ 2 dt 1.2310 s 31.973908g mol 8.279 1010 decays s 8.31010 decays s 51. (a) The decay constant is found from Eq. 41–8. ln 2 ln 2 1.3811013 s1 1.381013 s1 5 T1/ 2 1.59 10 yr3.156 107 s yr (b) The activity is the decay constant times the number of nuclei. N 1.3811013 s1 3.50 1018 4.8335 105 decays s 60s 1min 2.90 107 decays min
52. We use Eq. 41–7d, with R 16 R0 .
ln 2
R R et R e T1/2 0
0
t
T 1/ 2
ln 2
ln 2 9.8min 3.8min t 1 R ln 6 ln R0
53. We assume the initial sample of pure carbon is naturally occurring carbon, and so use the atomic weight of naturally occurring carbon to find the number of atoms in the sample. The activity is found from Eq. 41–7a, and the half-life is found in Appendix G. 315g 6.022 1023 atoms mol 1.579 1025 atoms N 12.0109g mol 1.3 N 1.579 1025 2.0531013 nuclei of 14 C 14 6 1012 ln 2 2.0531013 78.69decays s 79decays s N 5730 yr3.156 107 s yr 54. We find the mass from the activity. Note that NA is used to represent Avogadro’s number. The half-life is found in Appendix G. ln 2 mNA R N T1/ 2 A
Chapter 41
Nuclear Physics and Radioactivity
RT1/ 2 A 475decays s4.468 10 yr3.156 10 s yr238.050787 g mole NA ln 2 6.022 1023 nuclei moleln 2 9
m
7
3.8199 102 g 3.82 102 g
55. We assume that we currently have N nuclei of 87 Rb and 0.0260 N nuclei of 87 Sr. The sum of those 37
38
two would be the initial number of 8737 Rb nuclei, N . Thus, N 1.0260N . Use Eq. 41–6 and Eq. 0 0 41–8 to find the time for N nuclei of 87 Rb to decay to N nuclei of 87 Rb , which is the estimated 0
37
37
age of the fossils. N N0 et 1 N T1/ 2 N T1/ 2 N 4.97 1010 yr 1 t ln ln ln ln N N ln 2 ln 2 ln 2 1.026 0 0 1.0260N0 1.84 109 yr 56. We are not including the neutrinos that will be emitted during beta decay. 232 First sequence: Th 228 Ra 4 ; 228 Ra 228 Ac ; 228 Ac 228 Th ; 90
228
2
88
4
;
224
2
88
88
Th
90
Second sequence:
235 92
Th 42 ; U 231 90
227
89
Ac 227 Th
231 90
;
90
Ra 220 Rn 4
86
2
Th 231 Pa ; 231 Pa 227 Ac 4 ; 91
91
90
89
2
Th 223 Ra
227
90
89
89
4
88
2
57. Because the fraction of atoms that are 14 C is so small, we use the atomic weight of 12 C to find the 6
6
number of carbon atoms in 67 g. We then use the ratio to find the number of 14 6C atoms present when the club was made. Finally, we use the activity as given in Eq. 41–7c to find the age of the club. The half-life is found in gAppendix G. 67 6.022 N C 1023 atoms mol 3.362 1024 atoms of 126C 12g mol 6 12
N14 1.31012 3.362 1024 4.3711012 nuclei of 146C 6C
N N e N N T 1 t
14 6C
14 6C
today
14
t ln
0
14
C
6
today
N
1/ 2
ln
6
C
today
ln 2 14 N C T 1/ 2 6 0 5730 yr 7.0 decays s 7214 yr 7200 yr ln 2 ln ln 2 12 4.37110 nuclei 5730 yr3.156 107 s yr 0 14 C 6
0
ln 2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
N 58. The decay rate is given by Eq. 41–4b,
t
Instructor Solutions Manual
N . We assume equal numbers of nuclei decaying by
emission for the two isotopes. N T1/2 t 1.6 104 s N 218 218 218 218 7 214 8.6 10 T N 214 N214 214 1/2 3.1min60s min 218 t 214
59. The activity is given by Eq. 41–7a. The original activity is N 0 , so the activity 351.0 hours later is 0.945N0 . ln 2 t 0.945N 0 N 0 et ln 0.945 t T1/ 2 1d T ln 2 35.0 h 428.85 h 17.9d 1/ 2 ln 0.945 24 h 60. The activity is given by Eq. 41–7a. (a) Use Eq. 41–7d. 1 R T R 53d 25decays s R R0 et t ln 1/ 2 ln R ln 2 R ln 2 ln 320decays s 194.94d 190d 0 0 (b) We find the mass from the activity. Note that NA is used to represent Avogadro’s number, and A is the atomic weight. R N ln 2 m0 NA 0 0 T1/ 2 A
m
R0T1/ 2 A
320decays s53d86, 400s d7.016929g mole
6.022 10 nuclei moleln 2 23
NA ln 2
2.5 1014 g
61. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays becomes a daughter nucleus. N N 0et ND N0 N N0 1 e t
62. The activity is given by Eq. 41–7d, with R 0.01050R0 . R ln t ln 2 4.00 hln 2 0.6085h 36.5min R R et R0 ln 2 T 1/ 2 0 t T1/ 2 R ln 0.01050 ln R0
From Appendix G, we see that the isotope is
211 82
Pb .
Chapter 41
Nuclear Physics and Radioactivity
63. Because the carbon is being replenished in living trees, we assume that the amount of 146C is constant until the wood is cut, and then it decays. We use Eq. 41–6. The half-life is given in Appendix G. The N is 0.062. ratio N0 N ln N N et N0 ln 2 0 t T1/ 2 t
T1/ 2
ln
ln 2 64. (a)
N
5730 yrln 0.062 22,986 yr ln 2
N0
23, 000 yr
The mass number is found from the radius, using Eq. 41–1.
r 1.2 1015 m A1/3 3 3 7500 m r A 2.4411056 2.4 1056 15 15 1.2 10 m 1.2 10 m (b) The mass of the neutron star is the mass number times the atomic mass unit conversion in kg.
1.6605 1027 kg u 4.0531029 kg 4.11029 kg
m 2.4411056 u
Note that this is about 20% of the mass of the Sun. (c) The acceleration of gravity on the surface of the neutron star is found from Eq. 6–4 applied to the neutron star. Gm 6.67 1011 N m2 kg2 4.053 1029 g 4.806 1011 m s2 4.8 1011 m s2 2 r2 7500 m 65. Because the tritium in water is being replenished, we assume that the amount is constant until the N is 0.10. wine is made, and then it decays. We use Eq. 41–6. The value of N0 N ln N N 0et N0 ln 2 t T1/ 2 ln N 12.3 yrln 0.10 41yr t T ln 2 N ln 2
1/ 2
0
66. (a) We assume a mass of 70 kg of water, and find the number of protons, given that there are 10 protons in a water molecule. 3 N 70 10 g water 6.02 1023 molecules water 10 protons mol water water molecule protons 18g water mol water 28 2.34 10 protons We assume that the time is much less than the half-life so that the rate of decay is constant. ln 2 N t N T N 1/2 1033 yr N T 1 proton 4 t 1/2 6.165 10 yr 60,000 yr N ln 2 2.34 1028 protons ln 2 This is about 880 times longer than a life expectancy of 70 years.
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(b) Instead of waiting 61,650 consecutive years for one person to experience a proton decay, we could interpret that number as there being 880 people, each living 70 years, to make that 61,650 years (since 880 70 61,600). We would then expect one person out of every 880 to experience a proton decay during their lifetime. Divide 7 billion by 880 to find out how many people on Earth would experience proton decay during their lifetime. 7 109 7.95 106 8 million people 880 67. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy at the distance of closest approach. Note that the distance found is the distance from the center of the alpha to the center of the gold nucleus. The alpha particle has a charge of +2e, and the gold nucleus has a charge of +79e. 1 q q K U K U K 0 0 Au i i f f 40 r 2 2791.60 1019 C 1 q q 9 2 Au 8.988 10 N m 2.9511014 m C2 r 40 K 7.7MeV1.60 1013 J MeV
3.0 10
14
m
We use Eq. 41–1 to compare to the size of the gold nucleus. rapproach 2.9511014 m 4.2 rAu 1971/3 1.2 1015 m
So the distance of approach is about 4.2 the radius of the gold nucleus. 68. We find the number of half-lives from the change in activity. dN ln 0.0150 n dt 6.06 half-lives 21 0.0150 n dN ln 12 It takes 6.06 half-lives for a sample to drop to 1.50% of its original activity. 69. We can find the mass of 4019K from its activity. Note that NA is Avogadro’s number, N40 is the number of particles of 40 K , and A is the atomic mass of 40 K . Similar notation is used for 39 K . 19
40
19
19
R N ln 2 m40 N 40 40 T1/ 2 A40 A
m40
R40T1/ 2 A40 48decays s1.248 109 yr3.156 107 s yr39.963998g mole N A ln 2 6.022 1023 nuclei moleln 2
1.8101104 g 1.8 104 g We find the number of 39 K atoms from the number of 40 K atoms and the abundances given in 19
Appendix G. That is then used to find the mass of 3919K. R R T R N N 40 40 40 1/ 2 ; 40 40 ln 2
19
Chapter 41
Nuclear Physics and Radioactivity
0.932581 N 40 0.000117 m N A39 0.932581 N A39 0.932581 R40T1/ 2 A39 0.000117 40 N 0.000117 ln 2 N 39 39 N A A A 9 7 48decays s 1.248 10 yr 3.156 10 s yr 38.963706 g mole 0.932581 ln 2 0.000117 6.022 1023 nuclei mole N40 0.000117 NK ; N39 0.932581NK
1.407 g 1.4 g 70. We see from the periodic table that Sr is in the same column as calcium. If strontium is ingested, the body may treat it chemically as if it were calcium, which means it might be stored by the body in bones. We use Eq. 41–6 to find the time to reach a 1% level. N ln N N et N0 ln 2 0 t T1/ 2 29 yr ln 0.01 T N t 1/ 2 ln 192.67 yr 200 yr ln 2 ln 2 N0 The decay reactions are as follows. The strontium reaction is beta decay, as given in Appendix G. We assume the daughter also undergoes beta decay. 90 Sr 90 Y 0 e v ; 90 Y 90 Zr 0 e v 38
39
1
39
40
1
71. We find the mass from the initial decay rate and Eq. 41–7b. 6.022 1023 nuclei mole dN N0 m dt 0 atomic weight g mole dN 1 atomic weight dN T1/ 2 atomic weight m dt 0 6.022 1023 dt 0 ln 2 6.022 1023 3.94 104 s1
87.37 d86, 400s d 34.969032 g 2.49 1011 g 23 ln 2 6.022 10
72. (a) We find the daughter nucleus by balancing the mass and charge numbers: Z X Z Os Z e 76 1 77 A X AOs Ae 191 0 191 The daughter nucleus is 191 77 Ir . (b) See the included diagram. (c) Because there is only one energy, the decay must be to the higher excited state.
191 76Os
– (0.14 MeV)
(0.042 MeV) (0.129 MeV)
191 77 Ir* 191 77 Ir* 191 77 Ir
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73. The activity is the decay constant times the number of nuclei, as given by Eq. 41–7a. (a) We calculate the activity for 3215P. ln 2 ln 2 1.0g R N N 6.022 1023 nuclei mol T1/ 2 14.3d86, 400s d 31.973908g mol 1.057 1016 decays s 1.11016 decays s (b) We calculate the activity for 23290Th. ln 2 R N T1/ 2
1.0 g 23 6.022 10 nuclei mol 10 7 232.038054 g mol 1.40 10 yr 3.156 10 s yr ln 2
4.1103 decays s
74. From Fig. 41–1, the average binding energy per nucleon at A 63 is ~8.6 MeV. We use the mass average atomic weight as the average number of nucleons for the two stable isotopes of copper. That gives a binding energy of 63.5468.6 MeV 546.5MeV 550 MeV . The number of copper atoms in a penny is found from the atomic weight. 3.0g N 6.022 1023 atoms mol 2.8431022 atoms 63.546g mol Thus the total energy needed is the product of the number of atoms times the binding energy.
2.8431022 atoms546.5MeV atom1.60 1013 J MeV 2.5 1012 J
75. (a)
24 He m 24 He A24 He 4.002603u 4 0.002603u
(b) C m C A C 12.000000 u 12 0 for both units. (c) Sr m Sr A Sr 85.909261u 86 0.090739 u 0.090739 u931.5MeV uc2 (d) U m U A U 235.043928u 235 0.043928u 0.043928 u931.5MeV uc2 0.002603u 931.5 MeV uc2
12 6
12 6
12 6
86 38
86 38
235 92
235 92
86 38
235 92
(e) From Appendix G we see that 0 for 0 Z 7 and Z 86; 0 for 8 Z 85.
0 for 0 A 15 and A 214; 0 for 16 A 214.
76. The reaction is 11 H 10n 21H. If we assume the initial kinetic energies are small, then the energy of the gamma is the Q-value of the reaction. Q m 11H m 10n m 2 1H c2
1.007825u 1.008665u 2.014102 u c2 931.5 MeV uc2 2.224 MeV
Chapter 41
Nuclear Physics and Radioactivity
77. The mass of carbon 60,000 years ago was 1.0 kg. Find the total number of carbon atoms at that time, and then find the number of 146 C atoms at that time. The text says that the fraction if C-14 atoms is 1.31012 . Use that with the half-life to find the present activity, using Eq. 41–7d and Eq. 41–8. 1.0 103 g 6.022 1023 atoms mol N 5.014 1025 C atoms 12.0109g mol N14 5.014 1025 1.31012 6.518 1013 nuclei of 146C N 0 ln 2
t ln 2 ln 2 T N R N 0 e 1/2 N T1/ 2 T1/ 2
ln 2
6.518 10 e 13
ln 260,000 yr
5730 yr
0.1759decays s 0.2decays s Since the time only has 1 significant figure, we only keep 1 significant figure in the answer. 78. The energy to remove the neutron would be the difference in the masses of the 24 He and the combination of 23 He n. It is also the opposite of the Q-value for the reaction. The number of electrons doesn’t change, so atomic masses can be used for the helium isotopes. 931.5MeV c2 Q m m m c 2 3.016029 u 1.008665u 4.002603u c 2 He He-3 n He-4 u 20.58 MeV
Repeat the calculation for the carbon isotopes. 931.5MeV c2 Q m m m c 2 12.000000 u 1.008665u 13.003355u c 2 C C-12 n C-13 u 4.946 MeV The helium value is 20.58MeV 4.946 MeV 4.161 greater than the carbon value. 79. (a) Take the mass of the Earth and divide it by the mass of a nucleon to find the number of 1/3 nucleons. Then, use Eq. 41–1 to find the radius. 5.98 1024 kg r 1.2 1015 m A1/3 1.2 1015 m 183.6 m 180 m 27 1.67 10 kg (b) Follow the same process as above, just using the Sun’s mass. 1/3 1.99 1030 kg 4 r 1.2 10 15 m A1/3 1.2 10 15 m 12700 m 1.3 10 m 27 1.67 10 kg 80. (a) The usual fraction of 14 C is 1.3 1012. Because the fraction of atoms that are 14 C is so small, 6
6 12 6
we use the atomic weight of C to find the number of carbon atoms in 84 g. We use Eq. 41–6 to find the time. 84g 23 24 N12 6.022 10 atoms mol 4.215 10 atoms 12g mol N14 4.215 1024 atoms1.31012 5.4795 1012 atoms 1 N T N 5730 yr ln 1 N N 0 et t ln 1/ 2 ln 2.4 105 yr N ln 2 N ln 2 5.4795 1012
0
0
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Instructor Solutions Manual
(b) We do a similar calculation for an initial mass of 340 grams. 340g 23 12 13 N14 6.022 10 atoms mol1.310 2.218 10 atoms 12g mol 1 N T N 5730 yr ln 1 N N0 et t ln 1/ 2 ln 5 N ln 2 N ln 2 2.218 1013 2.5 10 yr
0
0
5
This shows that, for times on the order of 10 yr, the sample amount has fairly little effect on the age determined. Thus, times of this magnitude are not accurately measured by carbon dating. 81. (a) This reaction would turn the protons and electrons in atoms into neutrons. This would eliminate chemical reactions, and thus eliminate life as we know it. (b) We assume that there is no kinetic energy brought into the reaction, and solve for the increase of mass necessary1to make 1the reaction energetically possible. For calculating energies, we write the reaction as H n v, and we assume the neutrino has no mass or kinetic energy. 1 0 Q m 1 H m 1 n c2 1.007825u 1.008665u c2 931.5 MeV uc2 0 1 0.782 MeV This is the amount that the proton would have to increase in order to make this energetically possible. We find the percentage change. 0.782 MeV c2 m 100 m 938.27 MeV c2 100 0.083%
82. We assume the particles are not relativistic, so that p 2mK . The radius is given in Example 27–7 mv as r . Set the radii of the two particles equal. Note that the charge of the alpha particle is twice qB that of the electron (in absolute value). We also use the “bare” alpha particle mass, subtracting the two electrons from the helium atomic mass. r r m v m v m v 2m v p 2 p 2eB eB 2 4 p2 p 40.000549 u K 2m 2m 4m 4 K p2 p2 m 4.002603u 20.000549 u 5.48 10 2m
2m
83. Natural samarium has an atomic mass of 150.36 grams per mole. We find the number of nuclei in the natural sample, and then take 15% of that to find the number of 147 Sm nuclei. We first find the 62 number of 14762Sm nuclei from the mass and proportion information. N147
0.15 N natural Sm
0.151.00g6.022 1023 nuclei / mol
150.36g mol The activity level is used to calculate the half-life. 62
6.008 1020 nuclei of 14762Sm
Chapter 41
Nuclear Physics and Radioactivity
Activity R N
T 1/ 2
ln 2
N
T1/ 2 ln 2
N
R
ln 2
6.008 10 3.44069 10 s 20
11 3.156 107 s 1.110 yr
18
120decays s
1yr
84. Since amounts are not specified, we will assume that “today” there is 0.720 g of 23592U and 100.000 0.720 99.280g of 23892U. We use Eq. 41–6. (a) Relate the amounts today to the amounts 1.0 109 years ago. t
N Net NeT1/2
N N et 0
0
ln 2
ln 2
1.010 yr 7.0410 yr ln 2 1.927 g 0.720g e
ln 2
1.010 ln 2 4.46810 115.94g 99.280g e
9
t
N0 235 N 235 eT
1/2
8
9
t
N0 238 N 238 eT
1/2
9
1.927
The percentage of 235 U was
1.927 115.94
92
100% 1.63%
(b) Relate the amounts today to the amounts 100 106 years from now. 10010 yr t ln 2 ln 2 t N N e T1/2 0.720ge 7.0410 yr 0.6525g NN e 6
8
0
235
N N e T
t
ln 2
1/2
238
0 235
99.280ge
0 238
The percentage of 235 U will be
100106 9yr 4.46810 yr ln 2
0.6525 0.6525 97.752
92
97.752g
100% 0.663%
85. (a) The decay constant is calculated from the half-life using Eq. 41–8. ln 2 ln 2 12 1 3.83 10 s T1/2 5730 yr 3.156 107 s yr (b) In a living organism, the abundance of 146C atoms is 1.31012 per carbon atom. Multiply this abundance by Avogadro’s number and divide by the molar mass of carbon-12 to find the number of carbon-14 atoms per gram of carbon. 23 12 14 14 C atoms N C atoms 6.022 10 atoms 6 C mole 12 6 12 6 A N 1.310 12 1.310 12 C atoms M C atoms 12.000000 g 12 C mole
6
6
6
6.524 atoms 146 C g 126C 6.5 10 atoms 6 C g 6 C 10
14
12
(c) The activity in natural carbon for a living organism is the product of the decay constant and the number of 146 C atoms per gram of 126 C . Use Eq. 41–7d and Eq. 41–4b.
R N 3.831012 s1 0.25 decays s g 6 C 12
6.524 10 atoms C g C 0.2499 decays s g C 10
14 6
12 6
12 6
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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(d) Take the above result as the initial decay rate (while Otzi was alive), and use Eq. 41–7d to find the time elapsed since he died. t
R R0 e
1 0.121 1.894 1011 s ln 1yr 7 t 1 ln R 12 1 0.2499 3.156 10 s 6000 yr 3.8310 s R
0
Otzi lived approximately 6000 years ago. 86. (a) We use the definition of the mean life given in the problem. We use a definite integral formula from Appendix B-5. t t 1 1 tN t dt tN e dt te dt
0
N t dt
0
0
0
Ne
0
dt
0
0
t
0
(b) We evaluate at time t
1 1 2
1
e dt
t
2
1
.
N t N e 0 e e 1 0.368 N t 0 N 0 e0 87. The mass number changes only with decay, and changes by 4. If the mass number is 4n, the new number is 4n 4 4 n 1 4n. There is a similar result for each family, as shown here. 4n 4n 4 4 n 1 4n 4n 1 4n 4 1 4 n 1 1 4n 1 4n 2 4n 4 2 4 n 1 2 4n 2 4n 3 4n 4 3 4 n 1 3 4n 3 Thus, the daughter nuclides are always in the same family.
