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Contents 3.1
Measurements and their errors 3.1.1 Use of SI units and their prefixes 3.1.2 Limitations of physical measurements 3.1.3 Estimation of physical quantities
4 4 6 14
3.2
Particles and radiation 3.2.1 Particles 3.2.2 Electromagnetic radiation and quantum phenomena
15 15 40
3.3
Waves 3.3.1 Progressive and stationary waves 3.3.2 Refraction, diffraction and interference
49 49 61
Examination preparation Practical and mathematical skills Data and formulae Practice exam-style questions Answers
76 78 81 90
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Mathematical skills Data handling Interpreting graphs Geometry and trigonometry
96 96 100 109
Glossary
114
Index
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3.1
Measurements and their errors
3.1.1 Use of SI units and their prefixes Physics relies on the measurement of physical quantities like mass, speed or force. These measurements are always made with reference to a standard value, known as a unit. For example, the width of this book could be measured in terms of your hand-widths, or in units of feet and inches. To ensure that the numerical value of any specific measurement is the same, no matter who makes the measurement or where it is done, an agreed set of standard units is used throughout physics. These agreed standards are known as SI (Système Internationale) units.
Fundamental or base units Some physical quantities, such as mass or length, are said to be fundamental. SI units are based on seven fundamental quantities (see Table 1). These fundamental quantities are measured in units, known as base units, which are carefully defined so that measurements can be replicated exactly. Each unit has a name and a standard abbreviation. For example, the base unit of time, the second, is defined as the time taken for 9 192 631 770 oscillations of a particular electromagnetic wave from a caesium-133 atom. (You are not expected to know these definitions at A-level.) Table 1
Essential Notes Note that the unit name begins with a lower case letter, even if it is named after a scientist. The candela is not required for A-level.
Quantity
Unit
Abbreviation
Mass
kilogram
kg
Length
metre
m
Time
second
s
Quantity of matter
mole
mol
Temperature
kelvin
K
Electric current
ampere
A
Light intensity
candela
Cd
Derived units Essential Notes At GCSE you will have written the units of speed as m/s, but at AS and A level it is written as m s−1.
All other physical quantities have units that can be expressed in terms of these base units. These are known as derived units. For example, speed is defined as the distance travelled in a specified time and so has units of metres per second. In some cases, the derived unit is given a name such as joule (energy), newton (force) or pascal (pressure). These can always be expressed in terms of the base units.
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Section 1
Example Force is measured in newtons. Express the newton in terms of SI base units.
Answer Force can be calculated from the equation, force = mass 3 acceleration, where acceleration is the change in velocity per second. The units of acceleration are m s−1 / s = m s−2. Mass is measured in units of kg. So in terms of base units, force is measured in kg m s−2.
SI prefixes Measurements of a physical quantity, such as length, can span an enormous range. For example, the diameter of our galaxy is approximately 1030 times larger than the diameter of an atom. SI units use standard prefixes to express very large and very small numbers (see Table 2). Multiplier
SI prefix
Abbreviation
10−15
femto-
f
10−12
pico-
p
10
nano-
n
10−6
micro-
μ
10
milli-
m
103
kilo-
k
10
mega-
M
109
giga-
G
tera-
T
−9
−3
6
12
10
Table 2
Essential Notes −15
For example, the diameter of a uranium nucleus is 1.75 3 10 m, which can be written as 1.75 fm. The power output of the Three Gorges Dam hydroelectric power station is 22 500 000 000 W, which can be written as 22 500 000 kW, or 22 500 MW, or 22.5 GW.
Example Re-write the following values using SI standard prefixes. (a) 0.00125 A (b) 725 000 000 N (c) 0.000 12 kg
Answers Write the answers in scientific notation. Choose a power of 10 that is a multiple of 3. (a) 0.00125 A = 1.25 3 10−3 A = 1.25 mA (b) 725 000 000 N = 725 3 106 N = 725 MN (c) 0.000 12 kg. Dividing by 1000 gives 0.12 g. Dividing by 1000 again gives 120 mg.
The SI base unit for mass is the kilogram rather than the gram. This means that a mass of 2.5 kg would be entered into an equation as 2.5 kg rather than 2.5 3 103 g and 2.0 mg would be entered as 2.0 3 10−6 kg, rather than 2.0 3 10−3 g.
Essential Notes SI standard prefixes represent powers of 10 which are multiples of 3; 103, 106, 109, etc. You will come across non-standard multipliers, such as centi (3 10−2), but it is good practice to express your answers in terms of the standard prefixes. 5
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3.1.2 Limitations of physical measurements
Uncertainty Essential Notes Engineers often refer to the uncertainty in a value, particularly of a manufactured item, as the tolerance.
A measurement of a physical quantity, such as mass or speed, can never be exact. There is some uncertainty associated with every measurement. For example, a recipe may call for 200 g of sugar, which you could measure using digital kitchen scales (a type of electronic top pan balance). These scales typically show readings to the nearest gram. We say the resolution of the measuring instrument is 1 g, since this is the smallest change that can be registered on the device. If the reading is 200 g, the actual mass of sugar could be as low as 199.5 g or as high as 200.4999… g. The value of the measurement cannot be known exactly. Instead it falls within a range of values, which can be written as: 200 g ± 0.5 g
or
(200 ± 0.5) g
The ± 0.5 g is known as the absolute uncertainty in the reading. It is measured in the same units as the reading itself. An absolute uncertainty of ± 0.5 g is unlikely to be a problem if you are weighing yourself, but would be vital in the manufacture of medicines. It is the fractional uncertainty, often expressed as a percentage, which is important. percentage uncertainty = (absolute uncertainty / measured value) 3 100%
Essential Notes Precision and accuracy are terms which are often misused. The precision of a reading is determined by its percentage uncertainty. The accuracy of a reading refers to how close it is to the actual value.
