Physics secondary stage 2 by Mohammed Talaat

Book Sector

2011 - 2012

™HÉ£e

Arab republic of egypt Ministry of Education Book Sector

Table of Contents Foreword: Unit 1: Waves: Chapter 1 : Wave Motion Chapter 2 : Sound Chapter 3 : Light Unit 2: Fluid Mechanics: Chapter 4 : Hydrostatics Chapter 5 : Hydrodynamics Unit 3 : Heat: Chapter 6 : Gas Laws Chapter 7 : Kinetic Theory of Gases Chapter 8 : Cryogenics (Low Temperature Physics) Unit 4 : Dynamic Electricity and Electromagnetism: Chapter 9 : Electrical Current and Ohm's Law Chapter 10 : Magnetic Effects of Electric Current and Measuring Instruments. Chapter 11 : Electromagnetic Induction. Unit 5 : Introduction to Modern Physics: Chapter 12 : Wave Particle Duality Chapter 13 : Atomic Spectra Chapter 14 : Lasers Chapter 15 : Modern Electronics General Revision : Appendixes : Appendix 1 : Symbols and Units of Physical Quantities Appendix 2 : Fundamental Physical Constants Appendix 3 : Standard Prefixes Appendix 4 : Greek Alphabet Appendix 5 : Gallery of Scientists Appendix 6 : Selected Physics Sites on the Internet

1 : 79 2 23 47 81 : 132 82 117 134 : 182 135 159 173 184 : 271 185 202 231 273 : 382 274 306 324 350 383 : 404 407:418 406 409 411 412 413 418

intensively. It is targeted to use genetics, atoms and lasers in the computer of the future. It is a limitless world, enriched by imagination, where sky is the limit. The scientific progress is a cumulative effort. This collective endeavor has led to where we are today. A scholar of physics must be acquainted with such accumulated knowledge in a short time, so that he could add to it within the limited span of his life. In studying what others have found, we must skip details and trials, and extract the end results and build on them.A global view is, therefore, more important at this stage than being drowned in minute details that could be postponed to a later stage of study. This book is divided into 5 units. Unit 1 deals with waves, which are the basis of communication in the universe. (Chapter 1) deals with wave motion, (chapter 2) with sound, and (chapter 3) with light. Unit 2 deals with fluid mechanics, : hydrostatics (chapter 4) and hydrodynamics (chapter 5). Unit 3 deals with heat, where (chapter 6) deals with gas laws, (chapter 7) with the kinetic theory of gases and (chapter 8) deals with low temperature physics. Unit 4 treats electricity, where (chapter 9) covers the electric current and Ohmâ&#x20AC;&#x2122;s law, (chapter 10) covers the magnetic effects of electric current and measuring instruments, while (chapter 11) covers electromagnetic induction. Unit 5 gives an introduction to modern physics, where (Chapter 12) deals

Foreword Physics is the cornerstone of basic sciences. It deals with the understanding of nature and what goes around us, big and small in this universe. It is the root of all sciences. Interwined with it is chemistry which focuses on reactions between materials, biology which deals with living creatures, geology which is involved with the layers of the Earth, and astronomy which treats celestial objects. But in the end, physics remains the mother of all sciences and the basis for the tremendous present scientific and technological progress. Understanding physics means understanding the laws governing this universe. Such understanding has led to the current industrial development spearheaded by the West. The Arabs and Moslems were once the pioneers of civilization in the world when they realized the importance of understanding the laws of this universe. We owe them the discovery of most laws of physics centuries before the West. The foundations of medicine, physics, chemistry, astronomy, mathematics and music were all laid by Arab and Moslem scientists. In fact, understanding physics and its applications converts a poor, and underdeveloped society into an affluent and developed one. This has taken place in Europe, US, Japan and South East Asia. Computers, satellites, cellular (mobile) phones, and TV are all byproducts of physics. Genetics is currently being looked into

with wave particle duality, (Chapter 13) deals with atomic spectra, and (chapter 14) deals with lasers and their applications, while (chapter 15) covers modern electronics. Suzanne Mubarak Science Exploration Center has carried out the preparation, and the typing of manuscript as well as the design of the artwork. In the end, we want the student to take liking to physics. For this is the way to the future. We want the teacher to teach the subject of physics in an innovative way, to arouse the interest of the students by constantly referring to the use and applications of physics in the daily life. We hope that one day we will have great inventors and industrialists among today's students.

Committee for the preparation of this new version of the textbook. Prof. Mustafa Kamal Mohammad Yussef,Ph.D. Prof. Mohammad Sameh Said ,Ph.D. Mustafa Mohammad El-Sayed ,Ph.D. Tarik Mohammed Tala'at Salama,Ph.D. Karima Abdel-Alim Sayed Ahmad

electromagnetic waves spreading in space and the surrounding medium . When received by electrical signals and then to sound or even to an image. We can see water waves but we cannot see the radio, TV or mobile waves. However, we can detect them. Water waves are mechanical waves, so are sound waves and waves in

Unit 1:

the mobile antenna at the receiver, electromagnetic waves are transformed back into

vibrating strings. But radio, TV, and mobile waves are electromagnetic waves. Among are used in radiology. Mechanical waves require a medium to propagate through, while (em) waves do not require a medium. They can propagate in space.

Waves

these electromagnetic (em) waves, there are, for example, light waves and X-rays which

Mechanical Waves Mechanical waves require the following :2 ) a disturbance transmitted from the source to the medium. 3 ) a medium that carries a vibration. There are many forms of vibrating sources : 1 ) a simple vibrating pendulum (Fig 1 - 2). 2 ) a tuning fork (Fig 1 - 3).

Chapter 1:

1 ) a vibrating source.

3 ) a vibrating stretched wire (or string) (Fig 1- 4).

Wave Motion

4 ) a plumb (bob) attached to a vibrating spring (Yoyo) (Fig 1 - 5).

Wave Motion Chapter 1:

Wave Motion

Chapter 1 Overview :

Many of us enjoy watching waves on the surface of water pushing a fishing float or a boat up and down, or even making waves by throwing a pebble in a pond or still water. Each pebble becomes a source of disturbance in the water, spreading waves as concentric circles (Fig 1-1). Hence, waves are disturbances that spread and carry along energy.

Waves

Fig (1 -1)

Waves spreading from a point source

Unit 1:

Waves are not only water waves. There are, for example, radio waves. We often hear the announcer say: "This is Radio Cairo on the medium wave 366.7 m". Also, TV stations transmit both sound and image in the form of waves which are received by the aerial (antenna) . Such waves are transformed into electrical signals in the receiver, where they are eventually converted back to sound (audio) and image (video). Also, the mobile phone runs on waves. Sound signals are transformed into electrical signals then into 2

A pendulum

Fig (1 -3)

A tuning fork

Fig (1 -4)

A vibrating string

Amplitude (A) (meter): is the maximum displacement of the vibrating object or the distance between two points along the path of the object, where the velocity at one point is maximum and zero at the other.

Complete Oscillation: is the motion of a vibrating body in the interval between the

Unit 1:

Chapter 1:

Fig (1 -2)

instants of passing by one point along the path of its motion twice successively with

Waves

amplitude

Frequency (Î˝) (Hertz or Hz): is the number of complete oscillations made by a vibrating body in one second.

Fig (1 -5) Yoyo

Periodic Time (T) (seconds): is the time taken by a vibrating body to make one

To studyoscillation, vibrations, orwetheneed define relevantbody physical quantities such as: complete timetotaken by some the vibrating to pass by the same point displacement, oscillation, periodic time and follows:and the along the pathamplitude, of motioncomplete twice successively with motion in frequency the same as direction

same displacement.

(1-1) body at any instant from its rest Displacement (meter): is the Î˝distance = 1 of a vibrating

T position or its equilibrium origin. It is a vector quantity.

Simple Harmonic Motion: called a simple harmonic motion, e.g., a swing (Fig 1-6) or a simple pendulum (Fig 1-7). The vibration starts from point "a" then increases to a positive maximum at "b" then to zero at "a" then to negative maximum at "c" then to zero at "a". and the cycle is repeated continually (Fig 1-7 a ).

Fig (1-6)

Wave Motion

A vibrational motion in its simplest form is

Chapter 1:

Unit 1:

starting point of motion.

Waves

rest position

motion in the same direction and same displacement, i.e., at the same phase, relative to the

A swing as an example of a simple harmonic motion

Wave Motion Chapter 1:

Fig (1 -3)

A pendulum

Fig (1 -4)

A tuning fork

Fig (1 -2)

A vibrating string

Fig (1 -3)

A pendulum

A tuning fork

Fig (1 -4)

A vibrating string

rest position amplitude

rest position

Waves

Wave Motion Chapter 1: Waves Unit 1:

Fig (1 -2)

Fig (1 -5) Yoyo

amplitude

To study vibrations, we need to define some relevant physical quantities such as: displacement, amplitude, complete oscillation, periodic time and frequency as follows:

Fig (1 -3) Fig (1 -4) Fig (1 -5) Displacement of astring vibrating body at any instant from its rest A tuning fork (meter): is theAdistance vibrating

nit 1:

position or its equilibrium origin. It is a vector quantity. Yoyo

To study vibrations, we need to define some relevant physical quantities

displacement, amplitude, complete oscillation, periodic time and frequency as follo

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Wave Motion

Collapse of Tacoma bridge (USA) due to the wind causing the vibration of the bridge at the natural (resonant) frequency of the bridge

Chapter 1:

Fig (1-9)

Waves

At a certain frequency, the amplitude of the mechanical vibration may get out of hand, e.g., the crushing of a glass cup due to nearby sound waves (Fig 1-8), and the collapse of Tacoma bridge (USA) due to strong winds in November 1940 (Fig 1-9). This condition is called resonance. It is the cause of the collapse of many buildings. It occurs when a simple harmonic motion is set at the natural (or resonant) frequency of the building. Similar to mechanical resonance, there is also Fig (1-8) electrical resonance which is the basis of tuning a radio A glass cup crushed due to nearby sound waves or a TV receiver to a certain station, where one out of many electrical signals picked up by the aerial is amplified and made to coincide with the resonant frequency of the amplifier in the receiver when tuned to that particular station.

Unit 1:

Resonance

Complete Oscillation: is the motion of a vibrating body in the interval between the instants of passing by one point along the path of its motion twice successively with motion in the same direction and c same displacement, b i.e., at the same phase, relative to the a starting point of motion.

rest position

Frequency (Î˝) (Hertz or Hz):Fig is the of complete oscillations made by (1-7number a) a vibrating body in one second.

Displacement of a pendulum bob as time goes by

Periodic Time (T) (seconds): is the time taken by a vibrating body to make one complete oscillation, or the time taken by the vibrating body to pass by the same point along the path of motion twice successively with motion in the same direction and the same displacement. (1-1) Î˝=1 T

Simple Harmonic Motion: called a simple harmonic motion, e.g., a

Unit 1:

swing (Fig 1-6) or a simple pendulum (Fig 1-7). The vibration starts from point Fig "a" (1-7 thenb)

of a pendulum bobthen at different phases A, D are of the increases toDisplacement a positive maximum at "b" same phase (same displacement and direction)

to zero at "a"B,C then maximum at same direction but not the aretonotnegative of the same phase (the Fig (1-6) displacement) "c" then to zero at "a". and same the cycle is A swing as an example of repeated continually (Fig 1-7 a ).

Wave Motion

A vibrational motion in its simplest form is

Chapter 1:

maximum and zero at the other.

Waves

distance between two points along the path of the object, where the velocity at one point is

Unit 1:

Wave Motion

Amplitude (A) (meter): is the maximum displacement of the vibrating object or the

a simple harmonic motion

Unit 1:

Waves

Chapter 1:

Wave Motion

Longitudinal Waves:

Imagine a mass â&#x20AC;&#x153;mâ&#x20AC;? on a smooth horizontal surface attached to one end of a spring whose other end is attached to a vertical wall. If we pull the mass in the direction of the spring and let it go, the mass moves around its rest position in an oscillatory motion toward the spring and away (Fig 1-10). This is a simple harmonic motion. If we draw the curve that the center of gravity of the mass makes with respect to its rest position, we will obtain a sine wave (Fig 1-11). This is what distinguishes a simple harmonic motion from any other type of motion.

rest position

pulling the spring

releasing the spring

Fig (1-10) A vibrating spring

Fig (1-11) A sine wave resulting from a simple harmonic motion

vertical displacement

Unit 1:

distance

Waves

Fig (1-13 b) Vertical displacement as a sine wave

Fig (1-14)

A pulse resulting from part of a simple harmonic motion

Wave Motion

Chapter 1:

You can do this experiment yourself by using a long stretched rope. The far end is attached to a vertical wall while the near end is in your hand. When you move your hand up and down in the form of a pulse, you note that the wave spreads in a pulse form along the rope. This is known as a traveling wave (Fig 1-14).

spreading along a stretched rope

Wave Motion

Thus, a vibrating source making a simple harmonic motion may generate a wave propagating at velocity “v”. Each particle of the medium performs, in turn, a simple harmonic motion about its equilibrium position. An example of this motion is the longitudinal waves of sound in air.

Transverse Waves: Imagine a mass “m” attached to a vertical spring. A long horizontal taut (stretched) rope is also attached to this mass at the near end, while the other (far) end of the rope is attached

Chapter 1:

to a vertical wall. When the mass “m” performs a simple harmonic motion in the vertical direction, then the near end of the rope performs the same motion. Consequently, the following parts of the rope do the same thing successively. Then the motion transfers horizontally along the rope in the form of a wave at velocity “v”, while the other parts of the rope oscillate vertically in a simple harmonic motion about their rest positions. This wave is called a

Unit 1:

Waves

transverse wave (Fig 1-13 ).

Fig (1-13 a) Vertical displacement in a simple harmonic motion

longitudinal wave

A vibrating spring forming a transverse wave

In conclusion, we may classify mechanical waves into two types: 2 ) Longitudinal waves In transverse waves, the particles of the medium oscillate about their equilibrium positions in a direction perpendicular to the direction of the propagation of the wave. In longitudinal waves, the particles of the medium oscillate about their equilibrium positions along the direction of the propagation of the wave. The work done by the oscillating source is converted to the particles of a string (or a

Wave Motion

1 ) Transverse waves

Chapter 1:

Fig (1-17)

Waves

A vibrating spring forming a

Unit 1:

Fig (1-16)

stretched rope) in the form of potential energy stored as tension in the string and kinetic 13

Wave Motion Chapter 1:

Fig (1-15) A train wave spreading in a taut (stretched) rope due to a continuous simple harmonic motion at the near end

Unit 1:

Waves

A wave may also be continuous (called a traveling wave train) as long as the simple harmonic motion of the source keeps on(Fig 1-15). The stretched rope may be replaced by a spring in which a longitudinal wave (Fig 1-16) or a transverse wave (Fig 1-17) may be generated . We conclude that as a source oscillates , the particles of the medium oscillate successively in the same way. The vibration transfers first from the source to the particle of the medium next to it, then into the one connected to it, then into the following ones and so on. Thus, the vibration or disturbance forms a wave, since the wave is nothing but a disturbance (or energy) on the move along which energy is carried through . 12

wavelength amplitude

crest rest position

direction of vibration

Wavelength in a transverse wave

Thus, the wavelength is the distance between two successive points of the same phase

Waves

Fig (1-19)

Unit 1:

direction of propagation

(Fig 1-21). Alternatively, it is the distance which the wave travels during one periodic time (Fig 1-22).

Wavelength in a longitudinal wave

The number of waves passing by a certain point along the wave path in one second is

Wave Motion

displacment

Chapter 1:

Fig (1-20)

called frequency. 15

Chapter 1:

Wave Motion

energy manifested in the vibration of the particles of

direction of wave propagation

the string. Referring to (Fig 1-18), the points at maximum upward displacement in the positive direction are called crests, while the points of maximum downward displacement are called troughs. Observing any part of a vibrating string carrying a transverse wave, we find that it has one crest and one trough during one complete oscillation .

Frequency (谓) (Hertz) and wavelength (位) (meter): The distance between two successive crests or two successive troughs in a transverse wave is called

Waves

wavelength (Fig 1-19). Similarly, the distance between two successive contractions (compressions) or two successive rarefactions in a longitudinal wave is called wavelength (Fig 1-20).

Fig (1-18)

A piece of foam floating on the top of

Unit 1:

Thus, we may represent the wavelength by either a wave (crest) or at bottom (trough) of the two distances (AC) and (BD)(Fig 1-19 ). It is to be noted that the two successive pairs of points (A,C) and (B, D) move in the same way at the same time. We say they have the same phase, i.e.,the same displacement in the same direction. 14

Unit 1: Waves

Fig (1-22b)

A train of waves at velocity “v” generated by a vibrator

h T

pν = 1 T 1 pν

v = λν This is a general relation for all types of waves. In all cases , within a periodic time “T” a wave travels a wavelength. Frequency is the number of oscillations in one second or the number of wavelengths traveled by a wave propagating in a certain direction in one second.

Wave Motion

Chapter 1:

The relation between frequency,wavelength and velocity of propagation: If a wave travels at velocity “v”, a distance equal to the wavelength “λ” , then the wave takes time equal to the periodic time “T” to travel this distance.

Waves

Chapter 1:

Wave Motion

rest position

at the end of 1/4 period

at the end of 1/2 period

at the end of 3/4 period

amplitude

Fig (1-21)

The distance after each one full vibration completed in one period T is the wavelength

at time t

Unit 1:

at time t + T

at the end of one period

Fig (1-22a)

The distance which a wave moves in a periodic time T is the wavelength

In a Nutshell

Frequency =

1 Periodic time

. . .

Wave Motion

Chapter 1:

It is also the number of waves passing by a certain point along the path of a wave in one second. Periodic time “T” is the time taken by a continuously vibrating body to perform one complete oscillation, or the time taken by a continuously vibrating body ( e.g. a simple pendulum ) to pass by a point along its path twice successively with motion in the same direction. Mechanical waves are either: 1 ) transverse waves. 2 ) longitudinal waves. Transverse waves are waves in which the particles of a medium oscillate about their equilibrium positions in a direction perpendicular to the direction of propagation of the wave. Longitudinal waves are waves in which the particles of a medium oscillate about their equilibrium positions along the same path of propagation of the wave. Transverse waves comprise crests and troughs in succession.

Waves

from its rest position, or the distance between two points along the path of the oscillating object where the velocity at one point is maximum and at the other is nil. A complete oscillation is the movement of a continuously vibrating body ( e.g. a simple pendulum) is the interval between the instants of time as it passes by a certain point along its path twice successively with motion in the same direction. Frequency “ν” is the number of complete oscillations produced by a vibrating object in one second and is equal to the inverse of the periodic time.

Unit 1:

.A wave is a disturbance which spreads and carries energy along. .Displacement is the distance of an object at any instant from its rest(equilibrium) position. . The amplitude of oscillation “A” is the maximum displacement of an oscillating object

Chapter 1:

Wave Motion

Examples:-

1) If the wavelength of a sound wave produced by a train is 0.6 m and the frequency is 550 Hz,what is the velocity of sound in air? Solution:v=λν v = 0.6 x 550 = 330 m/s

2) If the number of waves passing by a certain point in one second is 12 oscillations and the wavelength is 0.1 m, calculate the speed of propagation. Solution:v=λν v = 12 X 0.1 = 1.2 m/s

3) Light waves propagate in space at speed 300 000km/s (3x108m/s), and the wavelength of light is 5000 A˚ .What is the frequency of this light ? 0

Waves

1 Angstrom(A ) =10-10 m

Solution:c = v = 3 x 108 m/s λ = 5 x 103 x 10-10 = 5 x10-7 m

Unit 1:

c= λν

3 x 108 = 5 x 10-7 x ν

pν =

3 x 10 8 14 Hz = 6 x 10 -7 5 x 10

Questions and Drills

III) Essay question: Deduce the relation between frequency, wavelength and velocity of wave propagation.

Chapter 1: Wave Motion

IV) Put a tick sign (√) next to the right choice in the following : 1) The relation between the velocity of propagation of the waves “v” in a medium , its frequency and wavelength is : a) v = λν b) v = ν / λ c) v = λ d) none (there is no correct answer) ν 2) Transverse waves are waves consisting of : a) Compressions and rarefactions b) Crests and troughs c) Crests and troughs, where the particles of the medium move short distances about their equilibrium positions in a direction perpendicular to the direction of propagation. d) Compressions and rarefactions, where the particles of the medium move short distances about their equilibrium positions along the direction of propagation of the wave . 3) If the wavelength of a sound wave produced by an audio ( sound producing) source is 0 . 5 m , the frequency is 666 Hz ,then the velocity of propagation of sound in air is : a) 338 m / s b) 333 m / s c) 330 m / s d) 346 m / s

Waves

II) Complete: a) Displacement is ...... b) Amplitude of oscillation is ...... c) Complete oscillation is ...... d) Periodic time is ...... e) Frequency is ......

Unit 1:

I) Define Wave - Transverse Wave - Longitudinal Wave - Wavelength

Wave Motion Chapter 1: Waves Unit 1: 20

. Longitudinal waves comprise compressions and rarefactions in succession . . Wavelength is the distance between two successive points along the direction of .

propagation of the wave, where the phase is the same (same displacement and same direction). The relation between frequency, wavelength and velocity of a wave is given by: v = 位谓

Wave Motion Chapter 1: Waves Unit 1: 22

4) If the velocity of sound in air is 340 m/s, for a sound of frequency (tone) 225 Hz , the wavelength(m) is : a) 4/3 b) 3/4 c) 20 d) 3/2 5) Light of wavelength 6000 AË&#x161;(1AË&#x161; = 10-10m ) propagates in space at velocity 300 x 103km/s, its frequency is: 10

(b) 4 x 10 Hz

(d) 5 x 10 Hz

(a) 4 x 10 Hz (c) 5 x 10 Hz

14 12

6) Two waves whose frequencies are 256 Hz and 512 Hz propagate in a certain medium , the ratio between their wavelengths is a) 2/1 b) 1/2 c) 3/1 d) 1/3

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A bat (ultrasonic radar)

A bat (Fig 2-1) does not see by its eyes. It transmits ultrasonic waves and detects the

Waves

Fig (2-1)

Unit 1:

How does a bat see ?

surroundings by the echo. It works in this fashion as a radar or rather sonar (radar using ultrasonic waves). Learn at Leisure

Ultrasonic waves (not heard by human ear) are used to image an embryo. They are

considered the safest. Ultrasonic imaging depends on the reflections of sound (Fig 2-2 ).

Sound

Fig (2-2)

Chapter 2:

Imaging the embryo

Use of ultrasonic waves for imaging an embryo

Sound

Chapter 2:

Chapter 2

Sound

Overview :

Sounds and tones are produced due to the vibration of objects. Such vibrations travel through air or any other medium in all directions. When these vibrations reach the ear, they are transmitted through the auditory nerve, then the brain translates them to sounds and tones

Reflection and Refraction of Sound: Firstly: Reflection of Sound:

Unit 1:

Waves

When a loud sound is produced at a suitable distance from a wall or a mountain, a sound is heard back resembling the original one. It is generated due to the reflection from the wall or the mountain, appearing as if it were coming from behind the wall or the

mountain. This sound is called echo. Hence, echo is the repetiton of sound produced due to reflection. Sound waves propagate in air in the form of concentric spheres of successive

compressions and rarefactions, whose center is the source of the sound . If these waves are obstructed by a large obstacle, they are reflected back also in the form of concentric spheres of compressions and rarefactions,whose center appears as if it lay behind the reflecting surface and at a distance equal to the distance of the original source from that surface. According to the laws of reflection: 1) the angle of reflection is equal to the angle of incidence. 2) the incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in one plane normal to the reflecting surface. Note that: the sound ray is a straight line indicating the direction of propagation of the sound wave.

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source

Waves

The velocity of sound in the air depends on the temperature of the air, since sound waves propagate in hot air faster than in cold air. When sound travels between two layers of air of different temperatures, it undergoes refraction (Fig 2 - 4). The figure illustrates the decrease of sound intensity as heard by an observer at a certain distance due to the refraction of sound upwards (leaking away), due to the increase of temperature on a hot day. At night, the sound is heard for longer distances due to the decrease of the temperature of the air adjacent to the surface of the Earth compared to layers above. Therefore, sound at night is refracted more toward the surface of the Earth.

Unit 1:

Why does sound travel easier at night than during the day ?

observer cold air

at daytime

ground hot air

at night

ground

Fig (2-4)

Sound

cold air

Chapter 2:

hot air

Sound travels easier at night

Sound

Secondly: Refraction of sound

When sound falls on a surface between two media, part of it is reflected back to the first

medium according to the laws of reflection, while the rest is transmitted to the second medium, deviating from its original path (Fig 2-3 ). Refraction of sound - upon transmitting

Chapter 2:

from one medium to another - depends on the velocity of sound in these two media. In other words :

sinφ v1 = sinθΘ v2

(2-1)

This means that when the velocity of sound in the first medium v1 is greater than the

velocity of sound in the second medium v2, the sound refracts nearer to the normal, i.e., φ > θ

and vice versa. It is to be noted that the velocity of sound in gases decreases as their density increases, while in liquids and solids the velocity is affected by another factors,which is more effective than denisty.

Waves

Normal

v1 Hot air

Unit 1:

Cold air

Fig (2-3)

Refraction of sound

Normal

Such sources may be obtained, for example, using two speakers for the same electrical

compressions, while the dashed arcs represent the maxima of rarefactions. The distance between any two successive arcs or any two successive dashed arcs is the wavelength.

Due to the combination of two waves of equal frequency and amplitude (Fig 2-6), some

Unit 1:

source (Fig 2-5). The connected arcs in the figure represent the positions of the maxima of

points or regions exist where the compressions of the first source intersect the

compressions of the second source, and the rarefactions of the first source intersect the m is an integer . Therefore, at such positions we have constructive interference, where the

intensity of the wave increases (Fig 2-6 a). There are also points or regions where the compressions of the first source intersect rarefactions of the second source or vice versa.

Waves

rarefactions of the second source. In both cases, the path difference must equal m 位, where

Therefore, the path difference equals (m + 1 ) 位 , where m is an integer, then we have 2

destructive interference and the intensity of the wave diminishes to zero (Fig 2-6 b). Learn at Leisure

The experiment of sound interference is similar to the ripple tank experiment, where two sources vibrate and produce mechanical waves. These waves interfere producing regions of constructive interference and regions of destructive interference ( Fig 2-7).

Interference fringes between two waves

Fig (2-7 b)

Sound

Fig (2-7 a)

Chapter 2:

Can we see the interference of sound waves ?

Interference pattern

Chapter 2:

Sound

Interference and diffraction of sound Firstly: Interference of sound

Interference is a combination of two waves or more of the same frequency, amplitude, and direction of propagation. Interference may be constructive (strengthing the intensity) or destructive (weakening the intensity or eliminating it altogether). Fig(2-5) demonstrates the interference of sound waves, which are longitudinal waves consisting of compressions and rarefactions. The two sources S1 and S2 emit waves of the same frequency and amplitude. source 1

Waves

lines of constructive interference (maximum sound intensity)

Fig (2-5)

lines of destructive interference (minimum sound intensity)

Formation of constructive and destructive fringes due to two sound sources first wave second wave of the same phase

Unit 1:

source 2

resultant wave

Fig (2-6 a)

Constructive interference of the waves first wave second wave at 180Ë&#x161; phase shift

resultant wave

Fig (2-6 b)

Destructive interference of the waves

sinθΘ v2

Superposition of waves

(Fig 2-10 d).

displacement

resultant wave

first wave

second wave

Sound

distance

Chapter 2:

Waves combine such that the resultant wave has intensity equal to the sum of intensities of the individual waves (Fig 2-9 a). When the frequencies are slightly different while the amplitude is the same, this combination (superposition) leads to beats (Fig 2-9 b , c , d ). If we move a tight wire or rope such that one pulse is generated (Fig 2-10 a ), then this pulse continues until it reaches the far end. If this end is attached to a sliding ring, then the reflected wave is positive ,i.e., in the same direction as the incident pulse. Whereas if this end is fixed such that it cannot move, then the reflected wave is always reversed (Fig 2-10 b). When the reflected wave meets the incident wave, constructive interference is produced in the first case (Fig 2-10 c ),and destructive interference is produced in the second case

Waves

where φ is the angle of incidence and θ in the angle of refraction, v1 is velocity of sound in the first medium and v2 is the velocity of sound in the second medium. 4) Sound waves of equal frequency and amplitude interfere to produce regions of constructive interference (increase of intensity) and regions of destructive interference (decrease of intensity). 5) Sound diffracts when passing through a small aperture or a sharp edge, provided that the discontinuity is comparable to the wavelength.

Unit 1:

3) It refracts upon traveling from one medium to another due to the difference in velocity (Fig 2 - 4) : sinφ v

Fig (2-9 a)

The resultant wave is the sum of two waves

Chapter 2:

Sound

Secondly: Sound Diffraction What is meant by diffraction? Diffraction is a change (or bending) of the wave path when passing through a slit or an aperture, small enough compared to the wavelength, or when passing by sharp edges in the same medium. We observe sound diffraction in our daily life. For example, if you speak in one room and a window or a door is open, someone in the next room may overhear you , although he is not standing right next to the window or the door (Fig 2-8). The sound intensity in the neighboring room depends on the position of a listener in that room, being maximum of course directly next to the opening. The spreading of sound in the neighboring room is attributed to diffraction .

Waves

door

audio source

spreading of sound in the form of concentric spheres

room Fig (2-8)

Unit 1:

Diffraction of sound through a door opening to a nearby room

Sound as a wave motion From above, we conclude that sound has wave properties :

1) It propagates in a medium in straight lines in all directions. 2) It undergoes reflection when falling on a surface, and the angle of reflection equals the angle of incidence.

Fig (2-9d)

string and hand at rest far end free to move (attached to a sliding ring)

hand moves up pulling the string upwards

Unit 1:

reflected pulse

far end fixed to the wall

reflected pulse hand moves down

Waves

middle of the pulse

Fig (2-10b)

hand at rest

Reflection of a pulse

Fig (2-10a)

firstly

Unit 1:

Waves

Harmonic tones resemble two nearly equal interleaved combs incident pulse

Formation of an incident pulse

thirdly

32 secondly

incident pulse

fourthly

far end free to move (attached to a sliding ring)

firstly

Combination of two positive pulses propagating in opposite directions far end fixed to the wall reflected pulse

reflected pulse

secondly

Fig (2-10b)

Reflection of a pulse

thirdly firstly

fourthly

Fig (2-10d)

Unit 1: Waves Chapter 2: Sound

Fig (2-10c)

Combination of two pulses one positive and the other thirdly negative moving in opposite directions

secondly

fourthly

Fig (2-10c)

1 -1 1 0

-2

-1 -2

Fig (2-9c)

destructive interference

Fig (2-9b)

Formation of harmonic tones with distance Formation of harmonic tones with distance distance

time

distance

0 -1

Fig (2-9b)

2 1 0 -1 -2

Fig (2-9d)

1 -1

Harmonic tones resemble two nearly equal interleaved combs Harmonic tones resemble two nearly equal interleaved combs destructive interference

Fig (2-9b)

destructive destructive interference string and hand at destructive rest interference interference string and hand at rest Fig (2-9c)

Fig (2-9c)

Formation of harmonic tones with time Formation of harmonic tonesupwith time hand moves pulling the hand moves up pulling the string upwards

string upwards

Unit 1:

ation of harmonic tones with distance Formation of harmonic tones with distance

Fig (2-9d)

hand moves down

middle of the pulse

hand at rest

Fig (2-9d)

aves

Harmonic tones resemble two nearly equal interleaved combs Fig (2-10a) Harmonic tones resembleFig two(2-10a) nearly equal interleaved combs Formation of an incident pulse Formation of an incident pulse

Fig (2-9c)

Formation of harmonic tones Formation of harmonic tones with time

Fig (2-9b)

Unit 1:Chapter Waves 2:

Chapter 2:

destructive interference

-1

Waves

1 -1

Sound

-1

distance

Sound

-1

time distance 1

time

string and hand at rest

hand moves up pulling the hand moves up pulling the string upwards

Chapter 2:

Waves

A spiral spring oscillating from both ends in reciprocity

pulley

vibrator antinode

weights

node

Chapter 2:

( string, spiral spring, or rope) in one direction and a continuous train of waves reflected in the opposite direction. These two wave trains interfere, giving a pattern of particles of the waves reflected from tensionlocalized, i.e., medium which appears notfixed moving or to the left but moving both ends to the right, tension perpendicularly to the wire. This effect may be visualized by moving a spiral spring, (string Fig (2-11c) or rope) up and down in reciprocity from both ends (Fig 2-11 a), or fixing it from one end A string fixed at both ends and and moving the other end in a simple harmonic motion (Fig 2-11 b), or pulling a string pulled in the middle fixed from both ends - from the middle, such that waves are transmitted in both directions Melde's Experiment and get reflected, and hence interfere (Fig 2-11 c). Melde's experiment best illustrates standing waves on strings or wires. The apparatus is s in shown (Fig 2-12). It consists of a vibrating connected to a soft string whose length two wavesource, ctions e ir d e it s o p p o end of the string passes over a smooth pulley and is ranges from 2 to 3 meters. The other connected at its free end to appropriate weights. When the source vibrates, a wave train is node node produced in the string, which reflects upon reaching the pulley.The reflected and incident antinode node waves are combined (superposed orantinode superimposed) to form standing waves. These standing waves have nodes and antinodes (Fig 2-13), provided the source frequency has a certain Fig (2-11a) value compared to the string (wire) length.

Waves

generator

Fig (2-11b) Fig (2-12)

Sound

Unit 1:

waves generated at the middle and Standing waves are formed when there is a propagating continuous intrain waves in a tight wire bothofdirections

Unit 1:

Sound

Standing ( Stationary ) Waves:

A spiral spring oscillating in a simple harmonic motion at Melde's one end apparatus while fixed at the other

sliding ring)

reflected pulse

far end fixed to the wall

Chapter 2:

Sound

Standing ( Stationary ) Waves: Fig (2-10b)

Standing waves are formed when Reflection there is a of continuous a pulse train of waves in a tight wire Fig (2-10b) ( string, spiral spring, or rope) in one direction and a continuous train of waves reflected in Reflection of a pulsegiving a pattern of particles of the the opposite direction. These two wave trains interfere, medium whichfirstly appears localized, i.e., not moving to the right, or to the left but moving thirdly perpendicularly to the wire. This effect may be visualized by moving a spiral spring, (string firstly or rope) up and down in reciprocity from both ends (Fig 2-11 a), or fixingthirdly it from one end and moving secondly the other end in a simple harmonic motion (Fig 2-11 b), or pulling a string fixed from both ends - from the middle, such that waves are transmittedfourthly in both directions secondly and get reflected, and hence interfere (FigFig 2-11(2-10c) c).

Combination in positive pulses avofes two two wFig (2-10c) tions irecopposite propagating directions opposite din Combination of two positive pulses propagating in opposite directions node

Unit 1:

Waves

firstly firstly

node

antinode

Fig (2-11a)

secondly secondly

A spiral spring oscillating from both ends in reciprocity

thirdly thirdly

fourthly antinode Fig (2-10d) node

Combination of two pulses one positive and the other Fig in(2-10d) negative moving opposite directions Combination of two pulses one positive and the other negative moving in opposite directions Fig (2-11b)

A spiral spring oscillating in a simple harmonic motion at one end while fixed at the other

fourthly

Fig (2-11c)

Melde's Experiment

pulley

vibrator Fig (2-14)

Nodes and antinodes

weights

Fig (2-12)

The distance between two successive nodes or two Melde's apparatus successive antinodes is half a wavelength

Chapter 2: Sound

Fig (2-15)

generator

Waves Chapter 2:

Melde's experiment best illustrates standing waves on strings or wires. The apparatus is shown (Fig 2-12). It consists of a vibrating source, connected to a soft string whose length ranges from 2 to 3 meters. The other end of the string passes over a smooth pulley and is connected at its free end to appropriate weights. When the source vibrates, a wave train is produced in the string, which reflects upon reaching the pulley.The reflected and incident waves are combined (superposed or superimposed) to form standing waves. These standing waves have nodes and antinodes (Fig 2-13), provided the source frequency has a certain value compared to the string (wire) length.

Unit 1: Waves

A string fixed at both ends and pulled in the middle

Sound

A spiral spring oscillating from both ends in reciprocity

Chapter 2:

Waves

Unit 1:

Fig (2-11a)

antinode

node Fig (2-13a)

A vibrating string showing a standing wave pattern

Fig (2-11b)

A spiral spring oscillating in a simple harmonic motion at one end while fixed at the other

waves generated at the middle and propagating in both directions Fig (2-13b) waves reflected standing wavefrom patterns with the tension tension Variation of both fixed ends ratio of the string length to the wavelength as time goes on

Unit 1:

ranges from 2 to 3 meters. The other end of the string passes over a smooth pulley and is connected at its free end to appropriate weights. When the source vibrates, a wave train is produced in the string, which reflects upon reaching the pulley.The reflected and incident

Waves

The node is the position where theFig amplitude (2-11c)of the vibration is zero, while the antinode is the position where the amplitude of theatvibration is maximum. Nodes and antinodes are A string fixed both ends and pulled in the middle spaced at equal distances apart (Fig 2-14). The wavelength of a standing wave is twice the distance between any two successive Melde's Experiment antinodes or two successive nodes. As the weight in the experiment is increased, the tension Melde's best illustrates standingofwaves on strings or frequency wires. Theforapparatus in theexperiment wire is increased, so is the velocity propagation, and the the same is shownwire (Figlength 2-12).is Italso consists of (Fig a vibrating increased 2-15). source, connected to a soft string whose length

Unit 1:

Waves

(2-2)

where m is the mass per unit lengh of the sting material Since the arc small Vibration of isstrings:

Sound Chapter 2:

Sound

third overtone A string vibrating in this way may produce different tones. (4th harmonic) The fundamental tone (the lowest frequency the string may Fig (2-17) vibrate at on its own) will produce one antinode and two Formation of harmonics nodes (as above). The string may vibrate in many other ways. We may divide the wire into segments. For example, we may have two segments (3 nodes and 2 antinodes), or 3 segments (4 nodes and 3 antinodes) (Fig 2-17). When the string has one segment (λ = 2 l), it produces the fundamental frequency (first harmonic). When the string has two segments, it produces the first overtone

Chapter Waves 2:

If we have a tight stringθfixed = l/ 2on both ends and pulled R fundamental tone from the middle then released to vibrate freely, we note (1st harmonic) 2 2FT l/ 2 FT l that the particles of the ∴ string = Mv to =perpendicularly Fc = vibrate R R R its length, i.e., perpendicularly direction of FT l to Fthe 2= FT / m = T is = transverse, propagation of the wave. vSuch a vibration M /l M first overtone i.e., transverse waves propagate on both halves of the (2nd harmonic) v= /m (2-2) T fixed ends, string in bolh directions until they reachFthe then theymareis reflected backunit in the opposite direction. By where the mass per lengh of the sting material multiple reflections, the incident and reflected waves, second overtone . (3rd harmonic) interfere producing standing waves, where an antinode is Vibration of strings: formed at the middle, and a node is formed at each of the If we have a tight string fixed on both ends and pulled fixed ends. fundamental tone from the middle then released to vibrate freely, we note third overtone (1st harmonic) A string vibrating in this way may produce different tones. that the particles of the string vibrate perpendicularly to (4th harmonic) The fundamental tone (the lowest frequency the string may its length, i.e., perpendicularly to the direction of Fig (2-17) vibrate at on its own) will produce one antinode and two propagation of the wave. Such a vibration is transverse, first overtone Formation of harmonics nodes (as above). The string may vibrate in many other ways. i.e., transverse waves propagate on both halves of the (2nd harmonic) We may divide the wire into segments. string in bolh directions until they reach the fixed ends, For example, we may have two segments (3 nodes and 2 antinodes), or 3 segments (4 nodes and then they are reflected back in the opposite direction. By 3 antinodes) (Fig 2-17). When the string has one segment (λ = 2 l), it produces the fundamental multiple reflections, the incident and reflected waves, second overtone frequency (first harmonic). When the string has two segments, it(3rd produces the first overtone harmonic) interfere producing standing waves, where an antinode is formed at the middle, and a node is formed at each of the fixed ends.

Waves Unit 1:

FT / m

Sound

Chapter 2: Waves Unit 1: 36

Fig (2-13a)

A vibrating string showing a standing wave pattern

Fig (2-13b)

Variation of standing wave patterns with the ratio of the string length to the wavelength as time goes on

The node is the position where the amplitude of the vibration is zero, while the antinode is the position where the amplitude of the vibration is maximum. Nodes and antinodes are spaced at equal distances apart (Fig 2-14). The wavelength of a standing wave is twice the distance between any two successive antinodes or two successive nodes. As the weight in the experiment is increased, the tension in the wire is increased, so is the velocity of propagation, and the frequency for the same wire length is also increased (Fig 2-15).