88. (a) If the initial nucleus is at rest when it decays, momentum conservation says that the magnitude of the momentum of the alpha particle will be equal to the magnitude of the momentum of the daughter particle. We use that to calculate the (non-relativistic) kinetic energy of the daughter particle. The mass of each particle is essentially equal to its atomic mass number, in atomic mass units. Note that classically, p . 2 2 2m K m A 4 p pD ; K D pD p K K K 2m 2m 2m m A A D
D
4 A K
D
D
D
D
K K D D 1 K KD 4 K AD K K 1 A K 4 A D 226 (b) We specifically consider the decay of 88Ra. The daughter has AD 222. KD 1 1 1 K KD 1 1 AD 1 222 0.017699 1.8%
4
4
Chapter 41
Nuclear Physics and Radioactivity
Thus, the alpha particle carries away 1 0.018 0.982 98.2% . 89. We determine the number of 4019K nuclei in the sample, and then use the half-life to determine the activity. 3 23 400 10 g6.022 10 atoms mol 7.208 1017 N 40 0.000117 N 0.000117 naturally 19 K 39.0983g mol occurring K ln 2 1yr 17 R N N ln 2 7.208 10 12.51decay s 13decay s 7 T1/ 2 1.265 109 yr 3.156 10 s
90. (a) The reaction is 23692U 23290Th 4 He. If we assume the uranium nucleus is initially at rest, 2 the magnitude of the momenta of the two products must be the same. The kinetic energy available to the products is the Q-value of the reaction. We use the non-relativistic relationship that p2 2mK. 2 2 2m K m p p ; K pTh pHe He He He He He Th Th 2m m K 2mTh 2m Th Th Th m Q KTh KHe mHe 1 KHe Th m 2 mTh K Th Q m m m c He m m m U Th He m He Th He Th 236.045566 u 232.038054 u 4.002603uc2 232.038054 u 4.002603u 232.038054 u
0.004826u 931.5 MeV uc2 4.495 MeV (b) We use Eq. 41–1 to estimate the radii.
rHe 1.2 1015 m4 1.9051015 m 1.9 1015 m 1/3
rTh 1.2 1015 m232 7.374 1015 m 7.4 1015 m 1/3
(c) The maximum height of the Coulomb barrier will correspond to the alpha particle and the thorium nucleus being separated by the sum of their radii. We use Eq. 23–10. 1 q q qHe qTh 1 He Th U 40 rA 40 rHe rTh 2 2901.60 1019 C 8.988 109 N m2 C2
27.898 MeV 28 MeV (d) At position “A”, the product particles are separated by the sum of their radii, about 9.3 fm. At position “B”, the alpha particle will have a potential energy equal to its final kinetic energy, 4.495 MeV. Use Eq. 23–10 to solve for the separation distance at position “B”. 1 qHe qTh UB 40 RB
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2901.60 1019 C
2
1 qHe qTh
R
40 UB
B
8.988 109 N m2 C2
4.495 MeV1.60 1013 J MeV
57.57 1015 m RB RA 57.6 fm 9.3fm 48.3fm Note that this is a center-to-center distance. 91. We take the momentum of the nucleon to be equal to the uncertainty in the momentum of the nucleon, as given by the uncertainty principle. The uncertainty in position is estimated as the radius of the nucleus. With that momentum, we calculate the kinetic energy using a classical formula. Iron has about 56 nucleons (depending on the isotope). px K
p
p p
x
r
2
2mr
2m
2
1.60 1013 J MeV 21.67 1027 kg561/3 1.2 1015 m 2
0.988MeV 1MeV 92. (a) From the figure, the initial force on the detected particle is down. Using the right-hand rule, the force on a positive particle would be upward. Thus, the particle must be negative, and the decay is decay. (b) The magnetic force is producing circular motion. Set the expression for the magnetic force equal to the expression for centripetal force, and solve for the velocity. 19 3 mv2 qBr 1.60 10 C0.012T4.7 10 m qvB v 9.9 106 m s r m 9.111031kg
93. If the original nucleus is of mass 256 u, the daughter nucleus after the alpha decay will have a mass of 252 u. Moreover, the alpha has a mass of 4 u. Use conservation of momentum – the momenta of the daughter nucleus and the alpha particle must be equal and opposite since there are only two products. The energies are small enough that we may use non-relativistic relationships. m K 4 4.5MeV 0.071MeV p pX 2m K K X 252 mX 94. We use Eq. 41–7a to relate the activity to the half-life. We estimate the atomic weight at 152 grams/mole since this isotope is not given in Appendix G. dN R N ln 2 N T1/ 2 dt 6.022 1023 nuclei 1yr ln 2 ln 2 7 T1/ 2 N 1.5 10 g 7 1.31021 yr 1decay s R 152g 3.156 10 s
95. We calculate the initial number of nuclei from the initial mass and the atomic mass. 1 nucleus 1.0187 1017 nuclei 1.02 1017 nuclei N 0 2.20 109 kg 27 13.005739 u1.660510 kg u See the adjacent graph. From the graph, the half-life is approximately 600 seconds. Note that the
Chapter 41
Nuclear Physics and Radioactivity
half-life as given in Appendix G is T1 2 9.965min
60s 597.9s . 1min
96. The emitted photon and the recoiling nucleus have the same magnitude of momentum. We find the recoil energy from the momentum. We assume the energy is small enough that we can use classical relationships. E p p 2mK KK K c 2 1.46 MeV E2 KK 2.86 105 MeV 28.6eV 2mK c2 931.5 MeV c2 2 239.963998 u c u 97. Section 41–1 states that the neutron, proton, and electron are all spin 1 2 particles. Consider the reaction n p e v. If the proton and neutron spins are aligned (both are 1 2, for example), the electron and anti-neutrino spins must cancel. Since the electron is spin 1 2, the anti-neutrino must be spin 1 , which means the magnitude of the neutrino spin must be 1 in this case. 2
2
The other possibility is if the proton and neutron spins are opposite of each other. Consider the case of the neutron having spin 1 and the proton having spin 1 . If the electron has spin 1 , the spins of 2
2
2 1 2
the electron and proton cancel, and the neutrino must have spin for angular momentum to be conserved. If the electron has spin 1 , the spin of the neutrino must be 3 for angular momentum to 2
2
be conserved. A similar argument could be made for positron emission, with p n e v.
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CHAPTER 41: Nuclear Physics and Radioactivity Responses to Questions 1.
Different isotopes of a given element have the same number of protons and electrons. Because they have the same number of electrons, they have almost identical chemical properties. Each isotope has a different number of neutrons from other isotopes of the same element. Accordingly, they have different atomic masses and mass numbers. Since the number of neutrons is different, they may have different nuclear properties, such as whether they are radioactive or not.
2.
With 88 nucleons and 50 neutrons, there must be 38 protons. The number of protons is the atomic number, and so the element is strontium. The nuclear symbol is 3888 Sr .
3.
Identify the element based on the atomic number. (a) Uranium (Z = 92) (b) Nitrogen (Z = 7) (c) Hydrogen (Z = 1) (d) Strontium (Z = 38) (e) Berkelium (Z = 97)
4.
The number of protons is the same as the atomic number, and the number of neutrons is the mass number minus the number of protons. (a) Uranium: 92 protons, 232 – 9 = 140 neutrons (b) Nitrogen: 7 protons, 18 – 7 = 11 neutrons (c) Hydrogen: 1 proton, 1 – 1 = 0 neutrons (d) Strontium: 38 protons, 86 – 38 = 48 neutrons (e) Berkelium: 97 protons, 252 – 97 = 155 neutrons
5.
The atomic mass of an element as shown in the periodic table is the average atomic mass of all naturally occurring isotopes. For example, chlorine occurs as roughly 75% 35 Cl and 25% 37 Cl , and 17
17
so its atomic mass is about 35.5 (= 0.75 35 + 0.25 37). Other smaller effects would include the fact that the masses of the nucleons are not exactly 1 atomic mass unit, and that some small fraction of the mass energy of the total set of nucleons is in the form of binding energy. 6.
The nucleus of an atom consists of protons (which carry a positive electric charge) and neutrons (which are electrically neutral). The electric force between protons is repulsive and much larger than the force of gravity. If the electric and gravitational forces are the only two forces present in the nucleus, the nucleus would be unstable as the electric force would push the protons away from each other. Nuclei are stable, and therefore, another force must be present in the nucleus to overcome the electric force. This force is the strong nuclear force.
7.
The strong force and the electromagnetic (EM) force are two of the four fundamental forces in nature. They are both involved in holding atoms together: the strong force binds quarks into nucleons and binds nucleons together in the nucleus; the EM force is responsible for binding negatively charged electrons to positively charged nuclei and for binding atoms into molecules. The strong force is the strongest fundamental force; the EM force is about 100 times weaker at distances on the order of 1017 m. The strong force operates at short range and is negligible for distances greater than about the size of the nucleus. The EM force is a long-range force that decreases as the inverse square of the distance between the two interacting charged particles. The EM force operates
Chapter 41
Nuclear Physics and Radioactivity
only between charged particles. The strong force is always attractive; the EM force can be attractive or repulsive. Both these forces have mediating field particles associated with them – the gluon for the strong force and the photon for the EM force. 8.
Quoting from Section 41–3, “… radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong chemicals.” Chemical reactions are a result of electron interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to electron interactions. Another piece of evidence is the fact that the -particle emitted in many radioactive decays is much heavier than an electron and has a different charge than the electron, so it can’t be an electron. Therefore, it must be from the nucleus. Finally, the energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron orbital transitions. All of these observations support radioactivity being a nuclear process.
9.
The resulting nuclide for gamma decay is the same isotope in a lower energy state: 64 Cu* 64 Cu γ . 29
29
The resulting nuclide for beta-minus decay is an isotope of zinc, 6430Zn : 64 Cu 64 Zn e . 29
30
The resulting nuclide for beta-plus decay is an isotope of nickel, 6428Ni : 64 Cu 64 Ni e . 29
10.
238
28
U decays by alpha emission into 234 Th, which has 144 neutrons.
92
90
11. The alpha particle is a very stable nucleus. It has less energy when bound together than when split apart into separate nucleons. In most cases, more energy is required to emit separate nucleons than an alpha particle. It is usually true that the emission of a single nucleon is energetically not possible. See Example 41–5. 12. Alpha (α) particles are helium nuclei. Each α particle consists of 2 protons and 2 neutrons, and therefore, it has a charge of +2e and an atomic mass value of 4 u. They are the most massive of the
or positrons . Electrons have a charge of –e and
three. Beta (β) particles are electrons
positrons have a charge of +e. In terms of mass, beta particles are much lighter than protons or neutrons, by a factor of about 2000, so are lighter than alpha particles by a factor of about 8000. Their emission is always accompanied by either an anti-neutrino (in decay) or a neutrino (in decay). Gamma (γ) particles are photons. They have no rest mass and no charge. 24 13. (a) Magnesium is formed: 24 11 Na 12 Mg e . 22
82
2
(b) Neon is formed: 22 11 Na 10 Ne e . 210 (c) Lead is formed: Po 206Pb 4 He . 84
14. (a) Sulfur is formed: 32 15 Pb 16 S e . 32
35 (b) Chlorine is formed: 16 S 35 17Cl e . (c) Thallium is formed: 211Bi 207Tl 4 He . 83
81
2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
15. (a)
45 20
Ca
(b)
58 29
Cu*
(c)
46 24
45 21
Sc e v
Scandium-45 is the missing nucleus.
Cu
Copper-58 is the missing nucleus.
58 29
Cr 46 V e v
(d)
234 94
(e)
239 93
Instructor Solutions Manual
The positron and the neutrino are the missing particles.
23
Pu
230 92
U
Np 239 Pu
Uranium-230 is the missing nucleus. The electron and the anti-neutrino are the missing particles.
94
16. The two extra electrons held by the newly formed thorium will be very loosely held, as the number of protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will be easy for these extra two electrons to escape from the thorium atom through a variety of mechanisms. They are in essence “free” electrons. They do not gain kinetic energy from the decay. They might get captured by the alpha nucleus, for example. 17. When a nucleus undergoes either or decay it becomes a different element, since it has either converted a neutron to a proton or a proton to a neutron. Thus, its atomic number (Z) has changed. The energy levels of the electrons are dependent on Z, and so all of those energy levels change to become the energy levels of the new element. Photons (with energies on the order of a few eV) are likely to be emitted from the atom as electrons change energies to occupy the new levels. 18. Alpha particles from an alpha-emitting nuclide are part of a two-body decay. The energy carried off by the decay fragments is determined by the principles of conservation of energy and of momentum. With only two decay fragments, these two constraints require the alpha particles to be monoenergetic. Beta particles from a beta-emitting nucleus are part of a three-body decay. Again, the energy carried off by all of the decay fragments is determined by the principles of conservation of energy and of momentum. However, with three decay fragments, the energy distribution between the fragments is not determined by these two constraints. The beta particles will therefore have a range of energies. 19. In electron capture, the nucleus will effectively have a proton change to a neutron. This isotope will then lie to the left and above the original isotope. Since the process would only occur if it made the nucleus more stable, it must lie BELOW the line of stability in Fig. 41–2. 20. Neither hydrogen nor deuterium can emit an particle. Hydrogen has only one nucleon (a proton) in its nucleus, and deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither one has the necessary four nucleons (two protons and two neutrons) to emit an particle. 21. Many artificially produced radioactive isotopes are rare in nature because they have decayed away over time. If the half-lives of these isotopes are relatively short in comparison with the age of Earth (which is typical for these isotopes), there won’t be any significant amount of these isotopes left to be found in nature. Also, many of these isotopes have a very high energy of formation, which is generally not available under natural circumstances. 22. After two months, the sample will not have completely decayed. After one month, half of the sample will remain, and after two months, one-fourth of the sample will remain. Each month, half of the remaining atoms decay. 23. For Z > 92, the short range of the attractive strong nuclear force means that no number of neutrons is able to overcome the electrostatic repulsion of the large concentration of protons.
Chapter 41
Nuclear Physics and Radioactivity
24. There are a total of 4 protons and 3 neutrons in the reactants. The -particle has 2 protons and 2 3 neutrons, and so 2 protons and 1 neutron are in the other product particle. It must be 2 He . 6 1 4 3 Li p He 3
1
2
2
14
25. The technique of 6C could not be used to measure the age of stone walls and tablets. Carbon-14 dating is only useful for measuring the age of objects that were living at some earlier time. Stone walls and tablets were never alive. 26. The decay series of Fig. 41–12 begins with a nucleus that has many more neutrons than protons and lies far above the line of stability in Fig. 41–2. In a β+ decay, a proton is converted to a neutron, which would take the nuclei in this decay series farther from the line of stability and is not energetically preferred. 27. There are four alpha particles and four β– particles (electrons) emitted, no matter which decay path is chosen. The nucleon number drops by 16 as 222 Rn decays into 206 Pb , indicating the presence of four 86
82
alpha decays. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha decays would result in a decrease of eight protons. Four β– decays will convert four neutrons into protons, making the decrease in the number of protons only four, as required. (See Fig. 41–12.) 28. (i)
Since the momentum before the decay was 0, the total momentum after the decay will also be 0. Since there are only 2 decay products, they must move in opposite directions with equal magnitude of momentum. Thus, (c) is the correct choice: both the same. (ii) Since both products have the same momentum, the one with the smallest mass will have the greater velocity. Thus, (b) is the correct choice: the alpha particle. (iii) We assume that the products are moving slowly enough that classical mechanics can be used. In 𝑝2
that case, k = . Since both particles have the same momentum, the one with the smallest 2𝑚 mass will have the greater kinetic energy. Thus, (b) is the correct choice: the alpha particle. 29. Fig. 41–7 shows the potential energy curve for an alpha particle and daughter nucleus for the case of radioactive nuclei. The alpha particle tunnels through the barrier from point A to point B in the figure. In the case of stable nuclei, the probability of this happening must be essentially zero. The maximum height of the Coulomb potential energy curve must be larger and/or the Q-value of the reaction must be smaller so that the probability of tunneling is extremely low. 30. If the Earth had been bombarded with additional radiation several thousand years ago, there would have been a larger abundance of carbon-14 in the atmosphere at that time. Organisms that died in that time period would have had a greater percentage of carbon-14 in them than organisms that die today. Since we assumed the starting carbon-14 was the same, we would have previously underestimated the age of the organisms. With the new discovery, we would re-evaluate the organisms as older than previously calculated. 31. (a) An einsteinium nucleus has 99 protons and a fermium nucleus as 100 protons. If the fermium undergoes either electron capture or + decay, a proton would in effect be converted into a neutron. The nucleus would now have 99 protons and be an einsteinium nucleus. (b) If the einsteinium undergoes decay, a neutron would be converted into a proton. The nucleus would now have 100 protons and be a fermium nucleus.
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32. In decay, an electron is ejected from the nucleus of the atom, and a neutron is converted into a proton. The atomic number of the nucleus increases by one, and the element now has different chemical properties. In internal conversion, an orbital electron is ejected from the atom. This does not change the atomic number of the nucleus, nor its chemical properties. Also, in decay, both a neutrino and an electron will be emitted from the nucleus. Because there are three decay products (the neutrino, the particle, and the nucleus), the momentum of the particle can have a range of values. In internal conversion, since there are only two decay products (the electron and the nucleus), the electron will have a unique momentum and, therefore, a unique energy.
MisConceptual Questions 1.
(a) A common misconception is that the elements of the period table are distinguished by the number of electrons in the atom. This misconception arises because the number of electrons in a neutral atom is the same as the number of protons in its nucleus. Elements can be ionized by adding or removing electrons, but this does not change what type of element it is. When an element undergoes a nuclear reaction that changes the number of protons in the nucleus, the element does transform into a different element.
2.
(b) The role of energy in binding nuclei together is often misunderstood. As protons and neutrons are added to the nucleus, they release some mass energy, which usually appears as radiation or kinetic energy. This lack of energy is what binds the nucleus together. To break the nucleus apart, the energy must be added back in. As a result, a nucleus will have less energy than the protons and neutrons that constitute the nucleus.
3.
(c) Regarding (a), large nuclei (with many more neutrons than protons) are typically unstable, so increasing the number of nuclei does not necessarily make the nucleus more stable. Regarding (b), nuclei such as 14 O have more protons than neutrons, yet 14 O is not as stable as 16 O . 8
8
8
Therefore, having more protons than neutrons does not necessarily make a nucleus more stable. The large Coulomb repulsion between the protons helps lead to instability. Regarding (d), there are no electrons in the nucleus. Electrons do not affect the stability of nuclei. Regarding (e), large unstable nuclei, such as 238 94 Pu, have a much larger total binding energy than small stable nuclei such as 42 He, so the total binding energy is not a measure of stability. The correct answer is (c). Stable nuclei typically have large binding energies per nucleon, which means that each nucleon is more tightly bound to the others. 4.
(e) The Coulomb repulsive force does act inside the nucleus pushing the protons apart. Another larger attractive force is thus necessary to keep the nucleus together. The force of gravity is far too small to hold the nucleus together. Neutrons are not negatively charged. It is the attractive strong nuclear force that overcomes the Coulomb force to hold the nucleus together.
5.
(b) The exponential nature of radioactive decay is a concept that can be misunderstood. It is sometimes thought that the decay is linear, such that the time for a substance to completely decay is twice the time for half of the substance to decay. Radioactive decay is not linear but exponential. That is, during each half-life, 1/2 of the remaining substance decays. If half the original decays in the first half-life, then 1/2 remains. During the second half-life, 1/2 of what is left decays, which would be 1/4 of the initial substance. In each subsequent half-life, half of the remaining decays, so it takes many half-lives for a substance to effectively decay away. The decay constant is inversely related to the half-life.
Chapter 41
Nuclear Physics and Radioactivity
6.
(e) The half-life is the time it takes for half of the substance to decay away. The half-life is a constant determined by the composition of the substance and not on the quantity of the initial substance. As the substance decays, the number of nuclei decreases, the activity (number of decays per second) decreases, but the half-life remains constant.
7.
(d) A common misconception is that it would take twice the half-life, or 20 years, for the substance to completely decay. This is incorrect because radioactive decay is an exponential process. That is, during each half-life, 1/2 of the remaining substance decays. After the first 10 years, 1/2 remains. After the second ten-year period, 1/4 remains. After each succeeding half-life, another half of the remaining substance decays, leaving 1/8, then 1/16, then 1/32, and so forth. The decays stop when none of the substance remains. But after a time, there will be too few nuclei to reliably use statistics. Thus, the time cannot be exactly determined.
8.
(c) A common misconception is that after the second half-life, none of the substance remains. However, during each half-life, 1/2 of the remaining substance decays. After one half-life, 1/2 remains. After two half-lives, 1/4 remains. After three half-lives, 1/8 remains.
9.
(a) The decay constant is proportional to the probability of a particle decaying and is inversely proportional to the half-life. Therefore, the substance with the shorter half-life (Sr) has the larger decay constant and the larger probability of decaying. The activity is proportional to the amount of the substance (number of atoms) and the decay constant, so the activity of Tc will be smaller than the activity of Sr.
10. (d) The element with the largest decay constant will have the shortest half-life. Converting each of the choices to decays per second yields: (a) 100/s, (b) 1.610–7/s, (c) 2.510–9/s, (d) 1.2104/s. Answer (d) has the largest decay rate, so it will have the smallest half-life. 11. (b) The half-life of radium remains constant and is not affected by temperature. Radium that existed several billion years ago will have essentially completely decayed. Small isotopes, such as carbon-14, can be created from cosmic rays, but radon is a heavy element. Lightning is not energetic enough to affect the nucleus of the atoms. Heavy elements such as plutonium and uranium can decay into radium and thus are the source of present-day radium. 12. (a) The nature of mass and energy in nuclear physics is often misunderstood. When the neutron and proton are close together, they bind together by releasing mass energy that is equivalent to the binding energy. This energy comes from a reduction in their mass. Therefore, when the neutron and proton are far from each other, their net mass is greater than their net mass when they are bound together.
Solutions to Problems 1.
Convert the units from MeV c2 to atomic mass units. 1u m 775MeV c2 0.832 u 2 931.49 MeV c
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2.
The particle is a helium nucleus and has A = 4. Use Eq. 41–1.
15
r 1.2 10
3.
Instructor Solutions Manual
1
15
m A 1.2 10 3
1
m 4 3 1.9 10
15
m 1.9 fm
The radii of the two nuclei can be calculated with Eq. 41–1. Take the ratio of the two radii. 1/3 1/3 1.2 1015 m214 r 214 214 1.0128 1/3 15 r206 1.2 10 m206 206 206 So, the radius of 214 82 Pb is 1.28% 1% larger than the radius of 82 Pb.
4.
Use Eq. 41–1 for both parts.
15 5.9 10 m 5.9fm (a) r 1.2 10 m A 1.2 10 m 120 3 3 r 3.7 1015 m 15 1/ 3 29.3 29 A (b) r 1.2 10 m A 15 15 1.2 10 m 1.2 10 m
5.
15
1/3
15
1/3
To find the rest mass of an particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m mHe 2me 4.002603u931.5 MeV uc2 2 0.511MeV c2 3727 MeV c2
This is less than the sum of the masses of two protons and two neutrons because of the binding energy. 6.
Each particle would exert a force on the other through the Coulomb electrostatic force. The distance between the particles is twice the radius of one of the particles. The Coulomb force is given by Eq. 21–2. 2 9 2 2 19 8.988 10 N m C 21.60 10 C 1 F q q 63.41N 63 N 2 40 2r 2 24 1/3 1.2 10 15 m
The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha calculated in Problem 5. F 63.41N 9.5 1027 m s2 F ma a 27 kg m 2 1.6605 10 3727 MeV c 2 931.49 MeV c 7.
First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is approximately (A u) and its radius is r 1.2 1015 m A1/ 3. Calculate the density. m 27 27 A1.6605 10 kg A1.6605 10 kg 2.294 1017 kg m3
1.2 1015 m A We see that this is independent of A. The value has 2 significant figures. V
4 3
r3
4 3
3
Chapter 41
Nuclear Physics and Radioactivity
(a) We set the density of the Earth equal to the density of nuclear matter. Earth Earth nuclear 4 M R3 matter
3
Earth 1/3
1/3 5.98 1024 kg 183.9 m M Earth R 180 m 4 nuclear Earth 2.294 10 kg m 3 matter (b) Set the density of Earth equal to the density of uranium, and solve for the radius of the uranium. Then compare that to the actual radius of uranium, using Eq. 41–1, with A = 238. M M Earth mU 4 Earth 3 4 mU 3 R r 3 3 REarth rU 3 Earth 3 U 1/3 238u1.6605 1027 kg u 1/3 6 mU 6.38 10 m 2.58 1010 m r R 24 5.98 10 kg U Earth M Earth rU 2.58 1010 m 3.5 104 1/3 15 rU actual 1.2 10 238
8.
Use Eq. 41–1 to find the value for A. We use uranium-238 since it is the most common isotope. 1.2 1015 m A1/3 3 r unknown 0.5 A 2380.5 29.75 30 rU 1.2 1015 m2381/3 From Appendix G, a stable nucleus with A 30 is 3115P.
9.
The basic principle to use is that of conservation of energy. We assume that the centers of the two particles are located a distance from each other equal to the sum of their radii. That distance is used to calculate the initial electrical potential energy. Then, we also assume that since the Rf nucleus is much heavier than the alpha, the alpha has all of the final kinetic energy when the particles are far apart from each other (and so have no potential energy). 1 q qRf K 0 K U K U 0 i i f f 40 r rRf
21041.60 10 C 3.118 10 eV K 8.988 10 N m C 4 263 1.2 10 m1.60 10 J eV 19
9
2
2
7
2
1/3
1/3
15
19
31MeV 10. (a) The hydrogen atom is made of a proton and an electron. Use values from Appendix G. mp 1.007276 u 0.9994553 99.95% mH 1.007825u (b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by Eq. 41–1. For the atomic radius, we use the Bohr radius3 given in Eq. 37–12. 3 15 3 r 1.2 10 m Vnucleus 34 rnucleus nucleus 1.2 1014 3 4 Vatom 10 3 r atom ratom 0.5310 m
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.01 atomic mass unit. Therefore, in a 1.0-kg object, (1.0 kg) 6.022 1026 u kg N 5.96 1026 6 1026 nucleons 1.01u nucleon No , it does not matter what the element is because the mass of one nucleon is essentially the same for all elements. 12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha just touches the thorium nucleus. Assume the two nuclei are at rest when they “touch”. The distance between the two particles is the sum of their radii. 1 q qTh K U K U K 0 0 i i f f 40 r rTh K 8.988 10 N m C 9
2
2
2901.60 1019 C
2
2.790 107 eV
4 232 1.2 10 m1.60 10 J eV 1/3
1/3
15
19
28 MeV 13. From Fig. 41–1, we see that the average binding energy per nucleon at A = 63 is about 8.7 MeV. Multiply this by the number of nucleons in the nucleus. 638.7 MeV 548.1MeV 550 MeV
14. (a) From Fig. 41–1, we see that the average binding energy per nucleon at A 238 is 7.5MeV. Multiply this by the 238 nucleons. 2387.5MeV 1785MeV 1800 MeV (b) From Fig. 41–1, we see that the average binding energy per nucleon at A 84 is 8.7 MeV. Multiply this by the 84 nucleons. 848.7 MeV 730.8MeV 730MeV 15.