A reading with a smaller percentage uncertainty is said to be more precise.
Example Estimate the percentage uncertainty in these readings. (a) ‘Weighing’ yourself (finding your mass) using bathroom scales. (b) Measuring the width of A4 paper with a ruler marked in millimetres.
Answer (a) Typical mass of an A-level student = 60 kg. Bathroom scales have a resolution of 0.1 kg percentage uncertainty = (absolute uncertainty / measured value) 3 100% = (0.1 / 60) 3 100% or 0.2% (b) Width of A4 paper = 210 mm measured to the nearest millimetre, so the percentage uncertainty = 1 / 210 or 0.5% The resolution of a measuring instrument is the smallest scale division shown by the instrument: for example, ± 1 mm for a typical ruler. It tells us the smallest value that the uncertainty can take. There may well be other factors that lead to a larger uncertainty. These could be caused by variations in the measuring device itself. The manufacturer will often publish the tolerances for the resolution and the accuracy of the measuring device. We can reduce the uncertainty in a reading, perhaps by choosing a balance with a resolution of 0.1 g or 0.01 g, but the measurement would never be exact.
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Example The distance from the Earth to the Moon was first estimated by Hipparchus over 2000 years ago. His calculation gave the distance as 63 Earth radii, to the nearest whole number. Modern measurements using lasers are capable of measuring the distance to the nearest millimetre. The accepted mean value of the distance from the Earth to the Moon is 384 403 km. (a) Modern measurements of the Earth–Moon distance are both more accurate and more precise than those carried out by Hipparchus. Explain the terms that are written in bold type. (b) Compare the precision of Hipparchus’ measurements with modern measurements. (c) The average radius of the Earth is 6371 km. Calculate the accuracy of Hipparchus’ result.
Answer (a) A precise reading has a low fractional uncertainty. Repeated readings that are precise will have values that are close to each other. An accurate reading is close to the actual or accepted value. (b) Hipparchus has d = 63 ± 1, so the fractional uncertainty is 1 / 63 = 1.6%. Modern measurements = ± 1 3 10−3 / 6371 000 = 1.6 3 10−8%, 100 million times better! (c) 6371 km 3 63 = 401 373 ≈ 400 000, so Hipparchus was within 16 000 km, an uncertainty of less than 5%.
Experimental errors The method used to take measurements can introduce further uncertainties. For example, suppose that you are timing the oscillations of a pendulum using a stopwatch that displays a reading to the nearest 0.01 seconds. That is, the resolution of the stopwatch is 0.01 s. Your timings will not be as precise as this might suggest, since you will need to judge when the oscillation starts and finishes and then react by starting or stopping the stopwatch. Your reaction time may add another 0.1 to 0.2 s to the reading. A good experimental design would help to reduce the effect of these experimental errors. For example, in the pendulum experiment, you can use markers, known as fiducial markers, to help you to judge exactly when an oscillation starts and ends (Fig 1).
Essential Notes Experimental errors are not necessarily ‘mistakes’; rather they are factors which introduce some uncertainty into the results. There are ways of reducing errors, but it is not possible to eliminate all the uncertainty in your results.
Experimental errors fall into two categories. These are referred to as systematic errors and random errors.
Systematic errors Systematic errors are usually associated with the measuring instrument or technique and they tend to affect all the readings in a similar way.
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Fig 1 Viewing along the line of the two fiducial markers means that you always view from the same point and you are better able to judge the start and end of the oscillation. The fiducial markers should be placed in the centre of the oscillation, where the pendulum is moving at its highest speed.
Systematic errors may add a constant value to all the readings, or may have a multiplier effect, changing all the results by the same percentage.1 For example: • Using a meter or balance which is not zeroed correctly. This zero error will be added to each subsequent reading. • Misreading a scale due to poor alignment of the observer’s eye and the scale, often referred to as parallax error (see Fig 2). • Failing to take account of the resistance of the connections in an electrical circuit. • Using a poorly calibrated thermometer to measure the temperature of an object, or failing to make a good thermal contact between the thermometer and the object. Systematic errors reduce the accuracy of a measurement, so that it is further away from the actual value. In some cases, where the actual value is not known, it may be difficult to quantify, or even identify, systematic errors. However, systematic errors may come to light if the results are not reproducible: that is, they cannot be replicated by other experimenters, using different apparatus and possibly a different method.
Reducing systematic errors The effect of a systematic error cannot be reduced by repeating the readings, since the same error will affect all the readings in the same way. A change in the method is required. Examples that occur in A and AS-level experiments include: • Parallax errors. These occur when measurements are taken from an analogue meter or a scale such as a ruler or measuring cylinder. These can be reduced by observing from the correct angle. If a pointer is used, as in the case of an analogue meter, it should be kept close to the scale. 1 In some cases, the multiplier effect might not be linear; some readings may be increased by more than others. 8
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Section 1
Fig 2 Avoiding parallax errors a It is important to have your eyeline at 90° to the scale.