Questions and Drills 1) State the laws of reflection of sound.

2) Show how to demonstrate the interference of sound. 3) Explain : Sound is a wave motion. II) Define:

2) a node.

3) the wavelength of a standing wave.

III) Complete :

1) The velocity of propagation of a transverse wave in a string is given by:

Waves

1) an antinode.

Unit 1:

I) Essay questions

v = ......................................

2) The fundamental frequency produced in a string is given by:

Î˝ = .....................................

3) Keeping tension constant, the frequency of a string is.......................proportional to its length . proportional to .................................

5) Keeping the length of the string and tension constant, the fundamental frequency is inversely proportional to ................................

IV. Choose the right answer:

1) Standing waves are formed by the combination of two waves propagating

Chapter 2:

4) Keeping the length of a string constant, the fundamental frequency is directly

a) in the same direction.

c) in opposite directions without necessarily having equal frequency and intensity. d) in two perpendicular directions.

Sound

b) in opposite directions provided they have equal frequency and intensity.

Light

oscillating magnetic field

Chapter 3:

oscillating electric field

Fig (3 – 2) An electromagnetic wave consists of an electric field and a magnetic field perpendicular to each other and to the direction of propagation of the wave

Reflection and refraction of light

Light propagates in straight lines in all directions, unless met by an obstructing medium.

If so, it undergoes reflection, refraction and partial absorption depending on the nature of in optical density - then part of light is reflected and the rest is refracted, neglecting absorption. We note from Fig (3-3) that each of the incident ray, reflected ray and refracted incidence all lie in one plane perpendicular to the separating surface. In the case of reflection : the angle of incidence is equal to the angle of reflection In the case of refraction : the ratio between the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium is equal to the ratio

angle of incidence

incident ray

reflected ray

first medium (air) second medium (glass)

angle of θ refraction refracted ray

Fig (3 – 3)

Unit 1:

ray as well as the normal to the surface at the point of

Waves

the medium. When a light ray falls on a surface separating two media - which are different

Reflection and refraction of light

Unit 1:

Light

Chapter 3 Overview :

Light is an indispensible form of energy. The Sun is the main natural source of energy to

us. The energy from the Sun is almost divided between heat and light. Thanks to the light

Waves

from the Sun, the plants perform photosynthesis, hence make their own food. Man depends on plants and animals, which in turn feed on plants.

We have seen before that sound has a wave nature. It propagates from a source causing

mechanical waves in the medium. Light also has a wave nature. It is subject to the laws of

reflection, refraction, interference and diffraction. But light is different from sound in that it does not require a medium to propagate in. Light is part of an extensive range of waves

called electromagnetic waves, which all travel at a constant speed in space equal to 3 x 108 m/s, while varying in frequency. This range of waves is called the electromagnetic spectrum

Chapter 3:

(Fig 3-1). It includes, for example, radio waves, infrared, visible light, ultraviolet,

X- rays and Gamma rays. They all share common features. They are all transverse

electromagnetic waves, but of different frequencies (and wavelengths).

Electromagnetic waves consist of time varying electric and magnetic fields. Both

oscillate at equal frequency at the same phase, and are perpendicular to each other and to the direction of propagation (Fig 3-2), hence called transverse waves.

Light

Fig (3 â&#x20AC;&#x201C; 1) Electromagnetic spectrum

2) The refraction of light is attributed to the difference in the speed of light, when light is

Light

transmitted from one medium to another. c

v= n (3-3)

where v1 is the speed of light in the first medium, v2 is the speed of light in the second medium. Substituting equation (3-3) in (3-1), we have: n 2 sin φ = n1 sin θ

n1 sin φ = n2 sin θ

(3-4)

This is Snell’s law

Chapter 3:

v1 n = n2 v2 1

The absolute refractive index for the medium of incidence times the sine of the angle of

incidence is equal to the absolute refractive index of the medium of refraction times the sine of the angle of refraction.

wavelengths, since the absolute refractive index varies with wavelength. Therefore, white light may be decomposed into its components. This can be seen, for example, in soap bubbles.

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Why refraction ?

Waves

3) We can use refraction in analysing a bundle of light into its components of different

approaches the normal. This resembles a car in which one of its wheels goes through a muddy soil, while the other is free on the paved road. The wheel that goes into the mud

becomes slower. Therefore, the car changes direction (Fig 3-4). The opposite is also true, as in refraction from a more dense material to a less dense one. The refracted ray deviates

Unit 1:

If light falls from a less dense medium onto a more dense medium ,the refracted ray

away from the normal (Fig 3-5).

of the speed of light in the first medium to the speed of light in the second medium. This

Unit 1:

ratio is constant for these two media, and is called relative refractive index from the first medium to the second medium, denoted by 1n2 : Sin sin q = v 1 = n Sin sin Î¸O v 2 1 2

(3-1)

Waves

Important facts

1) The speed of light in space ''c'' is one of the physical constants of the universe and is equal to 3 x 108 m/s. It is larger than the speed of light in any medium ''v''. The ratio c = n is called the absolute refractive index for the medium and is always > 1. v c (3-2) n= v

The absolute refractive indices of some materials are listed below

Chapter 3: Light 50

refractive index

Air Water Benzine Carbon tetrachloride Ethyl alcohol Crown glass Rock glass Quartz Diamond

medium

1.00293 1.333000 1.501000 1.461000 1.361000 1.52000 1.660000 1.4850000 2.419000

incident reflected ray ray

paved road

Unit 1:

muddy soil

glass glass

air

muddy soil

Waves

Fig (3 – 4)

paved road

Fig (3 – 5)

Refraction from a less dense medium to a more dense medium

Refraction from a more dense medium to a less dense medium

Examples

1) If a light ray falls on the surface of a glass slab whose refractive index is 1.5 at an angle 30˚, calculate the angle of refraction. Solution

∴

Chapter 3:

Sin s q n= s Sin s θO

s 30 ∴ 1.5 = Sin

Sin s θO

sin θO =

0.5 = 0.333 1.5

∴

θ = 19˚ 28´

Light

2) If the absolute refractive index of water is 4 and glass 3 , find 3 2 a) the relative refractive index from water to glass b) the relative refractive index from glass to water. Solution

a) The relative refractive index from water to glass

Unit 1:

reaching the double slit have the same phase, hence, are coherent (having the same frequency, amplitude and phase). Waves emanating from S1 and S2 are cylindrical and spread toward the observation screen C. On such a screen, waves coming from S1 and S2 combine and produce an interference pattern, appearing as a sequence of bright and dark straight parallel regions, which are the interference fringes (Fig 3-7). The distance between two successive fringes ∆y is given by:

Waves

∆y ∆Χ=

λLR d

(3-5)

where λ is the wavelength of the monochromatic source, R is the distance between the double slit and the observation screen and d is the distance Fig (3 – 7) Interference between S1 and S2. fringes Therefore, this experiment may be used to determine the wavelength for any monochromatic light source. Learn at Leisure

Interpretation of interference in Thomas Young’s experiment

Chapter 3:

If light were not to manifest interference, we would

obtain fringes as in Fig (3-8a). We may interpret the

formulas of constructive and destructive interference

monochromatic Firest light soure slit

pattern in Young’s experiment as follows. If the distance of the screen from the double slit R is large relative to the distance d between the two slits of the

Light

double slit, then we may consider the two rays r 1, r2

emanating from the double slit on their way to the

observation screen as nearly parallel. If θ is the

inclination angle of the two rays, then the path

difference between these two rays is ∆r (Fig 3-9). This 54

dark fringes b

Second slit

bright fringes

Fig (3 – 8)

Fringes (a) resulting from

interference (b) if there were no interference

Airy’s disk central bright spot parallel rays first secondary bright spot

Chapter 3:

When a monochromatic light falls on a circular aperture in a screen, we expect that light should form a circular bright spot on an observation screen, considering that light propagates in straight lines. But careful examination of the bright spot (called Airy’s disk), i.e., studying the light intensity, reveals the existence of bright and dark fringes (Fig 3-12).

Light

Light Diffraction

Fig (3 – 12) Fig (3-13) demonstrates diffraction from a rectangular slit, while Fig (3-14) shows the diffraction pattern at a sharp edge of matter. In general, diffraction is evident when the wavelength of the wave is comparable to the dimensions of the aperture, and vice versa (Fig 3-15). In fact, there is no big difference between the mechanisms of interference and diffraction. In both cases, combination (superposition) of waves is involved (Fig 3-16). center of the central bright fringe

Waves

Diffraction in a circular aperture

light intensity

light ray

Fig (3 –13b)

Fig (3 – 13a)

Diffraction from a rectangular aperture

Distribution of light intensity on a screen with the succession of fringes resulting from diffraction from a rectangular aperture

Unit 1:

slit

Interpretation of diffraction

Fig (3-17) shows a plane wave advancing toward a screen in which there is a

rectangular slit. At an appropriate distance, there is a white parallel observation screen.

Light

Learn at Leisure

rays. In this case, wavelets have the same phase. Constructive interference results in a bright fringe (Fig 3-17 a).

Fig (3 â&#x20AC;&#x201C;17a) Formation of the central bright fringe

Fig (3â&#x20AC;&#x201C;17b)

Waves

observation screen at a point corresponding to the center of the slit as a lens collimates the

Unit 1:

sources of secondary wavelets. Light emanating from these secondary sources fall on the

Chapter 3:

According to the wave theory, points on the wave front at the slit may be considered as

Succession of the fringes

Unit 1:

plane edge

Waves

sphere

Chapter 3:

the dimensions of the obstacle are small incomparison with λ

Fig (3–14)

Diffraction patterns from different obstacles

the dimensions of the obstacle are medium in comparison with λ

the dimensions of the obstacle are large in comparison with λ

Fig (3 – 15)

circular aperture 0.8d

Light

incident wave

Fig (3 – 16) Diffraction is the interference of secondary wavelets from different points in the slit

0.2d

diffraction is more evident from a narrow slit in comparison with λ

observation screen

(a) 58

razor edge

(b)

Diffraction places a limit on resolving details in an image. This limit is called the limit of resolution. If we have two point sources, and light is emitted from each through a circular aperture, then each source forms separate fringes. When the distance between the two sources decreases, the fringes get closer, and it becomes difficult to identify one from the other. It is found that the angle between the centers of the two fringes under this condition is given by the approximate relation ∆θ = λ , where D is the aperture diameter. D Thus, the ability to resolve two small objects is inversely proportional to the aperture diameter and directly to the wavelength (Fig 3-19). In the case of a microscope, the lens takes the place of the aperture and the wavelength limits the ability of the microscope to distinguish between small objects. As λ decreases, we can discern details that were not seen before. This is the advantage of the electron microscope (Chapter 12).

Light

Resolving power

Chapter 3:

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Fig (3–19)

b) the resolution is nil as the objects get too small and too close together.

c) if the two sources are drawing near to the observer, then he can separate them visually

Resolving power

Unit 1:

a)as the two sources get closer to each other it becomes difficult to separate them visually because of diffraction

Waves

∆

d) bacteria appearing through an electron microscope and not appearing through an optical microscope

Light as a wave motion Unit 1: Waves

From above, we conclude that light 1) propagates in straight lines . 2) reflects according to the laws of reflection . 3) refracts according to the laws of refraction . 4) light interferes, and as a result, light intensity increases in certain positions (bright fringes)and diminishes to zero in other positions (dark fringes). 5) light diffracts if obstructed by an obstacle . These are the same general properties of waves. Hence, light is a wave motion

Total reflection and the critical angle

When a light ray travels from an optically dense medium (as water or glass) to a less dense medium (as air), then the refracted ray deviates away from the normal (Fig 3-20). As the angle of incidence increases in the more dense medium (of high absolute refractive index), the refraction angle in the less dense medium (of low absolute refractive index) increases.

Chapter 3:

ϕ θθ

(a)

(b)

(c)

(d)

Fig (3–20)

Incidence of light from a more optically dense medium to a less optically dense medium

Light

A point is reached when the angle of incidence in the more dense medium approaches

a critical value φc when the angle of refraction in the less dense medium reaches its

maximum, which is 90˚. Then,the refracted ray becomes tangent to the surface. 62

b) In the case of water

Unit 1:

Sin sin q c =

1 1 = = 0.7518 n1 1.33

φc = 48˚ 45´

2)Using the information in the example above, find the critical angle for light falling

from glass onto water

Solution Waves

Using Snell’s law, n2 sin 90˚ = n1 sin φc 1.33 x 1 = 1.6 sin φc

sin q c = 1 x 1.33 = 0.8313 1.6

Some Applications of Total Reflection Chapter 3:

I. Fiberoptics (Optical Fibers)

Fig (3-21) shows an optical fiber. It is a

thread-like tube of transparent material. When light falls at one end, while the angle of incidence is greater than the critical angle, it undergoes successive multiple reflections until it emerges from the other end (Fig 3-22). Fig (3-23) shows a bundle

Optical fibers contain the rays despite bending

Light

of fibers which can be bent while containing light so that light can be made to travel in non straight

lines to parts hard to reach otherwise.

Fibers can be used to transmit light without

much losses, and are widely used nowadays. 64

Fig (3–21)

Fig (3– 22)

Optical fibers

Light

The Bear’s fur

The bear’s fur does not provide the bear with thermal isolation only, but the fur hairs are massive optical fibers which reflect ultraviolet rays. The fur appears white (Fig 3-26). because visible light reflects inside the hollow transparent optical fibers, while the skin itself absorbs all rays reaching it. Therefore, it is actually black. Learn at Leisure

Fig (3–26) The bear’s fur

How an optical fiber works

Chapter 3:

Learn at Leisure

If we have a hollow tube and look through it to see a bright object on the other end, then the object is easily seen. If the tube is bent, then the object cannot be seen. Yet, we may be fibers, while the ray is incident at an angle greater than the critical angle, then multiple reflections take place, until the ray emerges from the other end, despite the bending of the fibers.

Waves

able to see it, if we use reflecting mirrors in the path of rays. Similarly, by using optical

The critical angle between glass (refractive index 1.5) and air is 42˚. Therefore, a glass prism whose angles are 90˚, 45˚, 45˚ is used to change the path of the rays by 90˚ or 180˚. Such a prism may be used in optical instruments, as periscopes in submarines and binoculars in the field (Fig 3-27).

Unit 1:

II. The reflecting prism

Unit 1: Waves

Fig (3–24b) Endoscopes

Chapter 3:

Fig (3–24c)

Fig (3–24d)

An image of esophogus by optical fibers

Endoscope lens

Light Fig (3–25)

Optical fibers used to carry electrical signals

of the Earth are heated, their density decreases. Hence, their refractive index becomes smaller than that of the upper layer. If we follow a light ray reflected off a palm tree, this

Light

This can be explained as follows. On very hot days, the air layers adjacent to the surface

normal, and keeps deviating taking a curved path. When its angle of incidence reaches more than the critical angle, it undergoes total reflection and the curve goes up. To the eye of the observer, the ray appears as if coming from under the surface of the Earth. The observer thinks that there is a pond.

Chapter 3:

ray is traveling from an upper layer to one below. Therefore,if refracts away from the

Deviation of light in a prism taking the path " bc ", until it falls on the surface xz and emerges in the direction "cd" (Fig 3-29). We notice from the figure that the light ray in the prism refracts twice. As a result, the ray deviates from its original path by an angle of deviation α . The angle of deviation α is the angle subtended by the directions of the extension of the

Waves

When a light ray such as " a b " falls on the surface xy of a prism, it refracts in the prism

incident ray and the emerging ray. If the angle of incidence is φ , the angle of refraction on the first surface is θ , the angle of incidence on the second surface is φ , the angle of 1

emergence is θ and the apex angle of the prism is A, we note from the geometry (Fig 3-29).

Unit 1:

b) a reflecting prism changing the light path by 180˚

Waves

a) a reflecting prism changing the light path by 90˚

A reflecting prism

Prisms are better for this purpose than reflecting surfaces, first, because light totally

reflects from such a prism, while it is seldom to find a metallic reflecting surface whose efficiency is 100%. Secondly, a metallic surface eventually loses its luster,and hence its ability to reflect decreases. This does not happen in a prism. The surface at which light rays fall on a prism or the surface from which the rays emerge may be coated with non reflective layer of material like cryolite (Aluminum fluoride and magnesium fluoride) whose refractive index is less than that of glass, to avoid any reflection losses on the prism, even little as they are.

Chapter 3:

III.Mirage

This is a familiar phenomenon observable on hot days, as paved roads appear as if wet (Fig 3-28 a). Also, an image of the sky is made on desert plains, where palm trees or hills appear inverted giving the illusion of water (Fig 3-28 b).

Light

Fig (3–28b)

Fig (3–28a)

Paved roads appear as if wet

Fig (3–27)

c) prisms in binoculars

Reflection of the sky in the desert gives the illusion of water

Substituting for φ and θ we find that the refractive index can be determined from the

relation

α sin | 0+ A Sin 2 n= A Sin sin 2

Light

sin q 0 n = Sin Sin sin e 0

(3 - 10)

Experiment to determine the ray path through a glass prism and its refractive index:

Tools:

An equilateral triangular prism (A = 60˚), pins, a protractor, a ruler.

Chapter 3:

But

Procedure:

and mark its position with a fine pencil line. Place two pins such that one of them (a) is very close to one side and the other (b) is about 10 cm from the first. The line joining them represents the incident ray. Look at the other side of the prism to incident ray see the image of the two pins, one behind the other. Place two other pins c and d between the prism and the eye such that they appear to be in line with the two pins a and b,i.e., the four pins appear to be in one straight line.Locate the positions of the four pins.

path of the ray (a b c d) from air to glass to air again.

emerging ray

Fig (3– 31)

Determination of light

Unit 1:

2) Remove the prism and the pins, join b and c to locate the

Waves

1) Place the glass prism on a sheet of drawing paper with its surface in a vertical position

ray path in a prism

minimum angle of deviation depends also on the wavelength. Thus, if a beam of white light falls on a prism set at the minimum angle of deviation, then the emerging light disperses into spectral colors as illustrated in Fig(3-32). From this figure, it is concluded that the

Light

Note also that the refractive index (n) depends on up the wavelength λ, then the

into which the white light is dispresed are arranged by the order: red, orange, yellow, green, blue, indigo and violet.

The thin prism: A thin prism is a triangular glass prism. Its apex angle is a few degrees and is in the position of minimum deviation:

Chapter 3:

violet ray has the largest deviation ( maximum refractive index). The visible spectral colors

Thus,

and

|α0+ A 2

and A 2

are small angles.

α α sin | 0+ A ≅= | 0+ A Sin 2 2

(radians)

sin A ≅= A (radians) Sin 2 2

Substituting from (3-10), we find that the refractive index of the material of the thin

prism is determined by

n = α 0 +A A

∝0 = A (n - 1)

(3 - 11)

Unit 1:

Since:

Waves

|α + A Sin sin 0 2 n= A Sin sin 2

(3 - 12)

3) Extend dc to meet extended ab .The angle between them is the angle of deviation α.

Unit 1:

4) Measure the angle of incidence φ1, the angle of refraction θ1, the inner incidence angle 5)

φ2, the angle of emergence θ2 and the angle of deviation (α).

Repeat the previous steps several times changing the angle of incidence and tabulate

the results.

Waves

angle of Angle of inner angle of Angle of the angle of Angle of θ emergence prism A incidence φ refraction θ incidence φ 2 deviationα 1 2 1

Find the minimum angle of deviation and the corresponding angles φ and θ . ˚ ˚ - Then obtain the refractive index from equation (3-10).

The dispersion of light by a triangular prism: It has been proven previously that in the case of

Chapter 3:

minimum deviation, the refractive index may be determined from the relation:

α Sin s | 0+ A 2 n= Sin s A 2 where (n) is the refractive index,αo is minimum angle of

deviation, and (A) is the angle of the prism.

Light

Since the angle of the prism is constant for a certain prism,

so the minimum angle of deviation changes by changing the

refractive index. As the refractive index increases, the minimum angle of deviation increases and vice versa.

Fig (3–32)

A prism disperses the spectrum

.Light refracts between two media because of the different velocities of light in the two

media v1 & v2

.Laws of refraction of light :

1) The ratio between the sine of the angle of incidence in the first medium, to the sine of the angle of refraction in the second medium is constant, and is known as the refractive index 1n2 sin φ 1n2 = sin θ

Light

1) Angle of incidence = Angle of reflection 2) The incident ray, the reflected ray, and the normal to the reflecting surface at the point of incidence, all lie in one plane perpendicular to the reflecting surface.

Chapter 3:

.Laws of reflection of light :

In a Nutshell

in the second medium 2) The incident ray, the refracted ray, and the normal to the surface of separation at the point of incidence, all lie in one plane normal to the surface of separation. • The relative refractive index between two media is the ratio between the velocity of light in the first medium v1 and the velocity of light in the second medium v2 1 n2 =

Vv1 Vv 2

Waves

where φ is the angle of incidence in the first medium, and θ is the angle of refraction

Cc Vv where c is the velocity of light in free space and v is the velocity of light in the medium. , • Snell s law : n=

n1sinφ = n2 sinθ

Unit 1:

• The absolute refractive index for a medium is given by :

ll 75

Dispersive Power Unit 1:

When white light falls on a prism, the light disperses into its spectrum due to the

variation of the refractive index with wavelength. (α0)r = A (nr - 1)

(α0)b = A (nb - 1)

where nr is the refractive index for red and nb for blue.

Waves

Subtracting,

(α0)b - (α0)r = A (nb - nr)

( 3 - 13)

The LHS represents the angular dispersion between blue and red. For yellow (middle

between blue and red), the angle is :

(α0)y = A (ny - 1)

Chapter 3: Light 74

(3 - 14)

where ny is the refractive index for yellow. If (α0)y is the average of (α0)r and (α0)b, then

nyis the average of nr and nb. We defineωα as : ωα= ( α0)b − ( α0)r = n b - n r (α0)y ny - 1

( 3 - 14)

where ωα is the dispersive power, and is independent of the apex angle.

|α 0+ A 2 A Sin s 2

s Sin

Chapter 3:

Light

• Refractive index of the prism material is given by:

where n is the refractive index, α is the minimum angle of deviation. ˚ • The minimum angle of deviation in a thin prism is : α = Α (n - 1)

˚ • The angular dispersion for a thin prism is :

(α0)b - (α0)r= Α (nb -nr)

• where (α )b is the minimum deviation angle of the blue ray, and (α )r is the minimum 0

(α ) (α ) ωtα= _ 0 bb< _ 0r r (α _ 0y)y

ωtα = n b - n r

n y -1

where (α0)y is the minimum angle of deviation of the yellow light, and ny is the refractive

index for the yellow light.

Unit 1:

• The dispersive power :

Waves

deviation angle of the red ray.

• The distance between two successive similar fringes (either bright or dark) is : Unit 1:

∆y =

λR d

where λ is the wavelength of light employed, R is the distance between the double slit

and the screen, and d is the distance between the two slits. • Light is a wave motion.

• The critical angle is the angle of incidence in the more dense medium, corresponding to

Waves

an angle of refraction in the less dense medium equal to 90˚.

• The absolute refractive index is equal to the reciprocal of the sine of the critical angle when light travels from this medium into air or vacuum. n =

1 Sin s qc

• Total internal reflection takes place when the angle of incidence in the more dense medium is greater than the critical angle.

• The mirage is a phenomenon that can be explained by total internal reflection.

Chapter 3: Light 76

• The angle of the apex of the prism is given by: • The angle of deviation is given by:

A = θ1 + φ2

α = (φ1 + θ2) - A

where φ1 is the angle of incidence θ2 is the angle of emergence

• In the case of minimum deviation :

φ1 = θ2 = φ0 θ1 = φ2 = θ 0

Unit 1:

Questions and Drills I) Essay questions

1) Explain why light is considered to be a wave motion . 2) Describe an experiment to demonstrate the interference of light. 3) Explain how mirage is formed.

Waves

II) Define :

a) the relative refractive index between two media. b) the absolute refractive index for a medium. c) the critical angle. d) the angle of deviation.

III) Complete :

Chapter 3:

a) The distance between two successive bright fringes is given by .................................... b) Snell’s law states that : ...................................................... c) The angle of deviation in a thin prism is given from relation :..................... d) The dispersive power is: ......................................................

Light

IV) Choose the right answer : 1) When light reflects : a) the angle of incidence is less than the angle of reflection. b) the angle of incidence is greater than the angle of reflection. c) the angle of incidence is equal to the angle of reflection. d) there is no right answer above . sin φ 2) When light refracts, the ratio ,where φ is the angle of incidence and θ is the angle sin θ of refraction is:

a) constant for the two media. b) variable for the two media.

Unit 1:

7) The angle of incidence in a medium is 60˚ and the angle of refraction in the second medium is 30˚. Then the relative refractive index from the first to the second medium is : a) 3 b) 2 c) 1 d) 2 2

8) An incident ray at an angle 48.5˚ on one of the faces of a glass rectangular block (n =1.5), the angle of refraction is :

Waves

a)20˚

b)30˚

c) 35˚

d) 40˚

9) In an experiment it was found that the minimum angle of deviation is 48.2˚ Given that the angle of the prism is 58.8˚, the refractive index of the material of the prism is : a) 1.5

b) 1.63

c) 1.85

d) 1/1.85

10) If the critical angle for a medium to air is 45˚, then the absolute refractive index is : a) 1.64

b)2

d) 2

Chapter 3:

11) A thin prism has an angle of 5˚. Its refractive index is 1.6. It produces a minimum angle of deviation equal to :

a) 5˚

b) 6˚

c) 8˚

d) 3˚

12) A ray of light falls on a thin prism at an angle of deviation 4˚ and its apex angle 8˚.Its refractive index is :

Light 80

c)1.7

a) 1.5

b)1.4

c) 1.33

d) 1.6

Unit 2:

Chapter 4

Hydrostatics

Fluids are materials which can flow. They are liquids and gases. Gases differ from

liquids in compressibility. Gases are compressible, while liquids are incompressible. Thus,

liquids occupy a certain volume, while gases can fill any volume they occupy, i.e., the volume of the container.

Density

Fluid Mechanics

Overview

Density is a basic property of matter. It is the mass per unit volume (kg / m 3) (4 -1)

where Vol is the volume Density varies from one element to another due to:

2) difference in interatomic or intermolecular distances or molecular spacings.

We know that bodies of less density float over more dense liquids. The following table shows density for different material.

Hydrostatics

1) difference in atomic weights

Chapter 4:

m lĎ = Vvol

densities. Normal blood density is 1040 kg / m3 – 1060 kg / m3. High density indicates

higher concentrations of blood cells and lower concentrations indicate anemia.

The normal urine density is 1020 kg / m3. In some diseases, salts increase and cause the

Pressure Pressure at a point is the average force which acts normal to unit area at that point. If

force F is normal to a surface of an area A, then the affected pressure P on the surface is determined by the following relation:

F A

(4 - 3)

Chapter 4:

Learn at Leisure

Fluid Mechanics

urine density to increase.

Unit 2:

2) Measuring density is used in clinical medicine, such as measuring blood and urine

Elephant’s foot vs human foot

Because the pressure is the force per unit area, the

pressure due to a pointed high heel is greater than the

pressure due to an elephant’s foot, since the area of the pointed heel is very small (Fig 4 – 1).

Meaning of pressure

Hydrostatics

Fig (4–1)

Hydrostatics

Material

Unit 2:

Fluid Mechanics

Chapter 4:

Solids: Aluminumâ&#x20AC; Brass Coper Glass Gold Ice Iron Lead Platinum Steel Suger Wax

Liquids: Ethyl Alchol Benzene Blood Gasoline

Density kg/m3 2700 8600 8890 2600 19300 910 7900 11400 21400 7830 1600 1800 790 900 1040 690

Material Kerosene Mercury Glycerin Water Gases: Air Ammonia Carbon dioxide Carbon mono oxide Helium Hydrogen Nitrogen Oxygen

Density kg/m3 820 13600 1260 1000 1.29 0.76 1.96 1.25 0.18 0.090 1.25 1.43

The ratio of density of any material to that of water at the same temperature is called the relative density of the material (no units). The relative density of a material, is equal to : =

the density of the material at a certain temperature the density of water at the same temperature

(4 - 2)

= the mass of a certain volume of matter at a certain temperature the mass of the same volume of water at the same temperature

Applications to density

1)Measuring density is of great importance in analysis, such as measuring the density of the electrolyte in a car battery. When the battery is discharged, the density of the electrolyte (dilute sulfuric acid) is low due to chemical reaction with the lead plates and the formation of lead sulfate. When the battery is recharged, the sulfate is loosened from the lead plates and go back to the electrolyte, and the density increases once more. Thus, measuring the density indicates how well the battery is charged.

pressure Po

Fluid Mechanics

pressure P

Fig (4 – 3) Pressure increases with liquid depth

Fig (4 – 4)

where g the acceleration due to gravity. The pressure due to the

liquid from under the plate x (acting upwards) must be : Ahρg P= F = A A ... P = hρg

(4 − 4)

Calculation of the pressure of a column of liquid

Hydrostatics

Fg = Ahρg

Chapter 4:

To calculate the pressure (P), we imagine a horizontal plate x of area A at depth h inside a liquid of density ρ (Fig 4 – 4). This plate acts as the base of a column of the liquid. The force acting on the plate x is the weight of the column of the liquid whose height is h and whose cross section is A. Because the liquid is incompressible, the force resulting from the liquid pressure must balance with the weight of the column(of the liquid. The volume of this column is Ah and its mass Ahρ, hence its weight Fg is given by :

Unit 2:

acted upon by two forces: its weight downwards and the force due to the pressure of the liquid around it. As the depth of the liquid increases, the pressure increases (Fig 4 – 3).

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 86

Pressure at a point inside a liquid and its measurement. If you push a piece of foam under water and let it go, it will rise and float. This indicates that water pushes the immersed foam by an upward force. This force is due to the pressure difference across this piece of foam. At any point inside a liquid, the pressure acts in any direction. The direction of the force on any surface is normal to that surface. The pressure on a body is the same as the pressure on a volume of the liquid that has the same shape of the body in case this body were removed. In other words, the liquid occupying the same size which a body would occupy is PdA PdA

PdA a) pressure inside a liquid is perpendicular to any surface inside the liquid

b) pressure is perpendicular to the surface of an immersed body at every point

c) pressure on the surface of an object is equal to the pressure on the surface of a similar size of the liquid of the same volume and shape

Fig (4 â&#x20AC;&#x201C; 2)

d) in a certain size of a liquid there is equilibrium between two forces: the weight of the liquid and the pressure due to the remainder of the liquid.

Pressure in a liquid

Let us take a U - shaped tube filled with an appropriate amount of water. Let us add a quantity of oil in the left branch of the tube, until the level of oil reaches level C at a height mix. Let the height of the water in the right branch be hw above level AD (Fig 4 – 7).

Because the pressure at A = pressure at D

... Pa + ρogho = Pa + ρwghw where Pa is the atmospheric pressure, ρo the density of oil, ρw the density of water. Thus, ho ρo = hwρw or ρ o hw ρw = h o

water

(4 -6) Fig(4 – 7)

Balance of liquids in a U - shaped tube

Atmospheric Pressure Torcelli invented the mercury barometer to measure the atmospheric pressure. He took a 1 m long glass tube and filled it completely with mercury and turned it upside down in a measured 0.76 m from the surface of mercury in the tank. The void above the column of mercury in the tube is vacuum (neglecting mercury vapor) is called Torcelli vacuum.

Hydrostatics

tank of mercury. He noticed that the level of mercury went down to a certain level that

Chapter 4:

Measuring ho and hw, we may determine practically the

density of oil, knowing the density of water.

oil

Fluid Mechanics

ho over the separating surface AD between water and oil, noting that both liquids do not

Unit 2:

Balance of liquids in a U - shaped tube

Hydrostatics Chapter 4:

Taking into consideration the fact that the free surface of the liquid is subject to atmospheric pressure Pa ,then the total pressure at a point inside a liquid it depth d is given

by:

P =Pa + h ρ g

(4 − 5)

Practical observations show indeed that the liquid pressure at a point inside it increases with increasing depth and with increasing density at the same depth. Thus, we conclude : 1) All points that lie on a horizontal plane inside a liquid has the same pressure.

Unit 2:

Fluid Mechanics

2) The liquid that fills connecting vessels rise in these vessels to

the same height, regardless of the geometrical shape of these vessels provided that the base is in a horizontal plane (Fig 4 – 5). Therefore, the average sea level is constant

Fig (4 – 5)

Water rises to the same level in connecting vessels

for all connected seas and oceans. 3) The base of a dam must be thicker than that the top to withstand the increasing pressure at increasing depths (Fig 4 – 6). Fig (4 – 6)

Dams must be thicker at the base to withstand the pressure at increasing depths

Hydrostatics Chapter 4:

From Fig (4 – 8), the height h of the mercury column in the tube is constant, wether the tube is upright or inclined. Taking two points A, B in one horizontal plane (Fig 4 – 9), such that A is outside the tube at the surface of mercury in the tank, while B is inside the tube. The pressure at B = the pressure at A. Thus: Pa = ρgh

(4 - 7) vacuum atmospheric pressure

mercury

Unit 2:

Fluid Mechanics

Fig (4 – 8)

Mercury height in a barometer is not affected by the tilting of the manometer

This means that the atmospheric pressure is equivalent to the weight of a column of mercury whose height is 0.76 m and cross sectional area 1m2 at OC° at sea level. This is known as S.T.P. (standard temperature and pressure). Since the density of mercury at O C° is 13595 kg/m3 and g = 9.8 m/s2

Pa = 1 Atm = 0.76 x 13595 x 9.81 = 1.013 x 105 N/m2

Fig (4 – 9)

A simple barometer

1) Blood is a viscous liquid pumped through a complicated network of arteries and veins turbulent flow (chapter 5), there is noise which can be detected by a stethoscope. There are two values for blood pressure: the systolic pressure, as blood pressure is maximum

(normally 120 Torr). This occurs when the cardiac muscle contracts and blood is pushed from the left ventricle to the aorta onto the arteries. The diastolic pressure is the minimum blood pressure (normally 80 Torr) when the cardiac muscle relaxes.

2) When a tire is well inflated (under high pressure) the area of contact with the road is as small as possible, while an underinflated (low pressure) tire has large contact area. As the area of contact with the road increases, friction increases and consequently, the tire

Fluid Mechanics

by the muscular effect of the heart. This is called steady flow (chapter 5). In the case of

Unit 2:

Applications to Pressure

is heated. Air pressure in a tire can be measured by a pressure gauge (Fig 4 â&#x20AC;&#x201C; 12).

Chapter 4:

graduated scale

Fig (4-12)

Measuring tire pressure with a pressure gauge

Hydrostatics

intake from the tire

Hydrostatics Chapter 4:

Manometer The manometer is a U - shaped tube containing a proper amount of liquid of a known density. One end is connected to a gas reservoir. The level of the liquid in the manometer may rise in one branch and go down in the other. Taking two points A,B in one horizontal plane in the same liquid (Fig 4 – 11 a), we have the pressure at B = the pressure A

a) when gas pressure > atmospheric pressure

Fig (4-11)

b)when gas pressure < atmospheric pressure

Unit 2:

Fluid Mechanics

Manometer

When P- the pressure of the gas enclosed in the reservoir - is greater than Pa, ρgh is the weight of a column the liquid in the free end of the manometer above point B and is the difference between P and Pa (Fig 4 -11a), P = Pa + ρgh In the case P < Pa (Fig 4 – 11 b), P = Pa - ρgh

i.e., the level of the liquid in the free end branch is lower than the level of the liquid in the end connected to the gas reservoir by a height h. In many cases, it suffices to measure the pressure difference, (4 - 8) ∆ P = P - Pa = ρgh Knowing the liquid density ρ in the manometer and the height difference h between the liquid levels in the two branches and the acceleration due to gravity g,we can calculate ∆P. Knowing Pa ,we may determine P of the gas enclosed in the reservoir.

Hydrostatics Chapter 4:

Examples

1) A solid parallelepiped (5cm x 10cmx 20cm) has density 5000kg /m3 is placed on a horizontal plane. Calculate the highest and lowest pressure. (g = 10 m/s2) Solution

For the highest pressure it is placed on the side with the least area (5 cm x 10 cm), where the force is the weight. P=

FFg 5 x 10 x 20 x 10 -6 x 5000 x 10 4 N/ m 2 ) = = 10 A 5 x 10 x 10 -4

For the lowest pressure, it is placed on the side of the greatest area (10 cm x 20cm) P=

FFg 5 x 10 x 20 x 10 -6 x 5000 x 10 = = 2500 N/m 2 -4 A 10 x 20 x 10

Unit 2:

Fluid Mechanics

2) Find the total pressure and the total force acting on the base of a tank filled with salty

water of density 1030 kg/m3. If the cross-section of the base is 1000 cm2 , the height of the water is 1cm and the surface of the water is exposed to air. Take g = 10 m/s2 and the

atmospheric pressure = 1 Atm = 1.013 x 105N/m2 Solution

Total pressure P = Pa + Ď g h

= 1.013 x 105 + 1030 x 10 x 1

Total force

= (1.013 + 0.103) x 105 = 1.116 x 105 N/m2 F = P x A = 1.116 x 105 x 1000 x 10-4 = 1.116 x 104 N

subsequently, to the surface of the large piston. If the force applied to the small piston is f

and the force affecting the large piston is F, and because the pressure on both pistons must be the same at equilibrium at the same horizontal plane, then :

generated on the large piston. The mechanical advantage of the hydraulic press Îˇ is given

by:

Îˇ = F = A f a (4 - 10)

Fluid Mechanics

P= f = F a A F = A ff (4 - 9) a From this relation, it is clear that if force f affects a small piston, a large force F is

Unit 2:

If pressure P is exerted to the small piston, this pressure is transmitted to the liquid and,

Thus, the mechanical advantage of a hydraulic press is determined by the ratio of the

moves down a distance y under the influence of f, then the large piston moves up a 1

distance y under the effect of F. According to the law of conservation of energy, the work 2

done in both cases must be the same (for 100% piston efficiency), f

fy =Fy 1

y 1 f y 2

y y

(4 - 11)

cylinder liquid

This shows that the mechanical advantage of the

piston may alternatively be expressed as the ratio

y /y

1 2

(Fig 4-15)

Mechanical advantage

Hydrostatics

f = y2 F y1

Chapter 4:

large piston to the small piston. Referring to Fig (4 â&#x20AC;&#x201C; 15), it is clear that if the small piston

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 96

Pascal’s principle Consider a glass container (Fig 4 – 13) partially filled with liquid and equipped with a piston at the top. The pressure at a point A inside the liquid at depth h is P =P + hρg where P1 is 1 the pressure immediately underneath the piston, which results from the atmospheric pressure, as well as the weight of the piston and the force applied on the piston. If we increase the pressure on the piston by an amount ∆P, by placing an additional weight on the piston. We note that the piston does not move inside because the liquid is incompressible. But the pressure underneath the Fig (4-13) piston must increase in turn by ∆P. This raises the pressure at Increase of weight on the piston increases the point A by ∆P. This make the pressure P =P1 + ρgh + ∆ P . pressure in the liquid Pascal formulated this result as follows : When pressure is applied on a liquid enclosed in a container, the pressure is transmitted in full to all parts of the liquid as well as to the walls of the container. This is known as Pascal’s principle or Pascal’s rule.

Application to Pascal’s rule : hydraulic Press The hydraulic press (Fig 4 – 14) consists of a small piston whose cross sectional area is “a” and a large piston whose cross sectional area is “A”. The space between the two pistons is filled with an appropriate liquid. cross sectional area a

Fig (4-14)

a) force to the left is transmitted to the right

Hydraulic Press

1kg

100kg

cross sectional area A

b) a weight of 1 kg to the left generates a force equivalent to 100 kg to the right if the ratio of the two cross sectional areas is 1:100

Examples

A hydraulic caterpillar

Fig (4-21)

Diving at large depths (500 m)

Solution

The force acting on the large piston : F

F A 3 F = 100 x 800 = 8 x 10 N 10

Hydrostatics

A hydraulic press has cross sectional area 10cm2 which is acted on by a force of 100N. The large cross sectional area is 800 cm2. Taking g = 10m/s2 ,calculate : a) the largest mass that can be lifted by the press b) the mechanical advantage of the press c) the distance traveled by the small piston so that the large piston moves a distance of 1cm

Chapter 4:

Diving at low depths

Fig (4-19)

Fluid Mechanics

Fig (4-20)

Unit 2:

3) The caterpillar also uses Pascal’s rule (Fig 4 – 19). 4) A diver wears a diving suit and a helmet to protect him from pressures at large depths. At low (shallow) depths, the diver - without the helmet - blows air in his sinuses to balance the external pressure (Fig 4 – 20). At large depths, the diving suit is appropriately inflated with air, and the helmet protects the diver’s head from crushing pressures (Fig 4 – 21).