18 8
O consists of 8 protons and 10 neutrons. We find the binding energy from the masses of the components and the mass of the nucleus, from Appendix G. Note that the electron masses from the hydrogens cancel out the electron masses for the oxygen. 2 8 c Binding energy 8m 1 1H 10m 1 n0 m 18 O
81.007825u 101.008665u 17.999160 u c2 931.5MeV c2 0.150090 139.81MeV u
Binding energy per nucleon 139.81MeV 18 7.767 MeV
Chapter 41
Nuclear Physics and Radioactivity
16. Deuterium consists of 1 proton, 1 neutron, and 1 electron. Ordinary hydrogen consists of 1 proton and 1 electron. Use the atomic masses from Appendix G. The electron masses cancel. 1 2 2 1 m 0n m H 1 c Binding energy m 1 H
931.5 MeV c2 1.007825u 1.008665u 2.014102 u c 2 u 2.224 MeV
17. We find the binding energy of the last neutron from the masses of the isotopes. Binding energy m 3115P m 1 0n m 3215P c2
30.973762 u 1.008665u 31.973908 u c2 931.5 MeV / c2 7.935MeV
18. (a)
7 4
Be consists of 4 protons and 3 neutrons. We find the binding energy from the masses, using hydrogen atoms in place of protons so that we account for the mass of the electrons. 1 7 2 1 3m n 0 m Be 4 c Binding energy 4m 1 H
41.007825 u 31.008665 u 7.016929 u c2 931.5 MeV / c2 37.60 MeV
Binding energy 37.60 MeV 5.372 MeV nucleon nucleon 7 nucleons (b) 197 79 Au consists of 79 protons and 118 neutrons. We find the binding energy as in part (a). Binding energy 79m 11H 118m 1 n0 m 19779Au c2
791.007825 u 1181.008665 u 196.966570 u c2 931.5 MeV / c2 1559.4 MeV 1559 MeV
Binding energy 1559.4 MeV 7.916 MeV nucleon nucleon 197 nucleons 19.
31 15
P consists of 15 protons and 16 neutrons. We find the binding energy from the masses:
Binding energy 15 m 11H 16m 10n m 3115P c2
151.007825 u 161.008665 u 30.973762 u c2 931.5MeV uc2
262.9 MeV Binding energy 262.9 MeV 8.481MeV nucleon nucleon 31 We do a similar calculation for 3215P, consisting of 15 protons and 17 neutrons. 1 32 2 Binding energy 15 m 1H 1 17m n 0 m 15P c
151.007825u 17 1.008665u 31.973908 u c2 931.5 MeV uc2 270.85 MeV
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Binding energy 270.85 MeV 8.464 MeV nucleon nucleon 32 31 By this measure, the nucleons in P are more tightly bound than those in 32 P. Thus, we expect 31 P 15 15 15 to be more stable than 32 P. Appendix G bears that out – 32 P is radioactive, while 31 P is stable. 15
15
15
20. We find the required energy by calculating the difference in the masses. (a) Removal of a proton creates an isotope of nitrogen. To balance electrons, the proton is included as a hydrogen atom: 168 O 11H 157 N. 2 Energy needed m 157N m 11H m 16 O 8 c
15.000109 u 1.007825 u 15.994915 u 931.5 MeV uc2 12.13MeV
(b) Removal of a neutron creates another isotope of oxygen: 168O 10n 158O. 2 Energy needed m 158O m 10n m 16 O 8 c
15.003066 u 1.008665 u 15.994915 u 931.5 MeV uc2 15.66 MeV
The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the proton because there is also the repulsive electric force from the other protons “helping” to remove the proton. 21. (a) We find the binding energy from the masses. Binding Energy 2m
He m Be c2 4
2
8
4
2 4.002603u 8.005305u c2 931.5 MeV uc2 0.092 MeV Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy state as two alphas instead of a beryllium. (b) We find the binding energy from the masses. 2 Binding Energy 3m 42He m 12 C 6 c
34.002603u 12.000000 u c2 931.5 MeV uc2 7.3MeV Because the binding energy is positive, the nucleus is stable. 22. The wavelength is determined from the energy change between the states. c hc 2.6 1012 m E hf h E 0.48 MeV1.60 1013 J MeV
0 23. The nuclear decay is 31 H 32He 1 e v. When we add one electron to both sides in order to use
atomic masses, we see that the mass of the emitted particle is included in the atomic mass of 3 He. 2 The energy released is the difference in the masses. Energy released m 31H m 32He c2
3.016049 u 3.016029 u c2 931.5 MeV uc2 0.019 MeV
Chapter 41
Nuclear Physics and Radioactivity
24. For the decay 116C 105B 11p, we find the difference of the final and initial final masses. To balance the electrons (so that we can use atomic masses), we assume the proton is an atom of 11 H. m m 105B m 11 H m 116C
10.012937 u 1.007825 u 11.011433u 0.009329 u Since the final masses are more than the original mass, energy would not be conserved. 0 25. The decay is 01 n 11p 1 e v. The electron mass is accounted for if we use the atomic mass of 1 H as the combination of the proton and electron. If we ignore the recoil of the proton and the 1 neutrino, and any1 possible 1mass of the neutrino, we get the maximum kinetic energy. K m n m H c2 1.008665u 1.007825u c2 931.5 MeV uc2 max 1 0
0.782 MeV 26. For each decay, we find the difference of the initial and the final masses. If the final mass is more than the initial mass, the decay is not possible. 1 233 (a) m m 232 92 U m0 n m 92 U 232.037155u 1.008665u 233.039634 u 0.006816 u Because an increase in mass is required, the decay is not possible.
(b) m m 137 N m 01 n m 147 N 13.005739 u 1.008665u 14.003074 u 0.011330 u Because an increase in mass is required, the decay is not possible. 39 (c) m m 19 K m01 n m1940 K 38.963706 u 1.008665u 39.963998u 0.008373u
Because an increase in mass is required, the decay is not possible. 24 27. (a) From Appendix G, 11 Na is a emitter .
(b) The nuclear decay reaction is
. We add 11 electrons to both sides in
order to use atomic masses. Then, the mass of the beta is accounted for in the mass of the magnesium. The maximum kinetic energy of the corresponds to the neutrino having no kinetic energy (a limiting case) and no mass. We also ignore the recoil of the magnesium. K m 24Na m 24Mg c 2 11
12
23.990963u 23.985042 u c2 931.5 MeV uc2 5.515 MeV 28. (a) We find the final nucleus by balancing the mass and charge numbers. Z X Z U Z He 92 2 90 A X A U AHe 238 4 234 Thus, the final nucleus is 234 90 Th . (b) If we ignore the recoil of the thorium, the kinetic energy of the particle is equal to the Qvalue of the reaction. The electrons are balanced. K Q m
U m Th m He c2 238 92
234 90
4 2
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m 234 Th m 238 U m 4 He 90
92
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K
2
c2 4.20 MeV 1u 238.050787 u 4.002603u 2 c2 931.5 MeV c 234.043675u
This answer assumes that the 4.20 MeV value does not limit the sig. fig. of the answer. 60 29. The nuclear reaction is 60 27 Co 28Ni . We add 27 electrons to both sides of the equation so that we can use atomic masses. The kinetic energy of the will be maximum if the (essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel.
K m 2760 Co m 6028Ni c2
931.5MeV c2 59.933816 u 59.930785u c 2 2.823MeV u
30. We add three electron masses to each side of the nuclear reaction 74 Be 10 e 73Li v. Then for the mass of the product side, we may use the atomic mass of 73 Li. For the reactant side, including the three electron masses and the mass of the emitted electron, we may use the atomic mass of 74Be. The energy released is the Q-value. Q m47 Be m 73Li c2 2 2 931.5 MeV c 0.863MeV 7.016929 u 7.016003u c u
31. For alpha decay, we have 21884 Po 21482Pb 42He. We find the Q value, which is the energy released. 214 4 2 Q m 218 84 Po m 82Pb m 2He c
218.008972 u 213.999804 u 4.002603uc 2 931.5MeV c u 2
6.115 MeV 218 0 For beta decay, the nuclear reaction is 218 84 Po 85 At 1e v. We add 84 electrons to both sides of the equations so that we can use the atomic masses. We assume the neutrino is massless, and find the Q value. 218 2 84 Po m 85At c Q m 218 2 218.008972 u 218.008694 uc 2 931.5 MeV c 0.259 MeV u
32. (a) We find the final nucleus by balancing the mass and charge numbers. Z X Z P Z e 15 1 16 A X A P Ae 32 0 32 Thus, the final nucleus is 3216S.
Chapter 41
Nuclear Physics and Radioactivity
(b) If we ignore the recoil of the sulfur and the energy of the neutrino, the maximum kinetic energy 32 of the electron is the Q-value of the reaction. The reaction is 15 P 32 16S v . We add 15 electrons to each side of the reaction, and then we may use atomic masses. The mass of the emitted beta is accounted for in the mass of the sulfur. 32 K Q m 1532 P m 16 S c2 K 1.71MeV 1u m 32 S m 32 P 31.973908 u 2 2 2 c 931.5 MeV c 16 15 31.97207 u c
33. We find the energy from the wavelength. 6.63 1034 J s3.00 108 m s E hc 11.8MeV 13 19 1.05 10 m1.602 10 J eV This has to be a ray from the nucleus rather than a photon from the atom. Electron transitions do not involve this much energy. Electron transitions involve energies on the order of a few eV.
34. The nuclear decay reaction is
. We add 10 electrons to both sides in order to
use atomic masses for the calculations. Then the mass of the beta is accounted for in the mass of the sodium. The maximum kinetic energy of the corresponds to the neutrino having no kinetic energy (a limiting case) and no mass, and it is the Q-value of the reaction. We also ignore the recoil of the sodium. K Q m 23 Ne m 23 Na c2 22.9945 u 22.9898 u c2 931.5 MeV uc2 max 11 10 4.4 MeV If the neutrino were to have all of the kinetic energy, the minimum kinetic energy of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be the Q-value, and so the neutrino energies are 0 and 4.4 MeV, respectively. 35. A emission can be modeled as a proton changing into a neutron and a positron. The nuclear reaction here is 137 N 136C 01 v . If we add 7 electrons to each side, the nitrogen atomic mass can be used, and the carbon atomic mass can be used, but there will the mass of an electron and the mass of a positron as products that must be included. Thus, the atomic reaction is 13 13 0 0 1 v . See Problem 36 for a general discussion. 7 N 6 C 1 1 e The kinetic energy of the particle will be maximum if the (almost massless) neutrino and the electron have no kinetic energy. We also ignore the recoil of the carbon. 0 2 K m 137 N m 136 C m 10 e m 10 c2 m 137 N m 136 C 2m 1e c
13.005739 u 13.003355 2 0.00054858 u c2 931.5 MeV uc2 1.199 MeV
If the and electron have no kinetic energy, the maximum kinetic energy of the neutrino is also 1.199 MeV . The minimum energy of each is 0 when the other has the maximum. 36. For the positron-emission process, A N Z
A Z 1
N e v. We must add Z electrons to the nuclear
mass of N to be able to use the atomic mass, and so we must also add Z electrons to the reactant side. On the reactant side, we use Z 1 electrons to be able to use the atomic mass of N. Thus, we
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have 1 “extra” electron mass and the -particle mass, which means that we must include 2 electron masses on the right-hand side. We find the Q-value given this constraint. Q M M 2m c2 M M 2m c2. P D e D e P 37. We assume that the energies are low enough that we may use classical kinematics. In particular, we will use p 2mK . The decay is 23892U 23490Th 42He. If the uranium nucleus is at rest when it decays, the magnitude of the momentum of the two daughter particles must be the same. p 2 p 2 2m K m K 4u 4.20 MeV p p ; K Th 0.0718MeV Th Th 2m m 234 u 2m 2m Th
Th
Th
Th
The Q-value is the total kinetic energy produced. Q K KTh 4.20MeV 0.0718MeV 4.27 MeV 38. Both energy and momentum are conserved. Therefore, the momenta of the product particles are equal in magnitude. We assume that the energies involved are low enough that we may use classical kinematics; in particular, p 2mK . m 2m K 4.002603 2 2 p p p pRn ; K Rn Rn 2m m K 222.017576 K 2m 2m Rn Rn Rn Rn The sum of the kinetic energies of the product particles must be equal to the Q-value for the reaction. 4.002603 226 222 4 2 K K K K m Ra m Rn m He c 88 86 2 Rn 222.017576 m 226 Ra m 222 Rn m 4 He c2 88 86 2 K 4.002603 1 222.017576
226.025409 u 222.017576 u 4.002603u c2 931.5 MeV uc2 4.79 MeV 4.002603 1 222.017576 2 The rest energy (mc ) of an alpha particle is about 4000 MeV, so our assumption of classical kinematics is valid.
39. (a) The decay constant can be found from the half-life, using Eq. 41–8. ln 2 ln 2 1.5 1010 yr1 4.9 1018 s1 T1/ 2 4.5 109 yr (b) The half-life can be found from the decay constant, 1h using Eq. 41–8. 18240s ln 2 ln 2 T 5.1h 1/ 2 3.8105 s1 3600s
40. We find the half-life from Eq. 41–7d and Eq. 41–8. ln 2 t R R et R e T1/2 T ln 2 t ln 2 4.2 h 1.4 h 0 0 1/ 2 R 140 ln ln 1120 R0 We can see this also from the fact that the rate dropped by a factor of 8, which takes 3 half-lives.
Chapter 41
Nuclear Physics and Radioactivity
41. We use Eq. 41–6 to find the fraction remaining. ln 22.2 yr12mo yr N t 9 mo t e e N N0e 0.1309 0.13 13% N0 42. The activity at a given time is given by Eq. 41–7b. The half-life is found in Appendix G. ln 2 dN N ln 2 N 5.6 1020 nuclei 2.1109 decays s 7 T1/ 2 dt 5730 yr3.16 10 s yr 43. Every half-life, the number of nuclei left in the sample is multiplied by one-half. N n 5 1 1 0.03125 1 32 3.125% 2 2 N0 44. We need the decay constant and the initial number of nuclei. The half-life is found in Appendix G and is about 8 days. Use Eq. 41–8 to find the decay constant. ln 2 ln 2 9.99668 107 s1 T1/2 8.0252 days24 h day3600s h 846 106 g 6.022 1023 atoms mol 3.8918 1018 nuclei. N 0 130.906126g mol (a) We use Eq. 41–7b to evaluate the initial activity.
Activity N N 0 et 9.99668 107 s1
3.8918 1018 e0 3.8905 1012 decays s
(b) We evaluate Eq. 41–7d at t 1.50 h. This is much less than one half life, so we expect that most of the initial nuclei remain, and the activity is essentially unchanged. 3600s R R e 3.8905 10 decays se 9.9966810 s 1.50 h 3.8696 10 decays s 7
t
1
12
h
12
0
3.87 1012 decays s (c) We evaluate Eq. 41–7d at t = 3.0 months. We use a time of 1/4 year for the 3.0 months – other approximations (like 1 month = 30 days) for that time period will give a slightly different final answer. In any event, this time period is more than 10 half-lives, so we expect the activity to be significantly lower than the initial value. 9.99668107 s1 0.25yr3.156107 s yr R R 0et 3.8905 1012 decays se 1.4607 109 decays s 1.46 109 decays s 45. The activity of a sample is given by Eq. 41–7a. There are two different decay constants involved. Note that Appendix G gives half-lives, not activities. t N N N e I N e Cot I e I Co t I I Co Co I 0 Co 0 Co
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I
5.2712 y365.25d y ln T I 8.0252d Co 1/2 1/2 63.715d ln 2 ln 2 1 1 ln 2 TI TCo 8.0252d 5.2712 y365.25d y 1/2 1/2
ln T
ln t
Instructor Solutions Manual
Co
I Co
46. We find the number of nuclei from the activity of the sample and the half-life. The half-life is found in Appendix G. dN N ln 2 N T1/ 2 dt T1/ 2 dN 4.468 10 yr3.156 10 s yr370decays s 7.5 1019 nuclei ln 2 dt ln 2 9
N
7
47. Each emission decreases the mass number by 4 and the atomic number by 2. The mass number changes from 235 to 207, which is a change of 28. Thus, 7 particles must be emitted. With the 7 emissions, the atomic number would have changed from 92 to 78. Each emission increases the atomic number by 1. Thus, to have a final atomic number of 82, 4 particles must be emitted. 48. We will use the decay constant frequently, so we calculate it here. ln 2 ln 2 0.022432s1 T1/ 2 30.9s (a) We find the initial number of nuclei from an estimate of the atomic mass. 9.6 106 g N0
124g mol
6.022 1023 atoms mol 4.662 1016 4.7 1016 nuclei.
(b) Evaluate Eq. 41–6 at t 2.6min. N N 0et 4.662 1016 e
0.022432s1
2.6 min60s min 1.409 1015 1.4 1015 nuclei
(c) The activity is found by Eq. 41–7a. N 0.022432s1 1.409 1015 3.1611013 (d) We find the time from Eq. 41–7a, relative to the time when we had the initial 9.6 g of 12455Cs .
N N 0et
N 1 1decay/s 16 ln 0.022432s 4.662 10 decay/s ln N0 1542s 26 min t 1 0.022432s This is about 50 half-lives.
49. We find the mass from the initial decay rate and Eq. 41–7b. 6.022 1023 nuclei mole dN N0 m dt 0 atomic weight g mole
Chapter 41
Nuclear Physics and Radioactivity
m
dN
1 atomic weight
dt 0 6.022 1023
dN
T1/ 2 atomic weight
6.022 10 1.6 10 s 1.24810 yr3.156 10 s yr 39.963998g 0.6034g 0.60g ln 2 6.022 10 5 1
23
dt 0 ln 2
9
7
23
50. The number of nuclei is found from the mass and the atomic weight. The activity is then found from number of nuclei and the half-life, using Eq. 41–7b.6 7.8 10 g ln 2 ln 2 dN 6.022 1023 atoms mol N N 6 T1/ 2 dt 1.2310 s 31.973908g mol 8.279 1010 decays s 8.31010 decays s 51. (a) The decay constant is found from Eq. 41–8. ln 2 ln 2 1.3811013 s1 1.381013 s1 5 T1/ 2 1.59 10 yr3.156 107 s yr (b) The activity is the decay constant times the number of nuclei. N 1.3811013 s1 3.50 1018 4.8335 105 decays s 60s 1min 2.90 107 decays min
52. We use Eq. 41–7d, with R 16 R0 .
ln 2
R R et R e T1/2 0
0
t
T 1/ 2
ln 2
ln 2 9.8min 3.8min t 1 R ln 6 ln R0
53. We assume the initial sample of pure carbon is naturally occurring carbon, and so use the atomic weight of naturally occurring carbon to find the number of atoms in the sample. The activity is found from Eq. 41–7a, and the half-life is found in Appendix G. 315g 6.022 1023 atoms mol 1.579 1025 atoms N 12.0109g mol 1.3 N 1.579 1025 2.0531013 nuclei of 14 C 14 6 1012 ln 2 2.0531013 78.69decays s 79decays s N 5730 yr3.156 107 s yr 54. We find the mass from the activity. Note that NA is used to represent Avogadro’s number. The half-life is found in Appendix G. ln 2 mNA R N T1/ 2 A
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
RT1/ 2 A 475decays s4.468 10 yr3.156 10 s yr238.050787 g mole NA ln 2 6.022 1023 nuclei moleln 2 9
m
Instructor Solutions Manual
7
3.8199 102 g 3.82 102 g
55. We assume that we currently have N nuclei of 87 Rb and 0.0260 N nuclei of 87 Sr. The sum of those 37
38
two would be the initial number of 8737 Rb nuclei, N . Thus, N 1.0260N . Use Eq. 41–6 and Eq. 0 0 41–8 to find the time for N nuclei of 87 Rb to decay to N nuclei of 87 Rb , which is the estimated 0
37
37
age of the fossils. N N0 et 1 N T1/ 2 N T1/ 2 N 4.97 1010 yr 1 t ln ln ln ln N N ln 2 ln 2 ln 2 1.026 0 0 1.0260N0 1.84 109 yr 56. We are not including the neutrinos that will be emitted during beta decay. 232 First sequence: Th 228 Ra 4 ; 228 Ra 228 Ac ; 228 Ac 228 Th ; 90
228
2
88
4
;
224
2
88
88
Th
90
Second sequence:
235 92
Th 42 ; U 231 90
227
89
Ac 227 Th
231 90
;
90
Ra 220 Rn 4
86
2
Th 231 Pa ; 231 Pa 227 Ac 4 ; 91
91
90
89
2
Th 223 Ra
227
90
89
89
4
88
2
57. Because the fraction of atoms that are 14 C is so small, we use the atomic weight of 12 C to find the 6
6
number of carbon atoms in 67 g. We then use the ratio to find the number of 14 6C atoms present when the club was made. Finally, we use the activity as given in Eq. 41–7c to find the age of the club. The half-life is found in gAppendix G. 67 6.022 N C 1023 atoms mol 3.362 1024 atoms of 126C 12g mol 6 12
N14 1.31012 3.362 1024 4.3711012 nuclei of 146C 6C
N N e N N T 1 t
14 6C
14 6C
today
14
t ln
0
14
C
6
today
N
1/ 2
ln
6
C
today
ln 2 14 N C T 1/ 2 6 0 5730 yr 7.0 decays s 7214 yr 7200 yr ln 2 ln ln 2 12 4.37110 nuclei 5730 yr3.156 107 s yr 0 14 C 6
0
ln 2
Chapter 41
Nuclear Physics and Radioactivity
N 58. The decay rate is given by Eq. 41–4b,
t
N . We assume equal numbers of nuclei decaying by
emission for the two isotopes. N T1/2 t 1.6 104 s N 218 218 218 218 7 214 8.6 10 T N 214 N214 214 1/2 3.1min60s min 218 t 214
59. The activity is given by Eq. 41–7a. The original activity is N 0 , so the activity 351.0 hours later is 0.945N0 . ln 2 t 0.945N 0 N 0 et ln 0.945 t T1/ 2 1d T ln 2 35.0 h 428.85 h 17.9d 1/ 2 ln 0.945 24 h 60. The activity is given by Eq. 41–7a. (a) Use Eq. 41–7d. 1 R T R 53d 25decays s R R0 et t ln 1/ 2 ln R ln 2 R ln 2 ln 320decays s 194.94d 190d 0 0 (b) We find the mass from the activity. Note that NA is used to represent Avogadro’s number, and A is the atomic weight. R N ln 2 m0 NA 0 0 T1/ 2 A
m
R0T1/ 2 A
320decays s53d86, 400s d7.016929g mole
6.022 10 nuclei moleln 2 23
NA ln 2
2.5 1014 g
61. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays becomes a daughter nucleus. N N 0et ND N0 N N0 1 e t
62. The activity is given by Eq. 41–7d, with R 0.01050R0 . R ln t ln 2 4.00 hln 2 0.6085h 36.5min R R et R0 ln 2 T 1/ 2 0 t T1/ 2 R ln 0.01050 ln R0
From Appendix G, we see that the isotope is
211 82
Pb .
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63. Because the carbon is being replenished in living trees, we assume that the amount of 146C is constant until the wood is cut, and then it decays. We use Eq. 41–6. The half-life is given in Appendix G. The N is 0.062. ratio N0 N ln N N et N0 ln 2 0 t T1/ 2 t
T1/ 2
ln
ln 2 64. (a)
N
5730 yrln 0.062 22,986 yr ln 2
N0
23, 000 yr
The mass number is found from the radius, using Eq. 41–1.
r 1.2 1015 m A1/3 3 3 7500 m r A 2.4411056 2.4 1056 15 15 1.2 10 m 1.2 10 m (b) The mass of the neutron star is the mass number times the atomic mass unit conversion in kg.