(a) incorrect eye position
correct eye position
0
1
2
3
4
5
6
7
0 cm
8
9
10
8.2 cm
(b)
b The curved surface of a liquid in glass, known as the meniscus, can lead to errors in measurements.
Low Rea ding Correct Line of Volume
h Hig ing d Rea
(c) PERCENT
50
.3
60
.2
70
.1
80
.05
TRANSMITTANCE TR
90
c A mirror can be placed behind the scale to ensure that the observation is made from the correct angle. The reading is taken when the pointer is aligned with its own image.
100
0 ABSORBANCE
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Reducing the uncertainty in measurements of length A meter ruler marked in millimetres is useful for measuring objects which are between 10 and 100 cm long. The uncertainty is ± 1 mm, so at 10 cm the fractional uncertainty is 1%. For shorter lengths we need an instrument with better resolution. There are two such instruments that are commonly used in school labs. Electronic versions of these instruments are also available. • Vernier callipers typically have a range of 0 to 10 or 20 cm and a resolution of 0.1 mm. Fig 3 Vernier calliper The scale gives the distance between the jaws of the calliper. The first part of the reading is given by the point where the zero on the sliding scale meets the upper scale. In this case that point lies between 1.1 and 1.2 cm. The next significant figure is given by the mark on the sliding scale that is most closely aligned with a scale mark on the upper scale. In this case the mark is 7. The distance between the jaws is therefore 1.17 cm, or (11.7 ± 0.1) mm
cm
0
1
2
3
4
0 1 2 3 4 5 6 7 8 9 10
• Micrometers have a smaller range than vernier callipers, typically only a few cm, but they are capable of measuring to the nearest 0.01 mm.
Notes
35
There are a couple of things to be aware of here: • T he barrel on some micrometers goes round twice to alter the gap by 1 mm. The 0.5 mm marks are above the line on the scale. If one is showing, you will need to add 0.5 mm to your reading. • Y ou should use the ratchet at the end of the micrometer to turn the screw. It is designed to slip when it meets a resistance. Otherwise, you may crush the object that you were trying to measure.
0
5
30 25
Fig 4 The gap between the jaws is found by reading the scale at the point where the rotating barrel meets it, in this case just beyond the 5. The next significant figures are given by the rotating scale at the point where it crosses the line, 34 in this case. So the distance shown here is 5.34 mm or (5.34 ± 0.01) mm
Random errors Random errors affect measurements in an unpredictable way, causing the readings to fluctuate. Random errors may arise due to changing environmental conditions, such as temperature variations, or they may be
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Section 1
due to changes inside the measuring instrument itself. Some examples of random errors are: • Fluctuations in timing the oscillations of a pendulum using a stopwatch, due to the observer’s lack of consistency in starting and stopping the stopwatch. • Random electrical noise may affect the readings taken on an oscilloscope. • The diameter of a wire may vary along its length due to variations in the manufacturing process. Random errors increase the range of values that a reading may take. Random errors therefore decrease the precision of a reading. As random errors fluctuate above and below a mean value, the uncertainty may be reduced by repeating the readings and finding the mean. Large random errors can make it difficult to be sure that the measurement is repeatable. A repeatable measurement is one where the same result can be obtained by the same experimenter using the same method and apparatus2. The fractional, or percentage, uncertainty can be reduced either by: • Decreasing the absolute uncertainty: for example by using meters with a better resolution. OR • Increasing the magnitude of the measurements involved, for example: ❍
❍
❍
find the period of oscillation of a pendulum, measure the time for To 20 oscillations, then divide by 20. find the mass of a small ball bearing, ‘weigh’ a 100 of them and To divide by 100. find the thickness of a sheet of paper, find the thickness of a large To number and divide by the number of sheets.
Uncertainty and significant figures The uncertainty in a measurement will determine how many significant figures should be recorded. The number of significant figures should not go beyond that justified by the precision of the measurement. For example, suppose that you found the mass of some sugar using a balance with a resolution of 0.1 g. The balance reading was 201.0 g. It would not be correct to write the reading as 201.00 g since the significant figures imply a higher precision than is justified. Nor would it be correct to write 201 g ± 0.1 g, since that underestimates the precision. The reading should be written as 201.0 ± 0.1 g, with the uncertainty in the last significant figure.
2 Statistical analysis may be necessary to decide whether variations are due to random errors, or whether they represent real variations in the variable being measured. This is not required at AS level.
Notes Students often make this mistake when finding the mean of several readings. Suppose an event has been timed using a stopwatch with an uncertainty of ± 0.1 s. If the reading was repeated three times and the results were 11.2 s, 11.5 s and 11.0 s then the mean is 11.2333 s. But writing this answer would be penalised in an exam for the inappropriate use of too many significant figures. The answer should be recorded as 11.2 ± 0.1 s.
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Notes
Combining uncertainties
In a mathematics problem, an answer of ½ or 1/3 is perfectly acceptable. This is not the case in a physics answer. Numbers like ½ or 1 /3 are ideal concepts and are generally not realisable in the physical world. It is not possible to cut a cake exactly in half. The most modern apparatus (a laser cake cutter?) might be able to achieve 0.500000 ± 0.000001, but there will always be some uncertainty. Answers to physics questions should always be expressed using decimal or scientific notation, and should be given to the correct number of significant figures.