Hydrostatics Chapter 4:

Learn at Leisure

Applications to Pascal’s rule 1) The hydraulic brake in a car uses Pascal’s rule as the braking system uses a brake fluid. Upon pushing on the brake pedal with

a small force and a relativeley long stroke (distance), the

pressure is transmitted in the master brake cylinder, hence, onto

the liquid and the whole hydraulic line, then to the piston of the wheel cylinder outwardly, and finally to the brake shoes and the

brake drum. A force of friction results, which eventually stops the car. This type of brakes is called drum brake (rear brake) (Fig

Fig (4-16)

Rear brakes

4 – 16). In the case of the front (disk) brake (Fig 4 – 17), the forces

resulting from the braking action press on the brake pads which

Unit 2:

Fluid Mechanics

produce friction enough to stop the wheel. It should be noted that

the distance traveled by the brake shoes in both cases is small because the force is large.

2) In another application to Pascal’s rule, a hydraulic lift uses a liquid to lift up cars in gas stations (Fig 4 – 18).

hydraulic liquid

Fig (4-18)

A hydraulic lift

Fig (4-17)

Front brakes

∆P = P1 - P2

= h1ρg - h2ρg

= (h1 - h2) ρg

∆P = hρg

Fb = Ahρg

Fluid Mechanics

Fb = ∆P X A

Unit 2:

1) Horizontal forces cancel each other out , because each two opposite forces are equal in magnitude and opposite in direction. 2) As to forces in the vertical direction, we find that the weight of the liquid enclosed in volume Vol in which (Fg) = Vol ρg acts downwards. Since this liquid is static, the liquid must exert L on the enclosed liquid an equal force Fb upwards , which is equal to the weight of the enclosed liquid. This force results from the difference of the pressure on the upper and lower surfaces of the parallepiped which is ∆P x A

Substituting, noting that Ah is the volume:

(4 - 12)

where (Fg) is the weight of the displaced liquid. l We see that Fb is equal to the weight of the parallepiped of the enclosed liquid. Equilibrium requires that the force Fb works upwards, and it is named buoyancy (buoyant force-upthrust).

If we substitute the virtual parallepiped by a solid parallepiped of the same shape and volume and of density ρs (Fig 4 – 22 b), the buoyancy (upthrust or buoyant force) which the liquid exerts on the solid Fig (4-22b) parallepiped remains the same Fb acting upwards. The liquid is displaced a distance ∆h. The weight of the parallepiped representing Archimedes’ rule using a real parallepiped instead of a virtual parallepiped the immersed body (Fg)s acts downwards. The resultant force on

Hydrostatics

The relation between the weight of a body in air and the weight when immersed in a liquid

Chapter 4:

Fb = Volρg = (Fg)

101

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 100

a)To calculate the largest mass that can be lifted by the large piston, 3 F 8 x 10 m= = = 800 kg Kg g 10 b)To calculate the mechanical advantage, ¡η = F = A = 800 = 80 f a 10 c)To calculate the distance traveled by the small piston, fy1 = F y 2

y = 8000 x 1 = 80 cm Cm 1 100

Buoyancy and Archimedes’ Principle

We are familiar with the following observations: 1)An object can be easily lifted if immersed under water level, whereas it might be difficult to lift at in air. 2) A piece of foam floats when immersed in water. 3) An iron nail sinks in water while a large steel ship floats. 4) Balloons filled with helium rise up. We can interpret the above observations as follows : When an object is immersed under the liquid surface then the object exerts a force on the liquid. Consequently, the fluid (liquid or gas) pushes back by an equal and opposite force (Newton’s third law). This force is called buoyancy. It acts in all directions,but the net effect is upwards. The buoyancy is given by the weight of the liquid displaced by the immersed body. To show this, let us imagine the existence of a volume Vol of the Fig (4-22a) liquid as a virtual parallepiped whose cross sectional area is Archimedes’ rule considering A and height h.This parallepiped is acted upon by forces in all a virtual parallepiped directions (Fig 4 – 22 A). This part of the liquid (like any other part of stable liquid) does not move, so it is in equilibrium:

Archimedes’ rule and Newton’s law

Chapter 4: Hydrostatics

The immersed body replaces an equal size of the liquid. Because liquids are incompressible, such a body displaces a volume of the liquid equal to the volume of the immersed body (or part of it that is immersed). This displaced liquid acts as a mass placed on the surface of the liquid. Its weight presses on the surface of the liquid. Thus, the pressure on each point of the liquid increases by this amount. Because we calculate the buoyancy on the body as the difference between two forces acting on the surface of the immersed body, then this difference stays the same. Hence, buoyancy on the immersed body is unchanged by the displacement. It is equal to the weight of the displaced liquid (Fig 4 – 12). We can understand what happens as that the weight of the immersed body acts on the liquid as a whole, then the liquid acts back with buoyancy as a reaction equal in magnitude and opposite in direction (Newton’s third law). In the case when the weight balances out with buoyancy, we have equilibrium, and the body remains suspended in the liquid. If the weight of the body exceeds buoyancy, the body sinks to the bottom where it settles there. If the weight of the body is less than buoyancy, the body floats on top of the surface where it settles afloat, while the weight of the displaced liquid whose volume is equal to the volume of the immersed part of the body, which is then equal to the weight of the floating body.

Fluid Mechanics

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Unit 2:

difference between the body’s weight in air and the buoyancy of the liquid (Fg)s = (Fg)s - Fb where (Fg)s is the weight of the body while totally immersed in the liquid, (Fg)s is its weight in air and Fb is the buoyancy (Fig 4 – 24). Concluding from above, we may formulate Archimedes’ principle as follows: A body partially or fully immersed in a fluid (liquid or gas) is pushed upwards by a force equal to the weight of the volume of the fluid displaced partially or fully by the body.

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The story of Archimedes and the crown Archimedes was one of the most celebrated scientists of Ancient Greece. There is an interesting story of Archimedes’ discovery of his rule. When Heron-king of Syracuse (one of ancient Greek cities)- doubted that his new crown might not have been made of pure gold, he summoned Archimedes to seek his counsel as to whether or not the crown was rigged, without destroying the crown of course. Archimedes was first at a loss. But one day, he was bathing in a tub. He noticed that as he dipped himself in the tub the water level rose.

concluded that the maker cheated and used less dense and hence cheaper materials. It is often told that Archimedes was euphoric with joy as he got this idea. He came out from the tub, and ran out naked shouting: “Eureka Eureka (I found it – I found it)". This expression

was the displaced water for the rigged crown more or less than that for equal mass of pure gold? (Fig 4-28).

Fig (4-28)

The crown’s story

Hydrostatics

has been ever since coined as a motto for scientific discovery. Question:

Chapter 4:

Archimedes brought the crown and submerged it in a water filled tub and measured the displaced (overflown) water. He then calculated the density of the material of the crown. He then repeated the experiment on a similar block of pure gold of the same mass, and measured the volume of the displaced water. He found that this volume was different. He

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Unit 2:

(displacement of the liquid). In the case of gases, there is buoyancy force as well, and it acts upwards. But it is not mandatory that the volume of the displaced gas is equal to the volume of the immersed body, since gases are compressible. But buoyancy must be equal to the weight of the displaced gas. This is why balloons filled with helium rise up (Fig 4 – 27).

105

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 104

Application to buoyancy

1) Hydrotherapy technique is prescribed to patients who are unable to lift limbs because of disease or injury in the associated muscles or joints. When a body is immersed in water it becomes, in effect, nearly weightless. As a result, the force required to move the limb is greatly reduced, and the therapeutic exercise becomes possible. 2) Weightlessness experiments may involve immersion in containers filled with a liquid whose concentration is adjusted so that buoyancy cancels out the weight. 3) A submarine floats when its tanks are filled with air, and submerges when those tanks are filled with water. Fish and Wales also fill air sacs with air to enable them to float, and empty them from air when they go under. 4) A diver breathes air under compression when diving to shallow depths, to equate the pressure. At larger depths, the diver adjusts the pressure in the diving suit to control the buoyancy force (Fig 4 â&#x20AC;&#x201C; 26).

Fig (4-25)

A submarine floats and sinks by emptying or filling water tanks

Fig (4-26)

A diver floats or dives depending on the varying density of the diving suit

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Is there buoyancy for gases ?

In the case of liquids, the volume of the displaced liquid equals the volume of the immersed body, because liquids are incompressible. Then, the force acting on the body upwards is equal to the weight of the displaced liquid as a reaction to the immersion

Fig (4-27)

A balloon filled with helium

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 106

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Why does a ship float while an iron nail sinks ?

It is noted that a ship is of large mass but contains large voids and spaces, which cause it to displace plenty of water, hence large buoyancy pushes the ship upwards. Part of the ship (hull) remains submerged in water for the ship to be afloat. This submerged part is what causes the displacement of water of an equal volume causing buoyancy to balance out with the weight of the ship (4 – 29). As the cargo on the ship increases, the submerged part increases to build up more buoyancy enough to keep the ship afloat (Fig 4 – 30a). The iron nail sinks because the buoyancy force on it is small due to its small volume (Fig 4 – 30 b).

Fig (4-30a)

The part of the boat that is submerged depends on the weight

Fig (4-29)

A ship floats despite its massive weight

Fig (4-30b)

A ship floats despite its massive weight while a nail sinks

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The Dead Sea

Have you noticed the difference between swimming in the sea, the Nile, and swimming

pools? The Dead Sea in Jordan is enclosed. It is not connected to any sea or ocean. The salt concentration is very high. No one drowns in the dead sea (why ?)

Hydrostatics Chapter 4: Fluid Mechanics Unit 2: 106

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Why does a ship float while an iron nail sinks ?

Fig (4-30a)

The part of the boat that is submerged depends on the weight

Fig (4-29)

A ship floats despite its massive weight

Fig (4-30b)

A ship floats despite its massive weight while a nail sinks

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The Dead Sea

Have you noticed the difference between swimming in the sea, the Nile, and swimming

pools? The Dead Sea in Jordan is enclosed. It is not connected to any sea or ocean. The salt concentration is very high. No one drowns in the dead sea (why ?)

I.Definitions and Basic Concepts

mercury column of height about 0.76 m and base area 1m2 at 0°C

• The units of atmospheric pressure are : a) Pascal (1 N / m 2). 5

b) Bar (10 N / m 2).

c)cm Hg.

d) Torr (mm. Hg).

Fluid Mechanics

• The density (ρ) is the mass per unit volume (kg/m3) • The pressure P at a point is the normal force acting on a unit surface area (N/m2). • All points lying in the same plane have the same pressure. • The atmospheric pressure is equivalent to the pressure produced by the weight of a

Unit 2:

In a Nutshell

• The manometer is an instrument for measuring the difference in the pressure of a gas • Pascal’s principle :

The pressure applied on an enclosed liquid is transmitted undiminished to every portion of the liquid and to the walls of the container.

• Archimedes’ principle :

A body immersed wholly or partially in a fluid experiences an upthrust force

the body.

• The weight of the volume displaced = volume of the immersed body x density of the liquid x the acceleration due to gravity.

• The immersed body in the liquid is acted upon by two forces, the upthrust force Fb the weight of the body (Fg) . s

If (Fg)s = F the body will be suspended in the liquid. b

and

Hydrostatics

(buoyant force) in the vertical direction equal to the weight of the liquid displaced by

Chapter 4:

inside a container and the outer atmospheric pressure.

109

forces on the two pistons are :

a) 9:2

d) 81 : 4

b) 2: 9

e) 4:81

c) 4:18

acceleration due to gravity is 10 m / s2, then the upthrust force on the body is :

a) 10 kg d) 35 N

b) 10 N e) 90 N

c) 35 kg

7) A piece of wood floats on water such that 1/4 its volume appears over the water surface. If the water density is 1000 kg / m3 , the wood density is then :

a) 1333 kg / m3

d) 1000 kg / m3

b) 250 kg / m3 e) 500 kg / m3

c) 750 kg/m3

Fluid Mechanics

6) A body has mass 5 kg in air, its weight when immersed in liquid becomes 40 N. If the

Unit 2:

5) If the ratio between large and small piston diameters is 9:2. The ratio between the two

8) The lead density is greater than the copper density, the copper density is greater than that water and weigh them then, compared to their weights in air:

a) the decrease of weight of the lead cube is greater than the decrease of weight of the copper cube.

b) the decrease of weight of the aluminum cube is greater than the decrease of weight of the copper cube. lead cube.

d) the decrease of weight of the aluminum cube is less than the decrease of weight of the lead cube.

e) the decrease of weight of the lead cube is less than the decrease of weight of the copper cube.

Hydrostatics

c) the decrease of weight of the aluminum cube is equal to the decrease of weight of the

Chapter 4:

of aluminum. If we immerse a number of cubes with equal volumes of these metals in

113

Hydrostatics Chapter 4:

Questions and Drills I) Put mark ( ) to the correct statement:

1) The following factors affect the pressure at the bottom of a vessel except one, tick it: a) the liquid depth in the vessel.

c) the acceleration due to gravity. e) the area of the vessel base.

b) the density of the liquid

d) the atmospheric pressure.

2) Which of the following factors have no effect on the height of mercury column in a barometer?

a) the density of mercury c) atmospheric pressure.

e) the temperature of mercury.

b) the cross sectional area of the tube.

d) the acceleration due to gravity.

Unit 2:

Fluid Mechanics

3) When a ship moves from river water to sea water,

112

which of the following statements is right ?

a) the water density increases and the ship sinks slightly.

b) the water density increases and the ship floats upwards slightly c) the water density decreases and the ship sinks slightly.

d) the water density decreases and the ship floats slightly.

e) the water density does not change and nothing happens.

4) The water pressure at the bottom of the High Dam lake on the dam body depends on : a) the area of the water surface. b) the length of the dam.

c) the depth of the water.

d) the thickness of dam wall.

e) the density of wall substance.

top surface of the upper liquid. Describe what happens to the two liquids, the wooden cube, the bottom of the vessel and the table.

1) The pressure on the base of a cylinder containing oil with diameter 8 m is 1.5 x 103 N/m2. Find the total force on the base. (7.54 x 104 N)

2) A difference in pressure of 3.039 x 105 N/m2 is recommended for air in a car tire. If the

atmospheric pressure is 1.013 x 105 N/m2 ,calculate the absolute pressure of air in the tire in atmospheric units.

Fluid Mechanics

IV) Drills :

Unit 2:

5) A vessel on a table contains two liquids: oil and water. A wooden cube is placed on the

(4 Atm) pressure on its base.

(0.4 x 104 N/m2)

4) The large and small piston diameters of a hydraulic press are 24cm and 2cm respectively. Calculate the force that must be applied to the small piston to obtain a force (13.9 N, 144) 5) A piece of aluminum has a mass of 250 gram in air. When immersed in water it has an an apparent mass of 160 gram and it has 180 gram in alcohol. Calculate the densities of aluminum and alcohol if the density of water is 1000 kg /m3 and g = 9.8 m / s 2 (2777.8 kg / m3, 777.8 kg/m3)

Hydrostatics

of 2000 N on the large piston. Then calculate the mechanical advantage.

Chapter 4:

3) A fish tank of cross-sectional area 1000 cm2 contains water of weight 4000N. Find the

115

Hydrostatics Chapter 4:

9) If the ice density is, 900 kg / m 3 and the water density is 1000 kg /m3 , then the ratio of the floating part of an ice cube is :

a) 90 %

d) 80 %

Fluid Mechanics

e) 20 %

density,then the upthrust force exerted by the liquid on the body will be : a) equal to the mass of liquid displaced by the body. b) equal to the mass of the immersed body.

c) equal to the volume of the liquid displaced by the body. d) equal to the weight of the liquid displaced by the body.

e) greater than the body weight. II) Define each of the following : 3. Pascal’s principle

Unit 2:

c) 100 %

10) A body is immersed wholly in a liquid. If the body density is greater than the liquid

1. Density

114

b) 10 %

2. Pressure at a point

4. Archimedes’ principle

III) Essay questions :

1) Prove that the pressure (P) at depth (h) in a liquid is determined from the relation. P = Pa + ρgh

where Pa is the atmospheric pressure, ρ the liquid density and g is the acceleration due to

gravity.

2) Describe the manometer and show how it can be used for measuring a gas pressure inside a container.

3) What is meant by Pascal’s principle ? Describe one of its applications. 4) Show that the resultant forces on an immersed body is given by :F = (ρl – ρs) g Vol ,

where ρl and ρs are the densities of the liquid and the body respectively,g is the

acceleration due to gravity and Vol is the volume of the body. Explain the consequences of this relation.

Hydrostatics

6)The atmospheric pressure on the surface of a lake is 1Atm. The pressure at its bottom is

Chapter 4:

two floors if mercury density is 13600 kg / m3, the building height is 200m and g =

3 Atm. Calculate the depth of the lake (density of water 1000 kg/m3, 1 Atm = 1.013x105 N/m2, g = 9.8 m/s2 ). (20.673 m) 7) A man carries a mercury barometer with readings 76 cm Hg and 74.15 cm Hg at the lower and upper floors, respectively. Calculate the average density of air between the 9.8m/s2. (1.258 kg/m3) 8) A manometer containing mercury is attached to gas enclosed in a container. If the difference height in the manometer is 25 cm.

Unit 2:

Fluid Mechanics

Calculate the pressure difference and the absolute pressure of the enclosed air in units of

116

N/m2 (1Atm = 1.013 x 105 N/m2, mercury density = 13600 kg/m3 and g=9.8 ms2) (0.3332x105 N/m2, 1.3462x105 N/m2). 9) The volume of a huge balloon filled with hydrogen is 14x104m3. Find the maximum lifting force acting on it, if the hydrogen density is 0.092 kg/m3, air density is1.29 kg/m3 and the mass of the balloon with its accessories is 8x104 kg. (g = 10m/s 2). (87.72x104 N)

Unit 2 :

Fluid Mechanics

Chapter 5:

Hydrodynamics

3) The flow is irrotational, i.e., there is no vortex motion. 4)If no forces of friction exist between the layers of the liquid the flow is nonviscous .If there is friction it is viscous. If the velocity of flow of a liquid exceeds a certain limit, steady

flow changes to turbulent flow, which is characterized by the existence of vortices (Fig 5–2). The same thing happens to gases as a result of diffusion from a small space to a large space or from high pressure to low pressure (Fig 5 – 3).

Rate of flow and the continuity equation

3) the velocity of the liquid flow at any point in the tube does not change with time. There is a relation that ties the rate of flow of the liquid with its velocity and cross sectional area. Fig (5-2) This relation is called the continuity equation. To Vortices due to turbulent understand what the continuity equation entails, we choose flow or a violent motion of a body through a liquid two perpendicular planes normal to the streamlines at the two sections (Fig 5 - 4). The cross sectional area at the first plane is A1 and the cross sectional area at the second plane is A2. The volume rate of flow is the volume of the liquid flowing through area A1 in unit time.We have Qv = A1v1, where v1 is the velocity of the liquid at section A1. The mass of the liquid (of density ρ) flowing in unit time is called the mass rate of flow Qm ,which is given by:

Chapter 5:5: Hydrodynamics Chapter Hydrostatics

We shall focus on steady flow. Consider a flow tube such that: Fig (5-3) Smoke changes from steady 1) the liquid fills the tube completely. to turbulent flow 2) the quantity of the liquid entering the tube at one end equals the quantity of the liquid emerging out from the other end within the same time.

Fluid Mechanics Fluid Mechanics

Turbulent flow

Unit 2 Unit 2::

2) In steady flow, the velocity of the liquid at each point is independent of time.

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Unit 2: 2: Unit

Fluid Mechanics Fluid Mechanics

Chapter 5:Properties of Fluid dynamics

118 118

Chapter 5

Hydrodynamics

Overview Hydrodynamics (Fluid dynamics) deals, with fluids in motion. We must distinguish between two types of fluid motion, steady flow and turbulent flow.

Steady flow

If a liquid moves such that its adjacent layers slide with respect to each other smoothly, we describe the motion as a laminar flow or a streamline (steady) flow. In this type of flow, particles of the liquid follow continuous paths called streamlines.Thus, we may visualize the liquid as if it is in a real or virtual tube containing a flux of streamlines representing the paths of the different particles of the liquid (Fig 5 â&#x20AC;&#x201C; 1). These streamlines do not intersect, and the tangent at any point along the streamline determines the direction of the instantaneous velocity of each particle of the liquid at that point. The number of streamlines crossing perpendicularly a unit area at a point (density of streamlines) expresses the velocity of flow of the liquid at that point. Therefore, streamlines cram up at points of high velocity and keep apart at points of low velocity.

Conditions of Steady Flow

1)The rate of flow of the liquid is constant along its path, since the liquid is incompressible and the density of the liquid is independent of distance or time.

Fig (5-1)

Streamlines

Unit 2 :

Fluid Mechanics

Chapter 5:

Hydrodynamics

121

Chapter 5:5: Hydrodynamics Chapter Hydrostatics Unit 2: Fluid Mechanics

Unit 2: 2: Unit

Fluid Mechanics Fluid Mechanics

Chapter 5:Properties of Fluid dynamics

120 120

Qm=ρQv=ρ A1 v1 Similarly, the mass rate of flow through area A2 is ρQv = ρA2v2. Since the mass rate of flow is constant in steady flow ρ A1 v1 = ρ A2 v2

(5-1)

A1 v1= A2 v2

Fig (5-4)

Model for deducing the continuity equation

This is the continuity equation leading to

Vv11 A 2 = Vv22 A 1 (5-2) From this relation, we see that the velocity of the liquid at any point in the tube is

inversely proportional to the cross sectional area of the tube at that point. The liquid flows slowly where the cross sectional area A1 is large and flows rapidly, when the cross sectional

area A2 is small (Fig 5 – 5). To understand the continuity equation better, we consider a small amount of liquid ∆m = ρ∆V , where ∆V = A ∆x ,where ∆x is the distance traveled by the ol

liquid in time ∆t . Thus, ∆x1 = v1 ∆t. Then ∆Vol = A1v1 ∆t. This same value must emerge from the other side of the tube, since the liquid is incompressible ,i.e, ∆Vol = A2v2 ∆t . Thus, we

must emphasize that the rate of flow of the liquid is a volume rate Qv (m3/s), or a mass rate of

flow (kg/s). Both of these rates are constant for any cross section. This is called the conservation of mass, which leads to the continuity equation. A2 A1

Fig( 5-5)

Basis for the continuity equation

Hardening of the arteries The body controls the blood flow in the arteries by muscles surrounding these arteries.

equation (5 – 1), the blood velocity must hence increase. When they relax, the blood

velocity decreases. With age, these muscles lose that elasticity, and this is called hardening of the arteries, and hence these muscles lose the ability to control the blood flow. As

cholestrol(fats) precipitates on the inner walls of these blood vessels, the radius decreases further which increases the possibility of coagulation (formation of a clot) which blocks the

blood stream, leading to angina pectoris. The patient takes medication to ensure the

liquidity of the blood to prevent the coagulation. However, if the dose is excessive, he might end up with hemorrhage or (internal bleeding). It is known that one of the

Fluid Mechanics Fluid Mechanics

When these arteries contract, the radius of the artery decreases. From the continuity

Unit 2 Unit 2::

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constituents of blood is platelettes, which are responsible for normal coagulation to stop

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Measuring blood pressure Measuring blood pressure is one of the ways to check on the performance of the heart.

The sphygmomanometer is a type of manometer (Fig 5 – 9). It consists of a cuff in the form of an air bag wrapped around the patient’s arm. A hand pump is used to inflate the

bag and a mercury manometer is used to measure the pressure in the bag. The pressure is increased in the air bag, until the blood flow ceases momentarily in the brachial artery. A stethoscope is used to indicate the soundness of the artery’s muscle in pushing the blood.

Chapter 5:5: Hydrodynamics Chapter Hydrostatics

bleeding. Thus, viscosity of the blood and its composition play a vital role is man’s life.

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Unit 2: 2: Unit

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Chapter 5:Properties of Fluid dynamics

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Blood circulation in human body The circulatory system consists of a huge network of blood vessels including arteries,

veins, down to capillaries (Fig 5â&#x20AC;&#x201C; 8) .The heart pumps blood through this network at a rate of 5 liters per minute (or 8.33x10-5 m3/s) with a normal pulse rate of 70 pulse per minute.

The pumping rate may reach 25 liters per minute or 180 pulse per minute with excessive activity .Calculating the velocity of flow in the aorta (2 cm diameter),we find that the blood

velocity is 26.5 cm/s (check the calculation). If we add up all the capillaries, we find that the collective cross section is 0.25m2 (what is the velocity?). brain lungs

heart

arms and shoulders

veins

trunk and organs

veins

pulmonary artery aorta artery arteries

legs

veins

Fig (8-5)

A simplified diagram for blood circulation

Unit 2 :

Fluid Mechanics

Chapter 5:

Hydrodynamics

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Chapter 5:5: Hydrodynamics Chapter Hydrostatics Unit 2: Fluid Mechanics

The pressure in the air bag is decreased gradually, until the pressure exerted by the heart is sufficient to push the blood through, and open up the brachial artery. The sudden flow of the blood in the nearly closed artery causes a turbulent flow which produces a hiss, which the doctor can hear with the stethoscope, while the doctor monitors the reading on the manometer in mm Hg. This reading is the systolic pressure (normally

the hiss remains audible in the stethoscope, until the

Fluid Mechanics Fluid Mechanics

pressure in the air bag is equal to the lowest pressure exerted by the heart when the brachial artery is fully

Unit 2: 2: Unit

Chapter 5:Properties of Fluid dynamics

120 mm Hg). As the pressure in the air bag decreases,

cardiac arrest.

124 124

Fig (5 - 9)

Sphygmomanometer

open (diastolic pressure). Then,blood flows steadily and the hiss disappears. The reading on the manometer in this case (when the heart is in rest or relaxation) is normally 80 mm Hg. If the systolic pressure exceeds 150 mm Hg, the patient has hypertension which might cause brain hemorrhage, and hence stroke. If the diastolic pressure exceeds 90 mm Hg, the heart-which is supposedly then at rest -has extra pressure, causing fatigue and eventually fibrosis in the heart muscle, leading to heart failure or

The aorta cross section is given by

A 1 = / r 21 = / (0.007) 2 m 2

= / (0.0035) 2 x 30 m 2

A1 v1 = A2 v2

2 / (0.007) 2 (0.33) = / (0.0035) 2 (30) v2

v2 =

4 Ă&#x2014; 0.33 m/s = 0.044 0.44m/s 30

Thus, the velocity of the blood in the main arteries is 0.044 m/s. Consequently, the blood velocity in capillaries is much smaller, which gives time for the tissues to exchange

Fluid Mechanics Fluid Mechanics

The collective cross section for 30 main arteries is given by A 2 = / r 22 x 30

Unit 2 Unit 2::

Solution

oxygen and carbon dioxide as well as nutrients and excretion products. Divine wonder is Viscosity

We observe viscosity as follows : 1) We hang two funnels each on a stand and put a beaker under each. We pour alcohol in

one funnel and a similar volume of glycerine in the other, and observe the velocity of flow of each. We notice that the flow velocity of alcohol is higher than that of glycerine.

2) Take two similar beakers, one containing a certain volume of water and the other an equal volume of honey .Stir the liquid in both beakers with a glass rod .Which of the two

liquids is easier to stir ? Then, we remove the rod .We notice that :

Chapter 5:5: Hydrodynamics Chapter Hydrostatics

countless.

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Fluid Mechanics Fluid Mechanics

Chapter 5:Properties of Fluid dynamics

126 126

Examples 1) A water pipe 2 cm diameter is at the entrance of an apartment building. The velocity of the water in it is 0.1m/s. Then, the pipe tapers to 1cm diameter. Calculate

a) the velocity of the water in the narrow pipe. b) quantity of the water (volume and mass) flowing every minute across any section of the pipe (density of water = 1000 kg/m3) .

Solution

a) A1 v1= A2 v2 2

π (0.01m) (0.1 m/s) = π (0.005m) v2

π × 10-4 × 0.1 v2 = = 0.4m/s π × 2.5 × 10-5

b)The volume rate of flow (m3/s) is given by the relation = / x 10 -4 x 0.1 or / x 2.5 x 10 -5 x 0.4 Thevolume rate of flow (m3/min) is given by = 3.14 × 10-5 m 3 /s

Q x 60 = 3.14 x 10 -5 x 60 = 188.4 x 10 -5 m 3 /min The mass flowing per minute Qm = 3.14 x 10 -5 x 10 3 = 3.14 x 10 -2 The mass rate of flow (kg/min) is given by

Qm= 3.14 x 10 -5 x 10 3 x 60 = 1.884 K

2) The average velocity of blood in aorta ( radius 0.7 am ) for an adult is 0.33 m/s From the aorta, blood is distributed to main arteries (each radius 0.35cm). If we have 30 main arteries, calculate the velocity of blood in each.

force is due to the adhesive forces between the molecules of the solid surface and the stationary plate. Similarly, the upper layer moves at the same velocity of the upper plate.

b) The existence of another friction (shear) force between each liquid layer and the adjacent one, which resists the sliding of the liquid layers with respect to each other. This

produces a relative change in velocity between any two adjacent layers. Thus, viscosity is the property responsible for resisting the relative motion of liquid layers .This type of

flow is called nonturblent viscous laminar flow (or viscous steady flow), since no

vortices occur.

Coefficient of Viscosity velocity, a force F must exist. This force is directly proportional to both velocity and area of the moving plate A, and inversely proportional to the distance between the plates d. AV F| v ds Av vs d

(5 - 3)

Fd F = Av Av/d

(5 - 4)

F =d

where dvs (Eta) is a constant of proportionality called viscosity coefficient, given by dvs

The coefficient of viscosity ( Ns/m2 or kg/m s ) may be defined as : the tangential force acting on unit area, resulting in unit velocity difference between two layers, separated by unit distance apart .

Chapter 5:5: Hydrodynamics Chapter Hydrostatics

Referring to Fig (5 â&#x20AC;&#x201C; 6), we find that for the moving plate to maintains its constant

Fluid Mechanics Fluid Mechanics

contacting liquid molecules. This leads to zero velocity of the layer in contact with the

Unit 2 Unit 2::

a) Friction forces exist between the lower plate and the liquid layer in contact with it. This

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Unit 2: 2: Unit

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a) Water is easier to stir, which means that water resistance to the glass rod is less than the resistance of honey.

b) The motion in honey stops almost immediately after we remove the rod, while it continues for a little while longer in water .

3) We take two long similar measuring cylinders and fill them to the end, one with water and

the other with glycerine. Then , take two similar steal balls and drop one in each liquid, and record the time each ball takes in each liquid to hit the bottom. We observe that the time in water is less. Thus, the glycerine resistance to the ball motion is greater .

We, thus, conclude :1) Some liquids such as water and alcohol offer little resistance to the motion of a body in them, and are easy to flow. We say they have low viscosity

2) Other liquids such as honey and glycerine are not as easy to move through ,i.e.,they offer high resistance to body motion, and are said to have high viscosity.

To interpret viscosity, imagine layers of liquid trapped between two parallel plates, one

stationary and the other moving with velocity v (Fig 5 â&#x20AC;&#x201C;6). The liquid layer next to the stationary plate is stationary, while the layer next to the moving plate is moving at v. The layers in between move at velocities varying from o to v. The reason for this is as follow :

liquid

Layers of a liquid slide with respect to each other

Force acting on the upper layer of a liquid

Fig (6-5)

Friction between layers of a liquid

Unit 2 Unit 2::

Learn at Leisure

Syphon:can a liquid go up ? Suction of air from one end of a hose while the other to a certain head (vertical distance), then to flow downwards (Fig 5 â&#x20AC;&#x201C; 13). People often use this technique to draw gasoline from a car tank. This seems contrary to gravity. This phenomenon is called syphon. It can be explained as follows. The liquid molecules attract each other as beads in a chain (Fig 5 â&#x20AC;&#x201C;14). The molecules may

Fig (5-13) Syphon

go up to a certain distance overcoming gravity, then

Fluid Mechanics Fluid Mechanics

end is submerged in a liquid causes the liquid to rise up

come back down and flow from the free end of the hose. when the molecules of the liquid at the surface attract each other as a membrane. This is the same theory of the formation of air bubbles, in which the internal pressure balances out with the external pressure and the surface

Fig (5-13)

tension, so they do not blow up, unless this balance is Beads of liquid pulling each other disturbed. Another property is capillarity, where molecules of the container pull the molecules of the liquid by forces of adhesion. The surface of the liquid is curved due to the surface tension in the capillary. This property is responsible for drawing water in the stem of plants through the capillaries, so that the plant can obtain its water and nutrients from the soil and even out to the foliage.

Chapter 5:5: Hydrodynamics Chapter Hydrostatics

Liquids have another property, called surface tension,

131 131

Chapter 5:5: Hydrodynamics Chapter Hydrostatics Unit 2: Fluid Mechanics

Unit 2: 2: Unit

Fluid Mechanics Fluid Mechanics

Chapter 5:Properties of Fluid dynamics

130 130

Applications of Viscosity 1-Lubrication : Metallic parts in machines have to be lubricated from time to time. This process leads to:

a) reduction of heat generated by friction.

b) protecting machine parts from corrosion (wear).

Lubrication is carried out using highly viscous liquids. If we use water (low viscosity), it will soon seep away or sputter from the machine parts due its low adhesive forces. Therefore, we must use liquids with high adhesive (high viscosity), so they remain in contact with the moving machine parts.

2-Moving vehicles

When a car attains its maximum speed, the total work done by the machine which is supplied by the burnt fuel, acts most of the time against air resistance and the forces of friction between the tires and the road. At relatively low and medium velocities, air resistance to moving bodies resulting from air viscosity is directly proportional to the velocity of the moving body. When the velocity exceeds a certain limit, then the air resistance is proportional to the square velocity rather than the velocity,leading to a noticeable increase in fuel consumption. Therefore, it is advisable not to exceed such a limit ( 80 â&#x20AC;&#x201C; 90 km/h )

3-In medicine :

Blood precipitation rate : when a ball undergoes a free fall in a liquid, it is under three forces : its weight, buoyancy of the liquid and friction between the ball and the liquid due to viscosity. It is found that such a ball attains a final velocity which increase with its radius.

This is applied in medicine by taking a blood sample and measuring its precipitation rate. The doctor may then decide if the size of red blood cells is normal or not. In the case of rheumatic fever and gout, red blood cells adhere together, and therefore, their volume and radius increase and the sedimentation ( precipitation ) rate increases. In the case of anemia, the precipitation rate is below normal, since the red cells break up. Hence, their volume and radius decrease .

Unit 2:: Unit 2

Questions and Drills I) Define 2) viscosity

3) coefficient of viscosity

II) Essay questions 1) Prove that the velocity of a liquid at any point in a tube is inversely proportional to the cross sectional area of the tube at that point.

2) Explain the property of viscosity. 3) Illustrate some applications of viscosity. III) Drills 1- Water flows in a horizontal hose at a rate of 0.002 m3/s, calculate the velocity of the water in a pipe of cross sectional area 1cm2 .

(20 m/s)

diameter of the hose if the velocity of the emerging water is 27 m/s . (0.4cm) 3- A main artery of radius 0.035 cm branches out to 80 capillaries of radius 0.1 mm. If the velocity of blood through the artery is 0.044 m/s ,what is the velocity of blood in each (0.0067 m/s)

4- The cross sectional area of a tube at point A is 10 cm2 and at point B is 2cm2. If the velocity of water at A is 12 m/s, what is the velocity at B ? 5- The cross sectional area of a water pipe at the ground floor is

(60 m/s) 4x 10-4m2. The

velocity of the water is 2 m/s. When the pipe tapers to a cross sectional area of 2 x 10-4m2 at the end, calculate the velocity of the flow of water at the upper floor.

(4m/s)

Chapter Hydrostatics Chapter 5:5: Hydrodynamics

2- Water flows in a rubber hose of diameter 1.2 cm with velocity 3 m/s. Calculate the

of the capillaries?

Fluid Mechanics Fluid Mechanics

1) fluid

133 133

132

Unit 2 :

Fluid Mechanics

Chapter 5:

Hydrodynamics

It can be concluded that :

- In their motion, they collide with each other and collide with the walls of the container

- The distance between the molecules is called intermolecular distance (more or less

constant for different gases at the same conditions).

(NH3)

A cloud of Ammonium chloride diffuses to fill the two cylinders

diffusion of a white cloud of ammonium chloride

(NH4Cl)

paper

(HCl)

Fig (6 â&#x20AC;&#x201C; 2)

Presence of gas intermoleculor distances

When a graduated cylinder filled with ammonia gas is placed upside down on another cylinder filled with hydrogen chloride gas (Fig 6-2), a white cloud of ammonium chloride is formed, then it grows and diffuses until it occupies all the space within the two cylinders. This can be explained as follows. Hydrogen chloride gas molecules - in spite of their higher density - diffuse upwards, through spaces separating ammonia gas molecules, where

Gas Laws

The evidence of the existence of intermolecular distances can be shown as follows:

Chapter 6:

remove the paper

Hydrogen chloride gas

Heat

ammonia gas

Unit 3:

- Gas molecules are in a state of continuous random motion.

they combine together forming ammonium chloride molecules, which diffuse to fill the 137

Unit 3:

Chapter (6)

The Gas Laws

Overview

It can be shown that gas molecules are in continuous random motion called Brownian

motion as follows:

Heat

If we examine candle smoke through the microscope,we notice that the smoke

particles move randomly. The motion of the carbon particles is caIled Brownian motion, after Brown, an English botanist who discovered for the first time in 1827 that tiny pollen grains suspended in water moved randomly.

Chapter 6:

a. gas molecules undergo a random translational motion

b. liquid molecules undergo a translational and vibrational motion

c. solid molecules undergo a vibrational motion

Fig (6 â&#x20AC;&#x201C; 1)

Motion of molecules materials

Gas Laws

Interpretation of Brownian motion

Air (gas) molecules move in a haphazard (random) motion in all directions with

different velocities. During their motion, they collide with each other and collide with smoke particles. Due to the resultant force on a smoke particle, it will move in a certain

direction through a short distance and so on, always moving, colliding, and changing

direction. The reason for this is that the gas molecules are in a free motion (due to heat) and in constant collision, so they change their direction randomly (Fig 6-1).

136

To study the relation between the volume of a fixed mass of gas and its pressure at constant temperature, the apparatus shown in Fig (6- 3) is used. It consists of a burette (A) connected by a length of rubber tube to a glass reservoir (B) containing a suitable amount

Unit 3:

Firstly: the relation between the volume and pressure of a gas at constant temperature (Boyleâ&#x20AC;&#x2122;s law) :

of mercury. (A) and (B) are mounted side by side onto a vertical stand attached to a base movable along the stand either upwards or downwards and can be fixed at any desired

Heat

provided by three screws with which the stand is adjusted vertically. The reservoir (B) is position.

Procedure:

1- The tap (A) is opened and the reservoir (B) is

Chapter 6:

raised until the mercury level in burette A is about half full, taking into account that the mercury levels are the same in both sides. (Fig 6-3a) . 2- The tap (A) is then closed. The volume of the enclosed air is measured, let it be (Vol)1. Its equals the atmospheric pressure Pa (cmHg)

(b)

which may be determined using a barometer.

3- The reservoir (B) is then raised a few centimetres and the volume of the enclosed air is measured (Vol)2. The difference

(a) Fig (6 â&#x20AC;&#x201C; 3)

Boyleâ&#x20AC;&#x2122;s apparatus

(c)

Gas Laws

pressure is also measured, let it be P1, which

139

upper cylinder. Also, ammonia gas molecules - in spite of their lower density - diffuse

Unit 3:

downwards through spaces separating hydrogen chloride gas molecules, where they combine forming ammonium chloride molecules, which diffuse to fill the lower cylinder. Accordingly, we can conclude that there are large spacings separating the gas molecules, known as intermolecular spacings. This is to be tied to the compressibility of gases. These large intermolecular spacings allow gas molecules to get packed together when pressed.

Heat

Thus, a volume occupied by a gas decreases with increased pressure.

Gas Laws Experiments performed to evaluate the thermal expansion of a gas are complicated. The volume of a gas is affected by changes in pressure as well as by temperature. This difficulty does not arise in the case of solids or liquids, as these are very much less

Chapter 6:

compressible. In order to make a full study of the behavior of a gas, as regards volume, temperature and pressure, three separate experiments have to be carried out to investigate the effect of each pair, respectively, i.e., we study the relation between two variables only, keeping the third constant.These experiments are 1- The relation between the volume and pressure at constant temperature (Boyleâ&#x20AC;&#x2122;s law).

Gas Laws 138

2- The relation between the volume and temperature at constant pressure (Charleâ&#x20AC;&#x2122;s law). 3- The relation between the pressure and temperature at constant volume (Pessure law or Jollyâ&#x20AC;&#x2122;s law). We are going to study each of these three relations.