1.6605 1027 kg u 4.0531029 kg 4.11029 kg
m 2.4411056 u
Note that this is about 20% of the mass of the Sun. (c) The acceleration of gravity on the surface of the neutron star is found from Eq. 6–4 applied to the neutron star. Gm 6.67 1011 N m2 kg2 4.053 1029 g 4.806 1011 m s2 4.8 1011 m s2 2 r2 7500 m 65. Because the tritium in water is being replenished, we assume that the amount is constant until the N is 0.10. wine is made, and then it decays. We use Eq. 41–6. The value of N0 N ln N N 0et N0 ln 2 t T1/ 2 ln N 12.3 yrln 0.10 41yr t T ln 2 N ln 2
1/ 2
0
66. (a) We assume a mass of 70 kg of water, and find the number of protons, given that there are 10 protons in a water molecule. 3 N 70 10 g water 6.02 1023 molecules water 10 protons mol water water molecule protons 18g water mol water 28 2.34 10 protons We assume that the time is much less than the half-life so that the rate of decay is constant. ln 2 N t N T N 1/2 1033 yr N T 1 proton 4 t 1/2 6.165 10 yr 60,000 yr N ln 2 2.34 1028 protons ln 2 This is about 880 times longer than a life expectancy of 70 years.
Chapter 41
Nuclear Physics and Radioactivity
(b) Instead of waiting 61,650 consecutive years for one person to experience a proton decay, we could interpret that number as there being 880 people, each living 70 years, to make that 61,650 years (since 880 70 61,600). We would then expect one person out of every 880 to experience a proton decay during their lifetime. Divide 7 billion by 880 to find out how many people on Earth would experience proton decay during their lifetime. 7 109 7.95 106 8 million people 880 67. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy at the distance of closest approach. Note that the distance found is the distance from the center of the alpha to the center of the gold nucleus. The alpha particle has a charge of +2e, and the gold nucleus has a charge of +79e. 1 q q K U K U K 0 0 Au i i f f 40 r 2 2791.60 1019 C 1 q q 9 2 Au 8.988 10 N m 2.9511014 m C2 r 40 K 7.7MeV1.60 1013 J MeV
3.0 10
14
m
We use Eq. 41–1 to compare to the size of the gold nucleus. rapproach 2.9511014 m 4.2 rAu 1971/3 1.2 1015 m
So the distance of approach is about 4.2 the radius of the gold nucleus. 68. We find the number of half-lives from the change in activity. dN ln 0.0150 n dt 6.06 half-lives 21 0.0150 n dN ln 12 It takes 6.06 half-lives for a sample to drop to 1.50% of its original activity. 69. We can find the mass of 4019K from its activity. Note that NA is Avogadro’s number, N40 is the number of particles of 40 K , and A is the atomic mass of 40 K . Similar notation is used for 39 K . 19
40
19
19
R N ln 2 m40 N 40 40 T1/ 2 A40 A
m40
R40T1/ 2 A40 48decays s1.248 109 yr3.156 107 s yr39.963998g mole N A ln 2 6.022 1023 nuclei moleln 2
1.8101104 g 1.8 104 g We find the number of 39 K atoms from the number of 40 K atoms and the abundances given in 19
Appendix G. That is then used to find the mass of 3919K. R R T R N N 40 40 40 1/ 2 ; 40 40 ln 2
19
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
0.932581 N 40 0.000117 m N A39 0.932581 N A39 0.932581 R40T1/ 2 A39 0.000117 40 N 0.000117 ln 2 N 39 39 N A A A 9 7 48decays s 1.248 10 yr 3.156 10 s yr 38.963706 g mole 0.932581 ln 2 0.000117 6.022 1023 nuclei mole N40 0.000117 NK ; N39 0.932581NK
1.407 g 1.4 g 70. We see from the periodic table that Sr is in the same column as calcium. If strontium is ingested, the body may treat it chemically as if it were calcium, which means it might be stored by the body in bones. We use Eq. 41–6 to find the time to reach a 1% level. N ln N N et N0 ln 2 0 t T1/ 2 29 yr ln 0.01 T N t 1/ 2 ln 192.67 yr 200 yr ln 2 ln 2 N0 The decay reactions are as follows. The strontium reaction is beta decay, as given in Appendix G. We assume the daughter also undergoes beta decay. 90 Sr 90 Y 0 e v ; 90 Y 90 Zr 0 e v 38
39
1
39
40
1
71. We find the mass from the initial decay rate and Eq. 41–7b. 6.022 1023 nuclei mole dN N0 m dt 0 atomic weight g mole dN 1 atomic weight dN T1/ 2 atomic weight m dt 0 6.022 1023 dt 0 ln 2 6.022 1023 3.94 104 s1
87.37 d86, 400s d 34.969032 g 2.49 1011 g 23 ln 2 6.022 10
72. (a) We find the daughter nucleus by balancing the mass and charge numbers: Z X Z Os Z e 76 1 77 A X AOs Ae 191 0 191 The daughter nucleus is 191 77 Ir . (b) See the included diagram. (c) Because there is only one energy, the decay must be to the higher excited state.
191 76Os
– (0.14 MeV)
(0.042 MeV) (0.129 MeV)
191 77 Ir* 191 77 Ir* 191 77 Ir
Chapter 41
Nuclear Physics and Radioactivity
73. The activity is the decay constant times the number of nuclei, as given by Eq. 41–7a. (a) We calculate the activity for 3215P. ln 2 ln 2 1.0g R N N 6.022 1023 nuclei mol T1/ 2 14.3d86, 400s d 31.973908g mol 1.057 1016 decays s 1.11016 decays s (b) We calculate the activity for 23290Th. ln 2 R N T1/ 2
1.0 g 23 6.022 10 nuclei mol 10 7 232.038054 g mol 1.40 10 yr 3.156 10 s yr ln 2
4.1103 decays s
74. From Fig. 41–1, the average binding energy per nucleon at A 63 is ~8.6 MeV. We use the mass average atomic weight as the average number of nucleons for the two stable isotopes of copper. That gives a binding energy of 63.5468.6 MeV 546.5MeV 550 MeV . The number of copper atoms in a penny is found from the atomic weight. 3.0g N 6.022 1023 atoms mol 2.8431022 atoms 63.546g mol Thus the total energy needed is the product of the number of atoms times the binding energy.
2.8431022 atoms546.5MeV atom1.60 1013 J MeV 2.5 1012 J
75. (a)
24 He m 24 He A24 He 4.002603u 4 0.002603u
(b) C m C A C 12.000000 u 12 0 for both units. (c) Sr m Sr A Sr 85.909261u 86 0.090739 u 0.090739 u931.5MeV uc2 (d) U m U A U 235.043928u 235 0.043928u 0.043928 u931.5MeV uc2 0.002603u 931.5 MeV uc2
12 6
12 6
12 6
86 38
86 38
235 92
235 92
86 38
235 92
(e) From Appendix G we see that 0 for 0 Z 7 and Z 86; 0 for 8 Z 85.
0 for 0 A 15 and A 214; 0 for 16 A 214.
76. The reaction is 11 H 10n 21H. If we assume the initial kinetic energies are small, then the energy of the gamma is the Q-value of the reaction. Q m 11H m 10n m 2 1H c2
1.007825u 1.008665u 2.014102 u c2 931.5 MeV uc2 2.224 MeV
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
77. The mass of carbon 60,000 years ago was 1.0 kg. Find the total number of carbon atoms at that time, and then find the number of 146 C atoms at that time. The text says that the fraction if C-14 atoms is 1.31012 . Use that with the half-life to find the present activity, using Eq. 41–7d and Eq. 41–8. 1.0 103 g 6.022 1023 atoms mol N 5.014 1025 C atoms 12.0109g mol N14 5.014 1025 1.31012 6.518 1013 nuclei of 146C N 0 ln 2
t ln 2 ln 2 T N R N 0 e 1/2 N T1/ 2 T1/ 2
ln 2
6.518 10 e 13
ln 260,000 yr
5730 yr
0.1759decays s 0.2decays s Since the time only has 1 significant figure, we only keep 1 significant figure in the answer. 78. The energy to remove the neutron would be the difference in the masses of the 24 He and the combination of 23 He n. It is also the opposite of the Q-value for the reaction. The number of electrons doesn’t change, so atomic masses can be used for the helium isotopes. 931.5MeV c2 Q m m m c 2 3.016029 u 1.008665u 4.002603u c 2 He He-3 n He-4 u 20.58 MeV
Repeat the calculation for the carbon isotopes. 931.5MeV c2 Q m m m c 2 12.000000 u 1.008665u 13.003355u c 2 C C-12 n C-13 u 4.946 MeV The helium value is 20.58MeV 4.946 MeV 4.161 greater than the carbon value. 79. (a) Take the mass of the Earth and divide it by the mass of a nucleon to find the number of 1/3 nucleons. Then, use Eq. 41–1 to find the radius. 5.98 1024 kg r 1.2 1015 m A1/3 1.2 1015 m 183.6 m 180 m 27 1.67 10 kg (b) Follow the same process as above, just using the Sun’s mass. 1/3 1.99 1030 kg 4 r 1.2 10 15 m A1/3 1.2 10 15 m 12700 m 1.3 10 m 27 1.67 10 kg 80. (a) The usual fraction of 14 C is 1.3 1012. Because the fraction of atoms that are 14 C is so small, 6
6 12 6
we use the atomic weight of C to find the number of carbon atoms in 84 g. We use Eq. 41–6 to find the time. 84g 23 24 N12 6.022 10 atoms mol 4.215 10 atoms 12g mol N14 4.215 1024 atoms1.31012 5.4795 1012 atoms 1 N T N 5730 yr ln 1 N N 0 et t ln 1/ 2 ln 2.4 105 yr N ln 2 N ln 2 5.4795 1012
0
0
Chapter 41
Nuclear Physics and Radioactivity
(b) We do a similar calculation for an initial mass of 340 grams. 340g 23 12 13 N14 6.022 10 atoms mol1.310 2.218 10 atoms 12g mol 1 N T N 5730 yr ln 1 N N0 et t ln 1/ 2 ln 5 N ln 2 N ln 2 2.218 1013 2.5 10 yr
0
0
5
This shows that, for times on the order of 10 yr, the sample amount has fairly little effect on the age determined. Thus, times of this magnitude are not accurately measured by carbon dating. 81. (a) This reaction would turn the protons and electrons in atoms into neutrons. This would eliminate chemical reactions, and thus eliminate life as we know it. (b) We assume that there is no kinetic energy brought into the reaction, and solve for the increase of mass necessary1to make 1the reaction energetically possible. For calculating energies, we write the reaction as H n v, and we assume the neutrino has no mass or kinetic energy. 1 0 Q m 1 H m 1 n c2 1.007825u 1.008665u c2 931.5 MeV uc2 0 1 0.782 MeV This is the amount that the proton would have to increase in order to make this energetically possible. We find the percentage change. 0.782 MeV c2 m 100 m 938.27 MeV c2 100 0.083%
82. We assume the particles are not relativistic, so that p 2mK . The radius is given in Example 27–7 mv as r . Set the radii of the two particles equal. Note that the charge of the alpha particle is twice qB that of the electron (in absolute value). We also use the “bare” alpha particle mass, subtracting the two electrons from the helium atomic mass. r r m v m v m v 2m v p 2 p 2eB eB 2 4 p2 p 40.000549 u K 2m 2m 4m 4 K p2 p2 m 4.002603u 20.000549 u 5.48 10 2m
2m
83. Natural samarium has an atomic mass of 150.36 grams per mole. We find the number of nuclei in the natural sample, and then take 15% of that to find the number of 147 Sm nuclei. We first find the 62 number of 14762Sm nuclei from the mass and proportion information. N147
0.15 N natural Sm
0.151.00g6.022 1023 nuclei / mol
150.36g mol The activity level is used to calculate the half-life. 62
6.008 1020 nuclei of 14762Sm
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Activity R N
T 1/ 2
ln 2
N
T1/ 2 ln 2
N
R
ln 2
Instructor Solutions Manual
6.008 10 3.44069 10 s 20
11 3.156 107 s 1.110 yr
18
120decays s
1yr
84. Since amounts are not specified, we will assume that “today” there is 0.720 g of 23592U and 100.000 0.720 99.280g of 23892U. We use Eq. 41–6. (a) Relate the amounts today to the amounts 1.0 109 years ago. t
N Net NeT1/2
N N et 0
0
ln 2
ln 2
1.010 yr 7.0410 yr ln 2 1.927 g 0.720g e
ln 2
1.010 ln 2 4.46810 115.94g 99.280g e
9
t
N0 235 N 235 eT
1/2
8
9
t
N0 238 N 238 eT
1/2
9
1.927
The percentage of 235 U was
1.927 115.94
92
100% 1.63%
(b) Relate the amounts today to the amounts 100 106 years from now. 10010 yr t ln 2 ln 2 t N N e T1/2 0.720ge 7.0410 yr 0.6525g NN e 6
8
0
235
N N e T
t
ln 2
1/2
238
0 235
99.280ge
0 238
The percentage of 235 U will be
100106 9yr 4.46810 yr ln 2
0.6525 0.6525 97.752
92
97.752g
100% 0.663%
85. (a) The decay constant is calculated from the half-life using Eq. 41–8. ln 2 ln 2 12 1 3.83 10 s T1/2 5730 yr 3.156 107 s yr (b) In a living organism, the abundance of 146C atoms is 1.31012 per carbon atom. Multiply this abundance by Avogadro’s number and divide by the molar mass of carbon-12 to find the number of carbon-14 atoms per gram of carbon. 23 12 14 14 C atoms N C atoms 6.022 10 atoms 6 C mole 12 6 12 6 A N 1.310 12 1.310 12 C atoms M C atoms 12.000000 g 12 C mole
6
6
6
6.524 atoms 146 C g 126C 6.5 10 atoms 6 C g 6 C 10
14
12
(c) The activity in natural carbon for a living organism is the product of the decay constant and the number of 146 C atoms per gram of 126 C . Use Eq. 41–7d and Eq. 41–4b.
R N 3.831012 s1 0.25 decays s g 6 C 12
6.524 10 atoms C g C 0.2499 decays s g C 10
14 6
12 6
12 6
Chapter 41
Nuclear Physics and Radioactivity
(d) Take the above result as the initial decay rate (while Otzi was alive), and use Eq. 41–7d to find the time elapsed since he died. t
R R0 e
1 0.121 1.894 1011 s ln 1yr 7 t 1 ln R 12 1 0.2499 3.156 10 s 6000 yr 3.8310 s R
0
Otzi lived approximately 6000 years ago. 86. (a) We use the definition of the mean life given in the problem. We use a definite integral formula from Appendix B-5. t t 1 1 tN t dt tN e dt te dt
0
N t dt
0
0
0
Ne
0
dt
0
0
t
0
(b) We evaluate at time t
1 1 2
1
e dt
t
2
1
.
N t N e 0 e e 1 0.368 N t 0 N 0 e0 87. The mass number changes only with decay, and changes by 4. If the mass number is 4n, the new number is 4n 4 4 n 1 4n. There is a similar result for each family, as shown here. 4n 4n 4 4 n 1 4n 4n 1 4n 4 1 4 n 1 1 4n 1 4n 2 4n 4 2 4 n 1 2 4n 2 4n 3 4n 4 3 4 n 1 3 4n 3 Thus, the daughter nuclides are always in the same family.
88. (a) If the initial nucleus is at rest when it decays, momentum conservation says that the magnitude of the momentum of the alpha particle will be equal to the magnitude of the momentum of the daughter particle. We use that to calculate the (non-relativistic) kinetic energy of the daughter particle. The mass of each particle is essentially equal to its atomic mass number, in atomic mass units. Note that classically, p . 2 2 2m K m A 4 p pD ; K D pD p K K K 2m 2m 2m m A A D
D
4 A K
D
D
D
D
K K D D 1 K KD 4 K AD K K 1 A K 4 A D 226 (b) We specifically consider the decay of 88Ra. The daughter has AD 222. KD 1 1 1 K KD 1 1 AD 1 222 0.017699 1.8%
4
4
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
Thus, the alpha particle carries away 1 0.018 0.982 98.2% . 89. We determine the number of 4019K nuclei in the sample, and then use the half-life to determine the activity. 3 23 400 10 g6.022 10 atoms mol 7.208 1017 N 40 0.000117 N 0.000117 naturally 19 K 39.0983g mol occurring K ln 2 1yr 17 R N N ln 2 7.208 10 12.51decay s 13decay s 7 T1/ 2 1.265 109 yr 3.156 10 s
90. (a) The reaction is 23692U 23290Th 4 He. If we assume the uranium nucleus is initially at rest, 2 the magnitude of the momenta of the two products must be the same. The kinetic energy available to the products is the Q-value of the reaction. We use the non-relativistic relationship that p2 2mK. 2 2 2m K m p p ; K pTh pHe He He He He He Th Th 2m m K 2mTh 2m Th Th Th m Q KTh KHe mHe 1 KHe Th m 2 mTh K Th Q m m m c He m m m U Th He m He Th He Th 236.045566 u 232.038054 u 4.002603uc2 232.038054 u 4.002603u 232.038054 u
0.004826u 931.5 MeV uc2 4.495 MeV (b) We use Eq. 41–1 to estimate the radii.
rHe 1.2 1015 m4 1.9051015 m 1.9 1015 m 1/3
rTh 1.2 1015 m232 7.374 1015 m 7.4 1015 m 1/3
(c) The maximum height of the Coulomb barrier will correspond to the alpha particle and the thorium nucleus being separated by the sum of their radii. We use Eq. 23–10. 1 q q qHe qTh 1 He Th U 40 rA 40 rHe rTh 2 2901.60 1019 C 8.988 109 N m2 C2
27.898 MeV 28 MeV (d) At position “A”, the product particles are separated by the sum of their radii, about 9.3 fm. At position “B”, the alpha particle will have a potential energy equal to its final kinetic energy, 4.495 MeV. Use Eq. 23–10 to solve for the separation distance at position “B”. 1 qHe qTh UB 40 RB
Chapter 41
Nuclear Physics and Radioactivity
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Chapter 41
Nuclear Physics and Radioactivity
2901.60 1019 C
2
R B
1 qHe qTh 40 UB
8.988 109 N m2 C2
4.495 MeV1.60 1013 J MeV
57.57 1015 m RB RA 57.6 fm 9.3fm 48.3fm Note that this is a center-to-center distance. 91. We take the momentum of the nucleon to be equal to the uncertainty in the momentum of the nucleon, as given by the uncertainty principle. The uncertainty in position is estimated as the radius of the nucleus. With that momentum, we calculate the kinetic energy using a classical formula. Iron has about 56 nucleons (depending on the isotope). px K
p
p p
x
r
2
2mr
2m
2
1.60 1013 J MeV 21.67 1027 kg561/3 1.2 1015 m 2
0.988MeV 1MeV 92. (a) From the figure, the initial force on the detected particle is down. Using the right-hand rule, the force on a positive particle would be upward. Thus, the particle must be negative, and the decay is decay. (b) The magnetic force is producing circular motion. Set the expression for the magnetic force equal to the expression for centripetal force, and solve for the velocity. 19 3 mv2 qBr 1.60 10 C0.012T4.7 10 m qvB v 9.9 106 m s r m 9.111031kg
93. If the original nucleus is of mass 256 u, the daughter nucleus after the alpha decay will have a mass of 252 u. Moreover, the alpha has a mass of 4 u. Use conservation of momentum – the momenta of the daughter nucleus and the alpha particle must be equal and opposite since there are only two products. The energies are small enough that we may use non-relativistic relationships. m K 4 4.5MeV 0.071MeV p pX 2m K K X 252 mX 94. We use Eq. 41–7a to relate the activity to the half-life. We estimate the atomic weight at 152 grams/mole since this isotope is not given in Appendix G. dN R N ln 2 N T1/ 2 dt 6.022 1023 nuclei 1yr ln 2 ln 2 7 T1/ 2 N 1.5 10 g 7 1.31021 yr R 1decay s 152g 3.156 10 s
95. We calculate the initial number of nuclei from the initial mass and the atomic mass. 1 nucleus 1.0187 1017 nuclei 1.02 1017 nuclei N 0 2.20 109 kg 27 13.005739 u1.660510 kg u See the adjacent graph. From the graph, the half-life is approximately 600 seconds. Note that the
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
half-life as given in Appendix G is T1 2 9.965min
Instructor Solutions Manual
60s 597.9s . 1min
96. The emitted photon and the recoiling nucleus have the same magnitude of momentum. We find the recoil energy from the momentum. We assume the energy is small enough that we can use classical relationships. E p p 2m K K K K c 2 1.46 MeV E2 KK 2.86 105 MeV 28.6eV 2 2mK c2 931.5 MeV c 2 239.963998 u c u 97. Section 41–1 states that the neutron, proton, and electron are all spin 1 2 particles. Consider the reaction n p e v. If the proton and neutron spins are aligned (both are 1 2, for example), the electron and anti-neutrino spins must cancel. Since the electron is spin 1 2, the anti-neutrino must be spin 1 , which means the magnitude of the neutrino spin must be 1 in this case. 2
2
The other possibility is if the proton and neutron spins are opposite of each other. Consider the case of the neutron having spin 1 and the proton having spin 1 . If the electron has spin 1 , the spins of 2
2
2 1 2
the electron and proton cancel, and the neutrino must have spin for angular momentum to be conserved. If the electron has spin 1 , the spin of the neutrino must be 3 for angular momentum to 2
2
be conserved. A similar argument could be made for positron emission, with p n e v.
Chapter 41
Nuclear Physics and Radioactivity
CHAPTER 43: Elementary Particles Responses to Questions 1.
A reaction between two nucleons that would produce a is: p n p p .
2.
No, the decay is still impossible. In the rest frame of the proton, this decay is energetically impossible, because the proton’s mass is less than the mass of the products. Since it is impossible in the rest frame, it is impossible in every other frame as well. In a frame in which the proton is moving very fast, the decay products must be moving very fast as well to conserve momentum. With this constraint, there will still not be enough energy to make the decay energetically possible. In order for these products to be produced, the proton would have to interact with some other particle.
3.
Antiatoms would be made up of negatively charged antiprotons and neutral antineutrons in the nucleus with positively charged positrons surrounding the nucleus. If antimatter and matter came into contact, the particle–antiparticle pairs would annihilate, converting their mass into energetic photons.
4.
The photon signals the electromagnetic interaction.
5.
(a) Yes. If a neutrino is produced during a decay, the weak interaction is responsible. (b) No. For example, a weak interaction decay could produce a Z0 instead of a neutrino.
6.
The neutron decay process also produces an electron and an antineutrino; the antineutrino will only be present in a weak interaction.
7.
An electron takes part in the electromagnetic, weak, and gravitational interactions. A neutrino takes part in the weak and gravitational interactions. A proton takes part in the strong, electromagnetic, weak, and gravitational interactions.
8.
The chart here shows charge and baryon conservation checks for many of the decays in Table 43–2. Particle W
Decay W e v
e
(others are similar) Z0
Z0 e e (others are similar)
Higgs
H0 b b H0 W W H0 Z0 Z0
Charge conservation +1 1 0
Baryon conservation 000
0 1 1
000
0 1 1
0 1 1
0 1 1
3
3
3
3
000
000 000
muon
(others are similar) e v v
1 1 0 0
0000
tau
v v
1 1 0 0
0000
pion
(other is similar) v
+1 1 0
000
0
000
000
e
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
kaon
K v
000
0 1 1
000
K
000
000
0 1 1 0
0000
K L v
0 1 1 0
0000
K
0
0 1 1 0
0000
K
0
0000
0000
000 0000
000 0000
0 0
0 1 1 0
0000
0
0 1 1
000
000
000
+1 1 0
000
0 1 1 0
1 1 0 0
0 p
0 1 1
1 1 0
0 n 0
000 +1 1 0
1 1 0 1 1 0
n 0 0
+1 0 1 000
1 1 0 1 1 0
n 0 0 0
1 0 1
1 1 0
000
1 1 0
1 0 1
1 1 0
1 0 1
1 1 0
K
1 0 1
1 1 0
0
1 1 0
1 1 0
0
S
0
0
K e v
L 0
0 L
0
0
0
0
0
0
0
0
0
0 lambda sigma
xi
npe v
p
omega
9.
e
0 L
neutron
000
+1 1 0
0 S 0
rho
+1 1 0
K K0
eta
Instructor Solutions Manual
0
0
0
0
e
0
Since decays via the electromagnetic interaction are indicated by the production of photons, the decays in Table 43–2 that occur via the electromagnetic interaction are those of the Higgs boson, the 0 , the 0 , and the 0 .