Experimental results usually depend on several measurements that have to be combined in some way. For example, a measurement of a car’s average speed might depend on measuring the distance travelled by the car and the time taken for the journey. Each initial measurement has its own associated uncertainty. We need to combine these to find the overall uncertainty in the result.
Combining uncertainties when adding or subtracting measurements When measurements are added together, or subtracted from each other, we simply add the absolute uncertainties to find the overall uncertainty. For example, consider two lengths of wire, A and B, which are to be connected together. The length of each wire is measured with a metre ruler. Length of wire A = (120 ± 1 ) mm Length of wire B = (161 ± 1) mm The length of the combination is A + B = (281 ± 2) mm. Exactly the same approach is used when subtracting values; you must add the absolute uncertainties.
Combining uncertainties when multiplying or dividing measurements When measurements are multiplied together, or divided into each other, we add the fractional uncertainties to find the overall fractional uncertainty. For example, suppose we need to find the area of rectangular plate which has been measured using a metre ruler as (26 ± 1) mm long by (15 ± 1) mm wide. Fractional uncertainty in length = 1 / 26 = 0.038 Fractional uncertainty in width = 1 / 15 = 0.067 Fractional uncertainty in area = 0.038 + 0.067 = 0.105 or 10.5% So the area = 26 3 15 = 390 mm2 with an absolute uncertainty of 0.105 3 390 = 40.95 ≈ 41 mm2 Area = (390 ± 41) mm2 Exactly the same approach is used when dividing values; you must add the fractional uncertainties. Sometimes a calculation involves raising a measured variable to a power, such as (velocity)2 in the calculation of kinetic energy. In this case the uncertainty is dealt with as follows: Fractional uncertainty in An = n 3 fractional uncertainty in A
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Section 1
Example An experiment was carried out to find the density of a small metal cube. Measurements of the length of each of its edges were taken using vernier callipers. The readings were: Length / mm 15.0 15.2 14.9 14.8 15.0 15.1
The cube was weighed on a digital balance with a manufacturer’s statement of accuracy of ± 0.2 g. The result was 25.7 g. (a) Find the density of the cube. (b) Estimate the uncertainty in the result. (c) The metal is thought to be mild steel, which has a density of 7600 kg m−3. Do these measurements suggest that this metal could be mild steel? (d) What single improvement would you make to the method of measurement in order to reduce the uncertainty in the density?
Answer (a) Density = mass / volume, so the total fractional uncertainty is 4%. Mass has to be in kilograms and the volume in metres cubed. Mass = 25.7 g = 25.7 3 10−3 kg Volume = ((15 3 10−3) m)3 = 3.375 3 10−6 m3 So density = 25.7 3 10−3 kg / 3.375 3 10−6 m3 = 7615 kg m−3 (b) The resolution of the balance is 0.1 g and its accuracy is said to be ± 0.2 g so the mass of the cube = (25.7 ± 0.3) g. The fractional uncertainty is 0.01 or around 1%. Vernier callipers have a resolution of 0.1 mm, but the spread of results is rather larger, 0.4 mm, so the uncertainty is ± 0.2 mm. The mean length is (15.0 ± 0.2) mm. The fractional uncertainty is 0.2 / 15 = 0.01 or around 1%. Volume of cube = length3 = 15 3 15 3 15 = 3375 mm3. The fractional uncertainty = 3 3 1% = 3%. Since we are dividing mass by volume, we must add the fractional uncertainties, giving a total fractional uncertainty of 4%. (c) The density of steel is 7600 kg m−3, well within the possible range of this result, so the cube could be mild steel. (d) The measurements of mass and length both have a fractional uncertainty of 1%. But the length has to be cubed to find the volume and hence the density. It is therefore more important to reduce the uncertainty in the length measurement, perhaps by using an instrument with an improved resolution, such as a micrometer. 13
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3.1.3 Estimation of physical quantities
Estimation Being able to estimate the magnitude of physical quantities is a useful skill for a physicist; and it is one that develops with experience. Some quantities can be estimated directly; the mass of a lorry or the height of a building, perhaps. It is useful to have some similar known quantity to compare it to. For example, a typical family car has a mass of around 1500 kg or 1.5 tonnes, so a lorry’s mass might be estimated by comparison as roughly 10 cars or around 15 tonnes. Some other useful reference points are included in Table 3.
Essential Notes You do not need to recall the values in Table 1, but you should be able to recognise and suggest sensible values of the right order of magnitude (to the nearest power of ten) and to recognise differences in orders of magnitude. For example, the area of a football pitch can be estimated as about 100 m 3 70 m = 7000 m2, so the order of magnitude estimate is 10 000 m2. This is two orders of magnitude greater than the floor area of a typical house, and four orders of magnitude greater than the surface area of a human. Table 3 Orders of magnitude
Physical quantity
Example
Value
Value to nearest order of magnitude
Density
Air (at sea level and at 15°C)
1.2 kg m−3
1 kg m−3
Mass
Adult male
80 kg
100 kg 2
100 m2
Area
Typical semidetached house (floor area)
100 m
Area
Surface area of a human
2 m2
1 m2
Power
Kettle
2000 W
1000 W
Wind turbine
2 to 3 MW
106 W
Light bulb
100 W (incandescent)
100 W
30 W (fluorescent light)
1W
10 W
5 W (LED) Speed
Sound in air
330 m s−1
100 m s−1
Running
3 m s−1
1 m s−1
Derived estimates Some quantities can be estimated by breaking the problem down into steps, which can be estimated more easily.