The effect of temperature on the volume of a gas at constant pressure:

the same volume of different gases at constant pressure expand by the same amount? To show this, let us do the following experiment:

Unit 3:

We have already known that gases contract by cooling and expand by heating. But, does

1- Take two flasks of exactly equal volume, each fitted with a cork through which a tube

Heat

bent 90˚ is inserted. In each tube, there is a

thread of mercury of length 2 or 3 cm. Fill one of the flasks with oxygen and the other with carbon

dioxide or air. Submerge the two flasks in a vessel filled with water as shown in Fig (6 - 5).

CO2

2- Pour hot water into the vessel and notice the tubes. You will find that these distances are equal. This indicates that equal volumes of

Fig (6 – 5)

Effect of temperature on the volume of a gas at constant pressure

different gases expand equally when heated through the same temperature rise . In other words, they have the same volume expansion coefficient.

Volume expansion coefficient of a gas at constant pressure αv is defined as :

Chapter 6:

distance moved by the mercury thread in both

"It is the increase in volume at constant pressure per unit volume at 0˚C for 1˚C

Gas Laws

rise in temperature ".

141

between the two levels of mercury in both sides (h) is determined . In this case, the

Unit 3:

pressure of the enclosed air (cmHg) is P2 = Pa + h (Fig 6 - 3 b).

4- Repeat the previous step by raising the reservoir (B) another suitable distance and measure (Vol)3 and P in the same manner. 3

5- The reservoir (B) is then lowered until the mercury level in (B) becomes lower than its level in (A) by a few centimeters. Then, the volume of the enclosed air is measured

Heat

(Vol)4 and its pressure (P4) is determined P = P - h, where h is the difference between a 4 the two levels of mercury in both sides (Fig 6-3c).

6- The previous step is repeated once more by lowering (B) another suitable distance. Then (Vol)5 and P are measured in the same manner. 5

7- Plot the volume of the enclosed air (Vol) and the reciprocal pressure ( a straight line (Fig 6 - 4) Thus, we can conclude that:

Chapter 6: Gas Laws 140

Vol

1 ).We obtain P

1 P

,i.e.,the volume of a fixed mass of gas is inversely

proportional to the pressure, provided that the temperature remains constant. This is "Boyleâ&#x20AC;&#x2122;s law". Boyleâ&#x20AC;&#x2122;s law can be written in another form, as: VolV =

const const P

PVol = Const.

(6 - 1)

i.e., is : at a constant temperature, the product PVol of any given mass of a gas is constant.

Fig (6 â&#x20AC;&#x201C; 4) Relation between volume and reciprocal pressure of gas

Unit 3:

Secondly: the relation between the gas volume and its temperature at constant pressure (Charle’s law) : To investigate the relation between the gas volume and its temperature at constant pressure, the apparatus shown in Fig(6 - 6a) is used. It

steam inlet

Heat

consists of a capillary glass tube 30 cm long and about 1 mm diameter with one end closed. The tube contains a short pellet of mercury enclosing an amount of air inside it whose length is measured by a ruler stand. The apparatus is equipped with a thermometer inside a glass

Chapter 6:

envelope. We follow the folowing procedure: 1- The glass envelope is packed with crushed ice and water. It is then left until the air inside the

glass envelope pellet of mercury capillary tube

Cork steam outlet

Fig (6 – 6a)

Charle’s apparatus

glass tube has fully acquired the temperature of melting ice(0˚C). 2- The length of the enclosed air is then measured, and since the tube has a uniform cross-section, the length of the encloscd air is taken as being proportional to its volume

Gas Laws

(Vol )o˚c

3- The ice and water are removed from the envelope and steam is passed through the top and out at the bottom for several minutes to be sure that the temperature of air becomes 100˚C . Then, the length of the enclosed air is measured. It is taken as a measure of its volume (Vol )

100˚C

4- A relation between Vol and t˚C is plotted (Fig 6-6b). We see that such a relation is a straight line,which if extended will intersect the abscissa at -273˚C .

142

5- Repeating this experiment several times for different gases and measuring the amount of a- At constant volume, the pressure of a given mass of gas increases by increasing temperature.

b- At constant volume, equal pressures of gases increase equally, when heated through

Unit 3:

increase of gas pressure at constant volume for the same rise in temperature, we find:

the same range of temperatures.We define the pressure expansion coefficient of a gas

“lt is the increase in gas pressure at constant volume per unit pressure at 0˚C for

cm degree rise in temperature”. It is found to be the same for all gases.

Heat

at constant volume (βp) as :

Thirdly: the relation between the pressure and temperature of a gas at constant volume (pressure law or Jolly’s law): Chapter 6:

It was found experimentally that the increase in gas pressure is directly proportional to the initial pressure at 0˚C (P0˚C) as well as to the rise in its temperature, (∆t˚C). This is expressed as follows:

mercury

∝ P0˚C ∆ (t˚C)

∆P = βP P0˚C ∆ (t˚C)

ββP =

∆P∆P P0O˚∆COt∆ (t˚C)

(6-3)

where βp is a constant value. It is the pressure

Fig (6 – 8)

Gas Laws

∆P

thermometer

Jolly’s apparatus

145

The effect of temperature on the pressure of a gas at constant volume:

Unit 3:

1- To investigate how the pressure of a gas depends on temperature, the apparatus shown (Fig 6-7a) may be used. The gas under test is confined in a flask by mercury in a U tube. The flask is fitted with a cork. The surfaces of mercury in the two branches (A) and (B) have the same level at x,y. Thus, the pressure of the enclosed air is atmospheric. We then determine the temperature of air. Let it be t1˚C.

Heat

Chapter 6:

(a)

(b)

Fig (6 – 7)

(c)

Effect of the temperature on the pressure of a gas at constant volume

2- Submerge the flask in a vessel containing lukewarm water at t2˚C. You will notice that

Gas Laws

the level of mercury decreases in branch A, while it rises in branch B (Fig 6-7b).

3- We pour mercury in the funnel C, until the level of mercury in branch A returns to the

mark x then the volume of the enclosed air in the flask at t2˚C is equal to the volume at t1˚C (Fig 6-7c).

4- We notice that the surface of mercury in branch B exceeds that in branch A by an amount” h” (cm). This means that the pressure of the enclosed air has increased as a result of the temperature rise from t1˚C to t2˚C by an amount equal h (cmHg)( Fig 6-7c).

144

expansion cofficient of a gas with temperature, at constant volume.It is the same for all

Unit 3:

gases. To measure Î˛p of a gas at constant volume, Jollyâ&#x20AC;&#x2122;s apparatus shown in Fig (6-8) is used.

It consists of a glass bulb (A). The bulb is joined to a capillary tube (B) bent in the form of two right angles. The bulb and the tube are mounted on a vertical ruler attached to a board which is fixed on a horizontal base provided with 3 leveling screws.

Heat

The capillary tube (B) is connected to a mercury reservoir (C) by means of a rubber tube. We follow the following procedures: 1- Determine the atmospheric pressure (Pa) using a barometer.

2- Pour mercury in (A) to 1/7 of its volume to compensate for the increase in its volume when heated, so that the volume of the remaining part is still constant, (the volume

Chapter 6:

expansion coefficient of mercury is seven times the volume expansion coefficient of glass). 3- Submerge reservoir (A) in a beaker filled with water and pour mercury in the free end (C), until it rises in the other branch to mark (X). 4- Heat water in the vessel to the boiling point and wait until the temperature settles, and the mercury level in the branch connected to the reservoir stops decreasing.

Gas Laws 146

5- Move the free end (C) upwards until the the mercury level in the other branch rises to the same mark X. Then, measure the difference in height between the mercury levels in the two branches (h). From this, determine the pressure of the enclosed air P, which is equal to the atmospheric pressure (cm Hg) plus h, i.e., P=Pa+h

6- Move the branch (C) downwards and stop heating. Then let the reservoir cool down to nearly 90Ë&#x161;C. Then move the branch (C) upwards until the mercury level in the branch connected to the reservoir rises to mark X.

Other Forms of Charle’s and Jolly’s (pressure) laws :

Unit 3:

1- Refering to (Fig 6-9),

we note that the triangles ABC and ADE are similar. Therefore: BC = (Vol)1 DE = (Vol)2 AC = T1

Heat

AE = T2 Vol ∝ T

Vol = const. T ∴ (Vol)1 = (Vol)2 T1 T2

Chapter 6:

Thus, at constant pressure, the volume of a fixed mass of gas is directly proportional to its temperature on the Kelvin scale. This is another formulation of Charle’s law. 2- Using Fig.(6-10), the following relation can be obtained in a similar way:

That is

Gas Laws 150

(6 - 6)

P1 P = 2 Z1 Z2

(6 - 7)

P = const Z or P ∝ T Thus, at constant volume, the pressure of a fixed mass of gas is directly proportional to

its temperature on the Kelvin scale. This is another form of pressure (Jolly’s) Iaw.

2- The temperature of a normal human body on the Kelvin sca1e is about: d) 373˚K

b) 37˚K

e) 310˚K

c) 100˚K

3- The volume of a given mass of a gas is :

a) inversely proportional to its temperature at constant pressure.

Unit 3:

a) 0˚K

b) inversely proportional to its pressure at constant temperature. d) directly proportional to its temperature at variable pressure.

e) inversely proportional to its pressure at variable temperature.

4- The pressure of a gas at 10˚C is doubled if it is heated at constant volume to : a) 20˚C

d) 293˚C

b) 80˚C

e) 410˚C.

Heat

c) directly proportional to its pressure at constant temperature.

c) 160˚C

5- If we press a gas slowly to half of its original volume: b) its temperature is decreased to half its value. c) its pressure will be half of its original value. d) the velocity of its molecules is doubled. e) the pressure of the gas is doubled.

III) Eassy questions

constant pressure is the same for all gases?

2- Describe an experiment to find the pressure coefficient of a gas at constant volume and that it is the same for all gases.

3- How can you verify Boyle’s law experimentally?

4- How can you show that pressure of a gas increases by raising temperature at constant

Gas Laws

1- How can you show experimentally that the volume coefficient of expansion at

Chapter 6:

a) its temperature is doubled.

volume?

157

Questions and Drills Unit 3:

I) Complete (Fill in the spaces) :

Which phrase (a-e) completes each of the next following statements (1-3)? a) increases by a small value.

b) decreases by a small value.

Heat

c) remains constant. d) doubles.

e) dereases to its half value.

1- If the pressure of a gas is doubled at constant temperature. So its volume...............

2- If a barometer is transferred to the top of a mountain above the sea level, the length of mercury in the barometer .............

Chapter 6:

3- If the absolute temperature of a gas is decreased to be half its original value at constant pressure, so its volume ...............

II) Choose the correct answer:

1- The increase of the temperature of a carâ&#x20AC;&#x2122;s tire during motion leads to : 1) an increase in air pressure inside the tire. 2) an increase of air volume inside the tire.

Gas Laws

3) a decrease of the contact area of the tire with the road. Choose the correct letter (a-e)

a) (1, 2, 3) are correct.

b) (1, 2) only are correct. c) (1, 3) only are correct. d) 3 only is correct. e) 1 only is correct

156

5- How can you determine experimentally the absolute zero?

Unit 3:

6- Explain the meaning of zero Kelvin and the absolute temperature scale. 7- Deduce the general gas law.

IV) Drills:

1- The temperature of one liter of gas is raised from 10˚C to 293˚C at constant pressure, find its volume.

(2 liters)

Heat

2- A container containing air at 0˚C is cooled to (-91˚C). Its pressure becomes 40 cm Hg. Find the pressure of the gas at 0˚C.

(60 cm

Hg.)

3- The volume of a quantity of oxygen at 91˚C under 84 cm Hg is 760 cm3 (S.T.P). Find

Chapter 6: Gas Laws 158

its volume at 0˚C under a pressure of 76 cm.Hg.

(630 cm3)

4- A flask containing air is heated from 15˚C to 87˚C. Find the ratio between the volume of air that goes out from it to its original volume.

(25%)

5- A tire contains air under pressure 1.5 Atm at temperature (-3˚C). Find the pressure of

air inside the tire if the temperature is raised to 51˚C, assuming that the volume is constant.

(1.8 Atm)

6- An air bubble has a volume of 28cm3 at a depth of 10.13 m beneath the water surface. Find its volume before reaching the surface of the water, assuming that the temperature at a depth of 10.13 m, is 7˚C and that at the surface is 27˚C.

(Let g = 10ms-2 , Pa = 1.013 × 105 N/m2, ρ = 1000 kg/m3)

(60 cm3)

Different quantities of substances - even as small as 1cm3 - contain a huge number of atoms or molecules. It is convenient to express such numbers in terms of a unit called mole or gram mole. It is agreed upon that a mole (or gram mole) of any substance contains the number of atoms or molecules equal to the number of atoms in 12 gram of carbon .It is found experimentally that 12 gram of carbon contains 6.023x1023 carbon measuring unit of a quantity of matter in the international system of units. Although the original definition of mole was associated with carbon, yet the concept of the mole is generalized to any ensemble of particles, such that one mole contains Avogadro’s number of these particles. Thus, a mole of iron contains 6.023 x 1023 iron atom, and one mole of

water contains 6.023 x 1023 water molecules. In general, the mass of one mole of any

substance equals numerically the atomic or molecular mass (in grams) of this substance, i.e., the mass of one mole of carbon is 12 gram, one mole of oxygen is 32 gram and one mole of water is 18 gram. Each of these quantities contains the same number of atoms or molecules, which is Avogadro’s number. Thus, the mole of any substance is defined as the quantity of this substance in grams, which equals the atomic or molecular mass of the substance. Oxygen gas has atomic mass 16, i.e the molecular mass of a mole of Oxygen is 32. Avogadro’s law

Avogadro’s law states that: Equal volumes of different gases contain equal number of molecules under the same conditions of temperature and pressure. Alternatirely, Each mole of any gas at

Chapter 7: The KineticTheory of Gases

atoms. This is known as Avogadro’s number. NA The mole is, thus, introduced as a

Unit 3: Heat

Avogadro’s number

161

Chapter 7: The KineticTheory of Gases Unit 3: Heat

Chapter 7

The KineticTheory of Gases

Overview

To study the behavior of gases and explain their different laws, one can make use of the kinetic theory of gases. This theory is based on postulates given below: 1) A gas is composed of molecules which we shall regard as very minute perfectly elastic spheres obeying Newtonâ&#x20AC;&#x2122;s law of motion. 2) The intermolecular distances are relatively large, hence, the volume of the gas molecules is negligible compared to the volume occupied by the gas itself(the volume of the container). 3) The forces of intermolecular attraction between the gas molecules are very weak due to the large intermolecular distances, so they are negligible. Therefore, the potential energy of the molecules is zero. This means that the gas molecules do not interact with each other. Thus, the mean distances which a molecule moves before colliding with another (called mean free path) does not depend on the mass or type of molecule, and is statistically the same for all gases under the same conditions. Therefore, a certain volume of any gas at S.T.P. contains the same number of molecules regardless of the gas type. 4) Gas molecules are in continuous random motion due to the collisions between each other and due to their collisions with the walls of the container. The molecules move between any two successive collisions in straight lines. 5) The collisions between the gas molecules are perfectly elastic, i.e., the total kinetic energy of the gas molecules remains constant before and after the collisions. 6) The gas is in thermal equilibrium state with the walls of its container.

160

on the level of the molecules is called the microscopic point of view. This has led to the kinetic theory of gases. We start by the following postulates: 1) A gas contains a huge number of molecules in random motion. 2) The size of the molecule is much smaller than the total volume of the gas. lost. 4) Interactive forces among the molecules are negligible, except at collision. Hence, there is no potential energy involved. 5) Molecules obey Newtonâ&#x20AC;&#x2122;s laws. Consider one of the gas molecules in a box in the form of a cube whose side is l (Fig 7-1). The mass of the molecule is m, its average velocity is v and the x component of velocity is vx. The pressure exerted by the gas on the walls

of the box originates from the collision of the gas molecules with the walls. The

pressure P is the force per unit area P = F/A, where A= l2, and F is the force which the molecule applies to the wall. The linear momentum is PL .The change in the linear momentum for a molecule 6PL is the difference between the linear momentum before and after collision with the wall: F=

6P L 6t

Because the collision is elastic, the velocity after collision in the x direction is -vx . The change

in linear momentum transmitted to the wall 6PL is opposite to the change in the linear

Chapter 7: The KineticTheory of Gases

3) Collisions among the molecules and with the walls are elastic, and hence no energy is

Unit 3: Heat

In reality, a gas contains a huge number of molecules moving randomly. Studying the gas

163

Chapter 7: The KineticTheory of Gases Unit 3: Heat

0˚C and 1 Atm (S.T.P) occupies a volume of 22.4 liters. Therefore, each mole of any gas contains the same number of molecules at S.T.P. This number is given by NA = 6.023 x 1023 molecules/Mole. To understand this, let us say we have NA molecules of oxygen (molecular mass is 32 x mH where mH is the mass of the hydrogen atom). The Mass of one mole of oxygen is mH x 32 x NA. Thus, the mass is a constant (NAmH) times 32. If we have NA hydrogen molecules, then their mass is 2 x (NAmH). In other words, 32 gram oxygen (one mole oxygen) and 2 gram hydrogen (one mole hydrogen) have the same number of molecules (NA). Also, at the same temperature and pressure, the interatomic or intermolecular distances are the same on average, due to the random motion of gas molecules and non existence of attractive forces or potential energy between them, which are the postulates of the perfect gas. Consequently, any mole of any gas at S.T.P. occupies the same volume, which is found experimentally to be 22.4 liters. Gas density:

Knowing the number of molecules (N) in a given volume of gas Vol and the mass of one

molecule(m), we can calculate the density of the gas (ρ) from the relation: ρ=

Nm kg/m3 Vol

(7 - 1)

Gas pressure

Our study of the properties of gases in terms of pressure, volume and temperature have led to the deduction of the gas laws. Such a study is called the macroscopic point of view.

162

Chapter 7: The KineticTheory of Gases

momentum of the molecule.

Unit 3: Heat

between collisions tav as a substitute measure. In this case, the force is the average force

164

6PL = -mvx - (mvx) = - 2mvx

The change in the linear momentum delivered to the wall 6PL is opposite to the change in the linear momentum of the molecule.

6P = 2mv L

The force with which the molecule acts on the wall is given by : F=

l Fig (7 â&#x20AC;&#x201C; 1 )

6P L 2mv X = 6t 6t

Number of gas molecules in cube

where 6t , is the time of contact between the molecule and the wall upon impact. The impulse Iimp delivered by the molecule to the wall is given by: Iimp = F6t = 6PL = 2mvx Because the time interval 6t is very small and indeterminate, we can take the time acting all the time such that:

Fav tav = F6t where tav is the average time between for collisions of a molecule with the walls: t av =

2l vx

nRT =

 mv 22  2 nN A  av   2  3

mvav22 23  R  T = 23  N  2 A

(7 - 7)

where R is constant for every molecule and is called Boltzmann constant(k) : NA

(7 - 8)

32 22 kT = 1 mv uT mv av 23 2

(7 - 9)

From equations (7-7) and (7-8) ,

This relation ties the macroscopic theory of a gas to the microscopic model .It is to be noted that temperature T (a macroscopic quantity) measures the average kinetic energy of a molecule (a microscopic quantity). As the temperature increases, the kinetic energy increases. It is to be noted from equation (7-9) that each direction of motion (dimension or degree

of freedom) has associated with kinetic energy 12 kT. In fact this relation does not apply

solely to gas molecules but also to electrons in a metal , and even to any ensemble of particles in random motion. We might be tempted to believe that at T = 0 ˚K, the kinetic energy, and hence the velocity is zero, i.e., everything stops at absolute zero. In reality, we cannot claim so, because at absolute zero the equations of the ideal gas are no longer valid. It is known that gases are transformed in turn to liquids at low temperatures (chapter 8). Einstein showed that even at absolute zero, there is still energy called rest energy. In such a case, the above equations become inapplicable.

Chapter 7: The KineticTheory of Gases

ku = R = 1.338× γ 10 −23 J J/˚K / k˚ NA

Unit 3: Heat

167

Chapter 7: The KineticTheory of Gases Unit 3: Heat 166

Pp =

1 Nm 2 vav 3 Vv ol

Referring to equation (7-1),

∴ Pp = 1 ρv2 av 3

(7 - 2)

(7 - 3)

where ρ is the gas density and v2 is the mean square velocity of the gas molecules. The scientific concept of temperature :

From equation (7-2), multiplying the numerator and denominator by 2 :

2 PVol = 2 N mvav 3 2

We note that the number of gas molecules N is the number of moles times Avogadro’s number NA : N = nN A

mvav22   mv 2 PVPv ol = 3 nN A  2 

(7 - 4)

Boltzman

From the laws of the ideal gas, the macroscopic relation is given by : PV

= nR

PVPv == nRT nRT ol

(7 - 5) (7 - 6)

where R is the universal gas constant = 8.314 J/mole˚K, n is the number of moles in the substance. This relation is based on experimental observations, while the microscopic relation is based on theoretical deduction. We must equate the right hand side of both equations (7-4) and (7-6).

constant equals 1.38 x 10-23 J/˚K, and the atmospheric pressure is 1.013 x 105 N/m2 Solution

There are two methods to solve this problem. The first method: Po(0˚C)P =

1 lρ vo2 3 oo

1 MM 2 V 2 v 1 3 Vol o

3Po(0 C) (Vol) o

where M is the mass of one mole of the gas and Vol (0˚C), Po(0˚C) and the values of the

volume and pressure are those at S.T.P. and ρo is the density of the gas and vo is average

velocity at S.T.P.

v VV o

3 x 30.76 x 13600 9.8 xx 10 22.-3 = 493 m/s x 1.013 x 105 xx22.4 0.028 0.028

The second method: 1 m v22 = 3 KT k o 2 2 1 2 3KTN k NA = = o M

M 2 3 vo = KT kT NNAA 2 3 x 1.38 x 10 -23 x 273 x 6.02 x 10 23 0.028

= 493 m/s

Chapter 7: The KineticTheory of Gases

3Po(0 C) ρο

Unit 3: Heat

and occupies 22.4 liters at S.T.P., and that Avogadro’s number equals 6.02 x 1023, Boltzmann’s

169

I) Essay questions : 1. State the main postulates of the kinetic theory of gases. 2. On the basis of the postulates of the kinetic theory of gases, show how to prove that the gas pressure P is given by the relation:

1 Ď v2 3

where Ď is the gas density and v2 is the mean - square speed of its molecules. 3. Using the previous relation, show how to find expressions for each of the following: a) the root - mean - square speed of the gas molecules. b) the concept of the gas temperature. c) the average kinetic energy of a free particle. 4. A uniform cubic vessel of side length l has gas whose molecule has mass m moving in the x direction with velocity vx, and collides with the walls of the vessel in perfectly elastic collisions.

a) What is the linear momentum of the molecule before collision? b) What is the linear momentum of the molecule after collision? c) What is the change in linear momentum of the molecule on collision? d) What is the distance traveled by the molecule before the next collision with the walls of the vessel? e) What is the number of the collisions with the walls of the vessel per second made by

Chapter 7: The KineticTheory of Gases

Unit 3: Heat

Questions and Drills

171

Chapter 7: The KineticTheory of Gases Unit 3: Heat 170

In a Nutshell • The mole of any substance equals the molecular mass number in grams. • Avagadro’s number is the number of molecules in one mole and equals 6.023 x 1023 • The density of a gas is given by : ρ=

Nm kg/m3 Vol

where N is the number of gas molecules in a certain volume Vol and m is the mass of one molecule. • The pressure of a gas in a container is calculated from the relation: P= 1 lρ vav2 3 where ρ is the density of the gas and v2 is the mean - square speed of the molecules. • The average kinetic energy of one of the gas molecules is directly proportional to its absolute temperature in ˚K and the relation between them is: 1 3 m v 2 = KT k 2 2 where k is Boltzmann’s constant = 1.38 x 10-23 J/˚K

Chapter 7: The KineticTheory of Gases Unit 3: Heat 172

the molecule? f) What is the total change in linear momentum of one molecule per second due to its successive collisions with the walls of the vessel? g) What does the above quantity represent? h) If NA is the number of the gas molecules in the container, what will be the total force acting on the internal surface of the vessel?

II) Drills: 1. Hydrogen gas in a vessel is at S.T.P. Calculate the root mean square speed of its molecules, given that a hydrogen mole = 0.002 kg and Avogadro’s number = 6.02 x 1023, 1 Atm = 1.013 x 105N/m2

(1844 m/s)

2. What is the change in linear momentum of the hydrogen molecule in the above problem on each impact perpendicular to the walls of the vessel?

(1.224 x 10-23kg.ms-1)

3. Calculate the average kinetic energy of a free electron at 27˚C, given that Boltzmann’s constant k = 1.38 x 10-23 J˚K-1.

(6.21 x 10-21J).

4. Using the data given in the previous problem, find the root mean square speed of a free electron if its mass is 9.1 x 10-23 kg.

(1.168 x105m/s)

5. Find the ratio between the root mean square speed of the molecules of a certain gas at temperature 6000˚K (Sun’s surface) and that at temperature 300˚ K (Earth’s surface). (4.472)

Low temperatures may be achieved by drawing or removing energy out of the matter,

This can be done in various ways. The simplest is to establish contact with another precooled substance. Ice or dry ice ( solid CO2) or liquid air may be used. Temperatures of

77°K ( liquid nitrogen temperature) have been widely used. Liquid helium temperature (4.2°K) has even been reached . From the concept of latent heat of vaporization, the liquid

gas draws energy from the material to be cooled in order for the liquid gas to evaporate to be gaseous again. This results in the cooling of the substance required.

Some liquid gases have the property of superfluidity, i.e., they

can flow without resistance (or without friction) at temperatures

close to obsolute zero. Helium liquid has this property. In other words, it loses viscosity completely at such low temperatures. It can even flow upwards uninterruptedly against gravity or friction along

the walls of its container (Fig 8–1). It also has very low specific heat and is one of the best thermal conductors.

Fig (8-1)

Superfluidity

Dewar’s Flask It is a glass or metallic container evacuated to prevent heat

transfer. It is used to store liquid gases, since it is designed to

prevent heat losses by conduction, convection and radiation. It consists of a double walled pyrex container with silver plated walls

to minimize heat transfer by radiation. The spacing between the

walls is evacuated to prevent conduction and convection, e.g., as in

vacuum reflecting surfaces hot or cold liquid

Fig (8-2)

Dewar’s flask

Cryogenics (Low Temperature Physics

Superfluidity

Unit 3: Heat Chapter 8:

Mechanism of achieving low temperatures

175

Cryogenics (Low Temperature Physics Unit 3: Heat Chapter 8: 174

Chapter 8

Cryogenics (Low Temperature Physics)

Overview Cryogenics (or low temperature physics) is a branch of physics dealing with the cases when temperature approaches absolute zero (-237°C) or 0˚K . The temperature scale used in low temperature physics is the Kelvin temperature scale (the absolute temperature scale) which is based on the behavior of the ideal gas.

Van Der Waals’ Effect: One of the postulates of the kinetic theory of gases upon

which the ideal gas laws are based is neglecting the interactive (attractive) forces among the molecules of the gas, as well as

neglecting the size or the volume of the gas molecule in comparison with the volume of the container. The properties of

a real gas differ from those of the ideal gas, as the gas density increases. Interaction among gas molecules can no longer be neglected. This interaction is called van der Waals’ effect . It is

unlike chemical interaction between atoms leading to the

Van der Waals

formation of molecules. The attractive forces among the molecules become important as they lead to the liquefaction of the gas under high pressure. Due to the high pressure, van

der Waals’ interaction takes over, where two molecules approaching each other attract together, and eventually attract more molecules, until the gas switches to the condensed state of matter (liquid or even solid).

This mechanism explains the liquefaction of gases, which has led to achieving very low

temperatures approaching near absolute zero.

Cryogenics (Low Temperature Physics Unit 3: Heat Chapter 8: 176

a thermos bottle (Fig 8–2). It is used to store liquid nitrogen (boiling point 77°K) and liquid oxygen (boiling point 90°K). As to helium (boiling point 4.2° K and low specific heat), it is stored in a double Dewar’s flask, one inside the other. The spacing between the two flasks is filled with liquid nitrogen (due to the low specific heat and boiling point of helium).

How does a refrigerator work ? From the law of conservation of energy, if a gas acquires thermal energy Qth it is used up

in one of two ways:

1) an increase in internal energy U which is manifested by an increase in temperature. 2) work done by the gas molecules W. There are two types of heat transfer. One is at constant temperature with the surroundings, i.e. , at constant internal energy (6U = 0). In this case, the acquired energy is transformed in full into mechanical work done by the gas. This is called an isothermal process. The second type is performed when the gas is thermally isolated with its surroundings. So, it can neither acquire nor lose heat. In this case, Qth = 0 The work done by the gas must be at the expense of its internal energy. If W is positive, the gas does the work and the internal energy decreases (6U is

negative), i.e., the gas cools down. If work is done on the gas, then W is negative, so the internal energy increases and its temperature rises. The process when Qth = 0 (W is

positive or negative) is called adiabatic process. A refrigerator is an application to both isothermal and adiabatic processes , and the coolant (refrigerant) or the cryogenic liquid used is freon (boiling point- 30°C) or its substitutes.

resistance vanishes. This occurs at a critical (transitional)

temperature (Fig 8â&#x20AC;&#x201C;4). Materials having this property are called superconductors. If current happens to flow in a superconductor, it continues to flow even if the voltage

difference is removed. It will continue to flow for years, as it is met by nearly zero resistance.

critical temperature

Such a metal will not be heated by current flow. No energy is consumed in compensating electrical energy

Superconductivity

Superconductors can be used to pick up weak wireless signals. Therefore, they are used in the electric circuits of satellites. It is interesting to notice that if a permanent magnet is placed over a disk of a superconducting material, then the current in the superconductor generates a magnetic field, which is always repulsive with the external magnet, so the permanent magnet remains hanging in the air. This is called Meissner effect (Fig 8 â&#x20AC;&#x201C; 5).

Fig (8-5)

Meissner's effect

Fig (8-6)

Magnetically levitated train

Cryogenics (Low Temperature Physics

associated with an electric current, as in ordinary resistors.

Fig (8-4)

Unit 3: Heat Chapter 8:

metallic compounds) becomes very high, i.e., the electrical

179

Cryogenics (Low Temperature Physics Unit 3: Heat Chapter 8: 178

through a condenser (an apparatus outside the refrigerator). Heat exchange occurs in which

heat energy in the gas is radiated out to the surroundings being at a lower temperature. The high pressured gas condenses and becomes a liquid at constant temperature (isothermal

process). The liquid refrigerant is returned to the refrigerator once more. Before it enters into the freezer compartment, the refrigerant is forced to expand in an adiabatic process

through the expansion valve. In this case, the liquid molecules diffuse from a high pressure region into a low pressure region. The refrigerantâ&#x20AC;&#x2122;s volume increases and does work in

doing so against the spring in the valve. Thus, work is done at the expense of the internal

energy of the refrigerant. Since no external heat exchange is allowed (Qth= 0 or W is positive so â&#x2C6;&#x2020;U is negative). The internal energy of the liquid decreases so does its temperature. Thus, the refrigerant is now back to be a liquid as it was when we started the

cycle. The cycle repeats. The final result is the expulsion of thermal energy from the cabin to the condenser outside the refrigerator. Therefore, the refrigerator cools down. Of course,

the refrigerator must be well insulated. It should be noted that the electric energy needed by

the refrigerator throughout the cycle is the energy consumed in operating the piston. The internal energy one throughout complete cycle must remain unchanged (â&#x2C6;&#x2020;Unet = 0). The

compressor is, thus, a heat pump that transfers the heat from the refrigerator to the outside

through the work done by the compressor Wnet. Thus, the electric energy E required for

operation:

E = Wnet = (Qth)

net

Superconductivity In 1911, i.e., 3 years after the liquefaction of helium, Onnes and his assistants discovered superconductivity. When the temperature reaches a few degrees above absolute zero, the electrical conductivity of some metals (platinum, aluminum, zinc, lead, mercury and some

onnes

Magnetic resonance imaging ( MRI )

One of the most popular and safest tools in medical diagnosis nowadays is magnetic resonance imaging (MRI). In this method, a magnetic field affects the nuclei of hydrogen in the body. Exciting these nuclei with an alternating magnetic field, waves are radiated from the excited hydrogen nuclei, which are indicative of water ( hydrogen ) localization (oedemas and tumors).An internal image of the body can be made (Fig 8â&#x20AC;&#x201C;7), which helps identify such lumps. Superconducting magnets are used to counteract the huge energy losses in normal magnets, hence heating effects are reduced.

Cryogenics (Low Temperature Physics

Fig (8-7)

Unit 3: Heat Chapter 8:

Learn at Leisure

MRI image

181

Cryogenics (Low Temperature Physics Unit 3: Heat Chapter 8: 180

The reason for this is that superconductors belong to a class of materials called diamagnetic materials in which the magnetic field inside the material is zero. Therefore, an external magnet induces current in the superconductor which creates a magnetic field inside the superconductor in an opposite direction, so that the net magnetic field inside the superconductor is zero . This phenomenon has been put to use by designing a high speed (magnetically levitated) train. The train carries coils of a superconducting material. When the train moves, it induces current in fixed coils, which produces a magnetic field repelling the inducing field. The train is raised above the rails for a few centimeters . This levitation eliminates friction (Fig 8 â&#x20AC;&#x201C; 6), hence increases the train velocity. The levitated train may reach a velocity of 225 km/h . The discovery of room temperature superconductive materials will lead to expansion in the applications of superconductivity, since no cooling is then needed . Superconductors are also used in electric power plants and in transmission lines, where voltage losses are eliminated due to the vanishing resistance .

1) Explain each of the following phenomena : a) Van Der Waalsâ&#x20AC;&#x2122; effect . b) low temperature phenomena . c) superfluidity of some liquefied gases . d) superconductivity . 2 ) Give reasons : a)the use of two Dewar's flasks to store helium . c) helium can flow upwards along the walls of its container without stopping . d) a levitated train has been designed with a very high speed (225 km/hr ) . e) a magnet may remain hanging up above a superconductor regardless of the polarity . 3) State the most important applications for each of the following : a) Dewar's flask . b) superconductors . 4) Illustrate the difference between : a)chemical reaction and Van Der Waalsâ&#x20AC;&#x2122; reaction . b) the helium liquid and the nitrogen liquid .

Cryogenics (Low Temperature Physics

b) the spacing between the double walls in a Dewar's flask is evacuated .

Unit 3: Heat Chapter 8:

Questions

183

Cryogenics (Low Temperature Physics Unit 3: Heat Chapter 8: 182

In a Nutshell • Low temperature physics deals with the study of materials at temperatures near absolute zero • Van der Waals’ effect expresses the mutual interaction between molecules and is different from chemical interaction, which leads to the formation of molecules. • The mechanism for achieving very low temperatures depends on drawing energy from the material. This may be done by putting the material to be cooled in contact with a cooler material such as a liquefied gas. • Superfluidity : Some liquefied gases can flow without resistance or without friction at temperatures close to absolute zero. Helium is a superfluid , i.e. its viscosity vanishes. It can also flow up along the walls of the container against gravity and friction and has low specific heat. • Dewar's flask is a glass or metallic container evacuated to prevent heat transfer. It is used to store liquefied gases such as nitrogen, oxygen and helium and so on. • Superconductivity: Some metals have excessive electrical conductivity (zero resistance) at very low temperatures. • Meissner effect : If a permanent magnet is placed above a superconductor, the current induced in the superconductor generates a magnetic field which repels the permanent magnet so the permanent magent remains hanging in the air.

sectional area (m2) and (ρ ) is the resistivity (Ωm).The electrical conductivity of a certain e

is the reciprocal of the resistivity σ = ( ρ1 ). e

6) Ohm’s Law:

difference across its terminals at a constant temperature :

V = IR

7) as a convention, the direction of the electric current always goes from the positive terminal to the negative terminal outside the source into a closed electric circuit. It is opposite to the direction of motion of electrons. It is called the conventional direction of current.

Connection in series

Resistors are connected in series to obtain a higher resistance (Fig9–1). The equivalent resistance of a group of resistors connected in series can be obtained in connecting these resistors in an electric circuit comprising a battery,an ammeter,a rheostat (variable resistor) and a switch (Fig 9-2). The circuit is closed and the rheostat is adjusted so that an appropriate current I is passed. The voltage difference across each resistor is measured (V1

across R1, V2 across R2, V3 across R3) as well as the total voltage (V), which is equal to the

sum of the voltage differences across the resistors in the series circuit and this is called

Electric Current and Ohm’s Law

Fig (9 – 1)

Chapter 9:

Connecting resistors Firstly: series connection

Dynamic Electricity

The current intensity in a conductor is directly proportional to the potential

Unit 4:

material σ (Ω-1m-1)

Kirchhoff,s law

187

Secondly: Parallel connection bunch of large resistances (Fig 9 – 3). To obtain the equivalent resistance for a parallel ammeter and a rheostat all connected as shown (Fig 9 – 4).

Chapter 9:

Fig (9 – 3)

Connection in parallel

obtain an appropriate current in the main circuit of intensity I (A), which can be measured by the ammeter. The total voltage difference can then be measured across the terminals of the resistances by a voltmeter (V). The current in I3 in R3). In a parallel connection, the total current is determined by the smallest resistance. This case is similar to the flow of water in pipes.

Fig (9 - 4)

Measuring the equivalent resistance in a parallel connection

Electric Current and Ohm’s Law

We close the circuit and adjust the rheostat to

each branch is measured ( I1 in R1, I2 in R2, and

Dynamic Electricity

connection, the combination is included in an electric circuit comprising a battery, an

Unit 4:

The purpose of connecting resistors in parallel is to obtain a small resistance out of a

189

Electric Current and Ohmâ&#x20AC;&#x2122;s Law

Chapter 9: Dynamic Electricity

Unit 4:

Fig (9 â&#x20AC;&#x201C; 2)

Measuring the equivalent resistance in a series connection

V = V1 + V2 + V3

But ... V = IR

V1 = IR1 V2 = IR2 V3 = IR3 IR = IR1 + IR2 + IR3 R = R1 + R2 + R3

Thus, the equivalent resistance R of a group of resistors connected in series equals the sum of these resistances. It is to be noted that the largest resistance in the combination determines the total resistance in a series connection. If N resistances are connected in series each equal r then : R = Nr We conclude that if we want a large resistance out of a bunch of small resistances, we

simply connect them in series. 188

(9-1)

Ohm’s Law for a closed circuit and outside the cell to transfer an electric charge of 1C in the electric circuit. If we denote and the internal resistance of the cell by r, then VB = IR + Ir

VB = I (R + r) I=

VEB

R+r

intensity in a closed circuit is the emf of the total source divided by the total (external plus internal) resistance of the circuit.

Relation between emf and voltage across a source From Fig (9 – 5), we find

From this relation, we see that as I is decreased gradually in the circuit shown (Fig 9– 5),

-by decreasing the external resistance R- the voltage difference across the source increases.

When the current vanishes, the voltage difference across the source becomes equal to the

emf of the source. Hence, we may define the emf of a source as the voltage difference across it when the current ceases to flow in the circuit.

Electric Current and Ohm’s Law

V = VB - Ir

Chapter 9:

This is known as Ohm’s law for a closed circuit, from which we find that the current

Dynamic Electricity

the emf of a battery by VB, the total current in the circuit by I, the external resistance by R

Unit 4:

We know that the emf of an electric cell (battery - source) is the total work done inside

191

Unit 4:

Dynamic Electricity

Chapter 9:

Electric Current and Ohmâ&#x20AC;&#x2122;s Law

The smallest pipe ( the highest resistance) determines the flow rate in a series

connection, while the widest pipe (the least resistance) determines the rate of flow in a parallel connection, since it draws most of the water current. It is to be noted that : I=

V V V V , I1 = , I2 = , I3 = R R1 R2 R3

where R is the equivalent resistance, and V is the voltage difference across resistors

connected in parallel. The total current I is the sum of the branch currents. I1 + I2 + I3 . Thus:

1 1 1 1 = + + R R1 R2 R3 Hence, the reciprocal of the equivalent resistance R is the sum of the reciprocal of

resistances in the case of a parallel connection. In the case of two resistors in parallel, the equivalent resistance R is given by : R=

R1 R2 R1 + R2

(9 - 4)

When N resistances are connected in parallel each equal to r, 1 =N R r R= r (9 - 5) N Therefore, if we wish to obtain a small resistance out of a bunch of resistors, we simply

connect them in parallel. 190

V = V + V + V R R1 R2 R3

V1 = IR = 0.25 x 25 = 6.25V 1

V3 = IR3 = 0.25 x 85 = 21.25V

a) the current flowing in each resistor. b) the total resistance.

c) the current through the circuit.

solution :

R = 15.14 â&#x201E;Ś

c) The current flowing through the circuit I is : I = V = 45 = 2.972 A R 15.14

Electric Current and Ohmâ&#x20AC;&#x2122;s Law

I 3 = V = 45 = 0.529 A R 3 85 b)The total (equivalent or combined) resistance R is calculated as follows : 1 = 1 + 1 + 1 = 1 + 1 + 1 R R 1 R 2 R 3 25 70 85

Chapter 9:

a) the voltage difference across each resistor = 45V, since they are connected in parallel and the battery is of negligible internal resistance. The current flowing through each resistor is calculated separately as follows : I 1 = V = 45 = 1.8 A R 1 25 I 2 = V = 45 = 0.643 A R 2 70

Dynamic Electricity

2) If the resistors in the previous example are connected in parallel to the same battery, calculate:

Unit 4:

V2 = IR2 = 0.25 x 70 = 17.5V

193

Unit 4:

Dynamic Electricity

Chapter 9:

Electric Current and Ohm’s Law

It can be calculated also as the sum of the currents I1 , I2 , I3 flowing through all

resistors:

I = 1.8 + 0.643 + 0.529 = 2.972 A

3) In the figure shown above two resistors A and B are connected in parallel. The combination

is connected in series with a resistor C and a 18 volt battery of negligble internal resistance. If the resistances of A,B and C are 3Ω,6Ω,7Ω, respectively, calculate:

a) the total resistance. b) the current flowing through the circuit. c) the current through each of A and B.