10. All of the decays listed in Table 43–2 with a neutrino or antineutrino as a decay product occur via the weak interaction. These include the W, muon, tau, pion, kaon, K0L, and neutron. In addition, the Z particle and the Higgs boson both decay via the weak interaction. In each case, we could also list the corresponding antiparticle.
Chapter 43
Elementary Particles
11. Since the baryon has B = 1, it is made of three quarks. Since the spin of the baryon is 3/2, none of these quarks can be antiquarks. Thus, since the charges of quarks are either +2/3 or –1/3, the only charges that can be created with this combination are Q = –1 (= –1/3 – 1/3 – 1/3), Q = 0 (= +2/3 – 1/3 – 1/3), Q = +1 (= +2/3 + 2/3 – 1/3), and Q = +2 (= +2/3 + 2/3 + 2/3). There is no way to combine three quarks for a total charge of Q = –2. 12. Based on lifetimes shown in Table 43–4, the particle decays that occur via the electromagnetic interaction are J 3097 and 9460 , which have shorter lifetimes than the other mesons. 13. All of the particles in Table 43–4, except for J 3097 , 9460 , and the sigma particles decay via the weak interaction, based on their lifetimes. 14. Baryons are formed from three quarks or antiquarks, each of spin 12 or 2 , respectively. Any 1
combination of quarks and antiquarks will yield a spin magnitude of either 12 or 32. Mesons are formed from two quarks or antiquarks. Any combination of two quarks or antiquarks will yield a spin magnitude of either 0 or 1. 15. If a neutrinolet was massless, it would not interact via the gravitation force; if it had no electrical charge, it would not interact via the electromagnetic force; if it had no color charge, it would not interact via the strong force; and if it does not interact via the weak force, it would not interact with matter at all. It would be very difficult to say that it exists at all. However, a similar argument could be made for photons. Photons have no color, no mass, and no charge, but they do exist. 16. (a) No. Leptons are fundamental particles with no known internal structure. Baryons are made up of three quarks. (b) Yes. All baryons are hadrons. (c) No. A meson is a quark–antiquark pair. (d) No. Hadrons are made up of quarks and leptons are fundamental particles. 17. No. A particle made up of two quarks would have a particular color. Three quarks or a quark– antiquark pair are necessary for the particle to be white or colorless. A combination of two quarks and two antiquarks is possible, as the resulting particle could be white or colorless. The neutral pion can in some ways be considered a 4-quark combination. 18. In the nucleus, the strong interaction with the other nucleons does not allow the neutron to decay. When a neutron is free, the weak interaction is the dominant force and can cause the neutron to decay.
19. No, the reaction e p n v e is not possible. The electron lepton number is not conserved: The reactants have Le 1 0 1 , but the products have Le 0 1 1 . Thus, this reaction is not possible. If the product were an electron neutrino (instead of an antineutrino), the reaction would be possible. 20. The reaction 0 p e v e proceeds via the weak force. We know this is the case since an electron antineutrino is emitted, which only happens in reactions governed by the weak interaction.
21. (a) The two major classes of fundamental particles are quarks and leptons. (b) Quarks: up, down, strange, charm, bottom, and top.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
Leptons: electron, muon, tau, electron neutrino, muon neutrino, and tau neutrino. (c) Gravity, electromagnetic, weak nuclear, and strong nuclear. (d) Gravity is carried by the graviton; the electromagnetic force is carried by the photon; the weak nuclear force is carried by the W+, W–, and Z0 bosons; the strong nuclear force is carried by the gluon. The gravitational force is much weaker than the other three forces. 22. (a) Hadrons interact via the strong nuclear force (as well as the other three fundamental forces) and are made up of quarks. (b) Baryons are hadrons, are made of three quarks, and have a baryon number of either +1 or –1. (c) Mesons are hadrons, are made of one quark and one anti-quark, and have a baryon number of 0.
Responses to MisConceptual Questions 1.
(c) Quarks are NOT leptons, nor are they composed of leptons, so answer (a) is incorrect. They also are not components of electrons, but electrons are part of “ordinary matter”. Thus, answer (b) is also incorrect. Table 43–5 lists quarks as fundamental particles, so they have no internal structure. If they had internal structure, they would not be fundamental. Thus, answer (c) is correct.
2.
(a) Atoms interact with other atoms to form molecules by means of the electromagnetic force, which binds positive and negative (nuclei and electrons) charges together. The gravitational force is too weak to affect atomic structure, and the strong and weak forces are “short range” and only affect nuclei.
3.
(a, d) From Table 43–5, we see that electrons and quarks are considered to be fundamental. Protons, neutrons, pions, and mesons are all made of quarks and so are not fundamental.
4.
(a) A common misconception is that all six quarks make up most of the known matter. However, charmed, strange, top, and bottom quarks are unstable. Protons and neutrons are made of up and down quarks, so they make up most of the ordinary matter.
5.
(b, e) Quarks combine together to form mesons (such as the meson) and baryons (such as the proton and the neutron). The electron and Higgs boson are fundamental particles and therefore cannot be made of quarks.
6.
(c) All of the fundamental forces act on a variety of objects, including our bodies. Although the strong and weak forces are very short range, the electromagnetic force is a long-range force, just like gravity. One reason we notice the “weak” gravity force more than the electromagnetic force is that most objects are electrically neutral and so do not have significant net electromagnetic forces on them. It is true that the gravitational force between people and other objects of similar size is too small for us to notice, but due to the huge mass of the Earth, we are always aware of the influence of the Earth’s gravitational force on us.
7.
(d) A tau lepton has a tau lepton number of one. When it decays, the lepton number must be conserved, so it cannot decay into only hadrons, at least a tau neutrino would have to be one of the byproducts.
8.
(b, e, f, g) Atoms are not fundamental, because they are made up of protons, neutrons, and electrons. Protons and neutrons are not fundamental as they are made from quarks. The fundamental
Chapter 43
Elementary Particles
particles are leptons (including electrons), quarks, and gauge bosons (including the photon and the Higgs boson) 9.
(a) Unlike the electron, the positron has positive charge and a negative lepton number. Unlike the charge and lepton number, the mass of the positron has the same sign (and magnitude) as the mass of the electron.
10. (e) A common misconception is that the strong force is a result of just the exchange of mesons between the protons and neutrons. This is correct on the scale of the nucleons. However, when the quark composition of the protons, neutrons, and mesons are considered at the elementary particle scale, it is seen that the transfer is due to the exchange of gluons. Therefore, both answers can be considered correct at different scales. Students who answer (d) should be given credit for their answer as well. 11. (c) Pions are not fundamental particles and are made up of quark and anti-quark pairs. Leptons and bosons (including photons) are fundamental particles, but are not a constituent of protons and neutrons. Protons and neutrons are comprised of up and down quarks. 12. (d) Quarks, gluons, neutrons, and the Higgs boson interact through the strong force. Electrons and muons are charged particles and interact through the electromagnetic force. Neutrinos only interact through the weak force. 13. (a) Strangeness is conserved in strong interactions but not in weak interactions. The other quantities (energy, charge, and momentum) are listed in the Summary section as being conserved in all nuclear and particle reactions. 14. (c) Neutrinos, electrons, and positrons (antielectrons) are all leptons (and thus fundamental particles). Mesons are made of quark–anti-quark pairs and so are not leptons.
Solutions to Problems 1.
The total energy is given by Eq. 36–11a. E m0 c2 K 0.938 GeV 5.25 GeV 6.19 GeV
2.
Because the energy of the electrons is much greater than their rest mass, we have K E pc. Combine that with Eq. 43–1 for the de Broglie wavelength. h hc hc 5.2 1017 m E pc ; p E E
3.
Use Eq. 43–2 to calculate the frequency. The alpha particle has a charge of +2e and a mass of 4 times the proton mass. 2 1.60 1019 C1.7 T f qB 13 MHz 1.3107 Hz 2m 2 4 1.67 1027 kg
4.
The time for one revolution is the period of revolution, which is the circumference of the orbit divided by the speed of the protons. Since the protons have very high energy, their speed is essentially the speed of light.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
T
2r
v
27 103 m 3.0 10 m s 8
Instructor Solutions Manual
9.0 105 s
5.
The frequency is related to the magnetic field in Eq. 43–2. 27 7 qB 2mf 2 1.67 10 kg3.5 10 Hz f B 2.3T 19 2m q 1.60 10 C
6.
(a) The maximum kinetic energy is K q B R 1 mv . Compared to Example 43–2, the charge 2 2m has been doubled and the mass has been multiplied by 4. These two effects cancel each other in the equation, and so the maximum kinetic energy is unchanged. The kinetic energy from that example was 8.653 MeV.
2
2
2
2
K 8.7 MeV
2K
v
m
2
2
2
2.036 10 m s 2.0 107 m s 7
(b) The maximum kinetic energy is K q B R 1 mv . Compared to Example 43–2, the charge 2 2m is unchanged, and the mass has been multiplied by 2. Thus, the kinetic energy will be half of what it was in Example 43–2 (8.653 MeV). 2
K 4.3 MeV
2K
v
m
221 8.653 MeV1.60 10
13
2 1.66 1027 kg
J MeV 7 2.036 10 m s 2.0 107 m s
The alpha and the deuteron have the same charge to mass ratio and so move at the same speed. qB (c) The frequency is given by f . Since the charge to mass ratio of both the alpha and the 2 m deuteron is half that of the proton, the frequency for the alpha and the deuteron will both be half the frequency found in Example 43–2 for the proton. f 13 MHz 7.
From Eq. 41–1, the diameter of a nucleon is about d nucleon 2.41015 m. The 25-MeV alpha particles and protons are not relativistic, so their momentum is given by p mv 2mK . The wavelength is h h given by Eq. 43–1, . p 2mK 6.631034 J s h 2.88 1015 m 2 m K 2 41.66 1027 kg25 106 eV1.6 1019 J eV
p
h 2 mp K
6.631034 J s
2 1.67 10 kg25 10 eV1.6 10 27
6
19
J eV
5.751015 m
Chapter 43
Elementary Particles
We see that dnucleon and p 2dnucleon . Thus, the alpha particle will be better for picking out details in the nucleus. 8.
Because the energy of the protons is much greater than their rest mass, we have K E pc. Combine that with Eq. 43–1 for the de Broglie wavelength. That is the minimum size that protons of that energy could resolve. h hc hc 6.631034 J s3.0 108 m s E pc ; p E 1.9 1019 m E
9.
If the speed of the protons is c, the time for one revolution is found from uniform circular motion. The number of revolutions is the total time divided by the time for one revolution. The energy per revolution is the total energy gained divided by the number of revolutions. 2r 2r 2r t ct v T n T v c T 2r 12 9 3 E E 2r 1.0 10 eV 150 10 eV2 1.0 10 m Energy revolution n ct 8.9 105 eV rev 0.9 MeV rev
10. (a) At an energy of 6.5 TeV, the protons are moving at practically the speed of light. From uniform circular motion, we find the time for the protons to complete one revolution of the ring. Then the total charge that passes any point in the ring during that time is the charge of the entire group of stored protons. The current is then the total charge divided by the period. 2 R 2 R 2 R v T T v c Ne Nec I 0.355A 0.4 A 3 T 2 R 2 4.3 10 m
(b) The 6.5 TeV is equal to the kinetic energy of the proton beam. We assume the car would not be moving relativistically. K K K 1 mv2 beam
v
car
2Kbeam m
beam
2
1600 kg
509.9 m s
500 m s Our assumption that the car is not relativistic is confirmed. This is about 1200 mi/hr. 11. These protons will be moving at essentially the speed of light for the entire time of acceleration. The number of revolutions is the total gain in energy divided by the energy gain per revolution. Then the distance is the number of revolutions times the circumference of the ring, and the time is the distance of travel divided by the speed of the protons.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
N
6.510 eV 450 10 eV 12
E E rev
Instructor Solutions Manual
9
8.0 10 eV rev 6
7.56 105 rev
d N 2 R 7.56 10 2 4.310 m 2.04310 m 2.0 10 m 10 d 2.04310 m t 68s 8 c 3.00 10 m s 5
3
10
10
12. Because the energy of the protons is much greater than their rest mass, we have K E pc. A relationship for the magnetic field is given right before Eq. 43–2. We assume the LHC is like a cyclotron with a constant magnetic field around the entire ring. qBr mv qBr p qBr E qBr v m c E B 5.1T qrc 13. (a) The magnetic field is found from the maximum kinetic energy as derived in Example 43–2. 2 2 2 K q B R B 2mK 2m B
qR
14 10 eV1.60 10 J eV 0.7649 T
2 2.014 1.66 10 kg 27
19
6
(b) The cyclotron frequency is given by Eq. 43–2. 1.60 1019 C0.7649 T 5.826 106 Hz f qB 2m
2 2.0141.66 1027 kg
0.76 T
5.8 MHz
(c) The deuteron will be accelerated twice per revolution, and so will gain energy equal to twice its charge times the voltage on each revolution. 6 14 10 eV number of revolutions n 1.60 1019 J eV 19 3 2 1.60 10 C 22 10 V
318 revolutions 320 revolutions (d) The time is the number of revolutions divided by the frequency (which is revolutions per second). n t 318 revolutions 5.458 105 s 5.5 105 s 55 s f 5.826 106 rev s (e) If we use an average radius of half the radius of the cyclotron, the distance traveled is the average circumference times the number of revolutions. distance 12 2 rn 1.0 m318 999 m 1.0 km
14. Start with an expression from Section 43–1, relating the momentum and radius of curvature for a particle in a magnetic field, with q replaced by e. eBr v mv eBr p eBr m
Chapter 43
Elementary Particles
In the relativistic limit, p E c and so
E
eBr . To put the energy in electron volts, divide the
c energy by the charge of the object. E E B rc eBr e c 15. The energy released is the difference in the mass energy between the products and the reactant. E m c2 m c2 m c2 1115.7 MeV 939.6 MeV 135.0 MeV 41.1MeV 0
n
0
16. The energy released is the difference in the mass energy between the products and the reactant. E m c2 m c2 m c2 139.6 MeV 105.7 MeV 0 33.9 MeV
17. Use Eq. 43–3 to estimate the range of the force based on the mass of the mediating particle. 34 8 hc hc 6.6310 J s3.00 10 m s 2 mc d 3.981016 m 2 2d 2mc 18. The energy required is the mass energy of the two particles. 2 E 2mn c 2 939.6 MeV 1879.2 MeV
19. The reaction is multi-step, and can be written as shown here: 0 0 + The energy released is the initial rest energy minus the p final rest energy of the proton and pion, using Table 43–2. 2 E m 0 mp m c 1192.6 Mev 938.3 MeV 139.6 MeV 114.7 MeV
20. Because the two protons are heading towards each other with the same speed, the total momentum of the system is 0. The minimum kinetic energy for the collision would result in all three particles at 0 rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the . 0 Each proton will have half of that kinetic energy. From Table 43–2, the mass of the is 135.0 MeV c2 . 2Kproton m c2 135.0 MeV Kproton 67.5 MeV 0
21. Because the two particles have the same mass and they are traveling towards each other with the same speed, the total momentum of the system is 0. The minimum kinetic energy for the collision would result in all four particles at rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the K K pair. Each initial particle will have half of that kinetic energy. From Table 43–2, the mass of both the K and the K is 493.7 MeV c2 . 2 Kp or p 2mK c2 K p or p mK c 493.7 MeV 2
22. The energy of the two photons (assumed to be equal so that momentum is conserved) must be the combined rest mass energy of the proton and antiproton. c hc 1.32 1015 m 2m0c2 2hf 2h 2 m0c
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23. By assuming that the kinetic energy is approximately 0, the total energy released is the rest mass energy of the annihilating pair of particles. (a) E total 2m0 c2 2 0.511MeV 1.022 MeV (b)
E total 2m0c2 2 938.3 MeV 1876.6 MeV
24. The total energy is the sum of the kinetic energy and the mass energy. The wavelength is found from the relativistic momentum. E K mc2 12 109 eV 938 106 eV 1.294 1010 eV 13GeV
h p
h
hc
c
6.6310 J s3.00 10 m s 34
8
1
1.294 10 eV 938 10 eV 1.60 10 J eV 10
25. (a) 0 n (b) 0 p K
2
6
2
19
17
9.6 10 m
Charge conservation is violated, since 0 0 1 Strangeness is violated, since 1 0 0 Energy conservation is violated, since 1115.7 MeV c2 938.3 MeV c2 493.7 MeV c2 1432.0 MeV c2
(c) 0 +
Baryon number conservation is violated, since 1 0 0 Strangeness is violated, since 1 0 0 Spin is violated, since 12 0 0
26. The total momentum of the electron and positron is 0, so the total momentum of the two photons must be 0. Thus, each photon has the same momentum, so each photon also has the same energy. The total energy of the photons must be the total energy of the electron / positron pair. c 2 m c2 K 2hf 2h E E photons 0 e /e pair hc 1.380 1012 m 1.4 1012 m 2 m0 c K +
27. (a) For the reaction p n 0 , the conservation laws are as follows. Charge: 1 1 0 0 Charge is conserved. Baryon number is conserved. Baryon number: 0 1 1 0 Lepton number is conserved. Lepton number: 0 0 0 0 Strangeness: 0 0 0 0 Strangeness is conserved. The reaction is possible.
(b) For the reaction p n 0 , the conservation laws are as follows. Charge: 1 1 0 0 Charge is NOT conserved. The reaction is forbidden because charge is not conserved. (c) For the reaction p p e , the conservation laws are as follows.
Chapter 43
Elementary Particles
Charge: 1 1 1 1 Charge is conserved. Baryon number: 0 1 1 0 Baryon number is conserved. Lepton number: 0 0 0 1 Lepton number is NOT conserved. The reaction is forbidden because lepton number is not conserved. (d) For the reaction p e v e , the conservation laws are as follows. Charge: 1 1 0 Charge is conserved. Baryon number: 1 0 0 Baryon number NOT conserved. Mass energy is fine because mp me mv . The reaction is forbidden because baryon number is not conserved. (e) For the reaction e , the conservation laws are as follows. Charge: 1 1 0 Charge is conserved. Baryon number is conserved. Baryon number: 0 0 0 Electron lepton number: 0 1 0 Lepton number is NOT conserved. Mass energy is fine because m me mv . The reaction is forbidden because lepton number is not conserved. (f) For the reaction p n e ve , the conservation laws are as follows. Mass energy: 938.3 MeV c2 939.6 MeV c2 0.511MeV c2 Mass energy is NOT conserved. The reaction is forbidden because energy is not conserved. This reaction will not happen because charge is not conserved 2 0 ,
28. p p p p :
and baryon number is not conserved 2 0 . This reaction will not happen because charge is not conserved 2 1 ,
pp p p p :
and baryon number is not conserved 2 1 . pp ppp p:
This reaction is possible. All conservation laws are satisfied.
p p p e e p :
This reaction will not happen. Baryon number is not conserved 2 0 , and lepton number is not conserved 0 2.
29. Since the decays from rest, the momentum before the decay is zero. Thus, the momentum after the decay is also zero, so the momenta of the 0 and are equal in magnitude. Use energy and momentum conservation along with the relativistic relationship between energy and momentum. 2 2 m c E E E m c E
0
p c
2
0 2
2
E m c E m c m c E m c E m c m c 2E m c E m c p p
2
2
4
2 4
0
0 2 4
2 4 0
0
E 0
2
p c
0
2m c
0
2
0 2 4
m c m c m c 2
4
0
2
2
0
2
2
2
4
2
2
2
0
4
4
1321.7 MeV 1115.7 MeV 139.6 MeV 2
2
1124.4 MeV
2 1321.7 MeV
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E m c E 1321.7 MeV 1124.4 MeV 197.3 MeV 2
0
K E m c2 1124.4 MeV 1115.7 MeV 8.7 MeV
0
0
0
K E m c 197.3 MeV 139.6 MeV 57.7 MeV 2
30. Since the pion decays from rest, the momentum before the decay is zero. Thus, the momentum after the decay is also zero, so the magnitudes of the momenta of the positron and the neutrino are equal. We also treat the neutrino as massless. Use energy and momentum conservation along with the relativistic relationship between energy and momentum. 2 2 2 2 2 2 2 4 2 m c E E ; p p p c p c E m c E
e
2
4
e
e
2
e
2 e
e
2
2
e
e
e
2
e 2 e
e
2E m c m2 c4 m2 c4
2
e
2
e
K m c m c
2
2
4
K m c2 1 m c2 m c 2 e e 2m
1
2
2
E 1 m c2 m c 2 e 2m
e
m c 2E m c E 2
E m c m c E 2
e
me2 c2 1 139.6 MeV 0.511MeV
2
2m
69.3 MeV Here is an alternate solution using the momentum. Q m m mv c2 139.6 MeV 0.511MeV 139.1MeV
e
p c m c m c ; K E p c ; p p p
K E m c 2
e
2
e
e
2
2 2
e
v
e
e
v
e
v
0.511MeV pc
Q Ke Kv 139.1MeV
139.6 2 139.6 pc p c pc 0.511MeV
pc 0.511MeV
139.6 pc
v
2
2
2
2 2
2
139.6 MeV 0.511MeV pc 69.8 MeV 2 139.6 MeV 2
2
Ev Kv 69.8 MeV ; Q Kv K e 139.1MeV 69.8 MeV 69.3 MeV
31. (a) The Q-value is the mass energy of the reactants minus the mass energy of the products. 2 2 2 Q m c m c m c 1115.7 MeV 938.3 MeV 139.6 MeV 37.8 MeV
0
p
(b) Energy conservation for the decay gives the following. m c2 E E E m c2 E 0
p
0
p
Momentum conservation says that the magnitudes of the momenta of the two products are equal. Then convert that relationship to energy using E2 p2c2 m20c4 with energy conservation. 2 2 p p p c p c p
p
E 2 m2c4 E 2 m2 c4 m c2 E p
p
0
m c 2
p
2
4
2
Chapter 43
Elementary Particles
E2 m2c4 m2 c4 2E m c2 E 2 m2 c4 p
E
0
p
0 2 4 p
p 2 4 0
1115.7 MeV 938.3 MeV 139.6 MeV 2
m c m c m c
2 4
p
p
2
2
2
2 1115.7 MeV
2m c 0
943.7 MeV
E m c E p 1115.7 MeV 943.7 MeV 172.0 MeV 2
0
Kp E p m cp 2 943.7 MeV 938.3 MeV 5.4 MeV
K E m c 172.0 MeV 139.6 MeV 32.4 MeV 2
32. The two neutrinos must move together, in the opposite direction of the electron, in order for the electron to have the maximum kinetic energy. Thus, the total momentum of the neutrinos will be equal in magnitude to the momentum of the electron. Since a neutrino is (essentially) massless, we have Ev pvc. We assume that the muon is at rest when it decays. Use conservation of energy and momentum along with their relativistic relationship. pe p p
e
c E p c
m c Ee E E Ee p c p c E e p p 2
e
2
2
2
2
e
e
m c 2m c E E E m c E
4
2
2
e
e
e
2
2
e
4
e
e
2
2
e
e 2
e
4
m c me c 2
2
e
m c E p c E m c
m c Ee pe c
e
4
4
K m c2 e
2
2m c
e
2 4
2e 4
2
K m c m c m c e e 2m c2
105.7 MeV2 0.511MeV2
2 105.7 MeV
0.511MeV
52.3 MeV
33. A could NOT be produced by p p p n . The pion has a mass energy of 139.6 MeV, so the extra 130 MeV of energy could not create it. The Q-value for the reaction is as follows. Q 2m c2 m c2 m c2 m c2 m c2 m c2 m c2 p
p
n
p
n
938.3 MeV 939.6 MeV 139.6 MeV 140.9 MeV More than 140.9 MeV of kinetic energy is needed to create the pion. The minimum initial kinetic energy would produce the particles all moving together at the same speed, having the same total momentum as the incoming proton. We consider the products to be one mass M mp mn m
since they all move together with the same velocity. We use energy and momentum conservation, 2 2 2 2 4 along with their relativistic relationship, E p c m 0c . ; pp pM ppc pMc Ep mp c E M M c 2
Ep mp c EM 2
2
2
2 4
E 2 m2c4 E m c2 M 2c4 E 2 2E m c2 m2c4 M 2c4 2
p
p
p
p
M 2c4 2m2c4 Ep
p 2
2mp c
Kp mp c2
p
p
p
p
2
2 4
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M 2c4 2m2pc4 2 M 2c4 2m c2 m c Kp p p 2m c2 2m c2 p
p
938.3 MeV 939.6 MeV 139.6 MeV2 2 938.3 MeV 292.4 MeV 2 938.3 MeV 34. We use the uncertainty principle to estimate the uncertainty in rest energy. h 6.63 1034 J s E 9420 eV 0.009 MeV 2t 35. We estimate the lifetime from the energy width and the uncertainty principle. h h 34 6.6310 J s E t 2 1021 s 19 1.60 10 J 2t 2E 2 300 103 eV 1eV h . 36. Apply the uncertainty principle, which says that E 2t 34 h h 6.6310 J s E t 7.5 1021 s 19 1.60 10 J 2t 2E 2 88 103 eV 1eV
37. (a) For B b u , we have Charge:
1 13 23
Baryon number: Charm: Topness:
0 13 13 0 00 0 00
Spin:
0 12 12
Strangeness:
0 00 1 1 0
Bottomness:
(b) Because B is the antiparticle of B , B b u . The B must have an anti-bottom quark but 0
must be neutral. Therefore, B0 b d . Because B0 is the antiparticle to B0 , we must have B0 b d .