Example Estimate how much electrical energy is transferred as light in homes in the UK in a year.
Answer Estimate the number of light bulbs in the average house ≈ 15 Estimate the average power of bulb ≈ 30 W (assuming some LED lights) Estimate the average time each bulb is on each day ≈ 5 hours 14
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Section 2
Energy transferred per household per day = power 3 time = 0.03 kW 3 15 3 5 = 2.25 kWh Over 1 year, total energy = 2.25 kWh 3 365 = 820 kWh Estimate the population of the UK as 65 million Estimate the number of households in the UK as 25 million For the UK as a whole energy used is 820 kWh 3 25 3 106 = 2.1 3 1010 kWh This is intended to be an order of magnitude estimate. We have made some assumptions. For example, the 30 W figure assumes some older lighting together with some new, energy-efficient lamps. You may have other data which allows you to check the plausibility of your estimate. For example, the electricity demand in 2014 was 4000 kWh per household. This would mean that lighting was 21% of the domestic electricity demand, which sounds reasonable.
3.2
Particles and radiation
3.2.1 Particles
3.2.1.1 Constituents of the atom The electron The electron was first identified during experiments using electrical discharge tubes (Fig 5). When the voltage is turned on, the screen at the end of the tube emits a glow. The glow was said to be caused by ‘cathode rays’. When the rays hit the screen, their energy is converted into light. This energy conversion, known as fluorescence, is aided by coating the inside of the screen with a phosphor, such as zinc sulphide. 0V
+5 kV
cathode
anode
evacuated glass tube
Essential Notes Electrons are thought to be fundamental particles; there is no evidence that the electron can be broken down into any other particles. Fig 5 An electrical discharge tube
low-voltage heater fluorescent screen
In 1897, J. J. Thomson discovered that he could deflect the rays using electric or magnetic fields (Fig 6). He balanced the two deflections so that the rays moved in a straight line. This allowed him to calculate the charge– mass ratio for the particles making up the rays. Thomson found that the charge–mass ratio was constant and concluded that cathode rays were tiny, negatively charged particles, now called electrons. 15
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Definition The charge–mass ratio of any charged particle (an electron, a proton, a nucleus or an ion) is its charge in coulombs divided by its mass in kilograms. This is also called its specific charge and has units of C kg–1. Fig 6 Electrostatic and magnetic deflection: (a) cathode rays curve towards the positive plate, showing they carry a negative charge; (b) a magnetic field makes the cathode rays move in a circular path
magnetic field, out of plane of paper
electron
++++++ ++++++
circular path
electron
parabolic path straight path
(a)
(b)
Electrons are emitted from the negative electrode, or cathode. More electrons are emitted if the cathode is heated. This effect is known as thermionic emission. After they are emitted the electrons accelerate towards the anode, finally hitting the screen, where they cause fluorescence. Thomson realised that the electrons were torn away from atoms in the cathode’s surface by the electric field. He suggested that atoms were composed of many electrons, moving in various orbits inside a positively charged cloud. This model of atomic structure is often referred to as the ‘plum pudding’ model of the atom (Fig 7). Fig 7 Simple picture of ‘plum pudding’ model of atom
_ +
_ _
Essential Notes The charge on an electron is very small. One coulomb is the amount of charge that flows past a point when 1 ampere of current flows for one second. If this was all carried by electrons, there would be 6.25 3 1018 electrons flowing past each second.
+ _
+
_
+ _
+ +
_
+
Properties of the electron Mass of the electron me 5 9.1 3 10231 kg Charge of the electron e 5 21.6 3 10219 C e 1.76 1011 C kg 1 Charge–mass ratio me
The nuclear atom The plum pudding model of the atom had to be abandoned following Rutherford’s scattering experiments in which alpha particles (helium nuclei) were scattered off atoms of gold. These showed that atoms were almost all empty space but with a very small, very dense positively charged nucleus
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Section 2
(see in AQA Physics A-level Year 2). In the nuclear model, all the positive charge of the atom and nearly all the mass are located in the central nucleus. Tiny, negatively charged electrons orbit this nucleus, rather like planets orbiting the Sun (Fig 8). Fig 8 Nuclear atom
positively charged nucleus
negatively charged electrons
Definition Protons are the particles that carry the positive charge in the nucleus.
In a neutral atom, that is an atom with no net charge, the number of protons in the nucleus is balanced by the number of electrons orbiting the nucleus. A hydrogen atom has one proton and one electron. Helium has two protons and two electrons, and so on through the Periodic Table, until the heaviest naturally occurring element, uranium, which has 92 protons and 92 electrons.
Definition The neutron is a particle with a mass almost identical to that of the proton, but with no electric charge.