Solution :

The equivalent resistance for the combination (A, B) is : R´=

R1 R2 = 3x6 =21 R1 + R2 3 + 6

The equivalent resistance for the combination (A,B)and( C) is : R = R´ + R3 = 2 + 7 = 9 Ω

The current I flowing through the circuit is : 194

Unit 4:

• For N equal resistances each r :

R= r N I=

VEB R+r

resistance.

Electric Current and Ohm’s Law

where VB is the emf of the source, r is its internal resistance and R is the external

Chapter 9:

• Ohm’s law for a closed circuit:

1 = 1 + 1 + 1 R R1 R2 R3

Dynamic Electricity

• In a parallel connection:

197

6) In the circuit shown :

Unit 4:

a) the ammeter reading is ..................... b) the voltmeter reading is ..................... a) the ammeter reading A1 is .....................

b) the ammeter reading A2 is ......................

II) Choose the right answer: Four lamps 61 each are connected in parallel. The combination is connected to a 12V

battery with a negligible internal resistance: a) 8A

b) 6A

c) 4A

d) 2A

e) 72A

2) The total charge leaving the battery in 10s is............... b) 60C

c) 40C

d) 20C

e)2C

d) 1A

e) 2A

3) The current in each lamp is .........

a) 2 A b) 8A

c) 2 A

4) The voltage difference across each lamp is.......... a) 3V

b) 12 V

c) 6 V

d) 2 V

e) 4 V

5) The total resistance of the four equal lamps is...........

a) 3 1

b) 241

c) 2 1

d) 61

e) 121

6) If the 4 lamps are connected in series, the total resistance is.............

a) 2 1

b) 241

c) 3 1

d) 61 e) 121

Electric Current and Ohmâ&#x20AC;&#x2122;s Law

a) 80C

Chapter 9:

1) The current in the battery equals..........

Dynamic Electricity

7) In the circuit shown

199

and a switch. If the internal resistance of the battery is 0.31, determine : resistance is infinite.

b) the reading of the voltmeter when the switch is closed.

(15 V) (13.5)

4) A student wound a wire of a finite length as a resistor. Then, he made another of the

same material but half the diameter of the first wire and double the length. Find the ratio of the two resistances.

(1:8)

3V across. Calculate the current if the copper resistivity is 1.79 x 10-8 1.m

(11.17 A) 6) A 5.71 resistor is connected across the terminals of a battery of 121 emf and 0.31 a) the current in the circuit . b) the voltage difference across the resistor.

(2. A) (11.4 V)

Electric Current and Ohmâ&#x20AC;&#x2122;s Law

internal resistance. Calculate:

Chapter 9:

5) A copper wire 30 m long and 2x10-6m2 cross sectional area has a voltage difference of

Dynamic Electricity

a) the reading of the voltmeter when the switch is open, assuming that the voltmeter

Unit 4:

3) The circuit shown in Fig (9 â&#x20AC;&#x201C; 5) consists of a 15 V battery, an external resistance 2.71

201

Chapter 10

Magnetic Effects of Electric Current and Measuring Instruments

In 1819, Hans Christian Oersted- a Danish physicist-brought a compass near a wire carrying an electric current. He noticed that the compass was deflected. When he turned the current off, the compass assumed its original position. The deflection of the compasswhile current was flowing through the wire- indicated that it was being acted upon by an external magnetic field.This discovery started a chain of events that has helped shape our industrial civilization. In this unit we are going to study the magnetic field of current- carrying conductors in the form of: a) a straight wire. b) a circular loop. c) a solenoid.

Magnetic field due to current in a straight wire: We can examine the pattern of the flux density surrounding a long straight wire carrying a direct current using iron filings sprinkled on a paper surrounding the wire in a vertical position.It will be noted that they become aligned in concentric circles around the wire, as shown in Fig (10-1).

Fig (10 â&#x20AC;&#x201C; 1)

Pattern of iron filings around a wire carrying current

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

Overview

Oersted

203

The figure shows that the circular magnetic flux lines are closer together near the wire

and farther apart from each other as the distance from the wire increases. Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

As the electric current in the wire increases, the iron filings rearrange themselves after

gently tapping the board such that the concentric circles become more crowded.

This indicates that the magnetic field due to the electric current passing through a

straight wire increases with increasing the current intensity and vice verse.

The magnetic flux density β measured in Weber/m2 or Tesla(B = φm where φm is the A magnetic flux, A is the area) a point near a long straight wire carrying current I can be determined using the formula:

B= µI 2 /d

This relation is called Ampere's circuital law, where d is the normal distance between the point and the wire, and µ is the magnetic permeability of the medium (in air it is 4π

x 10-7 Weber/Am). Thus, B is inversely proportional to d and directly proportional to I.

This is why it is advisable to live away from high voltage towers.

Ampere's right hand rule: To determine the direction of the magnetic field resulting from an electric current in a wire, imagine that you grasp the wire with your right hand such that the thumb points in the direction of the current. The rest of the fingers around the wire give the direction of the magnetic field due to the current (Fig 10-2).

204

(10-1)

Fig (10 – 2)

Right hand rule

Thus circular loop carrying current may be considered as a bar magnet (Fig 10-3)

Fig (10 – 5)

Right hand screw

A circular loop carrying current in the direction

direction of screwing.

of screwing.

Examples: Determine the magnetic flux density at the center of a circular loop of radius 11cm

carrying a current of 1.4 A. if the wire loop consists of 20 turns and µ = 4 π x 10-7 air

Weber/Am

solution:

-7 µ NI 4 / x 10 x 20 x 1.4 B= = 2r 2 x 0.11

4 x 22 x 10 -7 x 20 x 1.4 = = 16 x 10 -5 Tesla 7 x 2 x 0.11

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

Fig (10 – 4)

207

To study the magnetic field due to a circular loop (or a coil),iron filings are sprinkled

on the board as shown in Fig (3 -10). Tapping it gently, the filings arrange themselves as Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

shown in figure, from which we can notice that:

1. the flux lines near the center of the loop are no longer circular. 2. the magentic flux density changes from point to point.

3. the magnetic flux lines at the center of the loop are straight parallel lines

perpendicular to the plane of the coil. This means that the magnetic field in this region

is uniform. The flux density at the center of a circular loop of N turns and radius r carrying current

I is given by :

B= µ N I 2r

(10-2)

From this relation, the magnetic flux density at the center of a circular loop depends

on three parameters:

1. number of turns of the circular loop where B ∝ N.

2. current intensity passing through the circular loop where B ∝ I.

3. radius of circular loop where B ∝ 1r .

- Right-hand screw rule: To determine the direction of the magnetic field at the center of a circular loop or coil, imagine a righthand screw being screwed to tie along the wire in the direction of the current. The direction of fastening of the screw gives the direction of the magnetic flux at the center of the loop (Figs.10 -4, 10 -5 ) Thus, a circular loop carrying current acts as a magnetic dipole or a bar magnet. It is to be noted that no single poles exist in nature. They always exist in N - S pairs.

206

Magnetic field due to current in a solenoid: When an electric current is passed through a solenoid ( a long spiral or cylindrical coil) as Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

shown in Fig(10-6), the resultant magnetic flux is very similar to that as a bar magnet. As shown in Fig(10-6A), the magnetic flux lines make a complete circuit inside and outside the coil,i.e., each line is a closed path. The side at which the flux emerges is the north pole, the other side where the magnetic flux reenters is the south pole. The magnetic flux density in the interior of a solenoid carrying an electric current depends on : 1) the current intensity passing through the coil where B∝ I. 2) the number of turns per unit length where B ∝ n :

a) field pattern

b) polarity of the field using

Fig (10 – 6)

Magnetic field due to a solenoid

∴ B ∝ nI B = µ nI

where µ is the permeability of the core material. In this case, it is air

This relation may be rewritten as follow: 208

Ampere’s right hand rule.

Force due to magnetic field acting on a straight wire carrying current. If we place a straight wire carrying current

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

between the poles of a magnet,a force results which acts on the wire and is perpendicular to

both the wire and the field (Fig 10-7). The

direction of the force is reversed if we reverse the current or the magnetic field. In all cases, the force is perpendicular to both electric

current and the magnetic field. In case the wire is allowed to move due to this generated

Fig (10 – 7)

field. The direction of the force with which a

denotes the direction into the paper.

force, the direction of motion is perpendicular Force due to a magnetic field acting on a to both the electric current and the magnetic straight wire carrying current, mark”x” magnetic field acts on a current- carrying wire perpendicular to the field can be obtained by applying Fleming’s left hand rule.

Fleming’s left hand rule Form your left hand fingers as follows: the pointer and thumb perpendicular to each other and to the rest of the fingers. Make the pointer point to the direction of the magnetic flux, and the rest of the fingers- except the thumb- in the

the pointer is in the direction of the magnetic flux

the rest of the fingers are in the direction of the current.

the thumb is in the direction of motion or force

Fig (10 – 8)

Fleming's left hand rule.

direction of the current. Then, the thumb points to the magnetic force or motion (Fig 10-8). It is found that the force acting on a wire carrying current flowing perpendicularly to a

210

You can imagine what the direction of the force will be in different cases. The

mark

means out of the page, and the mark x means into the page.

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

b) a force exists for θ other than zero

a) the force vanishes when θ = 0 (wire and magnetic field are parallel)

Fig (10 – 9)

A wire carrying current in a direction inclinded by an angle θ to the magnetic field.

The force between two parallel wires each carrying current.

When a current I1 passes in a wire and a current I2 passes in another parallel wire, a force results between the two wires. This force is attractive if the two currents flow in the same direction. The force is repulsive if the two currents flow opposite to each other. We can calculate this force as follows:

(b) a) the two currents are in the same direction.

b) the two currents are in opposite directions.

Fig (10 – 10)

Force between two parallel wires each carrying current

212

b) top view (plan)of the rectangle

magnetic field.

when the coil is parallel to the field.

c) top view (plan) of the rectangle when the magnetic dipole moment is perpendicular to the field.

e) top view (plan) when the rectangle is perpendicular to the field or the

d) top view (plan) of the rectangle when the magnetic dipole moment

magnetic dipole moment is parallel

makes an angle Î¸ with the field.

to the field and the couple is zero.

Fig (10 â&#x20AC;&#x201C; 11)

A torque acting on a coil carrying a

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

a) the coil is parallel to the

current

215

The essential parts of this device are shown in Fig(10-12). It consists of a rectangle of

a thin wire coil wrapped around a light aluminum frame mountend on a soft iron core.

shaped (horse shoe) magnet. Its rotational motion is restrained by a pair of spiral control springs, which also serve as current leads to the coil. Depending upon the direction of

the current being measured, the coil and pointer rotate either in clockwise or counterclockwise direction. The permanent magnet's poles are curved so that the magnetic flux lines are radially directed. Thus, the magnetic flux density is constant and

perpendicular to the side of the rectangle irrespective of the angle of the coil. the deflection of the pointer is proportional to the current in the coil.

The current flows in the coil from the right side upwards, and emerges from the other

side. Then the magnetic force generates a torque which makes the coil rotate clockwise.

The pointer deflects until it settles at a certain reading when the torque is balanced with the spring torsion which is counterclockwise. Thus, at balance, we can read the current value. When the current is reversed, the pointer deflects in the opposite direction.

The galvanometer sensitivity: The galvanometer sensitivity is defined as the scale deflection per unit current Î¸

intensity passing through its coil i.e, sensitivity = Î&#x2122; degree/micro ampere (deg/ÂľA) .

Direct current (DC) ammeter : An ammeter is a device which- through calibrated scales- is used to measure directly

the electric current. A galvanometer is an ammeter of limited range due to its moving

coil sensitivity. To extend the range of the galvanometer, it is necessary to add a very low resistance, called a shunt Rs to be connected in parallel with the galvanometer coil

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

The frame is pivoted on agate bearings. The assembly rotates between the poles of a U

Rg as shown in Fig (10-13).

217

Applications: Measuning Instuments The sensitive moving coil galvanometer Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

A sensutive moving coil galvanometer is an apparatus used to detect very weak

currents in a circuit, measure their intensities and determine their polarities. Its principle

of operation depends on the torque that is generated in a current -carrying coil moving in a magnetic field. graduated scale

pointer

spiral spring

pointer magnet

iron core

b) top view

a) a simplified view of a galvanometer when the pointer is in the middle of the graduated scale pointer

uniform scale

aluminum frame

central spring

control spring coil

falcrum

soft iron core aluminum frame

c) a galvanometer converted to

permanent magnet radial magnetic field

d) top view

a milliammeter

(Fig 10 â&#x20AC;&#x201C; 12)

A moving coil galvanometer

216

soft iron core

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

magnet

pointer coil

b) use of a shunt resistance a) a DC ammeter is a galvanometer whose pointer deflects in one direction

The DC ammeter

Placing the parallel shunt assures that the ammeter as a whole will have a very low resistance, which is necessary if the current in the circuit is to be unaltered after connecting the ammeter in series. Most of the current in the circuit passes through the shunt Rs, while only a small current Ig passes in the galvanometer coil Rg. If the maximum current to be measured is I, which is the full scale deflection (FSD), then

I = Ig + Is Is = I - Ig

Because Rs and Rg are connected in parallel, the voltage difference across each is the same. I s Rs = I g Rg I R Rs = g g . Is The two equations can be solved simultaneously to find Rs. Thus, Rs =

218

Fig (10 â&#x20AC;&#x201C; 13)

I g Rg I - Ig

Ohmmeter

Rg=250Ω

galvanometer standard resistance

6565Ω

3000Ω

battery instrument terminals

Fig (10 – 15)

A circuit for calibrating an ohmmeter

Measuring a resistance depends on measuring the current passing through it by an

ammeter and the voltage drop across it by a voltmeter. If the current is I and the voltage drop is V, the resistance R from Ohm's law is R =V/I If the voltage is fixed and known, we may remove the voltmeter from the circuit and calibrate the galvanometer to give the value of the resistance directly (Fig10-15).As the resistance is increased, the current in the circuit decreases,and consequently, the galvanometer reading.The Ohmmeter shown (Fig10-15) is actually a microammeter which reads 400µA as a full scale deflection (FSD).Its resistance is 250Ω connected in series with 3000Ω, a variable resistance whose maximum value is 6565Ω, and a 1.5 V battery of negligible internal resistance.When we short circuit (sc) the terminals of the instrument (RX =0), current flows in the circuit.For this current to give FSD, the resistance 1.5 = 3750 Ω in the circuit must be: -6 400 x 10 The variable resistance must be adjusted to give FSD, when the variable resistance is 500 since 250+3000+500=3750Ω. Now, if the unknown resistance is introduced into the circuit, the current flowing will be less, and the pointer will deflect short of FSD.

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

variable resistance

221

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

voltage,it must be converted to a high-resistance instrument.The voltmeter must draw a negligible current, so that it will not affect the voltage drop to be measured.To do this,a large multiplier resistor is connected in series with the galvanometer as shown in Fig. (10-14).The voltmeter is connected parallel across the two points between which the voltage difference is to be measured. Let us call the resistance of the galvanometer coil Rg and the multiplier resistance Rm which is connected in series parrallel with Rg. The maximum current that passes through it is Ig, which is the current needed for the full

scale deflection (FSD) voltage V.

The voltage difference across the coil is :

Vg = Ig Rg

The maximum voltage drop to be measured is:

V = Ig Rg + Ig Rm = Vg + Ig Rm

Rm =

(10-8)

Example A galvanometer has an internal resistance of 0.1Ω and gives a full scale deflection

for a current of 1mA. Calculate the multiplier resistance necessary to convert this galvanometer to a voltmeter whose maximum range is 50V.

Solution

Vg = Ig Rg = 0.001 x 0.1 = 1 x 10-4 V Rm =

V - Vg 50 - 1 x 10-4 = 1 x 10-3 Ig = 49999.9 Ω

The total resistance of the voltmeter is : Rtotal = 49999.9 + 0.1 = 50000 Ω

220

V - Vg Ig

They depend on digital electronics (Chapter 15). All the above instruments measure

voltage or current in one direction (DC). Therefore, they are called DC/multimeters. But

Fig (10 â&#x20AC;&#x201C; 17)

Analog multimeter

Fig (10 â&#x20AC;&#x201C; 18)

Digital multimeter

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

if the current or voltage is AC, the instrument used is called AC/ multimeters.

223

Thus, we may calibrate the instrument in terms of the resistance to be measured. If Rx

= 3750Ω, the current in the instrument is 200 µA, which is 1/2 the maximum current, Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

222

and hence the deflection is 1/2 FSD.

1 3

If the resistance is doubled, i.e., 7500Ω., the deflection will be FSD. For three times 1 4

the total resistance, i.e., 11250Ω, the deflection will be FSD corresponding to a current

of 100µA. It is to be noted that the graduated scale used to measure the resistance (Fig 10-16) is opposite to the graduated scale for the current . This means that the maximum deflection corresponds to zero resistance (short circuit or sc). As the resistance increases, the deflection decreases. It is to be noted also that the scale is not linear. The spacings between the readings of the scale to the right are further apart than the readings to the left.

Rx(Ω)

IµA

400

3750

200

11250

100 0

Fig (10 – 16)

An ohmmeter has a nonlinear graduated scale

The instruments using a point, are called analog instruments. A combined instrument called multimeter can be switched around to measure voltage, curent and resistance (Fig 10-17). Another set of instruments now exist which depend on reading numerals, denoting voltage, current a resistance on a small LCD (liquid crystal display) without the need for a pointer. Such instruments are called digital multimeters (Fig 10-18).

a) the length of the wire. b) the current intensity.

d) the angle between the wire and the direction of the magnetic field.

• A moving coil galvanometer is an instrument used to detect, measure and determine the polarity of very weak electric currents.

• The operation of a moving coil galvanometer is based on the torque acting on a current loop in the presence of a magnetic field.

• The sensitivity of a galvanometer is defined as the scale deflection per unit current intensity flowing through its coil.

• The ammeter is a device which is used through a calibrated scale to measure directly the electric current.

• To extend the range of the galvanometer, a low resistor known as a shunt is connected in parallel with the coil of the galvanometer.

• The total resistance of the ammeter (with the shunt) is very small, therefore, it does not appreciably change the current to be measured in a closed circuit.

• The voltmeter is a device used to measure the potential difference across two points of

an electric circuit. It is basically a moving coil galvanometer having a very high resistance called a multiplier resistance connected in series with its coil.

• Since the total resistance of the voltmeter is very great, it does not affect much the flow of current through the element across which it is connected to measure its potential difference.

• The ohmmeter is an instrument which is used to measure an unknown resistance.

• An ohmmeter is basically a microammeter connected in series with a constant cell

resistance, a variable resistance and a 1.5 volt battery. If its terminals are in contact

(sc), the pointer gives full-scale deflection (FSD). If a resistor is inserted between its

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

c) the magnetic flux density.

terminals, the current flowing decreases. Hence, the pointer's deflection decreases, and indicates directly the value of the inserted resistor through a calibrated scale.

225

In a Nutshell Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

224

- Definitions and Basic Concepts:

• A megnetic field is produced around a current-carrying wire. • The intensity of the magnetic field produced around a current-carrying wire, increases, by : a) getting closer to the wire. b) increasing the current. • The direction of the magnetic field produced around a current-carrying straight wire is determined by Ampere’s right-hand rule. • The lines of force around a current-carrying wire forming a circular loop, resemble to a great extent those of a short bar magnet. • The magnetic flux density at the center of a current-carrying circular loop depends on: a) the number of loop turns. b) the current intensity in the loop. c)the radius of the loop. • The direction of the magnetic field at the center of a current-carrying loop is determined by the right-hand screw rule. • The magnetic field produced by a current flowing through a solenoid (coil of several closely spaced loops) resembles to a great extent that of a bar magnet. • The magnetic flux density at any point on the axis of a current-carrying a solenoid depends on : a) the current intensity. b) the number of turns per unit length. • Right-hand screw rule is used to determine the polarity of a solenoid carrying a current. • The unit of magnetic flux density is Web / m2, (Tesla or N/Am).

• The force exerted by a magnetic field on a current-carrying wire placed in the field depends on:

5) What is the magnetic flux density at a point on the axis of a solenoid of length 50 cm carrying a current of 2A and has 4000 turns?

(0.02 Tesla).

magnetic field of 0.4 Tesla flux density, such that the plane of the loop is parallel to the field. Calculate the torque acting on the loop.

(0.72 Nm)

7) A galvanometer's loop of 5 x 12 cm2 and 600 turns is suspended in a magnetic field of 0.1 Tesla flux density. Calculate the current required to produce a torque of 1 Nm. (2.78 A).

8) A loop of cross-sectional area 0.2 m2 and 500 turns,carrying a current of 10 A is placed at 30Ë&#x161; between the normal to its plane and a magnetic field of 0.25 Tesla flux density. Calculate the torque acting on the loop. (125 Nm). 9) The coil of an ammeter is capable of carrying current up to 40 mA. If the resistance of the coil is 0.51, and it is desired to use the ammeter for measureing a current of 1 A, What is the resistance value of the required shunt?

(0.0211)

10) A galvanometer gives full scale deflection at current 0.02 A, and its terminal voltage is 5 V. What is the value of the multiplier resistance required to make it valid to measure potential differences up to 150 V?

(72501)

11) A voltmeter reads up to 150 V at full scale deflection. If the resistance of its coil is

501 and the current flowing is 4 x 10-4 A. Calculate the resistance of the potential multiplier connected to the coil?

(3749501)

12) A galvanometer reads up to 5A and has a resistance of 0.1 1. If we want to increase its reading 10 times, what is the value of the required shunt resistor?

(0.01111).

Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

6) A rectangular loop (12 x 10 cm) of 50 turns, carrying a current of 3 A, is placed in a

13) An ammeter has resistance 301. Calculate the value of the required shunt resistor to 229

b) the coil of the moving coil galvanometer is attached to a pair of spiral springs. c) when the moving coil galvanometer is used as a voltmeter, a resistor of high Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

resistance is connected in series with its coil. d) an ammeter is connected in series with a circuit, but the voltmeter is connected parallel to it. e) connecting a constant resistor inside the ohmmeter. f) the cell connected to the ohmmeter should have a constant emf. 9) What is meant by each of: potential multiplier and shunt? What is the use of each? Deduce the rule related to each. 10) Explain how you can use the moving coil galvanometer to measure each of the electric current, the electromotive force and the electrical resistance.

II) Drills: 1) A coil of cross sectional area 0.2 m2 is placed normal to a regular magnetic flux of density 0.04 Weber/m2. Calculate the magnetic flux which passes through this coil. (0.008 Weber). 2) A wire of 10 cm length, carrying a current 5 A, is placed in a magnetic field of 1Tesla flux density. Calculate the force acting on the wire, when: a) the wire is at right angles to the magnetic field.

(0.5 N)

b) the angle between the wire and the field is 45Ë&#x161;.

(0.356 N)

c) the wire is parallel to the magnetic flux lines.

(0)

3) A straight wire of diameter 2 mm carries a current of 5A. Find the magnetic flux density at a distance of 0.2 m from the wire.

(5x10-6 Tesla).

4) A circular loop of radius 0.1 m carries a current of 10 A. What is the magnetic flux density at its center? (the loop has one turn). 228

(2/ x 10-5 Tesla)

decrease the ammeter FSD to one third (decrease the sensitivity), and determine also the total resistance of the ammeter and the shunt resistor. Unit 4: Dynamic Electricity Chapter 10: Magnetic Effects of Electric Current and Measuring Instruments

230

(15 1, 101).

14) A galvanometer of resistance 541, when connected to a shunt (a), the current flowing through the galvanometer is 0.1 of the total current. But if connected to a shunt (b), 0.12 of the total current flows through the galvanometer. Find the resistances of a and b.

(6 1, 7.3641)

15) A moving coil galvanometer of resistance 50 ohms gives full scale deflection at current 0.5A. How could it be converted to measure: a) potential differences up to 200V? b) electric currents up to 2A?

(350 1 in series). (16 2/3 1 in parallel)

16) A milliammeter of resistance 51 has a coil capable of carrying a current of 15 mA. It is desired to use it as an ohmmeter using an electric cell of 1.5V having internal resistance 11. Calculate the required standard resistor, and calculate the external resistance needed to make the pointer deflect to 10mA? Calculate the current that flows through it when connected to an external resistor of 4001 ? (941, 501, 3mA)

Faraday’s laws: From the above Faraday’s observations, one can conclude the following: 1) the relative motion between a conductor and a magnetic field in which there is time variation of the magnetic flux linked with the conductor, induces an electromotive force in the conductor. Its direction depends on the direction of motion of the conductor relative to the field. 2) the magnitude of the induced electromotive force is proportional to the rate by which the conductor cuts the lines of the magnetic flux linked with it, i.e., emf¡ _∝ 6 qm 6t where ∆φm is the variation in the magnetic flux intercepted by the conductor through the

time interval ∆t

3) the magnitude of the induced electromotive force is proportional to the number of turns N of the coil which cut (or link with) the magnetic flux., i.e., emf ∝ N

Thus, from the analysis of the above mentioned results, one can conclude the following relation: 6 qm emf¡ = - N (11 - 1) 6t

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

the coil, a deflection of the pointer was noticed in the opposite direction. This phenomenon is called "electromagnetic induction". According to this phenomenon, an electromotive force and an electric current are induced in the coil, when the magnet is plunged into or removed from the coil. As a result, Faraday concluded that the induced electromotive force and also the induced electric current were generatd in the circuit as a result of the time variation of the magnetic flux linked with the coil during the motion of the magnet. Moreover, the action of the magnet is met by a reaction from the coil.If the magnet is plunged into the coil, the induced magnetic field acts in a way to oppose the motion of the magnet. If the magnet is pulled out, the induced magnetic field acts to retain (or keep) the magnet in. Faraday concluded that the induced emf and current were generated in the circuit as a result of the time variation of magnetic field lines as they cut the windings of the coil while the magnet was in motion.

233

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 232

Chapter 11

Electromagnetic Induction

Overview It has been noticed that the passage of an electric current in a conductor produces a magnetic field. Soon after Oersted's discovery that magnetism could be produced by an electric current,a question arose, namely, could magnetic field produce an electric current ? This problem was addressed by Faraday through a series of experiments which led to one of the breakthroughs in the field of physics, namely, the discovery of electromagnetic induction. On the basis of such a discovery, the principle of operation and function of most of the electric equipment - such as the electrical generators (dynamos) and transformers depend.

Faradayâ&#x20AC;&#x2122;s Experiment: Faraday made a cylindrical coil of insulated copper wire, such that the coil turns were separated from each other. He connected the two terminals of the coil to a sensitive galvanometer having its zero reading at the mid point of its graduated scale, as shown in Fig (11-1) . When Faraday plunged a magnet into the coil, he noticed that the pointer of the galvanometer was deflected momentarily in a certain direction. On removing the magnet from

Fig (11-1a)

The magnet is plunged into the coil

Fig (11-1b)

The magnet is pulled out of the coil

above relation indicates that the direction of the induced etectromotive force or the induced current tends to oppose the cause producing it. This rule is known as Lenz's rule. Lenz’s rule The induced current must be in a direction such as to oppose the change producing it. Fig (11-2) illustrates a direct application of Lenz’s rule : The direction of the induced current in a straight wire: In one of his several experiments, Faraday showed that the induced current in a straight wire flowed in a direction perpendicular to the magnetic field. Many years later. Fleming concluded a simple rule: Fleming’s right hand rule Extend the thumb,pointer and the middle finger of the right hand, mutually perpendicular to each other. Let the pointer points to the direction of the field, and the thumb in the direction of motion, then the middle finger (with the thumb (motion)

rest of the fingers) will point to the direction of the induced current or voltage as shown in Fig (11-3).

point to (field)

rest of the fingers (induced current or voltage)

Fig (11 - 3)

Fleming’s right hand rule

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

This is known as Farady's law of electormagnetic induction. The negative sign in the

235

234

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

(S)

Fig (11 â&#x20AC;&#x201C; 2)

Lenz's law

6 I1 6t

(emf)ยก 2 = - M

6 I1 6t

(11 - 2)

where M is the coefficient of mutual induction (mutual inductance) of the two coils. Its unit is VsA-1 and is equivalent to what is calIed "Henry". Thus, the henry is the unit used to measure the inductance in general. The negative sign in equation (11-2) follows from Lenz's rule, namely, that the direction of the induced electromotive force (or the direction of the induced current) is such as to oppose the cause producing it. The coefficient of mutual inductanc between two coils depends on the following factors. 1. the presence of an iron core inside the coil. 2. the volume of the coil and the number of its turns. 3. the distance separating them. The transformer is considered as a clear example of mutual induction Experiment to study mutual induction One can study experimentally the mutual induction as follows: Connect one of the two coils in a circuit which contains a battery, a switch and a rheostat. One coil is called the "primary coil", while the other coil - connected to a sensitive galvanometer with its zero point at the middle of its scale - is known as the "secondary coil". Fig( 11-5). Let us do the experiment as follows: 1) Close the circuit of the primary coil, while plunging the primary coil into the secondary coil. One notices a deflection in the galvanometer in a certain direction, indicating the generation of an induced electromotive force in the secondary coil due to the variation of the number of magnetic flux lines linked with the turns of the secondary coil. On taking away the primary coil from the secondary coil, one notices that the pointer of the sensitive galvanometer is deflected in the opposite direction.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

(emf) _|

237

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 236

Mutual induction between two coils:

Fig (11 - 4A)

in case there is no current in the first coil, there is no emf in the second coil.

Fig (11- 4b)

at the instant of closing the ciruit of the first coil, an emf is generated in the second coil.

Fig (11 - 4c)

after the current in the primary is steady (the flux is steady) emf in the secondary coil: 0

If the two stationary coils are arranged such that one coil surrounds the other., i.e., one coil is plunged into the second one, or even one is placed in the neighborhood of the other as shown in Fig.(11-4), then the variation in the intensity of the electric current in one of the two coils (opening and closing the switch) will induce an electromotive force in the other coil, according to Faradayâ&#x20AC;&#x2122;s law. This induced electromotive force is proportional to the rate of change in the magnetic flux linked with the other coil. Since the magnetic flux is proportional to the intensity of current in the first coil.

II. The pointer of the galvanometer deflects in the opposite direction in the following cases: a) on the withdrawal of the primary, or taking it far away from the secondary coil. b) on decreasing the intensity of the current in the primary. c) on switching off the primary circuit. In the above cases, he intensity of the magnetic field affecting the secondary coil decreases and the magnetic flux linkage decreases. The induced emf in the secondary coil decreases as the affecting field decreases with time. The direction of the induced electromotive force (and the induced current) is in the forward direction, so as to produce a magnetic field in the same direction as the current in the primary. This in turn resists the decrease in the affecting magnetic field. All these observations clarify Lenz's rule, where the direction of the induced current is such as to resist (or to oppose) the time variation causing it. Self induction of a coil: One can understand what is meant by self induction of a coil by connecting the coil of a strong electromagnet (a coil of large number of turns) in series with a 6V

battery VB

switch

battery, and a switch as shown in Fig (11-6). Current passes in the considered coil, due to which a strong magnetic field is formed, since each turn acts as a small magnet. The magnetic flux links

neon lamp

with the neighboring turns.

electromagnet coil

On switching off the circuit, it is noticed that an electric spark is passed between the two terminals of the switch. This is explained as follows.

Fig (11 - 6)

Effect of self induction in a coil

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

magnetic field will be in a direction as to resist the increase in the affecting magnetic field.

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 238

2) Plunge the whole primary coil to reside in the secondary one, then increase the intensity of the current in the primary coil. Notice the deflection of the pointer of the galvanometer in a certain direction. Decrease the current in the primary, and notice that the deflection of the

battery switch

rheostat

primary coil

pointer takes place in the opposite direction. This indicates the generation of an induced electromotive force in the secondary coil on increasing or decreasing the intensity of the current in the primary coil. 3) With the primary coil inside the secondary one,

galvanometer secondary coil

close the circuit of the primary coil, a deflection is noticed in the galvanometer in a certain direction. Open the primary circuit, and notice that the deflection is in the opposite direction. This indicates that an electromotive force is

Fig (11 - 5)

Mutual inductance between two coils

induced in the secondary coil upon switching on or switching off the primary circuit. The analysis of the above mentioned observations leads to the following conclusions: I. The pointer of the sensitive galvanometer deflects in a certain direction in the following cases: a) bringing the primary coil close to the secondary coil or when the primary coil is plunged inside the secondary one. b) increasing the intensity of the current in the primary coil. c) switching on the primary circuit. In all cases above, there is a positive increase in magnetic flux linkage and the induced emf in the secondary coil increases as the affecting magntic field increases with time. The induced current is in opposing direction to that in the primary. In such a case, the induced

It is the self-inductance of a coil in which an emf of one volt is induced when the current passing through it changes at a rate of one Ampere per second (vsA)-1. The self inductance of a coil depends on: a) its geometry. b) its number of turns. c) the spacing between the turns. d) the magnetic permeability of its core. Among the applications of self induction is the fluorescent lamp, where magnetic energy is stored in the coil. This energy is discharged in an evacuated tube filled by an inert gas, causing collisions of its atoms and their subsequent ionization and collision with the

Henry

walls of the tube. The inner walls are coated with a fluorescent material which causes visible light to be emitted upon the collision of the inert gas ions with it. Electromagnetic induction is also used in Ruhmkorff coil, which is used as an ignition coil in internal combustion engines (such as a car). Eddy Currents: If the magnetic flux changes with time through a solid conductor ,currents will be induced in closed paths in the conductor. Such currents are called "eddy currents". The change in the intercepted magnetic flux is effected either by moving the solid in a suitable magnetic field or by subjecting the metallic solid to an alternating magnetic field( for example field due to an AC current). The eddy currents are associated with heating effects. Thus, they are useful in melting metals in what is called the induction furnaces.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

The Henry :

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 240

When the coil circuit is switched off, the current ceases to pass in it, and this is associated with a decrease of the magnetic field of the neighboring turns to zero. This in turn is accompanied by a time variation of the flux linkage, i.e., each turn cuts the diminishing lines of the magnetic flux, and thus, an induced electromotive force is generated in the coil. The induced electromotive force is formed in the turns of the coil as a whole as a result of the self induction of the coil itself. This induced electromotive force is generated due to the self induction of the coil on switching off or switching on the circuit following Lenz's rule. Thus, an induced electric current is generated in the same direction as the original current. When the circuit is switched off, to retain the existing current, a spark is formed between the two terminals of the switch. When the number of turns of the coil is large, the induced emf on switching off the circuit will be much larger than that of the battery, This causes a neon lamp connected in paralle1 between the two terminals of the switch to glow(a neon lamp requires a potential difference about 180V to glow). Since the induced electromotive force is proportional to the rate of change of the current in the coil, then the emf induced by self induction is directly proportional to the rate of change of the current in the coil. That is : 6 I1 (emf)1 ∝ 6t ∴ (emf) e ¡ 1 = - L

6 I1 6t

(11 - 3)

where L is a constant of proportionality known as the coefficiont of self induction (self inductance) of the coil, and the negative sign in equation (11-3) indicates that the induced electromotivc force opposes the change causing it (Lenz's rule). L=-

e¡ emf

6P/6t

Thus, the self inductance of a coil is defined as: It is the electromotive force induced in the coil when the current passing through it changes at a rate equals one Ampere per second. The self inductance is measured in the unit henry.

The AC generator (or the dynamo) is a device which converts the mechanical energy into electrical energy. In a generator, a coil rotates in a magnetic field, and the resulting induced current can be transferred (or transmitted) by wires for long distances. The simple electric generator consists as shown in Fig (11-8) of four main parts : a) a field magnet. b) an armature. permanent field magnetic

c)two slip rings. d) two brushes. The field magnet may be a permanent

direction of motion

magnet or an electromagnet. The armature consists of a single loop of wire or coil of many turns suspended between the two poles of the field magnet. A pair of slip rings are connected, one to each end of the loop. They rotate with the loop in the magnetic field. The induced current in the coil passes to the external circuit through two graphite brushes, each touching one of the two corresponding slip rings. Fig (11-9) shows

slip rings armature brushes

Fig (11 - 8)

A simplified schematic for an AC generator (dynamo)

the direction of rotation of the armature between the poles and the direction of the induced current at a certain instant. The loop rotates around its axis in a circle of radius r. Its linear velocity is v=Ď&#x2030;r where Ď&#x2030; is the angular velocity equal to 2Ď&#x20AC;f, (where f is the frequency). Substituting for

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

Alternating current generator:

243

armature

changes direction every half a revolution. It follows a sine wave. From figure, we can also understand the meaning of f. Throughout a complete revolution, the current

graphite brushes

increases from zero to a maximum, then decreases to zero, then reverses direction, and increases in the

slip rings

negative direction up to a negative maximum. Then, it heads back to zero. In one complete revolution, one complete oscillation has occurred. The number

Fig (11-10a)

AC generator

of oscillations per second is the frequency f. The frequecy of home use power is 50Hz

Example: The coil of a simple AC generator consists of 100 turns, the cross sectional area of each is 0.21 m2. The coil rotates with frequency 50 Hz (cycles/second) in a magnetic field of constant flux density B = 10-3 Weber/m2. What is the maximum induced emf generated? and what is the instantaneous value at θ = 30˚? Solution:

(emf)max = NBA ω = NBA (2 πf)

= 100 x 10 -3 x 0.21 x 2 x

22 x 50 = 6.6 V 7

Thus, the maximum induced emf generated equals 0.6 volts.

1 = 3.3 V 2 It is worth remembering that the induced current is directly proportional to the induced e= ¡max sin e = 6.6 x sin 30˚= 6.6 x emf = ¡(emf)

emf. Thus, the instantaneous value of the induced current is given by : I = Imax sin (2 πf t )

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

From Fig (11-10), we see that the induced current

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 246

This induced current reaches its maximum value when the induced emf reaches its maximum value, and it vanishes as the induced emf is zero. Effective value of the alternating current: It is worth mentioning that the average value

maximum positive current

of an AC current equals zero, because the AC current changes from (Imax) to (-Imax). Neverthless, the electric energy is consumed as thermal

Î¸

current

one complete revolution

energy due to the motion of electric charges, and the rate of the electric energy consumed is proportional to the square of the intensity of

maximum negative current

the current. The effective value of the intensity of the alternating current is the value of the direct

coil position

current which generates the same rate of thermal

Fig (11-10b)

effect in a resistance (or the same power) as that generated by the considered AC current.

The relation between current and angle of rotation(sine wave)

Ieff = 0.707 Imax

(11 - 11)

The value Ieff is called the "effective value of the alternating current". There is a similar relation for the effective electromotive force, that is : (emf)eff = 0.707 (emf)max

Example:

(11 - 12)

Veff = 0.707 Vmax

If the effective intensity of current in a circuit equals 10 A, and the effective voltage is 240 volts,what is the maximum value for current and voltage ?

negative pole of the dynamo. Accordingly, the current in the external circuit will be always in one direction as shown. It is noticed that using the commutator renders the induced emf in Fig (11-11d) in one direction, but its value changes from zero up to a maximum value, then decreases again to zero during each half cycle of the coil rotation, but it is always in one direction. To

obtain

uni-directional

current

approximately constant value, i.e., to obtain a nearly DC (value), many coils separated by small angles are used. A cylinder is used which is split into a number of segments, double the number of coils. Thus, the current in the external circuit is almost constant. This

Fig (11-12)

is the way to obtain a DC generator (Fig 11-12).