38. We find the energy width from the lifetime in Table 43–2 34 and the uncertainty principle. 19 h 6.6310 Js (a) : t 5.110 s ; E 1293eV 1300 eV 0 19 19 2t 2 5.110 s1.60 10 J eV h 6.631034 J s 24 1.499 108 eV 150 MeV (b) : t 4.4 10 s ; E 2t 2 4.4 1024 s1.60 1019 J eV
Chapter 43
Elementary Particles
Spin:
0 1 1 1 1 0 0 0 0 2 1 0 0
Energy:
1314.9 Mev c2 1189.4 Mev c2 139.6 Mev c2 1329 Mev c2
39. (a) Charge: Baryon number: Lepton number: Strangeness:
1 2
Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is NOT conserved.
1
Spin is conserved.
2
Energy is NOT conserved. The decay is not possible. Neither strangeness or energy is conserved.
1 0 1 0 1 1 0 0 0 0 0 1 3 1 0 0 0
(b) Charge: Baryon number: Lepton number: Strangeness:
3 2
Spin:
1
Charge is conserved. Baryon number is conserved. Lepton number is NOT conserved. Strangeness is NOT conserved.
1
2
Spin is NOT conserved.
2
Energy: 1672.5 Mev c2 1192.6 Mev c2 139.6 Mev c2 0 Mev c2 1332.2 Mev c2 Energy is conserved. The decay is not possible. Neither lepton number, strangeness, or spin is conserved.
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1
(c) Charge: Baryon number: Lepton number: Strangeness:
1 2
Spin:
Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is conserved.
1 2
Spin is conserved, if the gammas have opposite spins. 2 2 2 2 2 Energy: 1192.6 Mev c 1115.7 Mev c 0 Mev c 0 Mev c 1115.7 Mev c Energy is conserved. The decay is possible. 40. The Q-value is the mass energy of the reactants minus the mass energy of the products. 0 For the first reaction, p p p p : Q 2m c2 2m c2 m c2 m c2 135.0 MeV p
0
p
0
For the second reaction, p p p n : 2 2 2 2 2 2 2 Q 2m c m c m c m c m c m c m c a
p
p
n
p
n
938.3 MeV 939.6 MeV 139.6 MeV= 140.9 MeV
41. The expected lifetime of the virtual W particle is found from the Heisenberg uncertainty principle, as given in Eq. 38–2 and Example 43–11. Et
t
E
2 80.379 10 eV 9
1.60 10 J 19
8.20 1027 s 8 1027 s
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42. (a) The 0 has a strangeness of –2, so it must contain two strange quarks. In order to make a neutral particle, the third quark must be an up quark. So 0 uss . (b) The has a strangeness of –2, so it must contain two strange quarks. In order to make a particle with a total charge of –1, the third quark must be a down quark. So dss . 43. (a) The neutron has a baryon number of 1, so there must be three quarks. The charge must be 0, as must be the strangeness, the charm, the bottomness, and the topness. Thus, n udd . (b) The antineutron is the antiparticle of the neutron, and so n u d d . (c) The 0 has a strangeness of 1, so it must contain an “s” quark. It is a baryon, so it must contain three quarks. And, it must have charge, charm, bottomness, and topness equal to 0. Thus, 0 u ds . (d) The 0 has a strangeness of +1, so it must contain an s quark. It is a baryon, so it must contain three quarks. And, it must have charge, charm, bottomness, and topness equal to 0. Thus, . 44. (a) The combination u u d has charge = +1, baryon number = +1, and strangeness, charm, bottomness, and topness all equal to 0. Thus, u u d p . (b) The combination u u s has charge = –1, baryon number = –1, strangeness = +1, and charm, bottomness, and topness all equal to 0. Thus, u u s . (c) The combination us has charge = –1, baryon number = 0, strangeness = –1, and charm, bottomness, and topness all equal to 0. Thus, u s K . (d) The combination du has charge = –1, baryon number = 0, and strangeness, charm, bottomness, and topness all equal to 0. Thus, du (e) The combination c s has charge = –1, baryon number = 0, strangeness = –1, charm = –1, and bottomness and topness of 0. Thus, c s DS 45. To form the D0 meson, we must have a total charge of 0, a baryon number of 0, a strangeness of 0, and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must have a “c” quark, with a charge of 23 e . To have a neutral meson, there must be another quark with a charge of 23 e . To have a baryon number of 0, that second quark must be an antiquark. The only candidate with those properties is an anti-up quark. Thus, D0 c u . 46. To form the DS meson, we must have a total charge of +1, a baryon number of 0, a strangeness of +1, and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must have a “c” quark, with a charge of 23 e . To have a total charge of +1, there must be another quark
Chapter 43
Elementary Particles
with a charge of 13 e . To have a baryon number of 0, that second quark must be an antiquark. To have a strangeness of +1, the other quark must be an anti-strange. Thus,
DS c s .
47. Here is a Feynman diagram for the reaction p 0 n. There are other possibilities, since the 0 also can be represented as a dd combination or as a mixture of dd and uu combinations.
48. Since leptons are involved, the reaction n v p is a weak interaction. Since there is a charge change in the lepton, a W boson must be involved in the interaction. If we consider the neutron as having emitted the boson, it is a W , which interacts with the neutrino. If we consider the
neutrino as having emitted the boson, it is a W , which interacts with the neutron.
49. To find the length in the lab, we need to know the speed of the particle which is moving relativistically. Start with Eq. 36–10a. 2 1 1 1 K m c c 0.7522c 2 2 1 v c 1 1 1 v c 0 2 2 K 1 920 MeV 1 m c 2 1777 MeV 0 13 2.90 10 s t0 13 4.40110 s tlab 2 2 2 1 v c 1 0.7522
xlab vt lab 0.7522 3.00 10 m s
8
4.40110 s 9.9310 m 13
5
50. (a) For the reaction p K p , the conservation laws are as follows. Charge is NOT conserved. Charge: 1 1 01 1 0 Spin: 0 1 0 0 Spin is conserved. 2 2
0
Baryon number: 0 1 1 0 0 Lepton number: 0 0 0 0 0 Strangeness: 0 0 1 0 0
0
Baryon number is conserved. Lepton number is conserved. Strangeness is NOT conserved.
The reaction is not possible because neither charge nor strangeness is conserved.
Note that the reactants would need significant kinetic energy to be able to “create” the K0 . 0 0 (b) For the reaction K p , the conservation laws are as follows. Charge: 1 1 0 0 Charge is conserved. 1 1 Spin: 0 2 2 0 Spin is conserved.
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Baryon number: 0 1 1 0 Baryon number is conserved. Lepton number: 0 0 0 0 Lepton number is conserved. Strangeness: 1 0 1 0 Strangeness is conserved. The reaction is possible via the strong interaction. (c) For the reaction K n , the conservation laws are as follows. Charge: 1 0 1 0 0 Charge is conserved. 1 1 Spin: 0 2 2 0 1 Spin is conserved. Baryon number: 0 1 1 0 0 Baryon number is conserved. Lepton number: 0 0 0 0 0 Lepton number is conserved. Strangeness: 1 0 1 0 0 Strangeness is NOT conserved. The reaction is not possible via the strong interaction because strangeness is not conserved. It is possible via the weak interaction. 0 0 (d) For the reaction K , the conservation laws are as follows. Charge: 1 0 0 1 Charge is conserved. Spin: 0 0 0 0 Spin is conserved. Baryon number: 0 0 0 0 Baryon number is conserved. Lepton number: 0 0 0 0 Lepton number is conserved. Strangeness: 1 0 0 0 Strangeness is NOT conserved. Mass: 493.7 MeV/c2 > 135.0 + 135.0 + 139.6 = 409.6 MeV/c2 The reaction is possible energetically. The reaction is not possible via the strong interaction because strangeness is not conserved. It is possible via the weak interaction. (e) For the reaction e e, the conservation laws are as follows. Charge: 1 1 0 Charge is conserved. 1 1 Spin: 0 2 2 Spin is conserved. Baryon number is conserved. Baryon number: 0 0 0 Lepton number is conserved. Lepton number: 0 1 1 Strangeness is conserved. Strangeness: 0 0 0 0 0 2 Mass: 139.6 MeV/c > 0.511 MeV/c2 The reaction is possible energetically. The reaction is possible via the weak interaction.
0
51. (a) For the reaction p K , the conservation laws are as follows. Charge: 1 1 1 1 Charge is conserved. Spin: 0 12 0 12 Spin is conserved. Baryon number: 0 1 0 1 Baryon number is conserved. Lepton number: 0 0 0 0 Lepton number is conserved. Strangeness: 0 0 1 1 Strangeness is conserved. The reaction is possible via the strong interaction. (b) For the reaction p K , the conservation laws are as follows. Charge: 1 1 1 1 Charge is conserved. 1 1 Spin: 0 2 0 2 Spin is conserved. Baryon number: 0 1 0 1 Baryon number is conserved. Lepton number: 0 0 0 0 Lepton number is conserved. Strangeness is conserved. Strangeness: 0 0 1 1
Chapter 43
Elementary Particles
The reaction is possible via the strong interaction. (c) For the reaction p 0 K0 0 , the conservation laws are as follows. Charge: 1 1 0 0 0 Charge is conserved. 1 1 Spin: 0 2 2 0 0 Spin is conserved. Baryon number: 0 1 1 0 0 Baryon number is conserved. Lepton number: 0 0 0 0 0 Lepton number is conserved. Strangeness: 0 0 1 1 0 Strangeness is conserved. The reaction is possible via the strong interaction. (d) For the reaction p 0 0 , the conservation laws are as follows. Charge: 1 1 0 0 Charge is NOT conserved. 1 1 Spin: 0 2 2 0 Spin is conserved. Baryon number: 0 1 1 0 Baryon number is conserved. Lepton number: 0 0 0 0 Lepton number is conserved. Strangeness is NOT conserved. Strangeness: 0 0 1 0 The reaction is not possible because neither charge nor strangeness is conserved.
(e) For the reaction p p e v e , the conservation laws are as follows. Charge: 1 1 1 1 0 Charge is conserved. 1 1 1 1 Spin: 0 2 2 2 2 Spin is conserved. Baryon number is conserved. Baryon number: 0 1 1 0 0 Lepton number: 0 0 0 1 1 Lepton number is conserved. Strangeness: 0 0 0 0 0 Strangeness is conserved. The reaction is possible via the weak interaction. Note that we did not check mass conservation because in a collision, there is always some kinetic energy brought into the reaction. Thus, the products can be heavier than the reactants. 52. The is the anti-particle of the , so the reaction is as follows. Charge: 1 1 0 Baryon number: 0 0 0 Lepton number: 0 1 1 Strangeness: 0 0 0 Spin: 0 12 12
v . The conservation rules are
Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is conserved. Spin is conserved
53. Use Eq. 43–3 to estimate the mass of the particle based on the given distance. 6.631034 J s3.0 108 m s 1 1.981011 eV mc2 hc
200 GeV 19 1.60 10 J eV This value is of the same order of magnitude as the mass of the W , which is about 80 GeV. 2d
2 1018 m
54. The fundamental fermions are the quarks and electrons. In a water molecule, there are 2 hydrogen atoms consisting of 1 electron and 1 proton each, and 1 oxygen atom consisting of 8 electrons, 8 protons, and 8 neutrons. Thus, there are 18 nucleons consisting of 3 quarks each and 10 electrons. The total number of fermions is thus 18 3 10 64 fundamental fermions.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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55. As mentioned in Example 43–9, the 0 can be considered as either uu or d d. There are various models to describe this reaction. Four are shown here.
Another model that shows the 0 as a combination of both uu and dd is also shown.
56. (a) To conserve charge, the missing particle must be neutral. To conserve baryon number, the missing particle must be a meson. To conserve strangeness, charm, topness, and bottomness, the missing particle must be made of up and down quarks and antiquarks only. With all this information, the missing particle is 0 . (b) This is a weak interaction since one product is a lepton. To conserve charge, the missing particle must be neutral. To conserve the muon lepton number, the missing particle must be an antiparticle in the muon family. With this information, the missing particle is
v .
57. We assume that the interaction happens essentially at rest, so that there is no initial kinetic energy or momentum. Thus, the momentum of the neutron and the momentum of the 0 will have the same magnitude. From energy conservation, we find the total energy of the 0 . m c2 m c2 E m c2 K
0
p
n
n
E m c2 m c2 m c2 K 139.6 MeV 938.3 MeV 939.6 MeV 0.60 MeV 0
p
n
n
137.7 MeV From momentum conservation, we can find the mass energy of the 0 . We utilize Eq. 36–14 to relate momentum and energy. 2 2 E 2 m2c4 E 2 m2 c4 m2 c4 E 2 E 2 m2c4 p p p c p c 0
n
n
0
n
n
0
0
0
0
n
n
m c2 E 2 E n2 m2nc4 137.7 MeV 939.6 MeV 0.60 MeV 939.6 MeV 2
0
0
133.5 MeV m 133.5 MeV c2 0
The value from Table 43–2 is 135.0 MeV.
2
2
1/ 2
Chapter 43
Elementary Particles
58. (a) First, we use the uncertainty principle, Eq. 38–1. The energy is so high that we assume E pc, E and so p . c E h h x xp 2 c 2 34 8 9 6.63 10 J s3.00 10 m s 1GeV 10 eV E hc 2 1015 GeV 31 19 2x 2 10 m 1.60 10 J eV
Next, we use de Broglie’s wavelength formula. We take the de Broglie wavelength as the unification distance. h h p Ec 6.63 1034 J s3.00 108 m s 1GeV 109 eV E hc 11016 GeV 1031 m 1.60 1019 J eV Both energies are reasonably close to 1016 GeV . This energy is the amount that could be violated in conservation of energy if the universe were the size of the unification distance. (b) From Eq. 18–4, we have E 32 kT . E kT T 3 2
2E
3k
2 1025 eV1.6 1019 J eV 7.7 1028 K 1029 K 23 31.38 10 J K
59. The Q-value is the mass energy of the reactants minus the mass energy of the products. 2 2 2 2 Q m c m c m c m c 139.6 MeV 938.3 MeV 1115.7 MeV 497.6 MeV
0
p
K0
535.4 MeV 2 We consider the products to be one mass M m m K 1613.3 MeV c since they both have the 0
0
same velocity. Energy conservation gives the following: E mp c2 EM . Momentum conservation
says that the incoming momentum is equal to the outgoing momentum. Then convert that 2 2 2 2 4 relationship to energy using the relativistic relationship that E p c m c . 2 2 p p p c p c E 2 m2 c4 E 2 M 2c4
M
2
2
M
M
E m c E m c M c E 2E m c m2c4 M 2c4 4
K
2
2 4
2
2
4
2
p
M c m c mc 2 4
E
2
p
p
2 4
p
K m c
2
2mp c
M 2c4 m2 c4 m2c4
p
2
m c2
2
2mp c
1613.3MeV 139.6 MeV 938.3 MeV 2 938.3 MeV 2
2
2
139.6 MeV 767.8 MeV
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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60. Since there is no initial momentum, the final momentum must add to zero. Thus, each of the pions must have the same magnitude of momentum and therefore the same kinetic energy. Use energy conservation to find the kinetic energy of each pion. 2m c2 2K 2m c2 K m c2 m c2 938.3 MeV 139.6 MeV 798.7 MeV p p 61. The Q-value is the energy of the reactants minus the energy of the products. We assume that one of the initial protons is at rest, and that all four final particles have the same velocity and therefore the same kinetic energy since they all have the same mass. We consider the products to be one mass M 4mp since they all have the same speed. Q 2mp c2 4mpc2 2m pc2 Mc2 2m pc2 Energy conservation gives the following, where K th is the threshold energy. K m c2 m c2 E K Mc2 th
p
p
M
M
Momentum conservation says that the incoming momentum is equal to the outgoing momentum. Then convert that relationship to energy using the relativistic relationship that E2 p2c2 m02c4 . pp pM ppc pMc 2
K th mp c2 mp2c4 K M Mc2 M 2c4 2
2
2
Kth2 2K thm pc2 m2pc4 m2pc4 K 2th 4K thm pc2 4m2pc4 4m p c4 2 2 2 4 2 4 2 2 4 2 2K m c 4K m c 4m c 16m c 2K m c 12m c K 6m c 3 Q 2
th
p
th
p
p
p
th
p
p
th
p
62. We use 0 to represent the actual wavelength, and to define the approximate wavelength. The approximation is to ignore the rest mass in the expression for the total energy, E K mc2 . We also use Eqs. 36–11a, 36–14, and 43–1. 2 mc 2 2 2
K mc mc K 1 2 K
p c E mc 2 2
2
h 0 p
2
2
2
2
hc
K 1
hc ; ; 1.02 1/ 2 0 0 2mc2 K
K hc
2 9.38 10 eV 2mc2 hc 1.02 K 4.644 1010 eV 4.6 1010 eV 1/ 2 2 2 K 1.02 1 2mc 0.0404 K 1 K 8
hc
2.677 10 m 2.7 10 m 17
K
2 1 K _______ vc 2 2 1 mc 1 v c
17
2
2
K 1 c 2 mc
8 9.38 10 eV
0.9998c 1
Chapter 43
Elementary Particles
63. A relationship between total energy and speed is given by Eq. 36–11b. mc2 E 2 2 c 1 v 2 2 8 2 2 9.38108 eV 9.3810 eV mc v 8 1 1 1 1 2 12 11.2 10 0.999999988 12 E 6.0 10 eV c 6.0 10 eV This is about 3.6 m/s slower than the speed of light. 64. According to Section 43–1, the energy involved in the LHC’s collisions will reach 13 TeV. Use that with the analysis used in Example 43–1. From Eq. 41–1, nuclear sizes are on the order of 1.2 1015 m . hc 9.56 1020 m E
9.56 1020 m 1 5 8.0 10 15 1.2 10 m 12, 500
r nucleus
65. (a) The nucleus undergoes 2 beta-decays. Thus, in the nucleus, 2 neutrons change into protons. The daughter nucleus must have 2 more protons than the parent but the same number of nucleons. 96 96 0 Thus, the daughter nucleus is 96 Mo . The reaction would be Zr Mo 2 e . 40
42
42
1
(b) Since the reaction is neutrinoless, lepton conservation would be violated in this decay. Checking isotope masses at www.nist.gov/pml/data/comp.cfm shows that the reaction is energetically possible. 96 Zr has a mass of 95.908271 amu and 96 Mo has a mass of 95.904676 40
42
amu. 96 (c) If 40 Zr would undergo 2 beta decays simultaneously and emit 2 electron antineutrinos, it could decay to 4296 Mo without violating any conservation laws. 66. We write equations for both conservation of energy and conservation of momentum. The magnitudes of the momenta of the products are equal. We also use Eqs. 36–11a and 36–14. 2 p p ; E mc E E 1
2
0
1
2
mc E E p c m c p c m c E m c m c 2
2
2
1
2 2
2
2 4 2
2
2 2 1
m c 2mc E E E m c m c 2 4
2
2
1
1
2
1
2
1
1
2 4
1
2
m c m1 c m2 c 2 4
K E mc
2 4
1
2 4
2
2mc
mc m c m c 2
2mc
2 4 2
2 1
2 4 1
2 4 2
mc mc mc 2 4
E1
2 4 2
1
2 4
2 2
2mc 2 4 2 2 m c m1 c m2 c 2mc m1c 2 4
mc
2 4
1
2 4
2
2mc
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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67. The conservation laws that must hold are the conservation of momentum, angular momentum (spin), mass-energy, charge, baryon number, and lepton number. Momentum can be conserved by the two decay products having equal but opposite momentum in the rest frame of the parent particle. There are no measurements given to enable us to check conservation of momentum, so we assume that it holds. The other conservation laws are evaluated for each decay. Spin can of course be positive or negative, so our “check” means seeing if there is a way for spins to be equal, using either positive or negative values for the spin. For the mass-energy to be conserved the mass of the initial particle must be greater than the mass of the resulting particles, with the remaining mass becoming kinetic energy 2 of the particles. Those values are in GeV c , but units are omitted below. For the reaction t W b : 1 2
1 12
Spin: Mass-energy:
173 80.4 4.18
Charge:
2 3
Baryon number:
1 0 13 3
Baryon number is conserved.
Lepton number:
0 0 0
Lepton number is conserved.
1 13
Spin is conserved. Mass-energy is conserved. Charge is conserved.
For the reaction t W b : 1 1 Spin: 1 2 2 Mass-energy: 173 80.4 4.18 Charge: 23 1 13 Baryon number:
13 0 13 0 0 0
Lepton number: For the reaction W u d : Spin: Mass-energy: Charge: Baryon number:
1 12 12 80.4 0.0023 0.0048
1 23 13
0 13 13 0 0 0
Lepton number: For the reaction W v : Spin: Mass-energy:
1 21 12
Spin is conserved. Mass-energy is conserved. Charge is conserved. Baryon number is conserved. Lepton number is conserved. Spin is conserved. Mass-energy is conserved. Charge is conserved. Baryon number is conserved. Lepton number is conserved.
80.4 0.1057 0
Spin is conserved. Mass-energy is conserved.
Charge: Baryon number:
1 1 0 0 0 0
Charge is conserved. Baryon number is conserved.
Lepton number:
0 1 1
Lepton number is conserved.
68. (a) Each tau will carry off half of the released kinetic energy. The kinetic energy is the difference in mass-energy of the two tau particles and the original Higgs boson. Q m 2m c 2 125 10 3MeV 2 1777 MeV 121.4 10 3MeV 121.4 GeV
0
H
K 12 Q 60.7 GeV (b) The total charge must be zero, so one will have a positive charge and the other will have a negative charge. One of them must be an anti-tau particle.