Protons and neutrons are the only particles in the nucleus. They are referred to as nucleons. Electrostatic repulsion acts only between protons. The properties of protons, neutrons and electrons are summarised in Table 4. Proton
Neutron
Electron
symbol
p
n
e2
charge (C)
11.602 3 10219 0
mass (kg)
227
1.6726 3 10 6
1.6749 3 10
Table 4 Comparison of protons, neutrons and electrons
21.602 3 10219 227
9.109 3 10231
specific charge or charge–mass ratio (C kg21)
9.58 3 10
0
1.78 3 1011
charge (relative to proton)
1
0
21
mass (relative
1
1.0014 ≈ 1
5.45 3 1024
to proton)
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Nuclear composition The simplest atom is hydrogen. It has one proton in its nucleus and no neutrons. It has only one electron orbiting the nucleus (Fig 9). Helium has two protons and two neutrons in its nucleus, with two electrons in orbit around it. Fig 9 The hydrogen and helium atoms
_ electron
_ electron
+ proton
electron
proton neutron
_
hydrogen atom
helium atom
(not to scale)
Definition The proton number, or atomic number, is the number of protons in the nucleus and is given the symbol Z.
The atomic number of an atom is also the number of electrons in the neutral atom. This determines the chemical properties of the atom. The atomic number is used to place elements in the Periodic Table.
Essential Notes The mass number A and the atomic number Z are always whole numbers.
Definitions The nucleon number A is the total number of nucleons in the nucleus of an atom. It is also known as the mass number. The neutron number N is the number of neutrons in the nucleus.
The nucleon number is the number of protons plus the number of neutrons, so A Z N The nuclear composition of an atom can be described using symbols. For example, the most common form of carbon has six protons and six neutrons in its nucleus. It can be written as 126 C. The upper number is the nucleon number and the lower number is the atomic number. In general an element X with an atomic number Z and an atomic mass number A is written as ZA X. Using this system the hydrogen nucleus is represented as 11 H, and helium as 42He. The first five elements in the Periodic Table are listed in Table 5.
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Element
Symbol
hydrogen H
Atomic Neutron Atomic mass number, Z number, N number, A 1
0
1
helium
He
2
2
4
lithium
Li
3
4
7
beryllium Be
4
5
9
boron
5
6
11
B
Table 5 The first five elements in the Periodic Table
Isotopes Hydrogen usually has one proton, only, in its nucleus, but some hydrogen atoms have one or two neutrons as well (Fig 10). The different atoms are referred to as isotopes.
Definition Isotopes are forms of an element with the same proton number but with a different number of neutrons.
deuterium atom
hydrogen atom
tritium atom
electron
neutrons
proton
proton
Fig 10 Isotopes of hydrogen. The extra neutrons do not affect hydrogen’s chemical behaviour; for example, all three isotopes combine with oxygen to make water
electron
The different isotopes of an element have identical chemical behaviour because their atoms have the same number of electrons. Isotopes also have the same number of protons in their nucleus. The difference is simply in the number of neutrons. This makes some isotopes heavier than others. The isotope of carbon that has six protons and six neutrons in its nucleus, 12 6C, is referred to as carbon-12 (Table 6). Carbon-13 has six protons and seven neutrons; carbon-14 has six protons and eight neutrons.
Notes Isotopes are always the same element. For two isotopes, Z is the same, N and A are different.
Isotope
Atomic number, Z
Number of electrons
Neutron number, N
Mass number, A
% abundance
carbon-12
6
6
6
12
98.89
carbon-13
6
6
7
13
1.11
carbon-14
6
6
8
14
< 0.001%
Example
Table 6 Isotopes of carbon
Find the specific charge of the nucleus of an atom of carbon, 126 C.
Answer A nucleus of 126 C has 6 protons and 6 neutrons. To find the mass of the nucleus add together the masses of the constituents. (This is an approximation; the actual mass of a nucleus is slightly less than the 19
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total mass of its protons and neutrons. This is dealt with in AQA Physics A-level Year 2). Mass of 6 protons 5 6 3 1.6726 3 10–27 kg Mass of 6 neutrons 5 6 3 1.6749 3 10–27 kg Total mass 5 2.008 3 10–26 kg The charge on the nucleus is just due to the protons. Total charge 5 6 3 1.602 3 10–19 C 5 9.612 3 10–19 C Specific charge5 charge–mass ratio5
9.612 10 19 C
2.008 10 26 kg 4.787 4.787 107 C kg kg 1
Example The most commonly occurring isotope of oxygen is oxygen-16. This is written as 168O. Oxygen-15 is an isotope of oxygen that is used in positron emission tomography (PET) imaging to assist in medical diagnosis. (a) How many neutrons, protons and electrons are there in a neutral atom of oxygen-16? (b) How does a nucleus of the isotope oxygen-15 differ from a nucleus of oxygen-16? (c) Use ZAX notation to describe oxygen-15.
Answer (a) There are 8 protons and 16 − 8 = 8 neutrons in a nucleus of oxygen-16. Since the atom is neutral, the number of electrons must balance the number of protons, so there are 8 electrons. (b) Isotopes of the same element only differ in the number of neutrons in their nuclei. So an oxygen-15 nucleus must have one less neutron than an oxygen-16 nucleus. (c) 158O
3.2.1.2 Stable and unstable nuclei
Essential Notes The femtometre (fm) is a useful unit in nuclear physics. 1 fm 5 10215 m
What holds the nucleus together? Positive charges repel each other and at such short distances the electrostatic forces pushing the nucleus apart are very large. Another force acts inside the nucleus, known as the strong nuclear force or strong interaction. The strong nuclear force has a very short range; it has no effect at separations greater than about 5 fm (5 3 10215 m). The strong interaction is an attractive force until the separation is less than 1 fm, when the force becomes strongly repulsive (Fig 11). The overall effect of the strong interaction is to pull the nucleus together, but the repulsive action prevents it from collapsing to a point.