Nearly DC current

The transformer:

The electric transformer is a device whose function is based on the mutual induction between two coils, and is used to step up or to step down an AC voltage. Transformers are used to transfer the electric energy from generators at electric power stations. Such transformers are called step - up transformers, while the transformers used at the zones

primary coil

output

input secondary coil soft iron core ( laminas)

Fig (11-13a)

Step UP transformer

where the energy has to be distributed among buildings are called step-down transformers. The transformer as shown in Fig (11-13) consists of two coils: a primary coil and a secondary coil. The two coils are wound around a soft iron core made of thin iron sheets

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

Continuing the rotation, the brush F l acts as a positive pole, while F 2 acts as the

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 248

commutator consists of two halves 1 and 2 of a hollow metallic cylinder split in between, and are well insultated from each other as shown in Fig

direction of rotation

(11-11).Two brushes F and F touch the two halves 1

during the rotation of the coil. The external circuit is connected to the two brushes Fl and F2. It is necessary

that the two brushes F1 and F2 touch the insulator

split cylinder

between the two halves at the moment when the plane

brushes

of the coil is perpendicular to the magnetic field, i.e, at the instant when the generated electromotive force

Fig (11-11c)

in the coil is zero.

Use of a split cylinder

Let us consider that the coil starts rotation in the

rectifies the current

direction shown (Fig.11-11c).During the first half rotation, brush F1 touches the half cylinder (1),

current

while brush F2 touches the other half (2) of the

number of revolutions

cylinder. The current in such a case will pass in the coil in the direction w x y z. As a result, the current passes in the external circuit in the direction from Fl

to F2 during the first half of the cycle. In the second

half of the cycle, the electric current reverses its

Fig (11-11d)

Unidirectional current versus Î¸ (sine wave)

direction in the coil, i.e., the current passes in the coil in the direction z y x w. At the same time, brush F1 will be in contact with the half(2), while F2

will be in contact with the half(1), i.e., the two halves of the commutator reverse their position relative to the two brushes. In such a case, the current in the external circuit passes from, F1 to F2, which is the same direction as that in the first half of the cycle.

secondary coil will be larger than the emf in the primary one. For example, if the number of turns of the secondary coil is twice that for the primary coil, one gets Vs = 2VP.

While, for the case when Ns is less than Np one gets a step-down transformer, where, in

such a case Vs will be less than Vp.

The relation between the current intensities in the two coils of the transformer: Let us assume that there is no loss in the electric energy in the transformer (almost zero resistance), then according to the law of conservation of energy, the electric energy made available by the source in the primary coil must equal that delivered to the load in the secondary coil. Vp Ip t = Vs Ist

From which the input power is equal to the output power, i.e,

Thus,

Â&#x2018;

Vp Ip = Vs Is

From the equations (11-11) and (11-12), Is

Np Ns

This shows that the intensity of the electric current in either of the two coils is inversely proportional to the number of its turns. For example: if the number of turns of the secondary coil is twice that of the primary coil, then the intensity of current in the secondary coil equals half that in the primary coil. From this argument, we see the importance of the use of the step-up transformer at the

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

This equation shows the interrelation between the emf Vs in the secondary and Vp in the primary. If Ns is larger than Np, one has a step-up transformer, where the emf in the

251

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 250

(laminas) insulated from each other, to minimize the effect of eddy currents and to minimize the dissipated electric energy. When an electric current passes in the primary coil, a magnetic field is generated. The core makes the lines of such a field pass through the secondary coil. The relation between the two emfs in the two

coils of the transformer:

When the primary coil is connectcd to a source of AC

Fig (11-13b)

Transformer symbol

voltage, the variation in the magnetic field linked with the primary current generates an induced emf in the secondary coil having the same frequency. The induced emf in the secondary is determined from the relation: Vss==- N s 6 qm 6t 6 qm where, Ns is number of turns of the secondary coil and is the rate of change of the 6t mangetic flux linked between the primary through the secondary coil. The electromotive force in the primary is in turn related to the rate of change of the magnetic flux and is determined from the relation :

V p==-N p 6 q 6t where, Np is the number of turns of the primary coil. Assume that the wasted magnetic

energy is negligible, i.e., there is no considerable loss in the magnetic flux, i.e., the whole resulting magnetic flux passes through the secondary coil (no stray lines). Dividing the above two relations one can get the following formula : Vs N = s Vp Np

(11 - 13)

V η= s Vp η=

Is x100 Ip

V s Np x x100 Vp Ns

80 =

8 1100 x x100 220 Ns

Ns = 50 turns N Is = Np Ip s Is = 1100 50 0.1 Is = 2.2 A Learn at Leisure

AC/DC There are two types of current or voltage AC and DC .In the case of DC, Ohm’s law dictates that what determines the current is the resistance. In the case of AC, what determines the current are three elements,the resistor, the inductor and the capacitor. Household appliances use 220 V AC of frequency 50 HZ. Often, we need to convert this to a lower DC voltage, as in the case of the mobile charger and some other appliances.To do this, we use a down transformer and a rectifier. It is to be noted that a low AC current is more hazardous to man than a low DC current (why?).

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

Solution:

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 254

an audio signal (Fig 11-16). The same thing happens in the hard disk in the computer, where data is stored by magnetization. In this way, the data is not lost from the hard disk when the power of the computer is switched off. iron core

input audio

unmagnetized tape

magnetized tape

magnetic field

Fig (11-13b)

Use of electromagnetic induction in recording

Examples: 1- A trandformer connected to a 240 V AC power source gives 900 V output emf with current intensity 4A. What is the intensity of the source current assuming that the efficiency of the transformer is 100%?

Solution:

Vs I p = Vp I s I Â&#x2018; 900 = p 240 I4s ...

Â&#x2018; I p = 900 x 4 = 15 A 240 2) An electric bell is connected to a transformer of efficiency 80% which gives 8 V output, while the input household voltage is 220 volts. What is the number of turns of the secondary coil if the number of turns of the primary coil is 1100 ? and what is the intensity of current in the secondary coil if the current in the primary coil is 0.1 A ?

Starting from a position at which the plane of the coil is parallel to the lines of the magnetic flux, and the brush F - connected to the positive terminal of the battery - touches 1

the half cylinder x, while F - connected to the negative terminal of the battery - touches the 2

half cylinder y as shown in Fig (11-17). Thus, current passes in the coil in the direction dcba. Applying Fleming's left hand rule, one concludes that the wire ab is affected by a force in the upward direction, while the wire cd is affected by a force in the downward direction. The two produced forces (couple) form a torque, and the coil begins to rotate in the direction shown in the figure. As the coil rotates, the moment of the couple decreases gradually till it vanishes, when the coil plane becomes perpendicular to the lines of the magnetic flux. But the coil having gained a momentum will continue motion due to its inertia, which in turn pushes the coil to the other side. The two halves x and y of the commutator interchange position, such that the half cylinder x will be in touch with the brush F2,while the brush F

will touch to other half cylinder y. Thus, the current in the coil will reverse direction and pass in the direction abcd. Applying "Fleming's left hand rule" for the new position of the coil shows that the force acting on the wire ab will be downward, while the force acting on the wire cd will be upwards. The obtained torque enables the coil to continue rotation in the same circular direction. The torque increases gradually to its maximum value when the plane of the coil becomes parallel to the lines of the magnetic flux. Then, it decreases to zero when the plane of the coil is perpendicular to the lines of magnetic flux. The inertia of the coil then causes it to continue rotating to the other side. This permits the two halves to interchange positions and with respect to the two brushes F1 and F2, and thus, the current in the coil is reversed once more. The coil continues rotating in the same circular direction making one complete revolution, and so on. In order to increase the power of the motor, a number of coils may be used with equal

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

Operation of a DC motor through one complete revolution:

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 256

DC motor:

ficld magnet

It is a device which converts electric energy to mechanical energy. It operates on a DC source (battery) (Fig 11-17). It consists in its simplest

split cylinder

form of a rectangular coil abcd comprising a large number of turns of insulated copper wire wound around a soft iron core made of thin insulated sheets to cut down on eddy currents. The core and the coil can rotate between the

brushes variable resistance

two poles of a strong horseshoe (U-shaped) field magnet. The two terminals of the coil are connected to two halves of a split cylinder

Fig (11 - 17) DC motor

(commutator). The two halves (x,y) are insulated from each other and capable of rotating around the axis of the coil. The plane separating the two halves is perpendicular to the plane of the coil and the line connecting the two brushes is parallel to the lines of magnetic flux. To operate the motor, the two brushes must be connected to the battery. The motor and the galvanometer: The principle of operation of the electric motor and that of the moving coil galvanometer are alike. The main difference is that the electric motor must rotate continuously in the same direction. The design of the electric motor necessitates that the two halves x,y of the cylinder must interchange positions relative to the two brushes F1 and F2 each half cycle.

As a result, the electric current passing in the motor must reverse direction in the coil each half revolution.

Search for Metals A metal detector is used in the search for metals. Its operation depends on measuring the change in the self inductance L of a coil due to its proximity to a metal. The current in the detector changes giving away the hidden matal (Fig. 11-18). Fig (11 - 18)

A metal detector

Learn at Leisure

The microphone and the loud speaker The operation of a moving coil microphone depends

magnet

on the vibration of an armature according to the sound waves. The magnetic field in an iron core changes, which results in the generation of an emf in a coil wound around the iron core. This emf is of variable amplitude and frequency according to the individual sound. Thus, a sound signal (mechanical wave) is converted to an electrical (audio) signal. In a moving coil loud speaker, the reverse takes place.

Sound waves coil

Fig (11 - 19)

A microphone

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

Learn at Leisure

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 258

angles between their planes. The two terminals, of each coil are connected to two opposite splits of a cylinder. The cylinder is split into a number of segments twice that of the number of the coils. During rotation, each two opposite segments touch the two brushes F1 and F2 when their corresponding coil is in position of largest torque. Learn at Leisure

Motor and generator at the same time: While operating the motor, its coil cuts the line of magnetic flux of the field magnet. There is a rate of change of cutting magnetic lines. Therefore, an emf is induced opposite to the source, reducing the current, and hence, the speed.This back emf in the motor coil acts to the stabilize the speed of rotation of the coil. If the speed of the coil tends to increase, the back emf increases. The difference between the back emf and the external source is the voltage drop across the coil resistance. Therefore, as the back emf increases, the current decreases, and so does the speed of rotation. Conversely, if the coil speed tends to decrease, the back emf decreases, and the current increases. So the speed of rotation of a DC motor remains constant.But the speed can be changed by changing the source (battery) voltage. It should be noted that if we try to stop the motor by force while it is connected to the source, the motor coil burns out, because the back emf would disappear and the battery voltage is applied in full across the small coil resistance so it burns out. Also, if a transformer works on no load (secondary is open circuited), the secondary current is zero and the primary current should be zero. However, a back emf in the primary almost balances out with the input voltage due to self-induction. A small current in the primary exists even at no load to produce the flux linkage. So, an ideal transformer does not really exist. However, at full load, we may consider - as an approximation - that the ideal transformer model works.

point contact

iron core

copacitor cam

primary coil

spark distributer

secondary coil platinum point contacts

car battery

rotating distributer discharge arm

Fig (11 - 21a)

A schematic for the ignition circuit in a car

ignition coil

contact point contact

spark plug

Fig (11 - 21b)

A section in ignition circuit components

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

spark plug

261

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 260

flexible wires to the coil

The variable audio (electrical) signal produces variations in the coil current. The coil is connected to a diaphragm which vibrates due

paper core

to the force generated in the presence of a

soft iron cylinder

magnetic field. Mechanical (sound) waves

permanent magnet

diaphragm

result, resembling the original audio signal. Thus, sound is heard back (Fig 11-20).

moving coil soft iron sheet

Fig (11 - 20)

A loud speaker

Learn at Leisure

lgnition circuit in a car The ignition coil (Fig 11-21) consists of two coils one inside the other, both wound around a soft iron core. The primary coil and the core comprise an electromagnet. the primary circuit is opened regularly by a distributer cam as it rotates, thus, opening and closing the point contacts. The secondary coil contains thousands of turns. A large emf is generated from time to time at the same rate of opening the primary circuit. This large emf generates a spark across the air gap across the spark plugs. The spark plugs are connected alternately to the secondery coil as the distributer rotates. A capacitor is used to protect the points from corrosion due to the spark. Electronic ignition system works on the same principle, but the cam is replaced by transistors as a switch (Chapter 15).

Self-induction: It is the electromagnetic effect induced in the same coil when the intensity of the current increases or decreases.This effect acts to resist such a change in the intensity of current.

• Coefficient of self-induction :

It is measured numerically by the electromotive force

generated by induction in the coil when the intensity of the current passing through it changes at a rate of 1A/s.

• The unit of measuring the self induction (Henry): It is the self induction of a coil in which an emf of 1V is induced when a current passes through it which changes at a rate of 1A/s.

1H =

• The self-induction of a coil depends on :

1V.S s A

a) its geometry.

b) its number of turns.

c) the spacing between its turns.

d) the magnetic permeability of its core.

• The Dynamo (AC Generator): It is a device used to convert the mechanical energy to electric energy(AC current and voltage) when its coil rotates in a magnetic field. The simple dynamo (AC generator) consists of : a) field magnet (strong magnet). b) a coil of insulated copper wire suspended between the two poles of the magnet. c) two metallic rings in contact with two graphite brushes connected to an external circuit.

• A commutator: (cylinder split into a number of insulated segments) is used to obtain a DC current and voltage (DC generator).

•

The alternating current: It is current which changes periodically its intensity and

direction with time according to a sinusoidal curve.

•

The electric transformer: It is an electric device used to step up or step down an emf through mutual electromagnetic induction.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

•

263

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 262

In a Nutshell Definitions and Basic Concepts:Electromagnetic induction : It is a phenomenon in which an induced electromotive force and also an induced current are generated in the coil on plunging a magnet into orwithdrawing a magnet out of a coil.

• The presence of a soft iron core inside a coil concentrates the lines of magnetic flux that link with the coil. This in turn increases the induced electromotive force and also the

•

induced current. Faraday's law for the induced emf : The induced emf generated in a coil by electromagnetic induction is proportional to the time rate by which the conductor cuts the lines of magnetic flux and is also proportional to the number of turns of the coil.

• Lenz's rule: the direction of the induced current generated by induction is such that to oppose the change in the magnetic flux producing it.

• Fleming's right hand rule: Place the thumb, the pointer and the middle finger(with the rest of the fingers) of the right hand mutually at right angles. If the pointer points in the direction of the magnetic field and the thumb in the direction of motion then the middle finger (with the rest of the fingers) will point in the direction of the induced current.

• Mutual induction: It is the electromagnetic interaction between two coils kept close to each other (or one inside the other).An electric current with time varying intensity passing in one coil (primary coil)will produce in the second one (secondary coil) an induced current in a direction such that to oppose the variations of the current intensity in the primary coil.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

∆

264

• The efficiency of the transformer: It is the ratio between the output electric energy given in the secondary and that available to the primary.

• The electric motor: It is an electric device used to convert the electric energy into mechanical energy . Basic laws:

• The induced emf genrated in a coil of N turns as a result of time variation of magnetic flux φm linked with the coil in an interval of time is given by the relation: emf¡ = - N 6 q vV 6t The nagative sign indicates that the direction of the induced emf (and thus the current) is such as to oppose the cause producig it.

• The emf induced in a secondary coil due to the time variation in the lines of magnetic flux resulting from a primary coil linking with the secondary coil in a time interval t is given by the relation :

6q emf = - M I 6t where M is the coefficient of mutual induction.

• The emf induced by self induction as a result of the current ∆I passing through the coil in a time ∆t is given by the relation : where M is the coefficient of self induction of coil. emf = - L

6I 6t

whose terminals are connected to a galvanometer. The deflection of its needle will be in a direction: a) opposite to the current in the primary coil. b) points to zero reading c) increasing. d) same as the current in the primary coil e) variable 5) Opening the primary circuit while the primary coil is inside the secondary one, leads to the generation of : a) an induced forward current. b) an electric field c) an induced back current. d) an AC current. e) a magnetic field. 6) The slow rate of growth of the current in the solenoidal coil is due to the: a) production of forward current. b) production of a magnetic field. c) production of a back induced current opposing (resisting ) the original one. d) production of a magnetic flux. e) production of an electric field. 7) The ohmic resistors are made of double wound wires: a) to decrease the resistance of the wire. b) to increase the resistance of the wire.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

4) A current passes in the primary coil, then this coil is plunged into a secondary coil

267

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 266

Questions & Drills I) Put (3) against the right answer: 1) The pointer of a galvanometer whose terminals are connected to a solenoidal coil will be deflected if one withdraws the magnet quickly from the coil because: a) the number of the coil turns is very large. b) the coil intercepts the lines of the magnetic flux. c) the number of the turns of the coil is small. e) the number of turns of the coil is suitable. 2) The needle of the galvanometer whose terminals are connected to a solenoidal coil deflects on the withdrawal of the magnet in a direction opposite to that which occurs on plunging the magnet into the coil because: a) an induced current is generated in a direction opposite to that on plunging the magnet. b) an electric current is generated. c) the number of the lines of magnetic flux decreases. d) the number of the lines of the magnetic flux changes. e) the number of flux lines remains constant. 3) The emf induced in a coil on plunging a magnet into or withdrawing it out of a coil differs according to the difference in : a) [the intensity of the current - the length of the wire - the number of the lines of flux]. b) [magnet strength - the velocity with which the magnet moves- the number of turns of the coil]. c) [the cross sectional area of the coil - the mass of unit length - the material from which the wire is made]. d) [the length of the wire - the number of turns - the type of the magnet]. e) [the magnetic flux density - time - the intensity of the current].

e) a current rectifier. 12) The ratio between the electric energy in the secondary to that in the primary is called: a) the lost energy. b) the given energy. c) the efficiency of the transformer. d) the working strength of the transformer. e) the gained energy. II) Define the following : 1- Electromagnetic induction. 2- Faraday's law of induction 3- Lenz's rule. 4- Fleming 's right hand rule. 5- Mutual induction. 6- Unit of measuring the mutual inductance. 7- Self induction. 8- Coefficient of self induction. 9- The Henry. 10- The induction coil. 11- The AC current. 12- The dynamo. 13- The electric motor. 14 - The transformer. 15- The efficiency of the transformer. 16- The back emf in the motor.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

c) several magnets d) an insulated copper wire.

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 268

c) to avoid self-induction. d) to eliminate the resistance of the wire. e) to facilitate the connection process. 8) The direction of the current produced in the dynamo coil can be determined using: a) Fleming's left hand rule. b) Lenz's rule. c) Fleming's right hand rule. 9) The rate with which the coil intercepts the lines of magnetic field in the dynamo is maximum when: a) the plane of the coil is perpendicular to the flux lines. b) the plane of the coil is inclined to the lines by an angle 30Ë&#x161; c) the face area of the coil is minimum. d) the face area of the coil is maximum. e) the plane of the coil is parallel to the lines of the magnetic flux. 10) The intensity of the current in the two coils of the transformer is : a) directly proportional to the number of the turns. b) inversely proportional to the number of the turns. c) depending on the temperature of the wire. d) depending on the substance of the wire. e) depending on the temperature of the air (ambient temperature). 11) The power of an electric motor to rotate increases on using: a) larger number of turns. b) several coils with angles between their planes.

1) The core of an electic transformer is made of thin sheets insulated from each other. 2) A bar of soft iron will not be magnetized if a double wound wire carrying a current is wound around it. 3) A wire free to move in a magnetic field moves when a current passes through it. 4) The transformer is not suitable to convert DC voltage. 5) The electric motor rotates with uniform velocity. 6) The induced current dies out in a straight wire faster than in a coil with air core, and in a coil with air core faster than in a coil wound around an iron core. 7) The metallic cylinder used to obtain a unidirectional current in the dynamo is split into two halves completely insulated from each other. V) Drills 1) A coil of 80 turns, and cross sectional area 0.2 m2 is suspended in a perpendicular position to a uniform magnetic field. The average induced emf is 2 V when it rotates 1/4 revolution through 0.5 s. Find the magnetic flux density. (0.0625T) 2- If the magnetic flux density between the two poles of the magnet of a dynamo is 0.7 Tesla, and the length of its coil is 0.4 m, find the velocity of motion in such a field to obtain an induced emf in the wire equal to 1V.

(3.57m/s)

3) A coil of a dynamo consists of 800 turns each of face area 0.25 m2. It rotates at a rate of 600 revolutions per minute, in a field of magnetic flux density 0.3 Tesla. Calculate the induced emf when the angle made between the normal to the coil and the magnetic flux is 30Ë&#x161;.

(1885v)

4) A rod of copper of length 30 cm moves with at velocity 0.5 m/s in a perpendicular direction to a magnetic field of density 0.8 Tesla. Calculate the emf induced in such a rod.

(0.12v)

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction

IV) Give reasons

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Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 270

III) Essay questions: 1) What are the factors on which the emf induced in a conductor depends ? Mention the relation between the emf. and such factors. 2) State Faraday's law of the emf induced in a coil, then show how to verify this practically? 3) What is meant by mutual induction between two coils? and what is meant by the coefficient of mutual induction? How - using the mutual induction - one can verify Lenz's rule? 4) If a current passes through a coil, deduce an equation relating the induced emf in the coil and the rate of change of the current in the coil. From this, deduce a definition for the coefficent of self induction and the Henry. 5) When does the emf induced in a coil become maximum ? and when does it become zero? 6) Explain an experiment to show the conversion of the mechanical energy into electrical energy, and another experiment to show the opposite conversion. Then, state the rule used to define the direction of the current in the first case and the direction of motion in the second case. 7) Deduce the relation by which one can evaluate the instantaneous emf induced in an AC generator. 8) What are the modifications introduced to the AC generator to render it a unidrectional generator ? 9) Describe the structure of the electric transformer ? then explain the principle of its opertion. What is meant by saying that the efficiency of the transformer is 80%?. 10) What is meant by the efficiency of the transformer? What are the factors which lower such an efficiency and how to deal with them? 11) Draw a labelled diagram showing the structure of the motor and explain its operation.

Unit 4 : Electricity Dynamic Elictricity and Electromagnetism Chapter 11: Electromagnetic Induction 272

5) An antenna of length one meter fixed in a motor car, which moves at velocity 80km/hour in a direction perpendicular to the horizontal component of the Earthâ&#x20AC;&#x2122;s magnetic field. An emf of 4 x 10-4 V is induced in the antenna. In such a case, calculate the magnetic flux density of the considerd horizontal field. (18 x 10-6T) 6) Calculate the coefficient of self-induction for a coil in which an emf of 10 V is induced if the passing current changes at a rate of 40 A/s (0.25 Henry) 7) The mutual induction between two faces of opposite coils is 0.1 Henry and the intensity of current in one of them is 4 A. If this intensity drops to zero in 0.01s, find the emf induced in the other coil. (40V) 8) A rectangular coil of dimensions 0.4m x 0.2m and of 100 turns rotates with a uniform velocity 500 revolutions per minute in a uniform field of magnetic flux density 0.1 Tesla. The axis of rotation in the plane of the coil is perpendicular to the field. Calculate the emf induced in the coil. (41.89 V.) 9) A step-down transformer of efficiency 90% has a primary coil voltage of 200 V and that of the secondary is 9 V. If the intensity of the electric current in the primary is 0.5 A, and the number of turns of the secondary is 90 turns, what is the intensity of the current of the secondary coil, and what is the number of turns of the primary? (10 A, 1800 turns ) 10) A step-down transformer connected to an AC power source of 2500 V gives a current of 80 A.The ratio between the number of turns of the primary and the secondary coils is 20:1 Assuming that its efficiency is 80%, find the emf induced across the two terminals of the secondary, and find also the current in the primary coil. (100V,4A)

Blackbody Radiation: We are content so far to regard light as waves. Waves have common features, i.e, visible light

wavelength (m)

Fig (12-1) Electromagnetic spectrum

Wave Particle Duality

Chapter 12:

All what we have studied so far can be lumped under the title of classical physics. By classical, we do not mean outdated or obsolete. In fact, classical physics explains everything in our daily life and our common experiences. The present unit, however, entails some of the basic concepts of modern physics and a general view of a quantum physics. This branch of physics (modern or quantum) deals with a great collection of scientific phenomena which might not be directly observed in our daily life, but treat a number of situations in the universe which classical physics cannot explain, especially when we deal with atomic and subatomic systems, i.e, down to the subatomic scale . Also, this kind of physics explains all phenomena involved in electronics which is the basis for all modern electronic and communication systems. It also explains chemical reactions on the level of the molecule. Some of such reactions were photographed by Ahmed Zewail using a high speed laser camera. Such work entitled him to earn the Noble Prize in chemistry in1999.

Introduction to Modern Physics

Overview

Unit 5:

Wave Particle Duality

Chapter 12

275

12-4). We also note that the dominant color of light emitted from these sources varies. varies with wavelength. The distribution of the radiation intensity with wavelength is called Planck’s distribution (Fig 12-5). It was also found that the wavelength λm at which the peak of the curve occurs is inversely proportional to temperature. This is known as Wien’s law. Therefore, the higher the temperature,

We also note that as the wavelength tends to infinity (very large)or to zero (very small ) the intensity of radiation tends to zero. For

example, the

temperature at the surface of the Sun is

visable light

Radiation intensity

the smaller the wavelength of the peak .

peak is 5000˚A ( 0.5 µm). This is within the visible range. Thus, almost 40% of the total energy emitted by the Sun is in

Fig(12-5) The wavelength at the peak is inversely proportional to temperature

(infrared radiation), while the rest is distributed over the remaining spectrum. We practically obtain the same shape of radiation intensity distribution for a glowing incandescent lamp, except that the temperature is now 3000˚K which puts the wavelength at the peak at 1000 nm = 10-6 m = 10000˚A = 1 Micron. From such lamps we get nearly 20 % as visible light and most of the rest as heat. We cannot explain these observations using classical physics. It can be argued from

Wave Particle Duality

the visible range and almost 50% is heat

wave length (nm) nanometer)

Chapter 12:

6000˚K. Hence, the wavelength at the

Introduction to Modern Physics

Hence,an em source does not emit all wavelengths equally, but the intensity of radiation

Unit 5:

12-2) and other stars, a burning charcoal (Fig 12-3) and a glowing incandescent lamp (Fig

277

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 276

Fig (12-2) The Sun as a source of em radiation

Fig (12-4a) A glowing incandescent lamp emits em radiation

Fig (12-3) A burnig charcoal emits em radiation

Fig (12-4b) A lamp emitting less em radiation

reflection, refraction, interference and diffraction. We know also that visible light is but a small portion of the electromagnetic (em) spectrum (Fig 12-1). Electromagnetic waves may differ in frequency, and hence, in wavelength, but they propagate in free space at a constant speed c = 3 x 108 m/s. Electromagnetic waves do not need necessarily a medium to propagate in. We all observe that hot bodies emit light and heat. An example is the Sun (Fig

Unit 5:

tomography (tumor detection) (Fig 12-11), in embryology and in criminology, since the heat radiated has left. All these applications are called remote sensing. Egypt has been a pioneer in this field. How can we explain the bell shape of radiation? Planck in 1900 came up with the answer. Planck called this phenomenon black body radiation. The reason for naming it so is that a black body absorbs all radiation falling on it, regardless of the wavelength. It is, thus, a perfect absorber. It then re-emits this radiation wholly. It is therefore a perfect emitter. If we imagine an enclosed cavity with a small hole,

Fig (12-8) An image of southern Sinai taken by Land sat satellites

radiation within the cavity remains trapped due to multiple reflections. Only a small part of it leaks out, which is called blackbody radiation (Fig 12-12). Planck managed to explain this blackbody phenomenon with an interpretation that

Fig (12-10) An image taken by a night vision system

Wave Particle Duality

Fig (12-9) A night vision system

Chapter 12:

the inside of the cavity appears black because all of the

Introduction to Modern Physics

from a person lingers for a while even after the person

279

with frequency. Why then should the intensity of radiation go down at the high frequency end, (Fig 12-6)? This curve is repeated for all hot bodies which emit continuous radiation not only the Sun but also the Earth, and all bodies even living creatures. But the Earth- being a non glowing body it absorbs the radiation from the Sun and

radiation intensity

Wave Particle Duality Chapter 12:

classical physics that since the radiation is an em wave, the intensity of radiation increases

classical expectation

Planck's distribution

reemits it. But its temperature is far less than that of the Sun. Therefore, we find the wavelength at the peak to be nearly 10 Micron ,which is within the infrared region

detect and image moving objects in the dark due to the heat radiation which these objects

278

and photograph the surface of the Earth, using different regions of the spectrum including the infrared radiation emitted by the surface of the Earth ,in addition to the reflected visible light (Fig 12-8). Also, microwaves are used for the same purpose in radars. Scientists analyze such images to determine possible natural Earth resources.

radiation intensity

Introduction to Modern Physics

as well as terrestrial equipment which map

Unit 5:

(Fig 12-7).There are satellites, and airborne

Fig (12-6) Radiation decreases with increasing frequency in disagreement with classical expectations em radiation from the Sun

em radiation from the Earth 位碌m

Fig (12-7) Radiation from the Earth and from the Sun

This technique is also used for military purposes such as night vision systems, which re- emit (Figs 12-9,12-10). Thermal imaging is also used in medicine, particularly in

Unit 5:

(b)

(c)

(d)

(e)

(f)

Fig (12-13) An image where each shot has a different number of photons in increasing order from(a) to(f)

Introduction to Modern Physics

(a)

image taken for an object for different numbers of photons. It is worth mentioning though

Photoelectric Effect and thermoionic effect: Photoelectric Effect:

Wave Particle Duality

A metal contains positive ions and free electrons which can move around inside the metal but cannot leave it, due to the attractive forces of the surface which may be represented by a surface potential barrier. But some of these electrons can escape if given enough energy in the form of heat or light (Fig 12-14). This is the idea behind the cathode ray tube (CRT), which is used in TV and computer monitors (Fig 12 -15). This tube consists of metal surface called the cathode, which is heated by a filament. Electrons are, thus, emitted by the so called electron gun (E-gun). Due to heat, some electrons may overcome the forces of attraction at the surface. These electrons are then freed (liberated) from the metal and are then picked up by the screen, which is connected

Chapter 12:

that the human eye is so sensitive that it can detect as little as one photon falling on it.

281

Wave Particle Duality Chapter 12:

sounded weird at the time. He proposed that radiation was made up of small units (or packets) of energy, each he called quantum (or photon). Therefore, we may consider radiation from a glowing object as a flux of emitted photons. The photons’ energy increases with frequency, but their number decreases with increasing energy. The photons emanate from the vibrations of atoms. The energy of these vibrating atoms is not continuous but quantized (discrete or discontinuous) into levels. These

Fig (12-11) A thermal image for the face and neck

energy levels take values E=nhν, where h is Planck's constant h=6.625x10-34 Js, and ν is the frequency (Hertz

Unit 5:

Introduction to Modern Physics

- Hz). The atom does not radiate as long as it remains in

280

one energy level. But if the vibrating atom shifts from a high energy level to a lower energy level, it emits a photon whose energy E = hν. Thus, photons with high frequency have high energy and those with low frequency have low energy. Radiation consists

Fig (12-12a)

Radiation inside the cavity is trapped so it appears black

of billions upon billions of these photons. We do not see separate photons, but we observe the features of the stream of photons as a whole. These features express in, the stream of photons represent the classical properties of radiation Fig (12-13) shows an

Fig (12-12b) A small part of energy leaks out of the hole which is called blackbody radiation

Unit 5:

a photon

a freed electron

Fig (12-14c) A more tightly bound electron needs higher energy to escape

fluorescent screen cathode grid

anode plate Y plate X E-beam

filament heater

Fig (12-15) Light spot on a fluorescent screen (emits photons when struck by electrons)

light spot

Wave Particle Duality

E-gun

Chapter 12:

to a positive pole called the anode, thus causing current in the external circuit. When the electrons hit the screen, they emit light which varies in intensity from point to point on the fluorescent screen, depending on the intensity of the electrical signal transmitted. Such a signal controls the intensity of the electron beam emitted from the E -gun through a negative grid in its way. The E -beam can be controlled by electric or magnetic fields to sweep the screen point by

Introduction to Modern Physics

energy

283

uv radiation freed electrons

Chapter 12:

Wave Particle Duality

zinc plate

Fig (12-14a) Electrons may be freed from a metal if given sufficient energy

Unit 5:

Introduction to Modern Physics

a photon

282

An electron barely escaping

energy

Fig (12-14b) Minimum energy needed to free an electron is called work function

photocurrent

Chapter 12: light intensity

Fig (12-17b) Photocurrent versus light intensity for ν > νc

Wave Particle Duality

Fig (12-17a) Photocurrent versus light intensity for ν < νc

Introduction to Modern Physics

light intensity

Unit 5:

Einstein put forth an interpretation for all this, which led him to Nobel prize in 1921. He proposed that a photon with ν > νc falling on a metallic surface , has energy hν , while the energy needed to free an electron ( called the work function) is Ew = hνc ( Fig 12-14). Thus, the photon is barely able to free an electron, if it has energy hν=hνc= Ew. If the photon energy exceeds this limit, the electron is freed and the energy difference hν -Ew is carried by the electrons as kinetic energy, i.e., it moves faster as hν increases. Whereas if hν<Ew, the electron would not be emitted at all, no matter how intense the light might be. Also, the emission is instantaneous. There is no need for time to collect energy. The emission takes place instantly, once hν > hνc . It is to be noted that νc and Ew vary for different materials, and do not depend on the light intensity, the exposure time or the voltage difference between the anode and the cathode.

285

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 284

point generating the picture,so called raster until the frame is completed (Fig 12-15). When light falls on a cold cathode instead of heating a filament, a current flows in the circuit too. This means incident light electrons have been freed due to light. The emission of such electrons due to light falling on a metallic surface is a phenomenon called photoelectric effect (Fig 12-16). This emitted phenomenon cannot be explained by the classical theory of photoelectrons light. Considering light as a wave, part of the light falling vacuum on the surface of the metal is absorbed, giving some ammeter electrons enough energy to escape. We are then up aganist certain difficulties with the classical theory. If we attempt voltmeter using the classical model, the current intensity or the emission of such electrons (called photoelectrons) should battery depend on the intensity of the incident wave, regardless of Fig (12-16) its frequency. Also, the kinetic energy (or velocity) of the Photoelectric current emitted electrons should increase with increasing intensity achieved by absorbing of the incident radiation. Even in the case of low light photons on a metal surface intensity, giving sufficient time should give some electrons )Photo electeric cell( enough energy to be freed, regardless of the frequency of the incident light. But the practical observations are contrary to these classical expectations. It has been observed that the emission of electrons depends primarily on the frequency of the incident light not on its intensity. Such electrons are not emitted if the frequency is under a threshold (critical value) νc no matter how intense light may be. If the frequency exceeds νc, photocurrent increases with the intensity of light (Fig 12-17). Also, the kinetic energy (or velocity) of the emitted electrons depends on the frequency of the incident wave not its intensity.In addition, the emission of electrons occurs instantly as long as ν > νc. The electrons do not need time to collect energy if the light intensity is low, provided ν > νc .

Unit 5:

constant current

Fig (12-19a) Measuring KEmax for different frequencies at constant photon flux

Thus, Vs serves as a measure for KEmax. If the

Chapter 12:

photon energy is hν, we have: hν = Ew + KEmax KE max = hν - Ew Thus, KEmax is directly proportional to hν, number of photons/s). If ν becomes νc , we have voltage is needed to stop the current (Fig 12-19).

Fig (12-19b) A linear relation between KEmax and ν

Wave Particle Duality

regardness of light intensity φL (light flux is the hν=Ew, then KEmax = 0 and Vs = 0, i.e., no stopping

Introduction to Modern Physics

constant light intensity

287

Wave Particle Duality Chapter 12:

Learn at Leisure

Interpretation of the photo electric effect: To measure the velocity of photoelectrons (and their KE), we apply a negative retarding voltage between the anode and the

cathode

anode

light

cathode. The magnitude of the voltage. which causes the photocurrent to cease is called the stopping voltage Vs. At this voltage, electrons barely make it to the anode.Vs is the lowest voltage that does that. The kinetic energy of electrons at the anode

Unit 5:

Introduction to Modern Physics

become zero. The kinetic energy at the

286

cathode which would enable electrons to hardly reach the anode is the maximum kinetic energy at the cathode (most energetic

Fig (12-18a) A circuit for measuring photocurrent versus voltage

electrons), and hence, is called the maximum kinetic energy KEmax where e is the electron charge since the total energy at the cathode

constant light of φ

equals the total energy at the anode.

constant light of

-eVs + KE 1

max

Vs = e KEmax

constant

L2 current

constant

L1 current

ν > νc

stopping voltage

(12-1)

Fig (12-18b) Photocurrent versus voltage for different light intensity and constant frequency ν > νc

and it changed its direction (Fig 12-21). This observation could not be explained by the , wave (classical) theory of light. It can be argued based on Planck s hypothesis that electromagnetic radiation consists of photons which can collide with electrons as billiard balls collide. In this collision, linear momentum must be conserved (law of conservation of linear momentum), i.e., the linear momentum before collision must equal the linear momentum after collision. Also, the law of conservation of energy must apply, i.e., the the sum of the energy of the photon and the electron after collision must equal the sum of the energy of the photon and the electron before collision . We must, therefore, consider a photon as a particle with a linear momentum, i.e., it has mass and velocity as much as the electron is a particle which has mass and velocity, and hence a linear momentum. Photon Properties

incident photon electron

Fig (12-21) Compton effect

scattered electron

Wave Particle Duality

scattered photon

Chapter 12:

A photon is a concentrated packet of energy which has mass, velocity and linear momentum. Its energy E = hÎ˝, it always moves at the speed of light c regardless of its frequency. Einstein showed that mass and energy were equivalent E = mc2.A loss of mass

Introduction to Modern Physics

It was observed that when a photon ( X or Îł rays ) collided with a free electron, the photon frequency decreased and changed its direction. Also, the electron velocity increased

Unit 5:

Compton Effect

289

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 288

high light intensity low light intensity

Fig (12-20a) Relation of photocurrent with voltage for material A

high light intensity

low light intensity

Fig (12-20b) Relation of photocurrent with voltage for material B

Fig (12-20c) Relation between KEmax with ν for different materials

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 290

is translated to released energy as in the atomic bomb (Fig 12-22). Nuclear fission is associated with a small loss of mass which is converted to large amount of energy due to the c2(c2 = 9 x 1016 m2/s2) factor. Therefore, the law of conservation of mass and the law of the conservation of energy blend into the law of conservation of mass and energy. Thus, a photon whose energy is hν has a mass of hν/c2, while in motion . Since it is moving at velocity Fig (12-22) Atomic bomb c, its momentum which is the product of mass and velocity becomes hν/c . If a beam of photons is incident on a certain surface at the rate of φL (photons/s,), then each photon impinges on the surface and bounces off, and hence, suffers a change in linear momentum 2mc. The force which a beam of photons applies to the surface is the change in linear momentum per second: F = 2mcφL hν 2pP F = 2 ( c ) φL = c w

(12-2)

where Pw is the power in watts of the light incident on the surface. This force is too small to be noticed. But it is appreciable if it affects a free electron instead, due to its small mass and size, so it throws it off. This is the explanation of Compton effect. In the microscopic model, we can image a phaton as a sphere whose radius is roughly equal to λ and oscillates at frequency ν. The stream of photons collectively has a magnetic field and an electric field. These two fields are perpendicular to one another and to the direction of propagation.Both oscillate at ν. The photon flux (or stream) carries the energy of the wave. The wave properties are ,thus, manifested by the photon stream as a whole. The wave

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 292

eye

nasal cavity

brain tissue ear

skull

ear cavity

Fig (12-23b) A CT scan image of the head

intestines

lever

spine

kidney

Fig (12-23c)

Fig (12-23d)

An image of blood vessels using X- rays and a dye

A CT scan image of the abdomen

should be noted that when photons fall on a surface, a comparison is made between λ and photons sense the surface as a continuous one and get reflected from it as in wave theory. If the interatomic distance is comparable to λ, photons penetrate through the atoms. This is what happens in the case of X- rays.

Example

Calculate the photon mass and linear momentum if λ = 380 nm

Solution

(3x108 m/s) ν = c/λ = (380) (1x10-9m) = 7.89 x1014 Hz (6.625 x 10-34 Js) (7.89x1014 s-1) (3x108m/s)2

= 5.81x10-36 kg (6.625 x 10-34 Js) P = h/λ = L (380) (1x10-9m)

Wave properties of a particle In the universe, there is a great deal of symmetry. If waves have particle nature, could it be that particles might have wave properties ? a question posed by De Broglie in 1923 led to a hypothesis of wave particle duality applying to particles. The wavelength of a particle must be in analogy with a photon λ = h/PL where pL is the linear momentum of the particle.