Chapter 43
Elementary Particles
(c) No. The mass of the two Z bosons is greater than the mass of the initial Higgs boson. Thus, the decay would violate the conservation of energy. 69. (a) We work in the rest frame of the isolated electron, so that it is initially at rest. Energy conservation gives the following. m c2 K m c2 E K E K E 0 e
e
e
e
e
Since the photon has no energy, it does not exist and so has not been emitted. Here is an alternate explanation: Consider the electron in its rest frame. If it were to emit a single photon of energy E , the photon would have a momentum p E c . For momentum to be conserved, the electron would move away in the opposite direction as the photon with the same magnitude of momentum. Since the electron is moving (relative to the initial frame of reference), the electron will now have kinetic energy in that frame of reference, with a total energy given by Eq. 36–14, E
m c p c m c E . With the added kinetic 2
0
2
2
2
0
2
2
energy, the electron now has more energy than it had at rest, and the total energy of the photon and electron is greater than the initial rest energy of the electron. This is a violation of the conservation of energy, so an electron cannot emit a single photon. (b) For the photon exchange in Fig. 43–8, the photon exists for such a short time that the Heisenberg uncertainty principle allows energy conservation to be violated during the exchange. So if t is the duration of the interaction, and E is the amount by which energy is not conserved during the interaction; as long as Et , the process can happen.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
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CHAPTER 44: Astrophysics and Cosmology Responses to Questions 1.
Long ago, without telescopes, it was difficult to see the individual stars in the Milky Way. The stars in this region of the sky were so numerous and so close together and so tiny that they all blended together to form a cloudy or milky stripe across the night sky. Now using more powerful telescopes, we can see the individual stars that make up the Milky Way galaxy.
2.
If a star generates more energy in its interior than it radiates away, its temperature will increase, indicating a higher average kinetic energy of the particles. Consequently, there will be greater outward pressure opposing the gravitational force directed inward. To regain equilibrium, the star will expand. If a star generates less energy than it radiates away, then its temperature will decrease, reflecting a lower average kinetic energy of the particles. There will be a smaller outward pressure opposing the gravitational force directed inward, and, in order to regain equilibrium, the star will contract. In both situations a new equilibrium will be reached, which means the star will not produce either a runaway heating and explosion, or a runaway total collapse.
3.
Red giants are extremely large stars with relatively cool surface temperatures, resulting in their reddish colors. These stars are very luminous because they are so large. When the Sun becomes a red giant, for instance, its radius will be on the order of the distance from the Earth to the Sun. A red giant has run out of hydrogen in its inner core and is fusing hydrogen to helium in a shell surrounding the core. Red giants have left their main sequence positions on the H–R diagram and moved up (more luminous) and to the right (cooler).
4.
Although the H-R diagram only directly relates the surface temperature of a star to its absolute luminosity (and thus doesn’t directly reveal anything about the core), the H-R diagram does provide clues regarding what is happening at the core of a star. Using the current model of stellar evolution and the H-R diagram, we can infer that the stars on the main sequence are fusing hydrogen nuclei to helium nuclei at the core and that stars in the red giant region are fusing helium and beryllium to make heavier nuclei such as carbon and that this red giant process will continue until fusion can no longer occur and the star will collapse.
5.
The initial mass of a star determines its final destiny. If, after the red giant stage of a star’s life, its mass is less than 1.4 solar masses, then the star cools as it shrinks and it becomes a white dwarf. If its mass is between 1.4 and 2-3 solar masses, then the star will condense down to a neutron star, which will eventually explode as a supernova and become a white dwarf. If its mass is greater than 2-3 solar masses, then the star will collapse even more than the neutron star and form a black hole.
6.
When measuring parallaxes from the Moon, there are two cases: (1) If you did the measurements two weeks apart (one at full moon and one at new moon), you would need to assume that the Earth did not move around the Sun very far, and then the d shown in Fig. 44–11 would be the Earth-Moon distance instead of the Sun-Earth distance. (2) If you did the measurements six months apart and at full moon, then the d shown in Fig. 44–11 would be the Sun-Earth distance plus the Earth-Moon distance instead of just the Sun-Earth distance. From Mars, then the d shown in Fig. 44–11 would be the Sun-Mars distance instead of the Sun-Earth distance. You would also need to know the length of a Mars “year” so you could take your two measurements at the correct times.
7.
Measure the period of the changing luminosity of a Cepheid variable star. Use the known relationship between period and luminosity to find its absolute luminosity. Compare its absolute luminosity to its apparent brightness (the observed brightness) to determine the distance to the galaxy in which it is located.
8.
A geodesic is the shortest distance between two points. For instance, on a flat plane the shortest distance between two points is a straight line, and on the surface of a sphere the shortest distance is an arc of a great circle. According to general relativity, space–time is curved. Determining the nature of a geodesic, for instance, by observing the motion of a body or light near a large mass, will help determine the nature of the curvature of space–time near that large mass.
9.
If the redshift of spectral lines of galaxies were discovered to be due to something other than expansion of the universe, then the Big Bang theory and the idea that the universe is expanding would be called into question. However, the evidence of the cosmic background microwave radiation would conflict with this view, unless it too was determined to result from some cause other than expansion.
10. No. In an expanding universe, all galaxies are moving away from all other galaxies on a large scale. (On a small scale, neighboring galaxies may be gravitationally bound to each other.) Therefore, the view from any galaxy would be the same. Our observations do not indicate that we are at the center. (See Fig. 44–22.) Here is an analogy, frim Fig. 44–21. If we were sitting on the surface of a balloon and more air was put into the balloon causing it to expand, every other point on the balloon moves away from you. The points close to you are farther away because of the expansion of the rubber and the points on the other side of the balloon are farther away from you because the radius of the balloon is now larger. 11. If you were located in a galaxy near the boundary of our observable universe, galaxies in the direction of the Milky Way would be receding from you. The outer “edges” of the observable universe are expanding at a faster rate than the points more “interior”. Accordingly, due to a relative velocity argument, the slower galaxies in the direction of the Milky Way would look like they are receding from your faster galaxy near the outer boundary. Also see Fig. 44–22. 12. An explosion on Earth blows pieces out into the space around it, but the Big Bang was the start of the expansion of space itself. In an explosion on Earth, the pieces that are blown outward will slow down due to air resistance, the farther away they are the slower they will be moving, and then they will eventually come to rest. But with the Big Bang, the farther away galaxies are from each other the faster they are moving away from each other. In an explosion on Earth, the pieces with the higher initial speeds end up farther away from the explosion before coming to rest, but the Big Bang appears to be relatively uniform where the farthest galaxies are moving fastest and the nearest galaxies are moving the slowest. An explosion on Earth would correspond to a closed universe, since the pieces would eventually stop, but we would not see a “big crunch” due to gravity as we would with an actual closed universe. 13. To “see” a black hole in space we need indirect evidence. If a large visible star or galaxy was rotating quickly around a non-visible gravitational companion, the non-visible companion could be a massive black hole. Also, as matter begins to accelerate toward a black hole, it will emit characteristic X-rays, which we could detect on Earth. Another way we could “see” a black hole is if it caused gravitational lensing of objects behind it. Then we would see stars and galaxies in the “wrong” place as their light is bent as it passes past the black hole on its way to Earth.
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14. Both the formation of the Earth and the time during which people have lived on Earth are on the far right edge of Fig. 44–29, in the era of dark energy. 15. Atoms were unable to exist until hundreds of thousands of years after the Big Bang because the temperature of the universe was still too high. At those very high temperatures, the free electrons and nuclei were moving so fast and had so much kinetic energy, and they had so many high energy photons colliding with them, that they could never combine together to form stable atoms. Once the universe cooled below 3000 K, this coupling could take place, and atoms were formed. 16. (a) Type Ia supernovae are expected to all be of nearly the same luminosity. Thus they are a type of “standard candle” for measuring very large distances. (b) The distance to a supernova can be determined by comparing the apparent brightness to the intrinsic luminosity, and then using Eq. 44–1 to find the distance. 17. If the average mass density of the universe is above the critical density, then the universe will eventually stop its expansion and contract, collapsing on itself and ending finally in a “big crunch.” This scenario corresponds to a closed universe, or one with positive curvature. 18. (a) Gravity between galaxies should be pulling the galaxies back together, slowing the expansion of the universe. (b) Astronomers could measure the redshift of light from distant supernovae and deduce the recession velocities of the galaxies in which they lie. By obtaining data from a large number of supernovae, they could establish a history of the recessional velocity of the universe, and perhaps tell whether the expansion of the universe is slowing down. 19. Many methods are available. For nearby stars (from about 100 ly to as much as 1000 ly away) we can use parallax. In this method we measure the angular distance that a star moves relative to the background of stars as the Earth travels around the Sun. Half of the angular displacement is then equal to the ratio of Earth-Sun distance and the distance between the Earth and that star. See Fig. 44–11. Apparent brightness of the brightest stars in galaxies, combined with the inverse square law, can be used to estimate distances to galaxies, assuming they have the same intrinsic luminosity. The H-R diagram can be used for distant stars. Determine the surface temperature using its blackbody radiation spectrum and Wien’s law, and then estimate its luminosity from the H-R diagram. Using its apparent brightness with Eq. 44–1 will give its distance. Variable stars, like Cepheid variables, can be used by relating the period to its luminosity. The luminosity and apparent brightness can be used to find the distance. The largest distances are measured by measuring the apparent brightness of Type Ia supernovae. All supernovae are thought to have nearly the same luminosity, so the apparent brightness can be used to find the distance. The redshift in the spectral lines of very distant galaxies can be used to estimate distances that are further than 107 to 108 ly. The Cepheid variable method gives the most accurate distances for far-away stars. For the most distant galaxies, redshifts give the best measurements. 20. (a) All distant objects in the universe are moving away from each other, as indicated by the galactic redshift, indicating that the universe is expanding. If the universe has always expanded, it must have started as a point. The 25% abundance of He supports the Standard Big Bang Model. The Big Bang Theory predicted the presence of background radiation, which has since been observed.
Chapter 44
Astrophysics and Cosmology
(b) The curvature of the universe determines whether the universe will continue expanding forever (open) or eventually collapse back in on itself (closed). (c) Dark energy increases the total energy of the universe increasing the probability that it is an open universe. 21. About 380,000 years after the Big Bang, the initially very hot temperature of the universe would have cooled down to about 3000 K. At that temperature, electrons could orbit bare nuclei and remain there, without being ejected by collisions. Thus stable atoms were able to form. This allowed photons to travel unimpeded through the universe, and so the universe became transparent. The radiation at this time period would have been blackbody radiation at a temperature of about 3000 K. As the universe continued to expand and cool, the wavelengths of this radiation would have increased according to Wien’s law. The peak wavelength now is much larger and corresponds to the blackbody radiation at a temperature of 2.7 K.
Responses to MisConceptual Questions 1.
(c) At the end of the hydrogen-fusing part of the star’s life, it will move toward the upper right of the diagram as it becomes a red giant. Low mass stars, not large enough to end as neutron stars, then move to the lower left as they become white dwarf stars. The position of the star on the main sequence is a result of the size (mass) of the star. This does not change significantly during its main sequence lifetime and therefore the star does not change positions along the main sequence.
2.
(b) Parallax requires that the star moves relative to the background of distant stars. This only happens when the distance to the star is relatively small, less than about 100 light years.
3.
(b) If the universe were expanding in every direction, then any location in the universe would observe that all other points in the universe are moving away from it with the speed increasing with distance. This is observed when galaxy speeds are measured, implying that indeed the universe is expanding. The observation does not say anything about whether the expansion will continue forever or eventually stop.
4. (c)
5.
The Sun is primarily made of hydrogen, not heavy radioactive isotopes. As gravity initially compressed the Sun, its core heated sufficiently for the hydrogen atoms in the core to fuse to create helium. This is now the primary source of energy in the Sun. The Sun is too hot for molecules, such as water, to be formed by the oxidation of hydrogen. When fusion began in the core, the energy release balanced the force of gravity to stop the gravitational collapse. Since the Sun is no longer contracting, the gravitational potential energy is no longer decreasing and is not a significant source of energy.
(b, c, d) Parallax is only useful to find distances to stars that are less than 100 light years distant, much less than inter-galactic distances. The H-R diagram with luminosity and temperature of the star can tell us the absolute brightness of the distant star. The relationship between the relative brightness and absolute brightness tells us the distance to the star. Certain supernova explosions have a standard luminosity. A comparison of the apparent brightness and absolute brightness of the supernova can give the distance to the supernova. Finally, since the universe is expanding and the velocity of objects increases with distance, a measure of the redshift in light from distant stars can be used to find the speed of the stars and therefore the distance.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
6.
(b) Due to the speed of light, light takes a finite time to reach Earth from the stars, with light from more distant objects requiring more time to reach Earth. Therefore when an object is observed that is a million light years away, the observers sees the object as it was a million years ago. If the object is 100 million light years away, the observer sees it as it was 100 million years ago. The farther the observed object, the earlier the time that it is being observed.
7.
(d) At the time of the Big Bang, all of space was compacted to a small region. The expansion occurred throughout all of space. And none of the specific items mentioned, like the Earth, the Milky Way Galaxy, or the Andromeda Galaxy even existed at the time of the Big Bang.
8.
(d) In the first few seconds of the universe matter did not exist, only energy; the first atoms were not formed until many thousands of years later, and these atoms were primarily hydrogen. Stars on the main sequence and novae convert hydrogen into helium. They do not create the heavier elements. These heavy elements are typically created in the huge energy bursts of supernovae, or in neutron star mergers, as mentioned on page 1353.
9.
(c) The rotational period of a star in a galaxy is related to the mass in the galaxy. As astronomers observed the rotational periods, they were found to be much greater than could be accounted for by the visible matter (stars and dust clouds). Dust clouds are not dark matter, as they can reflect light and are therefore visible. The acceleration of the expansion of the universe is the foundation for the theory of dark energy, which is different from dark matter.
10. (a) If the Big Bang were a single large explosion that gave the universe an initial large kinetic energy and since currently the only large-scale force in the universe is the attractive force of gravity, we expect that the kinetic energy of the universe would be decreasing as that energy becomes gravitational potential energy. The expansion, however, is accelerating so there must be another energy source fueling the acceleration. This source has been dubbed “dark energy.”
Solutions to Problems Note – since the textbook says that the Hubble parameter is 23 0.4km s Mly , we will treat the Hubble parameter as having 3 significant figures. 1.
Convert the angle to seconds of arc, reciprocate to find the distance in parsecs, and then convert to light years. o 3600 3.5 104 1.26 1o
d pc
1
1 1.26
3.26 ly 0.794 pc 1pc 2.6 ly
3.26 ly 9.46110 m 16 2.4 10 m ly 1pc 15
0.794 pc 2.
Use the angle to calculate the distance in parsecs, and then convert to light years. 1 1 3.26 ly d pc 4.348 pc 4.348 pc 1pc 14.17 ly 14 ly 0.23
Chapter 44
Astrophysics and Cosmology
3.
The parallax angle is smaller for the further star. Since tan d D , as the distance D to the star increases, the tangent decreases, so the angle decreases. And since for small angles, tan , we have that d D . Thus if the distance D is doubled, the angle will be smaller by a factor of 2 .
4.
The apparent brightness of an object is inversely proportional to the square of the observer’s distance from the object, given by Eq. 44–1. To find the relative brightness at one location as compared to another, take a ratio of the apparent brightness at each location. L 2 2 1 2 d d b 4d 2 Jupiter Jupiter Earth 5.2 0.037 or only about 4%. L b d 2Earth d Earth
2 4d Earth
Jupiter
Jupiter
5.
The density is the mass divided by30the volume. M M 1.99 10 kg 3 3 2 10 kg m 3 4 3 V 10 4 3 r 3 6 10 m
6.
(a) The apparent brightness is the solar constant, 1.3103 W m2 . (b) Use Eq. 44-1 to find the intrinsic luminosity. L 26 2 11 2 2 3 2 b L 4d b 4 1.496 10 m 1.3 10 W m 3.7 10 W 2 4d
7.
The angular width is the inverse tangent of the diameter of our Galaxy divided by the distance to the nearest galaxy. According to Fig. 44–2, our Galaxy is about 5100,000 ly in diameter. Galaxy diameter 1.0 10 ly tan1 0.042 rad 2.4o Galaxy tan1 6 Distance to nearest galaxy 2.4 10 ly 6 Moon diameter 3.4810 m 1 tan1 9.1103 rad 0.52o Moon tan 8 Distance to Moon 3.84 10 m The Galaxy width is about 4.5 times the Moon width.
8.
The text says that there are about 4 1011 stars in the galaxy, and about 1011 galaxies, so that means about 4 1022 stars in the observable universe.
9.
The density is the mass divided by the30volume. M M Sun 1.99 10 kg 1.83109 kg m3 3 3 4 6 V REarth 4 6.38 10 m 3
3
Since the volumes are the same,30 the ratio of the densities is the same as the ratio of the masses. M 1.99 10 kg 3.33105 times larger 24 Earth M Earth 5.98 10 kg
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
10. The density of the neutron star is its mass divided by its volume. Use the proton to calculate the density of nuclear matter. The radius of the proton is taken from Eq. 41–1. 30 M 1.5 1.99 10 kg 5.354 1017 kg m3 5.4 1017 kg m3 neutron 3 V star 3 4 3 1110 m neutron neutron 5.354 1017 kg m3 5.354 1017 kg m3 star star 8 2.3 2.9 10 white nuclear 1.6731027 kg 1.83109 kg m3
dwarf
matter 4 3
3
11. The reciprocal of the distance in parsecs is the angle in seconds of arc. 1 1 (a) 0.02083 0.021 d pc o 48 pc 1 o 6 5.786 10 5.8 106 o (b) 0.02083 3600
1.2 1015 m
12. Convert the light years to parsecs, and then take the reciprocal of the number of parsecs to find the parallax angle in seconds of arc. 1 22.70 pc 1pc 74 ly 0.044 23 pc 3.26 ly 22.70 pc
13. Find the distance in light years. That value is also the time for light to reach us. 3.26 ly 77 pc 1pc 251ly 250 ly It takes light 250 years to reach us.
14. The Q-value is the mass energy of the reactants minus the mass energy of the products. The masses are found in Appendix G. 4 4 8 He He Be 2
2
4
Q 2m Hec m Bec 2 4.002603 u 8.005305 u c 2
4 2
2
2
931.5 Mev c 0.092 MeV 2
He 8 Be 12 C 4
6
Q mBe c mHe c2 m Cc2 4.002603 u 8.005305 u 12.000000c2 931.5 Mev c 2
2
7.366 MeV 15. The angular width is the inverse tangent of the diameter of the Moon divided by the distance to the Sun. Moon diameter 3.48 106 m o tan1 tan1 2.33105 rad 1.33103 4.79 11 Distance to Sun 1.496 10 m 16. A: The temperature increases, the luminosity stays the same, and the size decreases. B: The temperature stays the same, the luminosity decreases, and the size decreases. C: The temperature decreases, the luminosity increases, and the size increases.
Chapter 44
Astrophysics and Cosmology
17. Wien’s law (Eq. 37-1) says that the PT , where is a constant, and so P1T1 P2T2 . The Stefan–Boltzmann equation (Eq. 19–18) says that the power output of a star is given by P AT 4 , where is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by Eq. 44-1. It is given that b1 b2 , r1 r2 , P1 750 nm, and P2 450 nm. T T T2 P1 P1 1 P2 2 T1 P2 b1 b2
4
L1 L2 2 4d 4d 2 1
2
4 4 4r22T24 T2 T2 d22 L2 P2 A2T2 2 4 2 4 d L P AT 4r T T4 T
2
1
2
1
1
1 1
1
1
1
1
T 750 2.8 2 P1 d1 T1 P2 450
d
2
2
The star with the peak at 450 nm is 2.8 times further away than the star with the peak at 750 nm. 18. Wien’s law (Eq. 37-1) says that the PT , where is a constant, and so P1T1 P2T2 . The Stefan–Boltzmann equation (Eq. 19–18) says that the power output of a star is given by P AT 4 , where is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by Eq. 44–1. It is given that b1 b2 0.091 , d1 d2 , P1 470 nm, and P2 720 nm. L1 L 0.091 2 2 T T T2 P1 ; b 0.091b 2 1 2 P1 1 P2 2 4d 4d T
1
d2
1 d 22 1
0.091L 2
L
P2
0.091P 2
P
1
0.091A2T24 AT
1
4
0.091 4r2 T24 4 r T
1 1 2
2
1 2
2
4
1
1
2
4
2
0.091 T2 r2 T14 r12
T 470 nm 0.091 2 0.091 P1 0.091 0.1285 r2 T 1 P2 720 nm r1
2
The ratio of the diameters is the same as the ratio of radii, so
D1 0.13 . D2
19. The Schwarzschild radius is 2GM c2 . 2GM 2 6.67 1011 N m2 kg2 5.98 1024 kg Earth 8.86 103 m 8.9 mm REarth 2 8 c2 3.00 10 m s
20. The Schwarzschild radius is given y R
2GM c2
in Example 44–1. 2GM 2 6.67 1011 N m2 kg2 R
2
c
. An approximate mass for our Galaxy is calculated
2 10 kg 3 10 m 3.00 10 m s 8
41
14
2
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
21. The Schwarzschild radius is given by R 2GM c , so M 2
Rc2 2G
Instructor Solutions Manual
. The radius is given in Eq. 37–
12.
5.29 10 m3.00 10 m s 3.57 10 kg M 2G 2 6.67 10 N m kg 11
Rc2
2
8
16
11
2
2
22. The limiting value for the angles in a triangle on a sphere is 540o . Imagine drawing an equilateral triangle near the north pole, enclosing the north pole. If that triangle were small, the surface would be approximately flat, and each angle in the triangle would be 60o . Then imagine “stretching” each side of that triangle down towards the equator, while keeping sure that the north pole stayed inside the triangle. The angle at each vertex of the triangle would expand, with a limiting value of 180o . The three 180o angles in the triangle would sum to 540o . 23. (a) For the vertices of the triangle we choose the North pole and two points on a latitude line on nearly opposite sides of the Earth, as shown on the diagram. Let the angle at the North pole be 179°. (b) It is not possible to draw a triangle on a sphere with the sum of the angles less than 180°. To draw a triangle like that, a hyperbolic surface like a saddle (similar to Fig. 44–18) would be needed.
180° 90° 90o
90°o 90
24. To just escape from an object, the kinetic energy of the body at the surface of the body must be equal to the magnitude of the gravitational potential energy at the surface. Use Eq. 8–19.
vesc
RSchwarzchild
2GM c2
25. We find the time for the light to cross the elevator, and then find how far the elevator moves during that time due to its acceleration. 2 2 2 9.80 m s 3.2 m g x x 2 1 ; y g t t 5.6 1016 m 2 2 2 8 c 2c 2 3.00 10 m s
Note that this is smaller than the size of a proton. 26. Use Eq. 44–4, Hubble’s law. v 2250 km s 97.8 Mly 9.78 107 ly v H0d d H 23 km s Mly 0
27. Use Eq. 44–4, Hubble’s law. v 0.022 3.00 108 m s 287.0 Mly 2.9 102 Mly 2.9 108 ly v H0 d d H
0
28. (a) Use Eq. 44–6 to solve for the speed of the galaxy.
Chapter 44
Astrophysics and Cosmology
obs rest v v c 455 nm 434 nm 0.04839c c 434 nm 0.048 c rest (b) Use Hubble’s law, Eq. 44–4, to solve for the distance. z
d
v H0 d
v
H
631Mly 6.310 ly
0.04839 3.00 108 m s
8
0
29. We find the velocity from Hubble’s law, Eq. 44–4, and the observed wavelength from the Doppler shift, Eq. 44–3a. 4 v H0d (a) 2.310 m s Mly 6.0 Mly 4.6 104
c
c
1 v c
0
(b)
1 v c
656.3 nm
1 4.6 104
656.6 nm 0 0.3 nm
1 4.6 104
4 2.310 m s Mly 60 Mly 4.6 103 c c 1 v c 1 4.6 103 0 659.3 nm 0 3.0 nm 656.3 nm 1 4.6 103 1 v c
v
H0d
30. Use Eqs. 44–3a and 44–4 to solve for the distance to the galaxy. 2 2 rest 1 v c v c obs 2 2 obs rest 1 v c
d
v
H
obs
rest
3.00 10 m s 419.7 nm 393.4 nm H 2.30 10 m s Mly 419.7 nm 393.4 nm c
2
2
obs 2
rest 2
obs
rest
8
4
0
0
2
2
2
2
842.9 Mly 8.4310 ly 8
If instead we use Eq. 44–3b, we get the following. v v c 26.3 nm 2 c 6.685 10 c rest 393.4 nm rest c v c 3.00 108 m s 2 8 2 d 6.685 10 6.685 10 8.72 10 ly 4 H0 H0 2.310 m s Mly
31. Use Eqs. 44–3a and 44–5a to solve for the speed of the galaxy. z
obs rest obs 1 1 v c 1 rest rest 1 v c 2
z 1 1 1.068 1 0.06569 v 0.066 c c z 12 1 1.0682 1
v
2
The approximation of Eq. 44–6 gives v zc 0.068 c .