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Section 2
Force/kN
repulsive
proton–proton equilibrium separation
20
Fig 11 Force–distance graph for proton–proton pair (strong nuclear force plus electrostatic force)
500 N peak (note peak has been exaggerated)
10
Essential Notes
0 1
2
3
4
5
Separation/fm
–10
Electric force dominates at large separations Strong nuclear force dominates at small separations
–20 attractive
At distances of less than about 2 fm, the strong nuclear attraction between two protons is larger than the electrostatic repulsion (Fig 11) so the nucleus is held together.
For large nuclei there is a problem. The strong nuclear force acts over a much shorter range than electrostatic repulsion. It isn’t possible to get all the protons close enough together for the strong nuclear force to overcome the electrostatic repulsion. There has to be some other particle in the nucleus that helps to glue it all together and keep it stable. This is the neutron, discovered by James Chadwick in 1932. Neutrons exert a strong nuclear attraction on protons and on other neutrons. Because the neutron has no charge, it does not add any electrostatic repulsion. neutron–neutron or neutron–proton
Force/kN
repulsive 20 10
equilibrium separation
Essential Notes
0 1 –10
2
3
4
5
Separation/fm
–20 attractive
Fig 12 Force–distance graph for strong nuclear force without electrostatic repulsion
strong nuclear force only
Alpha decay Unstable nuclei decay by emitting high-energy particles and/or very high-frequency electromagnetic waves, known as gamma radiation. Some heavier unstable nuclei decay by emitting alpha particles. An alpha particle is a tightly bound group of two neutrons and two protons, exactly as in a helium nucleus. An alpha particle can therefore be represented as 4 2He. The alpha particle is emitted at high speed from the unstable nucleus, with an energy of a few MeV. The original (parent) nucleus is transformed into a different isotope by the decay, since it has lost two protons and two neutrons. The new (daughter) nucleus that remains may also be
Alpha particles tend to be emitted from large, unstable nuclei. An alpha particle is two protons and two neutrons bound together (just as in the helium nucleus, 42He). It escapes from the large (parent) nucleus, leaving a different (daughter) nucleus behind. For example when an isotope of uranium emits an alpha particle, leaving behind an isotope of thorium: 238 92 U
4 → 234 90 Th 1 2 He
This is sometimes written as: 238 92U
→ 234 90 Th 1 a 21
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Essential Notes Alpha emission is often accompanied by gamma emission.
Essential Notes Nuclear radiation has sufficient energy to remove one or more electrons from atom. This process is called ionisation.
unstable, which leads to further radioactive decays. An alpha decay can be represented as: A ZX
→
A −4 Z −2 Y
+ 42 He(+ )
An alpha particle has a positive charge due to its two protons. Because it is charged and relatively massive, an alpha particle travelling through a medium, such as air, has frequent collisions with the atoms of that medium. These collisions cause ionisation. Each ionisation requires an energy of between 5 and 25 eV, so the alpha particle rapidly loses energy and slows to halt in a few centimetres of air. The range is less in a denser medium: for example, an alpha particle is easily stopped by a thin solid such as a sheet of paper. An alpha particle can be deflected by a magnetic field. Because of its relatively large mass, a strong field is needed to create an observable deflection.
Beta decay Some radioisotopes decay by emitting a very high-speed electron, known as a beta particle. The beta particle is created when a neutron in the nucleus of an unstable atom decays into a proton and an electron. The proton remains in the nucleus but the electron is emitted at high speed. A particle called an antineutrino is also emitted.
Essential Notes An antiparticle is denoted by a horizontal line above the symbol for the particle, e.g. -p denotes an antiproton, and -n denotes an antineutron. The exception to this is the positron (the electron’s antiparticle), which is normally written as e1.
The nucleus has lost a neutron, but gained a proton, so the atomic (proton) number is increased by one but the nucleon number is unchanged. A ZX
→
A Z − 1Y
+
0 −1 e
+
Beta particles collide with atoms as they pass through a medium, like air. They travel further between collisions than alpha particles and so cause less ionisation. As a consequence they have a longer range than alpha particles, of up to several metres in air. Beta particles are negatively charged and so are also deflected by a magnetic field, but in the opposite direction to alpha particles. Some radioactive decays emit a positive beta particle, or positron. This involves the decay of a proton and can be written 1 p 1
→ 01 n 01 e e
Example Thorium-232 (Th-232) is a radioactive isotope with an atomic number of 90, which decays into radium-228 (Ra-228). Radium-228 is itself radioactive and decays into actinium-228 (Ac-228). Write decay equations for the two decays.
Answer The first decay reduces the mass (nucleon) number by 4, so it must be an alpha decay. 232 90 Th
→
228 88 Ra
+ 42 He
The second decay increases the atomic (proton) number by 1, so it must be a beta decay. 228 88 Ra
→
228 89 Ac
+
0 −1 e
+
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The tracks left by ionising radiation can be visualised by using a cloud chamber. Small droplets of liquid condense on the ions left in a particleâ&#x20AC;&#x2122;s wake as it travels through the chamber. The photograph shows cloud chamber tracks. Alpha particle tracks are short and thick, since the particles cause a lot of ionisation in a short distance. Beta tracks are thinner and longer.