(12 − 4)

Wave Particle Duality

= 1.74 x10-27 kgm/s

Chapter 12:

m= E/c2 = hν/c2 =

Introduction to Modern Physics

the interatomic distance of the surface. If λ is greater than the interatomic distance, these

Unit 5:

Thus, the wavelength is Planck’s constant divided by the linear momentum PL. It

295

Introduction to Modern Unit Physics 5: Introduction Chapterto12: Modern Wave Physics Particle Duality Chapter 12: Unit 5:

294

emit RF waves which are received by the detector. The computer reconstructs the images showing the concentration of hydrogen, and hence water collections, indicating tumors (Fig 12-24). Superconducting magnets are useful in reducing the heat due to eddy currents. MRI

Example

Magnetic (MRI) is another method of on tomography. Calculate theRessonance force appliedImaging by a beam of light whose power is 1W the surface Instead of a wall.of using X- ray with their possibly harmful side effects, radio(RF) waves are used. It has the Solution ability to detect tumors,2and computerized image slices. The Pw it depends 2 x 1 also on making -8 F= = ∴ 8 = 0.67 by x 10a N patient lies on a moving strong (superconductive) magnet. A c bed 3surrounded x 10 receiving detector is used to receiving RF waves from hydrogen nuclei in the body . The This force is too diminutive to affect the wall strong magnet orients the spins of the nuclei, RF waves produced by a source disturb this Relation between photon its linear spin orientation . Such waves arewavelength then stopped.and As nuclei relaxmomentum to their original state, they emit RF waves which are received by theλdetector. = c/ν The computer reconstructs the images showing the concentration and hence water collections, indicating tumors Multiplying the numeratorofbyhydrogen, h (Fig 12-24). Superconducting magnets are useful in reducing the heat due to eddy currents. h λ = hc = (12-3) hν hν/c

Example

P = mc Calculate the force applied by a beam Lof light whose power is 1W on the surface of a wall. hν = 2 c c Solution 2 Pw 2 x 1 hν F= = ∴ 8 == 0.67 x 10-8 N c 3 x 10 c h ∴ λthe= wall This force is too diminutive to affect PL Relation between photon wavelength and its linear momentum λ = c/ν Multiplying the numerator by h h λ = hc = hν hν/c PL = mc hν = 2 c c hν = c ∴ λ= h PL

(12-3)

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 296

intensity distribution

But what does this mean? We double slit source of consider light as a huge stream of electrons photons. Photons collectively have a wave property accompanying them, thus, manifesting reflection, refraction, interference and diffraction.The wave intensity screen describes the photon concentraction Fig (12-25a) as if a photon carries the genes of Diffraction of electrons through a double slit light, regarding frequency, speed and wavelength. By the same token, an electron ray (e-beam) is a huge stream of electrons. Collectively, there must be a wave accompanying them. An electron carries the genes of the stream, regarding charge, spin and linear momentum. The accompanying wave has wavelength 位 The intensity of the accompanying wave describes the electron concentration. Such a wave can disperse, reflect, refract, interfere and diffract, just as light does (Fig 12-25). But does this mean we can use an electron ray as much as we use a light ray in a microscope? The answer is yes. This answer is verified by the discovery of the electron microscope.

source of electrons

Fig (12-25b) Calculation of the path difference between two e-rays through a double slit

screen

Unit 5:

path which has an integral multiple of λ , or else the path over which a standing wave is formed (Fig 12-27), or the path over which ψ2 is maximum (Fig

Electron Microscope

electron beam might have a wavelength 1000 times or more shorter than visible light. Therefore, the electron-microscope can distinguish fine details. The lenses used in e-microscope are usually magnetic. These lenses focus the e-beam. Their design falls under the topic of electron optics.

optical microscope source

electron microscope

lens

object objective image image projection lens screen or photographic plate

Fig (12-28a) Electron microscope

Wave Particle Duality

the e-microscope uses an electron beam. The

Fig (12-27c) An orbit as a standing wave formed over a string fixed at both ends

Chapter 12:

Electron microscope is an important lab instrument which depends in its operation on the wave nature of electrons. It resembles an optical microscope in many ways. The important difference is in the resolving power. The e-microscope has a high resolving power, because the electrons can carry a high kinetic energy, and hence very short λ (equation 12-3). Thus, its magnification is so high that it can detect very small objects, so small that an ordinary optical microscope fails to observe (Fig 12-28). The optical microscope uses light, while

Introduction to Modern Physics

12-26 b).

299

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 298

Neuclus

Fig (12-27a) An orbit is an integer of 位 as a traveling wave in a closed path whose end coincides with its start

Fig (12-27b) The order of the orbit is determined by the integer value with the nucleus is zero (so the electron does not fall on the nucleus or else the universe

would vanish). Also, the probability for an electron to exist infinitely away from the nucleus is zero, or else the atom is ionized by itself. Ionization needs external energy. We conclude that the energy of the electron when trapped in an atom is less than that of a free electron by the magnitude of the ionization energy. Therefore, an electron remains trapped in the atom, unless acted upon by an external stimulus. Heisenberg showed also that we could not determine an exact path (orbit) for an electron in an atom. But we can say that an orbit is the

In a Nutshell

The interpretation of previous observations has paved the way to a new set of laws of

Chapter 12: Wave Particle Duality

extensive, then classical mechanics may be used.

Introduction to Modern Physics

• Classical physics explain many phenomena, in on which light ( or mechanics, namely, cannot Quantum Mechanics. This branchparticularly of science isthose based the following em radiation) interacts with electrons or atoms. assumptions formed by Schrodinger: • Light or any em radiation consists of a huge collection of photons, each photon having 1) An electron in an atom has energy values which belong to a set of allowable values energy hν, where h is Planck’s constant and ν is the frequency. called energyforlevels. Theisatom does not emit effect, energywhere unlessphotocurrent it falls from adepends high level to a • An evidence photons the photoelectric on the lower level. intensity of incident light as long as the frequency is greater than a critical value νc. But photocurrent flows.hνThe kinetictoenergy of the the emission frequencyis isin less than of νc ,a no 2) ifSuch the form photon whose energy is equal the difference electron the photoelectric effect depends the frequency not on the light between freed the twobyenergy levels. This process is calledon relaxation. intensity. 3) The absorption of energy by an atom occurs if the photon energy is exactly equal to the • A photon has a mass, a linear momentum and a constant speed which is the speed of energyIt difference twothe levels . In this Ifcase, the atom light. has a size between denoted by wavelength. a photon fallsisonexcited a wall,byit having applies its a electron move in energy to the level. the Thiselectron processwill is called excitation . to its small force on up it, but if it falls on higher an electron, be thrown off due andenergy size. is greater than the ionization energy of the atom, an electron is 4) small If the mass photon • Compton effect proves the particle nature photons, where a photon totally freed from the parent atom, and the of atom becomes ionized (ion). has mass, speed and linear momentum. 5) Relaxation and excitation are simultaneous processes. At thermal equilibrium, the atom • A wave describes the collective behavior of photons. is stable due to the simultaneity of these processes. • The wavelength of abalance photonand is Planck’s constant divided by the linear momentum. The 6) There is a function which is always positive thedescribes electron inthethewave atom. This same relation applies to a free particle, wherethat thedescribes wavelength nature of the particle wave theatparticle. functions tends,i.e.,to the zero at accompanying the nucleus and infinity (at the border of the atom). • The electronthemicroscope proves detrapped Brogliewithin relation particles. is used to detect Therefore, electron remains theforatom due toIt nuclear attraction diminutive particles. without falling onto the nucleus. To explain this, we may say that as the electron draws • Quantum mechanics is based on assumptions which agree with experimental near the nucleus, increases so muchis that it flies awaya (back It was also observations, for its thevelocity cases when an electron trapped within limitedoff). confinement. found that the assumptions of quantum with experimental observations While classical physics applies when themechanics electron isagree free to move or when the contining size extensive. in allis cases when electrons are tightly bound in a limited size. However, if the size is

Unit 5:

Quantum Mechanics

301 303

Wave Particle Duality Chapter 12: Introduction to Modern Physics Unit 5: 300

Fig (12-28b) Head of a fly as seen by an e-microscope

Fig (12-28c) Uranium atoms as seen by a special type of e-microscopes

Fig (12-28d) Rubella virus as seen by an e-microscope (white spots are on the surface of the infected cells)

II) Essay questions

0.67x10-3N

Introduction to Modern Physics Chapter 12: Wave Particle Duality

• Classical physics cannot explain many phenomena, particularly those in which light ( or 1) Show why the wave theory failed to explain the photoelectric effect, and how Einstein em radiation) interacts with electrons or atoms. managed to interpret the experimental results of this phenomenon. • Light or any em radiation consists of a huge collection of photons, each photon having 2) Show verifyh isthePlanck’s particleconstant nature ofand lightν isfrom blackbody radiation . where the the frequency. energyhow hν, to • An evidence for photons is the effect, where photocurrent 3) Explain the Compton effect andphotoelectric show how it proves the particle nature ofdepends light ? on the intensity of incident light as long as the frequency is greater than a critical value νc. But if the frequency is less than νc , no photocurrent flows. The kinetic energy of the electron freed by the photoelectric effect depends on the frequency not on the light intensity. • A photon has a mass, a linear momentum and a constant speed which is the speed of light. It has a size denoted by the wavelength. If a photon falls on a wall, it applies a small force on it, but if it falls on an electron, the electron will be thrown off due to its small mass and size. • Compton effect proves the particle nature of photons, where a photon has mass, speed and linear momentum. • A wave describes the collective behavior of photons. • The wavelength of a photon is Planck’s constant divided by the linear momentum. The same relation applies to a free particle, where the wavelength describes the wave nature of the particle ,i.e., the wave accompanying the particle. • The electron microscope proves de Broglie relation for particles. It is used to detect diminutive particles. • Quantum mechanics is based on assumptions which agree with experimental observations, for the cases when an electron is trapped within a limited confinement. While classical physics applies when the electron is free to move or when the contining size is extensive.

Unit 5:

whose mass is 10 kg , what happens object is an electron and why ? InifatheNutshell

303 305

Wave Particle Duality Chapter 12:

Learn at Leisure Questions and Drills Can you identify the contribution of each of the following scientists to modern physics ? I)Drills 1) Calculate the energy of a photon whose wavelength is 770 nm and find its mass and linear momentum? (2.58x10-19J , 0.29x10-35kg , 0.86x10-27kgm/s)

2) Calculate the mass of an X- ray photon and a γ ray photon if the wavelength of X-ray is 100 nm , and that of γ-ray is 0.05 nm Planck

Einstein

(mX=2.2 X10-35kg , mγ=4.4x10-32kg) de Broglie Bohr

3) Calculate the wavelength of a ball whose mass is 140 kg which moves at velocity 40 m/s. Also, calculate the wavelength of an electron if it has the same velocity.

Introduction to Modern Physics

λe=1.8 x10-5m)

4) A radio station emits a wave whose frequency is 92.4 MHz. Calculate the energy of

Unit 5:

(λ=1.18x10-37m ,

(velocity=0.725x106m/s , V=1.5Volt)

302 304

each photon emitted from this station. Also, calculate the rate of photons φL if the

power of the station is 100 kW. Schrodinger

Heisenberg

-28 29 -1 (E=612.15x10 J , φ =16.3 x10 s ) Compton Thompson L

5) An electron is under a potential difference 20 kV. Calculate its velocity upon collision with the anode from the law of conservation of energy. The electron charge is 1.6x10-19C, its mass is 9.1 x 10-31 kg. Then calculate λ and PL.

(v=0.838x108m/s , λ=0.868x10-11m , PL=7.625x10-23kgm/s)

6) If the least distance detected with an electron microscope is 1nm, calculate the velocity of the electrons and the potential of the anode. 7) Calculate the force by which an e-beam whose power is 100 kW affects an object

Chapter 13: Atomic Spectra

spectra of the atoms of all elements would have been continuous. This is contrary to all experimental observations.The spectra of the elements have a discrete nature, and are called line spectra, i.e., occurring at wavelengths characteristic of the element.

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

1310

screen

Atomic Spectra

Chapter 13

Bohr’s Model (1913)

The word atom goes back to a Greek origin, meaning the indivisible. Different models for the atom have been put forth since then by many great scientists based on many experimental evidences. Bohr’s Model (1913)

first shell

Bohr’sModel Model(1913) (1913) Bohr’s

first shell

a) Thompson’s Atom (1898)

first shell

1 – After Thompson conducted several experiments leading to the discovery of the e electron and the determination of the ratio m for the electron, he put forth a model of a e solid positively charged substance in which negative electrons were second immersed shell (Fig13–1). 2 – Since the atom is neutral, the sum of the negative charge is equal to the sum of the positive charge.

Fig ( 13-5a)

Bohr Bohr

second shell

Bohr’s Model second shell

Fig ( 13-5a)

Bohr studied the difficulties faced by Rutherford’s model,

second shell Bohr’s Model

and proposed a model for the hydrogen atom building on

Bohr’s Model

Bohr

free (electron Fig 13-5a)level continuous levels

Fig ( 13-5a)

Bohr faced by Rutherford’s model, Bohr studied the difficulties Rutherford’s findings : Fig (13-1)

free electron level Bohr’s Model continuous levels andBohr proposed a model for the hydrogen atom building on Thompson’s faced is byaRutherford’s model, free electron level 1) At thestudied centertheofdifficulties the atom there positively charged continuous levels Rutherford’s atomic findingsmodel : Rutherford

and proposed a model for the hydrogen atom building on nucleusstudied . thethe difficulties faced by Rutherford’s model, 1) AtBohr the center of Rutherford’s findings : atom there is a positively charged

free electron level

electrons around the nucleus equals the number of

in eachcharges shell (Fig 13 –nucleus. 5). positive in the

3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of

Energy

Energy Energy Energy

Rutherford’s Atom (1911) continuous levels 2)b) Negatively charged electrons move aroundatom the nucleus inon nucleus . and proposed a model for the hydrogen building 1) At the center of thehisatom a positivelybased charged Rutherford performed well there knownisexperiment on which he formulated a model shells. Each shell (loosely often called orbit) has an energy 2) Rutherford’s Negatively charged electrons move around the nucleus in findings : . fornucleus the structure of the atom and showed that it was not solid. Each charged shelldo(loosely often calledaround hasthey an energy value. Electrons not emit radiation asorbit) long as remain 2) 1)shells. Negatively electrons move the nucleus in -4 At the center of the atom there is a positively charged In his experiment, Rutherford bombarded a thin gold plate (10 cm) with a beam of alpha value. Electrons do not emit radiation as long as they remain in each shell (Fig 13 – 5). 4 shells. Each shell (loosely often called orbit) has an energy nucleus particles ( He.2) (Fig 13 – 2 a, b). in eachElectrons shell (Figdo13not – 5). emit neutral, radiationsince as longthe as they remain 3)2)value. The atom is electrically number of in Negatively charged electrons move around the nucleus 3) in The atom is electrically neutral, since the number of each shellaround (Fig 13the – 5).nucleus equals the number of electrons shells. Each shell (loosely often equals called orbit) has an energy the nucleus 3) electrons The atomaround is electrically neutral, since the the number number of of positive charges in the nucleus. Fig (13-5b) (13-5b) positive value. charges Electronsindothenotnucleus. emit radiation as long as they remain Fig Energy levels levels Energy Fig (13-5b)

Energy levels

Unit 5: modern physics Chapter 13: Atomic Spectra Unit 5:modern modern physics Chapter 13:Spectra Atomic Spectra Unit 5: physics Chapter 13: Chapter Atomic Unit 5: modern modern physics Chapter 13: Atomic Spectra Unit 5: physics 13: Atomic Spectra

first shell

Overview

307

304 305 311 311 311

Chapter 13: Atomic Spectra

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

310 305

screen

Bohr’s Model (1913) Emission of Light Emission from Bohr’s of Light Atom.from Bohr’s Atom.

first shell

Unit 5: modern physics

second shell

Fig ( 13-5a)

Bohr

Bohr’s Model

free electron level continuous levels

Electron transitions Electron transitions

Rutherford’s findings : between atomic levelsbetween atomic levels

1) At the center of the atom there is a positively charged

Fig (13 – 7)

Standing waves

Chapter 13: Atomic Spectra

Bohr studied the difficulties faced by Rutherford’s model, (13 – 6)for the hydrogen Fig (13 –atom 6) building on and proposedFig a model

Fig (13 – 7)

Standing waves

Energy

nucleus . 1) When hydrogen atoms 1) When are stimulated hydrogen(given atoms energy) are stimulated not all of (given themenergy) are excited not all theofsame them are excited th 2) Negatively charged electrons move around the nucleus in way. Thus, electrons inway. different Thus,atoms electrons move in from different the first atomslevel move K from (n = the 1) tofirst different level K (n = 1) to di shells. Each shell (loosely often called orbit) has an energy higher levels (n = 2, 3,higher 4..) levels (n = 2, 3, 4..) Electrons do not emit radiation as long as they 2) value. Electrons remain in2) excited Electrons levels remain (or states) in excited onlyremain levels for a (or short states) period only of for timea ,called short period of time -8 -8 inlifetime each shell (Fig 10 13 –s),5). then they revert10to the s), then lowest they level revert (ground to the state). lowest level (ground state). (nearly lifetime (nearly levelsince E1 , the the electron emits whose emits energya photon whose ene from3)level In going Eneutral, from level E2 to level , the electron 3)3) In Thegoing atomdown is electrically number of Ea1photon 2 todown c λ = ν = E E : where ν is ν the = E frequency E : where of the ν photon is the frequency and its wavelength of the photon is : and its h h 2 1 the nucleus equals 2 1 the number of electrons around ν wavelength is : λ

4) positive The linecharges spectrum 4)ofThe hydrogen line spectrum consists ofof hydrogen a particular consists energyofvalue, a particular and hence energy a value, and he in the nucleus. Fig (13-5b) particular frequency. particular frequency. Energy levels

307 313 311

absorbed photon

absorbed unabsorbed unabsorbed photon photon photon

gas

Fig (13-5c) Fig (13-5c)

modern physics

energy

potential difference

slit

energy

Chapter 13: Atomic Spectra

emitted photon prism

emitted photon screen

Fig (13 –Fig 4a(13 ) - 5d) Fig (13 - 5d)

Apparatus for studying the spectra of the elements Absorption ofAbsorption photons of photons Emissin of photons Emissin of photons

He then addedHethree thenmore addedpostulates: three more postulates: 1) If an electron 1) Ifmoves an electron from an moves outerfrom shellanof outer energyshell E2 to of an energy innerEshell of inner energyshell of energy 2 to an

E1 (E2 > E1 ),E1an(Eamount ofanenergy amount E2 of – Eenergy E2 – Ein released form ofinathe photon, form whose of a photon, whose 2 > E1 ), 1 is released 1 isthe

energy hν = Eenergy = E2ν–isEthe where frequency ν is the of frequency the emittedofphoton the emitted (Fig 13 photon – 6). (Fig 13 – 6). 2 – E1, hν 1, where

Unit 5:

2) The electric2)(Coulomb’s) The electric (Coulomb’s) forces and mechanical forces and(Newton’s) mechanicalforces (Newton’s) are at work forcesinare theat work in the atom.

atom.

3) We can estimate 3) We the canradius estimate of the shell radiusbyofconsidering the shell bythat considering the wave that accompanying the wave accompanying the the electron formselectron a standing forms wave a standing (Fig 13wave – 7 ).(calculate (Fig 13 – 7the).(calculate orbit radiusthefororbit n =radius 1, 2, 3,)for n = 1, 2, 3,) Fig (13 – 4b ) Spectra of some elements

310 312 306

Bohr’s level LModel (n = 2) (1913) from higher levels. This series lies in the visible range.

first shell

3) Paschen’s series: where the electron moves down to level M (n = 3) from higher 4) Bracket’s series: where the electron moves down to level N (n = 4) from higher levels. This series lies in the IR range.

Unit 5:

levels. This series lies in the infrared (IR) range.

5) Pfund’s series: where the electron moves down to level O (n = 5) from higher levels. among the line spectrum of hydrogen.

Spectrometer

second shell

To obtain a pure spectrum, a spectrometer is

Fig ( 13-5a)

Bohr of 3 parts : used (Fig 13 – 9). It consists

Bohr’s Model

1) a source of rays : a light source in front of which there is a slit whose width can be Bohr studied difficulties adjusted by the a screw. Thisfaced slit isbyatRutherford’s the focal model, and point proposed a modellens. for the hydrogen atom building on of a convex Rutherford’s findings : 1) At the center of the atom there is a positively charged

Fig (13 – 9a)

Spectrometer

nucleus . shells. Each shell (loosely often called orbit) has an energy source

value. Electrons do not emit radiation as long as they remain slit in each shell (Fig 13 – 5).

Energy

2) Negatively charged electrons move around the nucleus in red

prism

yellow

3) The atom is electrically neutral, since the number of

violet

electrons around the nucleus equals the number of Fig (13 – 9b) positive charges in the nucleus. Spectrometer schematic

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

This series lies in the far IR and is the longest wavelengths ( the lowest frequencies)

Fig (13-5b)

Energy levels

309 315 311

Chapter 13: Atomic Spectra

paschen’s series

Different series of atomic spectral lines for hydrogen are produced, and are arranged as spectra of the atoms of all elements would have been continuous. This is contrary to all follows (Fig 13 – 8): experimental observations.The spectra of the elements have a discrete nature, and are | spectra, i.e., occurring at wavelengths characteristic of the element. P n=line called n=6

Pfund’s series

Brackett’s series

N M

potential difference

Paschen’s series

gas

Balmer’s

slitseries

prism

screen

Apparatus for studying the spectra of the elements ultraviolet

Lyman’s series

Leyman’s series

Fig (13 – 8 a)

Atomic spectral series for hydrogen

1) Leyman’s series: where the electron moves down to level K (n = 1) from higher levels. This series lies in the ultraviolet range (short wavelengths and Fig (13 – 4b ) high frequencies).

n’s che pas ries se

Unit 5:

modern physics

Fig (13 – 4a )

Balmer’s series

Fig (13 – 8 b)

Atomic model for

Spectra moves of somedown elements 2) Balmer’s series: where the electron to hydrogen spectrum spect

310 314 308

is calledModel the emission spectrum. Bohr’s (1913)

first shell

- It was found experimentally that when white light passes through a certain gas, some those which appear in the emission spectrum of the gas (Fig 13 – 9) . This type of spectrum is called the absorption spectrum. Fraunhofer lines in the solar spectrum are

Unit 5:

wavelengths in the continuous spectrum are missing. These wavelengths are the same as

examples of the absorption spectrum of the elements in the Sun, basically helium and atomic hydrogen

second shell

barium

Fig ( 13-5a)

mercury Bohr

Bohr’s Model

sodium

Fig (13 – 10) Bohr studied the difficulties faced by Rutherford’s model,

free electron level

and proposed a model for the hydrogen atom building on

X-rays findings : Rutherford’s

1)What At theare center of the X-rays ? atom there is a positively charged

-13

nucleus . They are invisible electromagnetic waves of short wavelength (10

-8

– 10 m) between

value. Electrons not emit radiation as long as they remain Properties of do X-rays:

Energy

and gammacharged rays. They were first Rontgen.in He called it so (the unknown 2)uvNegatively electrons movediscovered around thebynucleus

- inThey eachcan shellpenetrate (Fig 13 –media 5). easily . 3)-

Theyatom can ionize gases . neutral, since the number of The is electrically They diffract in crystals . electrons around the nucleus equals the number of - positive They affect sensitive plates . charges in thephotographic nucleus.

Chapter 13: Atomic Spectra

Emission line spectra for some elements continuous levels

rays) because did(loosely not know whatcalled they orbit) were .has an energy shells. Each he shell often

modern physics

hydrogen.

Fig (13-5b)

Energy levels

311 317 311

Chapter 13: Atomic Spectra modern physics Unit 5:

Fig (13-9c)

Use of a spectrometer to measure the temperature of the stars and their gases potential

difference

2) a turntable on which a prism is placed. 3) a telescope consisting of two convex lenses (objective and eye piece). slit prism screen gas To use the spectrometer for obtaining a pure spectrum, the slit is lit Fig (13 4a ) at the minimum with bright light falling from the slit onto the â&#x20AC;&#x201C;prism for studying the spectra of thethe elements angle of deviation.Apparatus The telescope is directed to receive light passing through the telescope.The objective focuses the rays belonging to the same color at the focal plane of the lens . That is Fraunhofer how we obtain a sharp (pure) spectrum. From studying the line spectra of different elements whose atoms are excited , we notice different types of spectra (continuous and line) : - the spectrum consisting of all wavelengths in a continuous manner is called the continuous spectrum. - the spectrum occurring at specified frequencies and not continuously distributed is called the line spectrum. Alternatively, they may be divided as emission and absorption spectra : - the spectrum resulting from the transfer excited Figof(13 â&#x20AC;&#x201C; 4b )atoms from a high level to lower level Spectra of some elements

310 316 310

Bohr’s Model (1913)

first shell

Unit 5:

Fig ( 13-5a)

Bohr

Bohr’s Model

Bohr studied the difficulties faced by Rutherford’s model, and proposed a model for the hydrogen atom building on Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

second shell

Fig (13-5b)

Energy levels

313 311

Unit 5:

modern physics

spectra of the atoms of all elements would have been continuous. This is contrary to all Coolidge Tube experimental observations.The spectra of the elements have a discrete nature, and are cooling fins called spectra, i.e., occurring wavelengths characteristic of the element. Thisline is used to produce X-rays.at When the filament is

high energy, depending on the voltage difference between the target and the hot filament. When an electron collides

vacuum tube t

under the influence of the electric field,which gives them

copper rod

heated, electrons are produced and directed at the target

ta r

Chapter 13: Atomic Spectra

target

high dc voltage

potential

with the tungsten difference target, part- if not all- of its energy is converted to X-rays (Fig 13 – 11).

Spectrum of X-rays Analyzing a beam of X-rays generated from a target to gas

slit

prism

components of different wavelengths, we find that the Fig (13 – 4a ) spectrum consists of two parts :

X rays screen

hot filament

Apparatus for studying the spectra of the elements

a) the continuous spectrum of all wavelengths (within a certain range) regardless of the target material.

b) the line spectrum corresponding to certain wavelengths characteristic of the target material, called the

heating source

Fig(13 –11)

Coolidge tube

characteristic X-ray radiation.

Interpretation of X-ray generation a) characteristic radiation The line spectrum is generated when an electron collides with an electron close to the (13latter – 4b electron ) nucleus of the target material atom. IfFigthe receives sufficient energy, it Spectra of some elements jumps to a higher level, or leaves the atom altogether, and is replaced by another electron

310 318 312

Bohr’s Model (1913)

rib

first shell

spinal cord

Unit 5:

2) X- rays have a great penetrating power. This is why they are used to detect defects in metallic structures. 3) X- rays are used in imaging bones and fractures and second shell

Bohr

chest

heart lung Fig ( 13-5a) Fig (13 –Model 14) Bohr’s

An X-ray image for the chest

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

some other medical diagnosis (Fig 13 – 14).

Fig (13-5b)

Energy levels

315 321 311

Chapter 13: Atomic Spectra

energy brakingwould effecthave of the surrounding giving rise to spectra continually of the atomsdueof toallthe elements been continuous.electrons, This is contrary to all experimental observations.The spectra of the elements a discretesince nature, and are electromagnetic radiation covering all different possible have wavelengths, the electron called line spectra, i.e., occurring at wavelengths characteristic of the element. loses energy gradually. This is the origin of the continuous radiation of X-rays.

Important Applications of X-rays 1) One of the important features of X-rays is diffraction, as they penetrate materials. That is why X-rays are used in studying the crystalline structure of materials (Fig 13 – 13). The

potential difference atoms in the

crystal act as a diffraction grating (which is a generalization of

diffraction from a double slit). Bright and dark fringes form, depending on the difference in the optical path. gas

slit

prism

screen

Unit 5:

modern physics

Fig (13 – 4a )

aperture Apparatus for studying the spectra of the elements

X-ray tube

anode

X-rays high DC voltage

filament heating source

crystal vacuum

Fig (13 – 13) Figin(13 – 4b ) crystals Use of X-rays studying Spectra of some elements

310 320 314

Bohr’s Model (1913)

first shell

Unit 5:

Fig ( 13-5a)

Bohr

Bohr’s Model

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

second shell

Fig (13-5b)

Energy levels

317 311

Chapter 13: Atomic Spectra

spectra of the atoms of all elementsInwould have been continuous. This is contrary to all a Nutshell experimental observations.The spectra of the elements have a discrete nature, and are linepostulates spectra, i.e., at wavelengths characteristic of the element. ·•called Bohr’s and occurring model of the hydrogen atom : When an electron jumps from a high level to a lower level, it produces radiation in the form of a photon of frequency ν and energy hν, which is equal to the difference

between the two levels

hν = E2 – E1, E2 > E1

potential ·• The linedifference spectrum of hydrogen consists of 5 series. Each line corresponds to a definite

energy difference, frequency and wavelength Lyman uv Balmer gas visible

slit

prism

screen

modern physics

Paschen IR (infrared) Fig (13 – 4a )

Brackett IR for studying the spectra of the elements Apparatus Pfund far IR ·• The spectrometer is an apparatus used to decompose light to its components (visible and invisible) ·• X-rays are an invisible radiation of short wavelengths, first discovered by Rontgen (1895).He called them the unknown (X) rays ·• X-ray diffraction is used in studying the crystalline structure, and also in the industrial

Unit 5:

and medical applications.

Fig (13 – 4b )

Spectra of some elements

310 322 316

Chapter 13: Atomic Spectra

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

1310

screen

Laser

first shell

Overview Rarely has any discovery left an impact on applied science as the discovery of laser has

second shell

Spontaneous Emission and Stimulated Emission

The atom has energy levels, the lowest of which is called ground Fig state(E ( 13-5a) 1) in which the Bohr

atom initially exists. The atom may be excited to one of higher statesBohr’s E2, E3Model etc.

If we shine a photon with energy hν = E2– E1 on the atom, the atom absorbs this photon

Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain in each shellExcitation (Fig 13 –by 5).absorption of

Energy

2) Negatively charged electrons move around the nucleus in

Relaxation to a lower level after a lifetime and release of excitation

energy from an external 3) The atom is electrically neutral,source since the number of

energy

electrons around the nucleus equals the number of Fig (14-1) positive charges in the nucleus. Spontaneous emission

Chapter 13: Atomic Spectra Chapter 14: Laser

-8 andBohr getsstudied excited the to Edifficulties a lifetime (nearly getslevel rid of this facedafter by Rutherford’s model,10 s), the 2. Soon enough freeatom electron levels excitation energy in the form of ahydrogen photon andatom goesbuilding back to itsonoriginal continuous state (Fig 14 -1). and proposed a model for the

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done. Soon after its discovery, laser has been introduced into optics, biology, chemistry, medicine and engineering especially communications. The word laser is an acronym for Light Amplification by Stimulated Emission of Radiation. In 1960, Maiman built the first laser out of chromium-doped Ruby. Later, He-Ne laser was manufactured along with other types of lasers.

UnitUnit 5: 5:

Bohr’s Model Chapter 14(1913)

Fig (13-5b)

Energy levels

325 319 311

Chapter 13: Atomic Spectra

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

310 319

screen

Bohr’s Model (1913) remain unspread and unscattered, unlike photons emitted spontaneously.

first shell

Spontaneous emission

Stimulated emissions

Energy

1) At the center of the atom there is a positively charged 4 - The intensity of photons - The intensity remains constant over long nucleus . decreases according to the distances contrary to the inverse is thesquare law.inIt has been possible to 2) Negatively inverse charged square electronslaw. moveThis around nucleus called spreading. While send a laser beam to the Moon and shells. Eachcollisions shell (loosely called orbit) an energy withoften particles is has receive it back, without much loseses, calleddoscattering. In ordinary the long distance involved. value. Electrons not emit radiation as long asdespite they remain light sources both spreading Spreading effect is nil and limited in each shelland(Fig 13 – 5). occur. scattering scattering takes place. 3) The atom is electrically neutral, since the number of 5 - This is the dominant radiation - This is the dominant radiation in electrons around the nucleus equals the number of in ordinary light sources. laser sources. positive charges in the nucleus. Fig (13-5b)

Chapter 13:14: Atomic Spectra Chapter Laser

- Thetheemitted photons propagate - The emitted coherent Bohr3studied difficulties faced by Rutherford’s model, photonsfreeareelectron level randomly and propagate in onecontinuous directionlevels as a and proposed a model for the hydrogen atom building on collimated parallel beam. Rutherford’s findings :

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1 - occurs when the atom relaxes from - occurs where an external photon an excited state to a lower state, stimulates excited atoms to emit emitting spontaneously the energy the energy difference is the form difference in the form of a photon of a photon before the lifetime without the effect of an external interval is over. second shell photon. It occurs after the lifetime interval is over. Fig ( 13-5a) Bohr monochromatic 2 - The emitted photons have a - The emitted photons areBohr’s Model wide range of wavelengths. (single) wavelength.

Unit Unit 5:5:

The following table gives a comparison between spontaneous and stimulated emissions :

Energy levels

327 321 311

Chapter 13: Atomic Chapter 14: Spectra Laser modern physics modern physics Unit 5: Unit 5:

This type radiation is called spontaneous radiation. It is the typeThis of radiation common spectra of theof atoms of all elements would have been continuous. is contrary to all in ordinary light sources. The emitted the same energyandas are the experimental observations.The spectra photon of the has elements havefrequency a discreteand nature, called line characteristic of theareelement. photon thatspectra, causedi.e., theoccurring excitationat .wavelengths But the phase and direction arbitrary. In 1917, Einstein showed that in addition to spontaneous radiation, there is another type of radiation, called stimulated emission (the dominant emission in lasers). If a photon of energy E2-E1 falls on an excited atom at level E2 before the lifetime is over, this photon pushes the atom back to the ground state, and hence, the atom radiates the excitation energy

potential difference in the form

of a photon of the same frequency, phase and direction of the falling

photon. (Fig 14â&#x20AC;&#x201C;2).

gas

slit

prism

Fig (13 â&#x20AC;&#x201C; 4a ) incident photon Apparatus for studying the spectra of the elements Relaxation to a lower level

A photon passes by an

due to an external photon

excited atom

Fig (14-2)

before its lifetime is over

Stimulated emission

Thus, throughout stimulated radiation, there are two types of photons; the stimulating and the stimulated photons moving together at the same frequency, phase and direction. The emission of photons from the atoms of the material in this way renders these Fig (13 â&#x20AC;&#x201C; 4b ) photons coherent and collimated for long distances. They are highly concentrated, and Spectra of some elements

310 326 320

screen

Bohr’s Model (1913)

first shell

Unit Unit 5:5:

ordinary light source

an ordinary light source is scattered during propagation

laser light travels in parallel rays for long distances without much scattering

second shell

Fig (14-4a)

Fig ( 13-5a)

Bohr of an ordinary light source and a laser Scattering Bohr’s Model

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of Fig (14-4b) Launching laser beam from the Earth to a reflector on the positive charges in thea nucleus. Fig (13-5b) surface of the Moon, 380000 km away

Chapter 13:14: Atomic Spectra Chapter Laser

free electron level continuous levels

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laser source

Energy levels

329 323 311

light intensity

(φL)

max

light intensity

Chapter 13: Atomic Chapter 14: Spectra Laser modern physics modern physics Unit 5: Unit 5:

spectra of the atoms of all elements would have been continuous. This is contrary to all experimental observations.The spectra of the elements have a discrete nature, and are φL called line of the element. φL spectra, i.e., occurring at wavelengths characteristic

potential difference

gas

prism

screen

Fig (13 – 4a )

Fig (14-3a)

forordinary studying the spectra of the elements SpectralApparatus width for an monochromatic light source

Fig (14-3b)

Spectral width for a laser source

Properties of a laser beam 1) Monochromaticity: Each line in the visible spectrum in ordinary light sources includes a band of wavelengths (this is why the ordinary color appears to have different shades to the naked eye). The intensity of each wavelength in this band width is shown in Fig (14–3a). A laser source emits one spectral line with a very limited bandwidth and Fig (13 – 4bof) that spectral line Fig (14–3b), hence it the intensity is concentrated at the wavelength is called monochromatic.

310 328 322

slit

Spectra of some elements

Bohr’s Model (1913)

first shell

coherent light

Bohr

second shell

Coherence low light high light intensity intensity

Fig ( 13-5a)

Bohr’s Model

Bohr studied the difficulties faced by Rutherford’s model, and proposed a model for the hydrogen atom building on The intensity of ordinary light decreases with distance from the source inverse square law 1) At the center of the atom theredueistoa the positively charged

Rutherford’s findings :

2) Negatively charged electrons move around the nucleus in Fig (14-6) Lasercalled light maintains theenergy same shells. Each shell (loosely often orbit) has an intensityasas it propagates value. Electrons do not emit radiation long as they remain

Energy

nucleus .

Theory of the in each shell (FigLaser 13 – 5).Action:

3) The atom is electrically neutral, since the number of Laser action depends on driving the atoms or molecules of the active medium into a electrons around the nucleus equals the number of state of population inversion, while maintaining a form of dynamic equilibrium. In this positive charges in the nucleus. Fig (13-5b) state, the number of atoms in the excited state exceeds the number of atoms in the lower

Chapter 13:14: Atomic Spectra Chapter Laser

free electron level continuous levels

modern physics physics modern

Fig (14-5)

Unit Unit 5:5:

incoherent light

Energy levels

331 325 311

Chapter 13: Atomic Chapter 14: Spectra Laser modern physics modern physics Unit 5: Unit 5:

potential difference

Fig (14-4d)

Measuring the distance between the Moon

Fig (14-4c)

gas Measuring astronomical

distances by a laser beam

slit

and the Earth by the reflection of a laser prism

screen

beam from a reflector on the lunar surface

Fig (13 â&#x20AC;&#x201C; 4a )

for studying 2) Collimation : InApparatus ordinary light sources, the the spectra diameterofofthe theelements emitted light beam increases

with distance , where in lasers, the diameter stays constant for long distances without much unscattering. Thus ,energy is transmitted without much losses . 3) Coherence: Photons of ordinary light sources propagate randomly or incoherently. They emanate at different instants of time, and have inconsistent and varying phase. In lasers, however, photons emanate coherently both in time and place, since they come out together at the same time sequence, and maintains the same phase difference throughout, during propagation over long distances. This makes radiation intense and focused. 4) Intensity: Light produced by ordinary sources is subject to the inverse square law, since the intensity of radiation falling on unit area decreases, the further away from the light source, Fig (13 â&#x20AC;&#x201C; 4b ) due to spreading (Fig 14-4a). The laser rays falling on a unit surface are unspread. They Spectra of some elements

maintain a constant intensity and are not subject to the inverse squarelaw.

310 330 324

3) Resonant cavity is the container and the activating catalyst for amplification . It can be

Bohr’s Model (1913)

first shell

(a) external resonant cavity in the form of two parallel mirrors enclosing the active medium permitting multiple reflections leading to amplification as in gas lasers (Fig 14 – 7a).

as mirrors as in ruby laser (Fig 14 – 7b). One of the two mirrors is semitransparent to allow some of the laser radiation to leak out (Fig 14 – 8).

second shell

Fig ( 13-5a)

Bohr

Bohr’s Model

and proposed a model for the hydrogen atom building on

the active mediumlevels act as continuous mirrors

Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons not emit radiation as long as they remain Figdo(14-7a) in eachExternal shell (Figresonant 13 – 5). cavity

Energy

2) Negatively charged electrons move around the nucleus in Fig (14-7b)

Internal resonant cavity

3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13:14: Atomic Spectra Chapter Laser

Bohr studiedtwo the reflecting difficulties faced by Rutherford’s model, the two freepolished electronends levelof mirrors

modern physics physics modern

b) internal resonant cavity where the ends of the active material are polished so as to act

Unit Unit 5:5:

one of two types :

Fig (13-5b)

Energy levels

333 327 311

Chapter 13: Atomic Chapter 14: Spectra Laser modern physics modern physics Unit 5: Unit 5:

state. when stimulated emissionwould occurs,have it will amplified asThis photons are increased spectraThus, of the atoms of all elements beenbecontinuous. is contrary to all experimental observations.The of themedium, elementsduehave a discrete nature, between and are in number going back and forth spectra in the active to multiple reflections calledenclosing line spectra, i.e., occurring at wavelengths characteristic of theare element. two mirrors. In so doing, more and more excited atoms poised to generate stimulated emission, which is further amplified and so on. This is the origin of amplification of the laser (Fig 14–7), called laser action .

Main Components of a Laser potential

Despitedifference the variations in size, type and frequency, three common elements must exist in any laser: 1) Active medium: This can be a crystalline solid (e.g. ruby), semiconductor (chapter15) a liquid dye, gas atoms Ne laser),prism ionized gases (e.g. screenArgon laser), or gas (e.g. He – slit molecular gases (e.g. CO2 laser).