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
32. Use Eqs. 44–3a and 44–5a to solve for the redshift parameter. rest obs 1 1 1 0.07048 0.070 z obs rest rest Or, we use the approximation given in Eq. 44–6. v z 0.068 c 33. We cannot use the approximation of Eq. 44–6 for this circumstance. Instead, we use Eq. 44–3a combined with Eq, 44–5c, and then use Eq. 44–4 for the distance. 1 v c 1 v c 2 2 z 1 2.15 2.15 1 v c 2.15 1 v c 1 v c 1 v c 3.6225 c 0.6443c 2.152 1 v c 2.152 1 5.6225 v c 3.6225 v 5.6225 v 0.6443 3.00 108 m s d 8404 Mly 8.40 109 ly 3 H 2310 m s Mly 0
34. Use Eqs. 44–3a and 44–5a to solve for the speed of the galaxy. 625 nm 434 nm 191nm 2
191nm 1 1 1 1 434 nm v 1.0739 0 0.3494 z 1 v c 1 2 2 c rest 3.0739 1 v c 191nm 1 1 1 1 434 nm 0
2
v 0.349c or v = 0.3494 3.00 108 m s 1.05 10 m s 8
Use Hubble’s law, Eq. 44–4, to solve for the distance. v 0.3494 v c v H0 d d 4.557 103 Mly 4.56 109 ly 23000 m s Mly H0 H0 c 3.00 108 m s 35. For small relative wavelength shifts, we may use Eq. 44–6 to find the speed. We use Eq. 44–4 to find the distance. v v c ; v H d 0 c rest rest 8 cm v c 3.00 10 m s 0.10 23, 000 m s Mly 21cm d H H 62 Mly
0
0
rest
We also use the more exact relationship of Eq. 44–3a.
Chapter 44
Astrophysics and Cosmology 2
21.1cm obs 1 1 obs v 1 v c 21cm 2 4.75 102 c rest 2 21.1cm 1 v c rest obs 1 1 21cm rest 2
v H0 d d
36. Eq. 44–3a states
v H0
v c
H0 c
23000 m s Mly
61.96 Mly 62 Mly
3.00 108 m s
obs
4.75 102
.
rest
v 1/ 2 1/ 2 1 v 1 2 1 v v v 1 v c 1 1 1 1 rest 1 obs rest rest rest 2 2 2 c c c c c 1 v c v v v v 1 1 2 1 v obs rest 2 c rest rest rest rest rest c c c rest c
37. Wien’s law is given in Eq. 37–1.
T 2.90 10 m K 3
P
P
2.90 103 m K T
2.90 103 m K
3
1.110 m
2.7 K
38. (a) We approximate the temperature–kinetic energy relationship by kT K as given on page 1368. K 131012eV 1.60 1019 J eV 1.5 1017 K kT K T 1.38 1023 J K k
(b) From Fig. 44–29, this might correspond to a time around 1014 s . Note that this is just a very rough estimate due to the qualitative nature of Fig. 44–29. (c) According to Fig. 44–29, this is in the “Electroweak era” . 39. We use Wien’s law, Eq. 37–1. From Fig. 44–29, the temperature is about 1010 K. 2.90 103 m K 2.90 103 m K 3 PT 2.90 10 m K P 31013 m 1010 K From Fig. 31–12, that wavelength is in the gamma ray region of the EM spectrum. 40. We use the proton as typical nuclear matter. 1nucleon 26 kg3 10 m 1.67 1027 kg 6 nucleons m3 41. If the scale factor of the universe is inversely proportional to the temperature, then the scale times the temperature should be constant. If we call the current scale factor “1”, and knowing the current temperature to be about 3 K, then the product of the scale factor and the temperature is about 3. Use Fig. 44–29 to estimate the temperature at various times. For purposes of illustration, we assume the 10 universe has a current size of about 10 ly . There will be some variation in the answer due to reading the figure.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
(a) At t 106 yr , we estimate the temperature to be is about 1000 K. Thus the scale factor is found as follows. 3 3 3103 ScaleTemperature 3 Scale Temperature 1000
Size 3103
10 ly 310 ly 7
10
(b) At t 1 s , the temperature is about 1010 K . 3 3 Scale 10 31010 Temperature 10
Size 31010
10 ly 10
(c) At t 106 s , the temperature is about 1012 K . 3 3 Scale 3 1012 12 Temperature 10 14 2 Size 3 1012 1010 ly 3 10 ly 3 10 m 33
(d) At t 10 s , the temperature is about 10 K . 3 3 Scale 27 31027 Temperature 10 27
Size 31027
10 ly 310 ly 0.3 m 17
10
42. We approximate the temperature–energy relationship by kT E mc2 as suggested on page 1368. mc2 2 . kT mc T k mc2 500 MeV c2 c 2 1.60 1013 J MeV 6 1012 K (a) T 23 1.38 10 J K k 5
From Fig. 44–29, this corresponds to a time of 10 s . mc2 9500 MeV c2 c 2 1.60 1013 J MeV T 1 1014 K (b) 23 k 1.38 10 J K 11 From Fig. 44–29, this corresponds to a time of 10 s . mc2 100 MeV c2 c 2 1.60 1013 J MeV 1 1012 K (c) T k 1.38 1023 J K 6 From Fig. 44–29, this corresponds to a time of 10 s . There will be some variation in the answers due to reading the figure.
43. (a) According to the text, near Fig. 44–33, baryons (protons and neutrons) make up about 5% of the total mass-energy in the universe, in the form of stars and galaxies. M stars & 1011 galaxies 1011 stars galaxy 2.0 10303kg star galaxies 3 4 baryon stars & 4 R 14 109 ly 9.46 1015 m ly
galaxies
3
3
2.055 1027 kg m3 2.11027 kg m3 This is less than the critical density, suggesting a negative curvature.
Chapter 44
Astrophysics and Cosmology
(b) Again, according to the text, dark matter is about 27/5 = 5.4 times more plentiful than normal matter.
dark 5.4baryon 5.4 2.0551027 kg m3 This is somewhat larger than the critical density, suggesting a positive curvature. 44. Each black 2GMhole is twice the Schwarzschild radius R from the other, and is moving in a circular orbit, S 2 c
FG
FG
with the orbit radius RS . The gravitational attraction of one star on the other will produce uniform circular motion about their center of mass, the center of their orbits. M 2 RS GM M v2 M2 GM c2 2 G v G 2 2 R 2R 2R M 4RS 4 2GM 8 S S S 2 c
v
c
0.35 c
3.00 108 m s
8
8
1.1108 m s
45. The angular momentum is the product of the rotational inertia and the angular velocity. I initial I final 2 2 R 2 MR2 I 6 106 m initial final initial initial initial 5 initial initial 1rev month 3 2 2 I R 8 10 m final final 5 MRfinal 5.625 105 rev month 5.625 105 rev 1month 1d 1h 0.217 rev s month 30 d 24 h 3600 s
0.2 rev s 46. The rotational kinetic energy is given by 12 I . The final angular velocity, from Problem 45, is 2
5.625 105 rev month . 2
R final final final 5 final final 12 final final 2 2 2 2 K initial I MRinitial initial Rinitial initial 2 initial initial 5 K
1
I
2
2
MR 2
2
8 103 m5.625 105 rev month 5 5 5.625 10 6 10 6 6 10 m1rev month 2
47. The apparent luminosity is given by Eq. 44–1. Use that relationship to derive an expression for the absolute luminosity, and equate that for two stars.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
L
b
4d d
L 4d 2b ; L
distant star
d
11 bSun 1.5 10 m
distant star
Sun
Sun
4d 2
L
2
bdistant
Instructor Solutions Manual
4d 2 b
b
distant distant star star
Sun Sun
1
1ly 5 ly 1011 9.4611015m
star
48. The power output is the energy loss divided by the elapsed time. K K fraction lost 1 I2 fraction lost 1 2 MR22 fraction lost P initial 2 2 5 t t t t
8.0 10 m 2 rad s 1 10 1.74610 W 1.7 10 W
1 1.5 1.99 10 kg 30
2
3
9
2
1d24 h d3600 s h
5
25
25
49. Use Newton’s law of universal gravitation. FG
m1m2 2
r
6.67 1011 N m2 kg2
6
15
2
2.98 1028 N
28 3 10 N
50. We use the Sun’s mass and given density to calculate the size of the Sun. M M 4 3 r V 3
Sun
30 rSun 3M 31.99 10 kg 4
1/ 3
1/ 3
26 3 4 10 kg m
3.62 1018 m
r Sun
d
1.50 1011m 2 10
7
1ly 15 382 ly 9.46 10 m 400 ly
3.62 1018 m
r 382 ly 3 ; d Sun 100,000 ly 4 10
Earth-Sun
galaxy
51. The temperature of each star can be found from Wien’s law, Eq. 37–1. The peak wavelength is used as a subscript to designate each star’s properties. P T 2.90 103 m K 2.90 103 m K 2.90 103 m K T660 T480 4394 K 6042 K 660 109 m 480 109 m The luminosity of each star can be found from the H–R diagram.
L 660
7 1025 W
L
5 1026 W
480
The Stefan–Boltzmann equation says that the power output of a star is given by P AT , where is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L given in Eq. 44–1. Form the ratio of the two luminosities. 2 L A T4 4r2 T 4 r T2 L 5 1026 W 4394 K 1.413 480 660 480 480 480 480 480 480 2 L A T 4 4 r 2 T 4 r L T 2 7 1025 W 6042 K 660 660 660 660 660 660 660 480 The diameters are in the same ratio as the radii. 4
Chapter 44
Astrophysics and Cosmology
d480
1.413 1.4 d660 The luminosities are fairly subjective, since they are read from the H–R diagram. Different answers may arise from different readings of the H–R diagram.
52. (a) First calculate the number of parsecs. Then use the fact that the number of parsecs is the reciprocal of the angular resolution in seconds of arc. 1pc 1 100 ly =30.67 pc
3.26 ly 1 9.06 106 1 1 9 106 30.67 60 60
(b) We use the Rayleigh criterion, Eq. 35–10, which relates the angular resolution to the diameter of the optical element. We choose a wavelength of 550 nm, in the middle of the visible range. 1.22 550 109 m 1.22 D 1.22 4.24 m 4m 6 D 9.06 10 rad 180
The largest optical telescopes with single mirrors are about 10 m in diameter. 53. (a) To find the energy released in the reaction, we calculate the Q-value for this reaction. From Eq. 42–2a, the Q-value is the mass energy of the reactants minus the mass energy of the products. The masses are found in Appendix G.
Q 2mC c2 mMg c2 2 12.000000 u 23.985042 u c 2 931.5 Mev c2
13.93 MeV
(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching. The distance between the two nuclei will be twice the nuclear radius, from Eq. 41–1. Each nucleus will have half the total kinetic energy. 1 q2 1/ 3 1/ 3 15 15 r 1.2 10 m A 1.2 10 m12 nucleus U 40 2r 1 q2 nucleus K 12 U 12 4 0 2r 2 1MeV 2 6 1.60 1019 C 9 1 2 2 2 8.988 10 N n C 4.711MeV 13 1.60 10 J 4.7 MeV (c) We approximate the temperature–kinetic energy relationship by kT K as given on page 1368. kT K T
K
5.5 10 K 10
k
54. (a) Find the Q-value for this reaction. From Eq. 42–2a, the Q-value is the mass energy of the reactants minus the mass energy of the products.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
16 8
Instructor Solutions Manual
28 4 O 168 O 14 Si 2 He
2 Q 2m Oc2 m Sic2 m Hec2 2 15.994915 u 27.976927 u 4.002603c2 931.5 Mev c
9.594 MeV
(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching. The distance between the two nuclei will be twice the nuclear radius, from Eq. 41–1. Each nucleus will have half the total kinetic energy. 2 1 q 1/3 1/3 15 15 r 1.2 10 m A 1.2 10 m 16 nucleus U 40 2r 2 1 q nucleus K nucleus 12 U 12 40 2r 1eV 9 7.609 MeV 12 8.98810 N 1.60 1019 J
7.6 MeV (c) We approximate the temperature–kinetic energy relationship by kT K as given on page 1368. 1.60 1019 J 6 7.609 10 eV 1eV K 8.8 1010 K T kT K 23 k 1.38 10 J K 55. We must find a combination of c, G, and that has the dimensions of time. The dimensions of c are L L3 ML2 , T the dimensions of G are are , and the dimensions of . 2 T MT 2 3 L L ML t P c G T L 3 2 M T 2 T MT 2 T 3 2 0 ; 0 ; 2 1 5 0 ; 1 3 5 1 3 1 ; 1 ; 5 2
2
2
6.67 10 N m kg 21 6.6310 J s 5.38 10 s 3.00 10 m s 11
G t P c5/ 2G1/ 2 1/ 2 5 c
2
34
2
44
8
5
56. The radius of the universe is estimated as the speed of light (c) times the age of the universe (T). This radius is used to find the volume of the universe, and the critical density times the volume gives the mass. r cT ; V 43 r 3 43 cT ; 3
m V 4 cT 4 3.00 10 m s 13.8 10 yr 3.156 107 s 10 kg m 3 3 1yr 8 9 26 3 3
9.34 10 kg 1053 kg 52
3
Chapter 44
Astrophysics and Cosmology
57. (a) Assume that the nucleons make up only 2% of the critical mass density.
26
nucleon mass density 0.02 10 nucleon number density
3
kg m
0.02 1026 kg m3 1.67 10
27
0.12 nucleon m
3
kg nucleon
neutrino number density 10 nucleon number density 1.2 108 neutrino m3 9
0.98 1026 kg m3
8.17 10
kg
35
neutrino
1.66 1027 kg
(b) Assume that the nucleons make up only 5% of the critical mass density.
nucleon mass density 0.05 1026 kg m3 nucleon number density
0.05 1026 kg m3 1.67 10
27
0.30 nucleon m
3
kg nucleon
neutrino number density 109 nucleon number density 3.0 108 neutrino m3
0.95 1026 kg m3
kg
35
3.17 10
neutrino
9.315 108 eV c2 1.66 1027 kg
58. (a) From page 1358, a white dwarf with a mass equal to that of the Sun has a radius about the size of the Earth’s radius, 6380 km . From page 1358, a neutron star with a mass equal to 1.5 solar masses has a radius of about 20 km . For the black hole, we use the Schwarzschild radius formula. 2GM 11 2 2 30 R 2 6.67 10 N m kg 31.99 10 kg 8849m c2
3.00 10 m s
2
8
8.85 km
(b) The ratio is 6380 : 20 : 8.85 721: 2.26 : 1 700 : 2 : 1 . 59. Each helium atom requires 2 protons and 2 neutrons and has a total mass of 4u. Each hydrogen atom requires one proton and has a mass of 1u. If there are 7 times more protons than neutrons, then for every two neutrons there are 14 protons. Two protons combine with the two neutrons to produce a helium atom. The other 12 protons produce 12 hydrogen atoms. Therefore there is 12 times as much hydrogen as helium, by number of atoms. Each helium atom has a mass 4 times that of the hydrogen atom, so the total mass of the hydrogen is12 u, and the mass of the helium is 4 u. The mass ratio of hydrogen:helium is thus 12:4, or 3:1. 60. Because Venus has a more negative apparent magnitude, Venus is brighter. We write the logarithmic relationship as follows, letting m represent the magnitude and b the brightness. m k log b ; m2 m1 k log b2 log b1 k log b2 b1 5 m2 m1 k 2.5 log b2 b1 log 0.01 m m m m b m m k log b b 2 10 k bVenus 10 2.5 2 1 2 1 b1 bSirius 2
1
Venus
Sirius
4.41.4
10 2.5 16
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
61. We assume that gravity causes a centripetal force on the gas. Solve for the speed of the rotating gas, and then use Eq. 44–6 to estimate a z value for the gas that is moving away from Earth, near the receding outer edge of the cloud. mgas mblack m v 2 hole F F G gas gas gravity centripetal 2 r r mblack 2 109 1.99 1030 kg 6.42 105 m s vgas v z 2.14 103 2 103 c
62. We treat the energy of the photon as a “rest mass,” and so mphoton " " Ephoton
2
c . To just escape from
a spherical mass M of radius R, the energy of the photon must be equal to the magnitude of the gravitational potential energy at the surface. GMm GMm GM photon photon photon R Ephoton 2 c R Ephoton Ephoton
63. (a) The photon enters from the left, grazing the Sun, and moving off in a new direction. The deflection is assumed to be very small. Consider a small part of the motion in which the photon moves a horizontal distance dx cdt while located at (x,y) relative to the center of the Sun. Note that y R and
p
E
c
mc x
r yR F
GMm r2
r x y . If the photon has energy E, it has 2
2
2
a “mass” of m E c , and a momentum of p E c mc. To find the change of momentum in the y-direction, we use the impulse produced by the y-component of the gravitational force. GMm dx GMm R dx GMmR dx dp y Fy dt 2 cos 2 3/ 2 r c r r c c x 2 R2 2
To find the total change in the y-momentum, we integrate over all x (the entire path of the photon). We use an integral from Appendix B-4.
py
GMmR c
dx 3/ 2
x R 2
2
GMmR c
The total magnitude of deflection is py p .
x
R2 x 2 R2
1/ 2
2GMm cR
2GMp c2R
Chapter 44
Astrophysics and Cosmology
2GMp
2 c R
p p (b) We use data for the Sun. N m2 11 30 2 6.67 10 1.99 10 kg 2 kg 180 2GM 3600 0.87 2 2 c R 3.00 108 m s 6.96 108 m rad 1
64. Eq. 44–3b states
rest
and Eq. 44-6 also states that the ratio is z. Eq. 44–3a
. It is given that v 0.1c . Calculate v 0.1c
states obs
obs rest v v rest c
using both Eq. 44–3b
obs
rest
and Eq. 44–3a, and compare via percent difference. 44 3b : obs rest 0.1 obs rest
44 3a : % diff
1 v c
1 v c 1.10554rest 1.1rest obs
rest
1.10554rest
rest
1.1 1.10554 rest
0.9
100 5.01% 5%
65. If there are N nucleons, we assume that there are approximately 1 N neutrons and 1 N protons. Thus, 2
2
1
for the star to be neutral, there would also be 2 N electrons. (a) From Eq. 40–12 and 40–13, we find that if all electron levels are filled up to the Fermi energy E , the average electron energy is 3 E . 5
F
E N 3 E 1 N 3 e
e
5
F
2
F
h 3 Ne 2
2/3
5 8me V
(b) The Fermi energy for nucleons would be a similar expression, but the mass would be the mass of a nucleon instead of the mass of the electron. Nucleons are about 2000 times heavier than electrons, so the Fermi energy for the nucleons would be on the order of 1/1000 the Fermi energy for the electrons. We will ignore that small correction.
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
To calculate the potential energy of the star, think about the mass in terms of shells. We assume the star is spherical and has a uniform mass density. Consider the inner portion of the star with radius r < R and mass m, surrounded by a shell of thickness dr and mass dM. See the diagram. From Gauss’s law applied to gravity, the gravitational effects of the inner portion of the star on the shell are the same as if all of its mass were at the geometric center. Likewise, the spherically-symmetric outer portion of the star has no gravitational effect on the shell. Thus the gravitational energy of the inner portion–shell) combination is given by a form of Eq. 8–17, dU G
mdM
dr
R r
. The density of the star
r M 3 . We use that density to calculate the masses, and then integrate over the 3R
is given by 4
full radius of the star to find the total gravitational energy of the star. 3 M r m
r R r M R
3
4 3
3
4 3
4 3
3
3
2 4 r dr 3 dr 3 R R 3 3 2 r 3Mr M dr
3Mr
M
2
dM 4r 2 dr 4 mdM
2
3GM r 4 dr dU G G 6 r r R R 3GM 2 3GM 2 R 3GM 2 R5 2 4 4 r dr 3 GM U r dr 6 R 5 5 R R6 R6 0 0 R3
R3
(c) The total energy is the sum of the two terms calculated above. The mass of the star is primarily due to the nucleons, and so M Nmnucleon . h2 3 N 2/3 3 GM 2 Etotal Ee U 3 1 N 52
3h
M
5/3 320 4/3 me mnucleon equilibrium radius.
dE total
dR
5
R
2/3
92/33h2 M 5/3
Let a
8me 2V
3 2 3 GM 2 M 92/33h2 M 5/3 3 GM 3 2 5 R 80me mnucleon 2m nucleon 34 R 5 R 320 4/3me m5/3 nucleon R 2
2a
R3
b
3 2 dR 0 to find the and b GM , so E a b . We set dE total total 2 R R 5
2a
0 R eq b R2
2
92/33h2 M 5/3 320 4/3m m5/3 3
e
GM 2
5 We evaluate the equilibrium radius using the Sun’s mass.
nucleon
92/3 h2 nucleon 32 4/3GM 1/3m em5/3
Chapter 44
Astrophysics and Cosmology
Req
92/3 h2 32 GM m em nucleon 4/3
1/3
5/3
92/3 6.631034 J s
32
4/3
2
N m 1/3 5/3 11 30 31 27 6.67 10 2.0 10 kg 9.1110 kg 1.67 10 kg 2 kg 2
7.178 10 m 7.2 10 km The text states on page 1358 says that the radius of a white dwarf with a mass equal to that of the Sun would be about the size of the Earth. The Earth has a radius of about 6400 km, so the calculation gives a value in agreement with the text. 3
6
66. There are N neutrons. The mass of the star is due only to neutrons, and so M Nmn . From Eqs. 40– 12 and 40–13, we find that if all energy levels are filled up to the Fermi energy EF , the average energy is 3 E . We follow the same procedure as in Problem 65. The expression for the gravitational 5
F
energy does not change.
2/3 2/3 318 h2 M 5/3 M 3h2 3 2/3 2 h 3 N M E N E N n n 5 F 5 3 8mn V mn 40m n 34 R m n 160 4/3mn8/3 R2 2/3 2 5/3 2 318 h M 3 GM Etotal En U 4/3 8/3 2 160 mn R 5 R 3
3
318 h2 M 5/3 2/3
Let a
3 2 and b GM , so E total
n 160 4/3m8/3
5
equilibrium radius. dE total
dR
2a
a b . We set dE dR 0 to find the total 2 R R
318 h2 M 5/3 4/3 8/3 160 m n 2/3
R3
b
0 R eq
R2
2a b
2
3
2
18 h2 2/3
16 4/3GM 1/3m8/3 n
GM 5 We evaluate the equilibrium radius for a mass of 1.5 solar masses. 2/3 18 h2 Req 16 4/3GM 1/3m8/3 n 16 4/3
6.67 10
1822/3 6.631034 J s N m 11 30
kg2
2
27 1.52.0 10 kg 1.67 10 kg 1/3
8/3
1.086 104 m 11km 67. (a) Because the speed of the galaxy is small compared to the speed of light, we can use Eq. 44–6 instead of Eq. 44–3a. The velocity of the galaxy is negative. v 130 103 m s rest 656 nm 8 0.284 nm c 3.00 10 m s
Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition
Instructor Solutions Manual
(b) This is a blue shift because the wavelength has decreased, and because the galaxy is approaching. (c) The time is equal to the distance between the galaxies divided by their relative speed. This is assuming that they continue to move at the same present speed. (They would actually accelerate due to the gravitational force between them). 15 1yr d 2.5 106 ly 9.46 10 m 9 t 3.156 107 s 5.8 10 yr 3 ly v 130 10 m s
Chapter 44
Astrophysics and Cosmology