Example Study the alpha-particle and beta-particle tracks in the photographs above and to the right. (a) Why do the beta-particle tracks suffer large deviations at the end of their tracks? (b) How would you expect the tracks to look if a strong magnetic field was applied to the chamber (in a direction perpendicular to the page)? Explain your answer. (c) Gamma radiation can also cause ionisation, but the range of gamma rays is much longer than alpha or beta particles. How might the track left by a gamma ray look?
Answer (a) Beta particles are electrons, so they have low mass. At the end of their path, their velocity has dropped and they have less momentum, so they are easier to deflect. (b) The tracks would curve. Alpha tracks would curve slightly in one direction. Those made by beta particles would curve more markedly in the other direction. The tracks curve in opposite directions as the alpha and beta particles carry opposite charges. The beta tracks curve more because beta particles have much less mass, and so need less force to make them change direction. (c) The tracks would be less dense, as ionisation occurs less often. The direction would not be affected by a magnetic field because gamma rays are not charged.
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Neutrinos The existence of the neutrino was first predicted by Wolfgang Pauli in 1930 in an attempt to understand beta decay. Unlike alpha particles, which are always emitted with a well-defined energy, beta particles are emitted with a range of energies (Fig 13). This seemed to contravene the principle of conservation of energy. If a certain amount of energy is transferred by each radioactive decay, why did the emitted beta particle have a range of possible energies? Pauli suggested that another particle, the neutrino, is also emitted in beta decay. The neutrino carries away the balance of the energy, so that the total energy of the decay is always constant.
Fig 13 (a) Typical alpha spectrum (b) Typical beta spectrum
Energy (a)
Number of beta particles
Number of alpha particles
The neutrino is represented by the symbol ne. The subscript ‘e’ stands for ‘electron’; these neutrinos are more properly referred to as electron neutrinos, because other types of neutrinos exist.
maximum energy
Energy (b)
The neutrino is a fundamental particle which carries no charge. For some years it was believed to have zero mass, but experiments now suggest that it has a small mass, much less than that of an electron.
Definition The neutrino is a fundamental particle with no charge. It has a very small mass. It interacts with other matter very weakly.
3.2.1.3 Particles, antiparticles and photons Antimatter The British physicist Paul Dirac predicted the existence of a particle with exactly the same mass as the electron, but with a positive charge. Indeed he suggested that all particles must have such antiparticles.
Definition An antiparticle is a ‘mirror image’ of a particle, of identical mass but opposite charge.
The first antiparticle was discovered in 1932 by Anderson, who was observing tracks in a cloud chamber made by cosmic rays (high-energy particles from space). He used a strong magnetic field to curve the paths of 24
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Section 2
high-energy electrons. Some tracks seemed identical to the electron tracks but curved in the opposite direction. These were tracks of an antielectron, now known as a positron. This is an example of antimatter. Another example of antimatter is the antineutrino, which is emitted during beta decay. A free neutron is itself unstable and decays into a proton and an electron and an antineutrino. Beta particles are electrons emitted when a neutron decays into a proton and an electron. The proton stays inside the nucleus but the electron is emitted at high speed, together with an antineutrino: 1 n 0
→ 11 p 01 e e
Table 7 summarises the properties of the fundamental particles and their antiparticles.
Particle
Mass/kg
electron
9.109 3 10231
positron
231
9.109 3 10
proton
1.6726 3 10
antiproton
227
Charge/C
Rest energy/MeV
–1.602 3 10219
0.511
11.602 3 10219
0.511
219
11.602 3 10
938
1.6726 3 10227
–1.602 3 10219
938
neutron
1.6749 3 10227
0
939
antineutron
1.6749 3 10227
0
939
neutrino
≈ 1 3 10–36
0
< 0.0001
–36
0
< 0.0001
antineutrino
≈ 1 3 10
Example It seems likely that a large proportion of the mass in the Universe has not yet been observed. This ‘missing mass’ is known as dark matter, as we are unsure what it is. Some scientists have suggested that it is made up of particles known as neutralinos. These particles would be difficult to detect but would have a mass equivalent to an energy of at least 300 GeV. (a) Explain why mass can be specified in terms of GeV, which is a unit of energy.
Table 7 Mass, charge and rest energy of particles and antiparticles
Essential Notes The joule (J) is the SI unit of energy, but it is too large for measuring energy on an atomic scale. We use the electron volt (eV), a much smaller unit. 1 eV 5 1.602 3 10–19 J 1 keV (kilo electron volt) 5 1000 eV 1 MeV (mega electron volt) 5 1000 000 eV 1 GeV (giga electron volt) 5 109 eV 1 TeV (tera electron volt) 5 1012 eV
(b) What is 300 GeV in units of joules?
Answer (a) Einstein’s special theory of relativity showed that mass (m) and energy (E) are linked through the equation E = m c2, so that m = E / c2. Mass has the same units as (energy / c2). Particle masses are often quoted in MeV/c2. (b) 1 eV is equivalent to 1.6 3 10−19 J. 300 GeV = 300 3 109 3 1.6 3 10−19 J = 4.8 3 10−8 J.
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