Fig (13 – 4a ) 2) Sources of energy responsible for exciting the active medium as follows : Apparatus for studying the spectra of the elements (a) excitation by electrical energy, either by using radio frequency (RF) waves or by using electric discharge under high DC voltage gas lasers:( HeNe – Ar – CO2). (b) excitation by optical energy, also known as optical pumping, which can be done either by flash lamps (e.g. in ruby laser) or using a laser beam as a source of energy (liquid dye laser). (c) thermal excitation, by using the thermal effects resulting from the kinetic energy of gases to excite the active material (e.g. in He-Ne laser). (d) excitation by chemical energy as chemical reactions between giving gases energy to stimulate atoms toward lasing (e.g. the–reaction Fig (13 4b ) between hydrogen and fluorine or the reaction between Deuterium CO2 ) . Spectrafluoride of someand elements

310 332 326

Bohr’s Model (1913)

first shell

Unit Unit 5:5:

a unexcited condition

second shell

excited condition Fig ( 13-5a)

Bohr

Bohr’s Model

free electron level continuous levels

Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy d value. Electrons do not emit radiation as longincident as they photon remain

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5).

emitted photon 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of Fig (14-8c) positive charges in the nucleus.

Chapter 13:14: Atomic Spectra Chapter Laser

metastable state Bohr studied the difficulties faced by Rutherford’s model, c =E2 - Eon 1 and proposed a model for the incident hydrogenphoton atomhνbuilding

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Population inversion through a third Fig (13-5b) metastable state Energy levels

335 329 311

modern physics modern physics

Chapter 13: Atomic Chapter 14: Spectra Laser

relaxing electron

spectraincident of the photon atoms of all elements would have been continuous. This is contrary to all emittednature, photonand are experimental observations.The spectra of the elements have a discrete called line spectra, i.e., occurring at wavelengths characteristic of the element. excited electron

before excitation

incident photon

Fig (14-8a)

after excitation

Stimulated emission by an external photon potential difference

higher level

a normal condition

gas

ground state

slit

prism

inverted population

Fig (13 â&#x20AC;&#x201C; 4a )

Apparatus for studying the spectra of the elements

c a photon approaches which causes excitation

Unit 5: Unit 5:

stimulated emission is generated

e recurrence of stimulated emission

Fig (13 â&#x20AC;&#x201C; 4b ) (14-8b) Spectra Fig of some elements Laser action

310 334 328

screen

Helium Neon (He – Ne) laser Bohr’s–Model (1913)

mirror

window

vacuum tubefirst shell

near equality of the values of the same metastable excited energy levels in these two elements.

(a) Construction of He-Ne laser :

He – Ne laser schematic

neon in the ratio 10 : 1 at a low pressure of nearly 0.6 mm Hg (Fig 14 – 9).

second shell

2) At both ends of the tube there are two plane or concave Fig ( 13-5a)

parallel mirrors whichBohr are perpendicular to the tube axis.

Bohr’s Model

One has a reflection coefficient of nearly 99.5%, while

Fig (14-9b)

He – Ne laser

Energy

(b)shells. Operation Each shell: (loosely often called orbit) has an energy

free electron level continuous levels

1) The difference inside the tube leads to theremain excitation of the helium atoms to value.voltage Electrons do not emit radiation as long as they higher (Fig13 14––5). 10). in each levels shell (Fig 3) The The excited atom ishelium electrically sincethetheunexcited number neon of atoms inelastic collisions. 2) atoms neutral, collide with electrons around the nucleus equals the helium number of to the neon atoms due to the Thus, energy is transferred from the excited atoms positive charges in the nucleus. Fig (13-5b) near equality of the excited levels in both atoms. Neon atoms are, thus, excited.

Chapter 13:14: Atomic Spectra Chapter Laser

the other mirror is semitransparent with a reflection Bohr studied the difficulties faced by Rutherford’s model, coefficient of 98%. and proposed a model for the hydrogen atom building on 3) High frequency electric field feeding the tube from the Rutherford’s findings : outside to excite the helium and neon atoms, or a high 1) At the center of the atom there is a positively charged DC voltage difference inside the tube causing electric nucleus . 2) discharge. Negatively charged electrons move around the nucleus in

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1) A quartz tube including a mixture of helium and

Fig (14-9a)

Unit Unit 5:5:

These two elements have been selected due to the

laser beam

Energy levels

337 331 311

Chapter 13: Atomic Chapter 14: Spectra Laser

potential difference

Fig (14-8d) Multiple reflections between the two mirrors

mirror gas

glass tube

excited atom slit

prism

semitransparent mirror screen

Fig (13 â&#x20AC;&#x201C; 4a ) Figthe (14-8e) Apparatus for studying spectra of the elements Amplification by multiple reflections

Unit 5: Unit 5:

modern physics modern physics

spectra of the atoms of all elements would have been continuous. This is contrary to all mirror semitransparent mirror experimental observations.The spectra of the elements have a discrete nature, and are called line spectra, i.e., occurring at wavelengths characteristic of the element.

Fig (13 â&#x20AC;&#x201C; 4b ) Fig (14-8f) Spectra of some elements Output radiation from the semitransparent mirror

310 336 330

mirrors in its way, they bounce off inside the tube and cannot get out. Bohr’s Model (1913)

first shell

may well collide with some neon atoms in the excited metastable state, well before lifetime is over. Thus, they stimulate the neon atoms to emit photons of the same energy and direction as the colliding photon. Thus, the number of photons moving inside the

Unit Unit 5:5:

6) During the propagation of these photons inside the tube between the two mirrors, they

tube multiplies. action. This is how amplification takes place .

second shell

8) When radiation inside the tube reaches a certain level, we let it out partially through the Fig ( 13-5a) semitransparent mirror, Bohrwhile the rest of the radiation remains trapped inside the tube. Bohr’s Model

The stimulated emission and the lasing action go on.

continuous levels

and proposed a model for the hydrogen atom building on Helium atoms collide again with neon atoms, and the cycle repeats. Rutherford’s findings : 10) As to the helium atoms which have lost their energy by collision, they regain energy 1) At the center of the atom there is a positively charged through the electric discharge and so on. nucleus .

Laser applications

Energy

2) Negatively charged electrons move around the nucleus in Today there are different types and sizes of lasers. Laser light covers different regions of shells. Each shell (loosely often called orbit) has an energy thevalue. electromagnetic visible to uv and remain IR. Some laser systems can focus a Electrons dospectrum not emit from radiation as long as they laser beamshell in a small in each (Fig 13spot, – 5).where energy might get so high as to melt - and even evaporate - iron, pierce diamond. There are since lasers the which may have 3) The oratom is electrically neutral, number of enough energy to destroy electrons aroundinthe equals the number of missiles and planes whatnucleus is termed Star War. Some applications of lasers also include positive charges theand nucleus. holography (Fig 14 in - 11) medical applications.

Chapter 13:14: Atomic Spectra Chapter Laser

9) As to the neon atoms which have relaxed to a lower level, soon enough they lose further Bohr studied difficulties by Rutherford’s freeto electron level state. whatever leftthe of their energyfaced in different forms, andmodel, finally go back the ground

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7) The new stream of photons repeat the excursion, and thus, they remultiply by the lasing

Fig (13-5b)

Energy levels

339 333 311

Unit 5: Unit 5:

3) An ofaccumulation excited neon metastable spectra the atoms of of all elements wouldatoms have been continuous. This isstate contrary to all ensues. Theobservations.The excited level of aspectra neon atom haselements a experimental of the have a discrete nature, andlaser are -3 beam called line spectra, i.e., occurring relatively long lifetime (nearlyat10wavelengths s). Such acharacteristic of the element. level is called metastable state. Hence, 4) A group of neon atoms that are excited relax

energy transfer rapid decay by collisions between He and Ne atoms

excitation by collisions

energy

population inversion occurs in neon atoms. to a lower excited state. In so doing, they emit potential photons, which have energy spontaneous difference

equal to the difference in energy levels. Then, photons propagate randomly in all directions inside the tube.

5) Photons which propagate slit of the prism gas along the axis tube are reflected back by one of the two Fig (13 – 4a )

Figscreen (14-10a)

collisions of atoms rapid decay

excitation of electrons

rapid decay

collisions with the walls of the container ground state

Fig (13 – 4b ) Fig (14-10b)

Spectra of some Transitions between energyelements levels in He-Ne laser

310 338 332

ground state

He – Ne laser energy levels

Apparatus for studying the spectra of the elements He Ne

energy

modern physics modern physics

Chapter 13: Atomic Chapter 14: Spectra Laser

energy

Learn atModel Leisure (1913) Bohr’s

first shell

Unit Unit 5:5:

Types of holograms

A hologram is a kind of diffraction grating which is a generalization of a double slit, where interference occurs between the penetrating waves. To make a hologram, the are recorded on the holographic plate, while at the same time the reference beam is recorded. Interference pattern is

second shell

formed and a hologram is developed the same way as a

Fig ( 13-5a)

photographic plate. ToBohr read a hologram, a laser beam is

Bohr’s Model

used in the same direction as the laser beam during

Energy

theNegatively reference beam, the move same around side of the the nucleus observer,in if a screen or smoke is used, 2) charged.i.e.,on electrons where a 3D image be formed in space What we described above is the shells. Each shellmay (loosely often called orbit)(Fig has14–11). an energy transmission hologram, is lit from behind (Figremain 14–12). It may be lit also by an value. Electrons do notwhich emit radiation as long as they

in eachlight shellsource, (Fig 13but – 5).the image will have many colors. There are yet other types of ordinary 3) The atom holograms such isas electrically the reflectionneutral, hologram,since whichtheis litnumber up front,ofin which ordinary light may also the nucleus equalsLikethethenumber be electrons used. Therearound is also embossed hologram. reflectionofhologram, it may be lit up front,

Chapter 13:14: Atomic Spectra Chapter Laser

Fig (14-11) recording. The light rays read out the patterns formed on the Bohr studied the difficulties faced by Rutherford’s model, free electron level Hologram generates hologram, giving an imge as if the object is seen from that continuous levels a 3D image and proposed a model for the hydrogen atom building on angle, Looking through the hologram in the direction Rutherford’s findings : opposite to the reference beam, a virtual image is seen as if the object lay behind the 1) At the center of the atom there is a positively charged hologram, nucleus i.e., . on the beam side. We can also see a real image in the direction opposite to

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object is illuminated with a laser light. The reflected rays

positive charges in the nucleus. (13-5b) hologram i.e., on the observer side, and uses ordinary light. It may be considered as aFig transmission Energy levels

341 335 311

Unit 5: Unit 5:

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Chapter 13: Atomic Chapter 14: Spectra Laser

a) Holography:

spectra of the atoms of all elements would have been continuous. This is contrary to all Images of observations.The objects are formed by collecting rays reflected from them. experimental spectra of the elements have a discrete nature,Anandimage are represents in occurring the intensity of light fromcharacteristic point to point. Butelement. is light intensity all called line variations spectra, i.e., at wavelengths of the there is to it in the information about an image? If we have two rays leaving off an illuminated object at two points on it, there is a difference in intensity alright (proportional to the square of the amplitude of the wave). But in addition, there is a path difference between the two lit points and the corresponding points on the photographic plate where potential

the imagedifference is recorded due to the topology of the object. Thus, the waves leaving off the

object carry information in both amplitude and phase (phase difference = 2/ x path h

difference). The photographic plate records only the intensity (square of the amplitude) and does not record the phase. That is why a 2D image does not carry the 3D detail. In other gas

slit

prism

words, a plane image has only half the truth (only the intensity). In 1948, a Hungarian (13 â&#x20AC;&#x201C; 4a ) scientist Gabor (Nobel prize laurette) Fig proposed a method to obtain the component that is Apparatus for studying the spectra of the elements

missing from the information in the image and retrieve it from the beam, using another beam of the same wavelength called the reference beam. A laser beam is split into two beams. One is used to illuminate the object, and the other is used as the reference beam. The reflected beam and the reference beam meet at the photographic plate, and interference takes place. After the photographic plate is developed, resulting interference fringes appear coded, and we call such an image a hologram. Illuminating a hologram with a laser of the same wavelength and looking through it with the naked eye, we see an identical 3D image of the object without using any lenses. The full information (intensity and phase) is now retrieved due to the coherence nature of the laser. Fig (14 â&#x20AC;&#x201C; 12) shows the optical system used to obtain a hologram using a laser beam. Tens of photos may be stored in one Fig (13 â&#x20AC;&#x201C; 4b ) hologram. We may also obtain 3D images in holograms of moving objects. Spectra of some elements

310 340 334

screen

b) Lasers in medicine: Bohr’s Model (1913)

first shell

Unit Unit 5:5:

The retina contains light sensitive cells. In case of

retinal detachment, part of the retinal loses its function. Unless quickly treated, the eye may lose sight completely. In early stages, the eye may be treated by reconnecting the detached part with the layer underneath. Nowadays, lasers are used for that purpose Fig (14 – 14).

second shell

The operation takes less time and efford than before. The

Fig ( 13-5a)

thermal heat from theBohr laser cauterizes the points of

Bohr’s Model

detachment (endothermy). Lasers are also used to treat cases of far and near sightedness, so the patient can

treating retinal detachment Rutherford’s findings : optical fibers are used for diagnosis and even operative 1) At the center of the atom there is a positively charged surgery (Fig 14-16). nucleus .

Other applications of laser

Energy

2) Negatively charged electrons move around the nucleus in c) communications , where optical fibers carry information - loaded laser beam instead of shells. Each shell (loosely often called orbit) has an energy a wire carring electrical signals. value. Electrons do not emit radiation as long as they remain d)inindustry, each shellparticularly (Fig 13 – 5).fine industries. military include precision , smartofbombs and laser radar 3)e)The atomapplications is electrically neutral, sinceguidance the number (LADAR).around the nucleus equals the number of electrons

charges(Fig in the f) positive CD recording 14 –nucleus. 17).

Chapter 13:14: Atomic Spectra Chapter Laser

Bohr studied the difficulties facedOther by Rutherford’s free dispose with glasses Fig (14-15). applicationsmodel, of Fig electron (14-14)level continuous levels and proposed a model for the hydrogen atom building on Use of a laser beam in lasers in medicine include endoscopy, where lasers with

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This used to be a strenuous and delicate operation.

Fig (13-5b)

Energy levels

343 337 311

Chapter 13: Atomic Chapter 14: Spectra Laser modern physics modern physics Unit 5: Unit 5:

spectrareference of the atoms of all elements would have been continuous. This is contrary to all rays observations.The spectra of the elements have a discrete nature, and are experimental reference called line spectra, i.e., occurring at wavelengths characteristic of the element. mirror rays

hologram object

potential difference

virtual image

hologram

Fig (14-12b)

Fig (14-12a)

Hologram formation gas

slit

prism

Hologram as a grating

Fig (13 – 4a )

screen

Apparatus for studying thetospectra of theThere elements with a mirror behind. It is a cheap alternative a hologram. is also pulse hologram, which uses powerful laser pulses. Holograms can be made of people and moving objects at successive times, which may lead to the future 3D movies (Fig 14 – 13).

Fig (14-13)

Successive Fig stationary (13 – 4bshots ) giving an illusion of motion Spectra of some elements

310 342 336

real image

Bohr’s Model (1913)

first shell

Unit Unit 5:5:

Fig ( 13-5a)

Bohr

Bohr’s Model

free electron level continuous levels

and proposed a model for the hydrogen atomin building on Use of lasers endoscopy Rutherford’s findings :

pit

1) At the center of the atom there is a positively charged nucleus .

laser

pitless area

shells. Each shell (loosely lensoften called orbit) has an energy

Energy

2) Negatively charged electrons move around the nucleus in sideemit radiation as long as they remain value. Electronsback do not glass plate lens

in each shell (Fig 13 – 5).

detector

3) The atom is electrically neutral, since the number of mirror

electrons around the nucleus equals the number of positive charges in the nucleus.

Fig (14-17a)

Use of a laser in writing on CDs

Chapter 13:14: Atomic Spectra Chapter Laser

Bohr studied the difficulties faced by Rutherford’s Fig (14-16) model,

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second shell

Fig (13-5b)

Energy levels

345 339 311

Chapter 13: Atomic Chapter 14: Spectra Laser

(Fig 14-18).

h) arts and laser shows (Fig 14-19).

i) surveying (determining dimensions and areas). j) space research. potential difference

gas

slit

Fig (14-15)prism

Cornea treatment Fig (13 â&#x20AC;&#x201C; 4a ) by a

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

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screen

laser Apparatus for studying the spectra of the elements

Unit 5: Unit 5:

modern physics modern physics

g) laserofprinting where a laser beam would is used have to carry spectra the atoms of all elements beeninformation continuous. This is contrary to all experimental observations.The spectra of the have a discrete nature, and are from the computer to a drum coated by elements a photosensitive called line spectra, occurring wavelengths of the element. material. A toneri.e., is used to printat off from the characteristic drum onto paper

Bohr’s Model (1913)

In a Nutshell

first shell

It is the emission from one excited atom as it relaxes from a high energy level to a low energy level after its lifetime interval is over and under no external stimulus. • Stimulated emission:

Unit Unit 5:5:

• Spontaneous emission:

It is the emission from one excited atom as a result of a collision with an external end , come out in coherence ,i.e.,having the same phase, (direction and frequency). second shell • Properties of a laser beam :

Fig ( 13-5a)

1 ) spectral purity (monochromatic). Bohr

Bohr’s Model

2 ) collimation (parallel rays).

free electron level continuous levels

Energy

and proposed • Laser actiona: model for the hydrogen atom building on Rutherford’s findings : 1) the active medium must be in the state of population inversion . 1) At the center of the atom there is a positively charged 2) emission of radiation for the excited atom through the stimulated emission. nucleus . 3) amplification of stimulated emission through the resonant cavity 2) Negatively charged electrons move around the nucleus in • Basic elements of a laser : shells. Each shell (loosely often called orbit) has an energy 1 ) anElectrons active medium. value. do not emit radiation as long as they remain ) a shell source(Fig of 13 energy in 2each – 5). (pumping). 3 ) aatom resonant cavity. neutral, since the number of 3) The is electrically electrons nucleus equals the number of • He - Ne laseraround is a gasthelaser:

Chapter 13:14: Atomic Spectra Chapter Laser

3 ) coherence (same phase and direction). Bohr studied the difficulties facedand bysmall Rutherford’s model, 4 ) concentration (high intensity diameter).

modern physics physics modern

photon, which has the same energy as the one that caused it to be excited. Photons at the

in medium the nucleus. inpositive which charges the active is a mixture of helium and neon in the Fig ratio(13-5b) 10 : 1

Energy levels

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Chapter 13: Atomic Chapter 14: Spectra Laser

potential difference

gas

CDs

prism

screen

Apparatus for studying the spectra of the elements drum covered with a scanning with laser light sensitive material

laser light intensity controller

start

information

Fig (14-18) Use of a laser in printing

(Fig (14-19) Laser show

310 346 340

Fig (14-17b)

slit

Fig (13 â&#x20AC;&#x201C; 4a )

Unit 5: Unit 5:

modern physics modern physics

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

Bohr’s Model (1913)

Questions and Drills

first shell

2- Compare between spontaneous emission and stimulated emission operation - wise and feature - wise.

Unit Unit 5:5:

1- What is meant by laser ?

3- Laser light has special characteristics which distinguish it from ordinary light . 4- Discuss clearly the laser action . 5- What is meant by optical pumping and population inversion? 6- What is the role of the resonant cavity in laser operation ?

second shell

Fig ( 13-5a)

7- Lasers have 3 mainBohr components, what are they ?

Bohr’s Model

8- On what basis have helium and neon been chosen as an active medium in He - Ne laser ? Bohr studied model, 9What is thethe roledifficulties of helium faced in He by - NeRutherford’s laser ?

Explain clearly laserhydrogen beam is generated in He on - Ne laser . and10-proposed a modelhow fora the atom building

11- Explainfindings how holography works using lasers . Rutherford’s : are used extensively in medicine. Discuss one of its applications . 1) 12At Lasers the center of the atom there is a positively charged

shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

13Lasers nucleus . play an important role in missile guidance in modern warfare. Why is laser used as such? 2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13:14: Atomic Spectra Chapter Laser

free electron level continuous levels

modern physics physics modern

Discuss this statement .

Fig (13-5b)

Energy levels

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Chapter 13: Atomic Chapter 14: Spectra Laser

3 ) communications. 4 ) industry. 5 ) military applications. 6) CD recording potential 7) printing

difference

8) arts and shows 9) surveying 10) space research gas

slit

prism

Fig (13 – 4a )

Fig (13 – 4b )

Spectra of some elements

310 348 342

screen

Apparatus for studying the spectra of the elements

Unit 5: Unit 5:

modern physics modern physics

•spectra Laser of applications: the atoms of all elements would have been continuous. This is contrary to all 1 ) 3D photography (holography). experimental observations.The spectra of the elements have a discrete nature, and are called line spectra, occurring wavelengths characteristic of the element. 2 ) medicine (e.g.i.e., treating retinalat detachment).

Chapter 13: Atomic Spectra

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

1310

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Bohr’s Model15(1913) Chapter

Modern Electronics

first shell

The world witnesses a tremendous mushrooming in the field of electronics and communication to the point where they have become an insignia for this era. Electronics and communication are now indispensible in our life. TV, cellular (mobile) phone , computer, satellites and other systems are evidences for the vast progress in the second shell

applications of electronics and communications, whether in business, e-government, information technology(IT), entertainment or culture. They have become also an essential Fig ( 13-5a) Bohr ingredient in modern warfare. Weapons do not fare from the pointBohr’s of view of fire power Model only, but guidance, surveillance, monitoring jamming and deception ,called electronic

counter measures (or ECM) play an important role in combat. Also, in medicine whether in Bohr studied the difficulties faced by Rutherford’s model, free electron level continuous diagnois, prognosis, or operations, electronics plays a key role. In short, therelevels is no single and proposed a model for the hydrogen atom building on field in all walks of life Rutherford’s findings : where electronics has no part, starting from e –games to e– warfare.

2) Origin Negatively charged electrons move around the nucleus in of electronics:

Energy

you must of awareness about electronics – simplified as it 1)Therefore, At the center of theattain atoma certain there islevel a positively charged may be, yet nucleus . essential regardless of the prospective career you might end up with.

shells. Each shell (loosely often called orbit) has an energy The word electronics stems from the electron. Electronics describes the behavior of electrons. value. Electrons do not emit radiation as long as they remain There are two states for an electron: a free electron and a bound electron. The free electron – as in each shell (Fig 13 – 5). in the case of CRT- is subject to classical physics. A bound electron – however – is subject to 3) The atom is electrically neutral, since the number of quantum physics. The binding of an electron might be within an atom, a molecule or the bulk of electrons around the nucleus equals the number of matter. Matter has different forms : gas, liquid, solid or plasma (when the gases are ionized as positive charges in the nucleus. Fig (13-5b) in the fluorescent lamp). Matter in each form consists of molecules. What distinguishes states of

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

Overview:

Energy levels

351 345 311

Chapter 13: Atomic Spectra

potential difference

gas

slit

prism

Apparatus for studying the spectra of the elements

Unit 5:

modern physics

Fig (13 â&#x20AC;&#x201C; 4a )

Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

310 345

screen

level and the excited(1913) level. If the electron goes back to a lower level, it emits energy in the Bohr’s Model first shell

decreases as the energy of that level increases. There is a balance between the process of excitation and the process of relaxation, noting that the electron tends to go back to the ground state.

Pure Fig Semiconductors: (15-2a)

+4e Core

Fig (15-2b)

Fig (15-2c)

generation of 1) At the center of oftheatoms atom inthere a positively chargedatom has four electrons regular arrangement the issolid state. A silicon an electronin the free hole pair nucleusshell . (Fig 15 – 1).electron outermost Therefore, each silicon atom shares 4 electrons with 4 neighboring

2) Negatively charged moveis complete around theonnucleus atoms, so that the outer electrons shell of each sharing in basis to contain 8 recombination electrons each of an electron thermal called orbit) has an energy shells. Each shell (loosely often hole pair (Fig 15 – 2 a,b). We must distinguish energy here between two types of electrons in silicon. The first value. Electrons do not emit radiation as long as they remain type is the innermost (tightly bound) electrons, which are strongly attracted to their parents in each shell (Fig 13 – 5). atoms. The second type is the valence electrons, which have more freedom to move across 3) The atom is electrically neutral, since the number of energy interatomic distances. They exist in the outermost shell. Atthermal low temperatures (Fig 15 - 2c), all electrons around the nucleus equals the number of Fig (15-3b) bonds in the crystalFigare(15-3a) intact (unbroken). positive charges in the nucleus. Fig (13-5b) As temperature increases increases, Breaking a bond requires In this case – unlike metals – there are no free electrons. But as temperature Energy levels more bonds are broken energy Energy

Chapter 15: Modern Electronics

Covalent may of view ofSilicon crystal at T=0˚K There three types of materials frombonds. theWepoint Eachareatom shares Core represent a S atom (-14 e) i electrons with its all bonds are intact aroundconduct (+14 e) nucleus as a core and heat electrical neighbors conductivity. Conductors electricity second shell (+4e) and (-4e ) in the outer shell easily (as in metals). Insulators do not conduct electricity and heat some bonds, are broken and electrons are freed. Such an electron leaves behind a vacancy Fig ( 13-5a) (as in wood and plastics). Bohr Semiconductors are in between. At in the broken bond.This vacancy is called a hole (Fig 15 – 3). Because the atom is neutral, absolute zero, they act as insulators, whereas as temperature Bohr’s Model then the absence of an electron entails the appearance of a positive charge. We, thus, say increases ,their conductivity increases (as in silicon). that the hole has a positive charge. We do not call a silicon atom which loses an electron Fig (15-1) Bohr studied facedand by common Rutherford’s model,in the free electron level Silicon is onetheofdifficulties the important elements levels from its bond an ion, because soon enough, this atom may capturecontinuous aA freesilicon electron atomor an and proposed a model the) and hydrogen atom buildingcrust. on But universe. It exists in sandfor (SiO rocks of the Earth’s 2 electron from another bond to fill its own vacancy. Then, the atom returns neutral, and the Rutherford’s crystals of purefindings silicon :consist of silicon atoms bound together in covalent bonds. A crystal is a

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

Unit 5: modern physics 354

form of a photon. The probability of finding an electron in a particular excited level

353 347 311

ground state.

Pure Semiconductors:

Chapter 13: Atomic Spectra Unit 5:

modern physics

Chapter 15: Modern Electronics

outermost shell (Fig 1). Therefore, each slit silicon atom prism shares 4 electrons screenwith 4 neighboring gas An electron in 15 an– atom: atoms, so that the outer shellis ofconsidered each is complete on sharing basis to contain electrons eachIt An electron in an atom bound It cannot depart8 on its own. Figa (13 – 4aelectron. ) (Fig 15energy – 2 a,b). WeApparatus mustThis distinguish here between types electrons silicon. Theoffirst needs to do that. energy is called the two ionization i.e.,inthe energy an for studying the spectra of theofenergy, elements type is the innermostis(tightly bound) electrons,when whichfreearebystrongly attractedThat to their parents electron in bondage less than its energy this amount. is why the atoms. The second type is the valence have more freedom to move across electron remains in bondage in the first electrons, place. Thiswhich energy is called the binding energy. It is

interatomic They existstable. in theThe outermost At low 15 - 2c), all the cause of distances. keeping the atom electronshell. in the atomtemperatures has a set of(Fig discrete energy bondsaccording in the crystal are intactmodel. (unbroken). levels to Bohr’s It occupies one of the allowable levels and cannot have In this value case –inunlike metals there areinnothefreeatom electrons. But asbytemperature an energy between. The– electron is governed the laws ofincreases, quantum

mechanics. That is why the probability of having an electron fall onto the nucleus, or having the electron outside the atom (without external help) is zero. What binds the electron to the nucleus is the electric force of attraction. As long as the electron remains in one energy level, it does not gain or lose energy. But if Fig (13it is – 4b ) to a higher energy level, provided the electron acquires energy by absorption, excited

Spectra of some the energy absorbed is exactly equal to the energyelements difference between the original (ground)

310 352 346

Chapter 15: Modern Electronics

Unit 5: modern physics

matter apart is the intermolecular distance. In the case of a solid, this distance is very small. In Thereofare materialswould from have the point view of This is contrary toCore spectra thethree atomstypes of allofelements been of continuous. all the case of a gas,observations.The this distance is large. In the of a liquid,have it is somewhere in between. we experimental spectra of case theelectricity elements a discrete nature, and Ifare electrical conductivity. Conductors conduct and heat called line i.e.,the occurring wavelengths characteristic of the element. consider thespectra, solid state, atoms oratmolecules of matter get close enough to each other within a easily (as in metals). Insulators do not conduct electricity and heat certain distance due to the forces of attraction between them. If we imagine that they are made to (as in wood and plastics). Semiconductors are in between. At get close, then the forces of repulsion act in to prevent further proximity. Thus, the interatomic absolute zero, they act as insulators, whereas as temperature distance represents a point of equilibrium (or balance) between the forces of attraction and the increases ,their conductivity increases (as in silicon). Fig (15-1) forces of repulsion among the atoms. It is to be noted that these atoms oscillate around their Silicon potential is one of the important and common elements in the A silicon equilibriumdifference positions due to heat. But they are separated by space. We cannot see thisatom space by universe. It exists in sand (SiO2) and rocks of the Earth’s crust. But our naked eye, because the interatomic distance is much smaller than the wavelength of the crystals of pure silicon consist of silicon atoms bound together in covalent bonds. A crystal is a photons of visible light to which our eyes are sensitive. regular arrangement of atoms in the solid state. A silicon atom has four electrons in the

5: modern physics

excitation and the process of relaxation, noting that the electron tends to go back to the

353

first shell

As the temperature increases, the number of free electrons and holes increases, noting

that the number of free electrons equals the number of free holes in a pure semiconductor. But a state of dynamic equilibrium is reached (called thermal equilibrium) at which only a small percentage of bonds are broken. The number of bonds broken per second will be Fig (15-4b) equal to the number of bonds mended per second, so that a fixed Fignumber (15-4c)of free electrons Motion of holes is equivalent to motion of

At athe certain temperature, the and same and freeelectrons holes remains constant at every temperature. same electrons within bonds (in the opposite direction) But not number of free electrons and

holes remain free.They reshuffle ,but their number stays constant.holes is constant

second shell

Free electrons (a class of valance electrons) represent a third type of electrons in silicon. Doping: Such electrons in fact are stilltoconfined, but tothey are confined theFig full( size of thesilicon crystal Semiconductors are known be sensitive impurities and tototemperature. Since 13-5a) Bohr Bohr’s i.e., arethe limited by of theansoelement called surface of the crystal. Breaking a Model bond requires isitself, tetravalent, addition as phosphorus (P) or antimony (Sb) or any other a minimum energy optical). In theatom caseto replace of mending a atom bond in(called pentavalent element (thermal will causeorsuch an impurity a silicon the recombination), is released (thermal or optical). Bohr(Fig studied the faced by Rutherford’s model, crystal 15 – energy 5 a).difficulties Then, the phosphorus atom will try to do the same with the free bonding electron level continuous levels

As proposed the electrons moveatom infora random motion atom ,so dobuilding the holes,on since electrons in the bond move and model the hydrogen neighbors as theasilicon would do. around randomly to fill :in vacancies (voids) within the broken bonds (Fig 15 – 4 ). Rutherford’s findings 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of Fig (15-4a) Fig (15-5a) Holes move randomly between bonds positive charges in the nucleus. An antimony atom (pentavalent) replaces a silicon atom

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

356

broken bond is mended, and the hole shifts somewhere else, and so on. Bohr’s Model (1913)

Fig (13-5b)

Energy levels

355 349 311

Chapter 13: Atomic Spectra modern physics

+4e Core

small percentage of bonds are broken. The number of bonds broken per second will be equal to the number of bonds mended per of free electrons Fig second, (15-2b) so that a fixed number Fig (15-2a) Fig (15-2c) and free holes temperature. same Covalent bonds. We may But not the same Each atomremains shares constant at every Siliconelectrons crystal atand T=0˚K potential a Si atom (-14 e) with its reshuffle ,butrepresent all bonds are intact difference holes electrons remain free.They their around (+14 number e) nucleus asstays a core constant. neighbors (+4e) and (-4e ) in the outer shell Free electrons (a class of valance electrons) represent a third type of electrons in silicon. someelectrons bonds, are Suchconfined an electron behind a vacancy Such in broken fact areand stillelectrons confined,are butfreed. they are to theleaves full size of the crystal

in the i.e., broken is called a holeof(Fig – 3). Breaking Because the atomrequires is neutral, itself, are bond.This limited byvacancy the so called surface the 15 crystal. a bond a slit prism screen gas then the absence an electron the appearance of aofpositive charge. We, thus, say minimum energyof(thermal or entails optical). In the case mending a bond (called –call 4a ) a silicon atom which loses an electron that the hole hasenergy a positive charge.(thermal WeFig do(13 notoptical). recombination), is released or Apparatus forsoon studying the this spectra ofmay the capture elementsa free electron or an from its electrons bond an move ion, because enough, atom As the in a random motion ,so do the holes, since electrons in the bond move

electronrandomly from another to fill (voids) its ownwithin vacancy. Then, the atom and the around to fill inbond vacancies the broken bonds (Figreturns 15 – 4neutral, ). generation of an electron hole pair

free electron

recombination of an electron hole pair

thermal energy

Fig (15-4a) Fig (15-3b) Fig (15-3a) (13 – 4b ) Holes move Fig randomly between bonds As temperature increases Breaking a bond requires Spectra of some elements more bonds are broken energy

Chapter 15: Modern Electronics

Unit 5:

But a state of dynamic equilibrium is reached (called thermal equilibrium) at which only a

Unit 5: modern physics

310 354 348

broken bond is mended, and the hole shifts somewhere else, and so on. spectra of the atoms of all elements would have been continuous. This is contrary to all As the temperature increases, spectra the number free electrons holes increases, noting experimental observations.The of theof elements have and a discrete nature, and are called spectra, i.e., electrons occurringequals at wavelengths characteristic of in theaelement. that the line number of free the number of free holes pure semiconductor.

355

n = p + ND

(15 - 1)

take partdensity. in the In bonding scheme, one valence extra the hole this case, n > p sparing and the material is called n–type. Conversely, if anweexcess add electron (excess) electron. The force of attraction on the excess electron aluminum (Al) or boron (B) or any trivalent element, to which is left out is weak. Hence, it can easily be detached from pure silicon, the impurity atom replaces a silicon atom. its parent atom, which becomes a positive ion.This extra electron Since the impurity atom now has 3 electrons in the joins the stock of the free electrons in the crystal . In other outershell, it detaches electron a neighboring words, the crystal has ananadded sourcefrom of free electrons besides (Fig 15-5b) bond tobonds, complete its own bondatoms. creating extra hole, broken namely, impurity Suchanimpurity atoms are Doping with a pentavalent atom provides an extra free electron.

second shell called donors (givers).ion.AtAtthermal becoming a negative thermalequilibrium, equilibriumthe , sum of the A pentavalent atom has a core (+5 e) and 5 electrons (15 - 2) positive charge equals thep=sum charge. NA of + nthe negative Fig ( 13-5a) + Bohr n = p + N (15 concentration. - 1) where NA is the negative Dimpurity Bohr’s Model Fig (15-6a) + where NDSuch is theanpositive concentration, is the freeAelectron density Thus, p>n. atom is donor called ion acceptor (taker). Innall boron atom replacesand p is

a silicon atom

cases, have In this case, n > p and the material is called n–type. Conversely, if we add the holewestudied density. Bohr the difficulties2 faced by Rutherford’s model, free electron level np = ni (15 - 3) continuous levels aluminum (Al) aormodel boronfor (B)the or any trivalent element, to on and proposed hydrogen atom building pure silicon, the impurity Rutherford’s findings : atom replaces a silicon atom.

357

Thus, p>n. Such an atom called acceptor In all of electrons around the isnucleus equals (taker). the number cases, we have positive charges in the nucleus. np = ni2 (15 - 3)

Energy

1) At the center the atom positivelyin charged Since impurityof atom nowthere has is3 aelectrons the nucleus .it detaches an electron from a neighboring outershell, 2) Negatively charged electrons move around the nucleus in bond to complete its own bond creating an extra hole, shells. Each shell (loosely often called orbit) has an energy becoming a negative ion. At thermal equilibrium , value. Electrons do not emit- radiation as long as they remain (15 - 2) p= NA + n in each shell (Fig 13 – 5). where N is electrically the negativeneutral, impurity A 3) The atom is sinceconcentration. the number of

Unit 5: modern physics Chapter 13: Atomic Spectra Chapter Unit 5: 15: modern Modern physics Electronics Chapter 15: Modern Electronics

Because the impurity atomdonor has 5ion electrons, four of them willfree electron density and p is where NModel positive concentration, n is the Bohr’s D is the (1913) first shell

hysics

called donors (givers). At thermal equilibrium, the sum of the A pentavalent atom has a core (+5 e) and 5 electrons positive charge equals the sum of the negative charge.

Fig (15-6a) A boron atom replaces a silicon atom

Fig (13-5b)

Energy levels

357 351 311

provides an extra free electron. provi A pe ( (+5 e) and 5 electrons

called donors (givers). thermaltheequilibrium, the sum called donors (givers). At thermalAt equilibrium, sum of the A pentavalent atom of has athe core

Doping:

positive charge equals sum ofcharge. the negative charge. positive charge equals the sum ofthe the negative Semiconductors are known to be+sensitive to impurities and to temperature. Since silicon + n = p + NDn = p(15+-N (15 - 1) 1)D is tetravalent, the addition of an elementslitas phosphorus (P) or antimony (Sb) or any other prism screen gas + + where where N is Nthe positive the donor positive ion concentration, donor ionn concentration, is the free electron density n is the and pfree is ele D is pentavalent Delement will cause such an impurity atom to replace a silicon atom in the Figthe(13 –>4ap) and the thehole hole density. case, Inthe this n phosphorus > pcase, and natom material the material Conversely, is called if wethen–type. add crystal (Fig 15density. – 5Ina).this Then, willistrycalled to don–type. the same bonding with

Unit 5: modern physics

aluminum aluminum or (Al) boron or(B)boron or anydo.(B) trivalent or element, any trivalent to element, to neighbors as(Al) the silicon atom would g one valence extra an excess pure puresilicon, silicon, the impurity the impurity atom replaces atom a silicon replaces atom. a silicon atom. mpurity atom has 5 electrons, four of them will electron onbonding the scheme, excess electron Since Sincethe the impurity impurity atom nowatom has 3 now electrons hasin the 3 electrons in the sparing one valence extra an excess electron outershell, outershell, itfrom detaches it detaches an electronanfrom electron a neighboring from a neighboring asily beofdetached . The force attraction on the excess electron bond bondto to complete complete its own its bondown creating bond an extra creating hole, an extra hole,

is weak. Hence, it can easily be detached from ion.This extra electron

which becomes a becoming positive ion.This electron becoming a negative a extra negative ion. At thermal ion. At equilibrium thermal , equilibrium , Unit 5:

15In this - 1)case, n > p and the material is called n–type. Conversely, if we add

Chap

- - 2) thethe crystal (15 - 2) of free electrons. inInthe other crystal . Inp=other NA + n p= N(15 A +n l has anelectrons added source ofbesides freeelectrons besides (Fig concentration. 15-5b) free where the impurity negative impurity concentration. where NA N is Athe isnegative (Fig 15-5b) Fig (15-6a) mely, impurity atoms. Such impurity atoms are Doping with a pentavalent atom Thus,p>n.p>n. atom isancalled acceptor (taker). Inreplaces all Thus, Such Such an atoman is called acceptor (taker). In all A boron atom provides extra free electron. ch impurity atoms are Doping with a pentavalent atom ivers). At thermal equilibrium, the sum of the A pentavalent Fig(13 (15-5a) a silicon atom atom has Fig – 4b ) a core cases, cases, we we havehave provides an extra free electron. (+5 e) and 5 electrons An antimony quals the sum negative 2 (pentavalent) 2 brium, theof thesum ofcharge. the npA=pentavalent Spectra ofnatom some ni np =(15 3) elements - 3) has(15 a core i- atom replaces a silicon atom + (+5 e) and 5 electrons n = p + ND (15 - 1) tive350 charge. 310 356 he positive donor ion concentration, n is the free electron density and p is

Chapter 15: Modern Electronics

Chapter 13: Atomic Spectra

itsparent parent which abecomes a positive ion.This extra electron its atom,atom, which becomes positive ion.This extra electron Fig (15-4b) joinsthethe the freein electrons in other the crystal . In other joins stockstock of the of free electrons the crystal . In Fig (15-4c) ofhas holes equivalent toof motion of words, the has an added of free electrons besides words, theMotion crystalcrystal an isadded source free source electrons besides (Fig 15-5b) At a certain temperature, the electrons within bonds (in the opposite direction) number of free electrons and potential broken bonds, impurity atoms.atoms Such impurity broken bonds, namely,namely, impurity atoms. Such impurity are holes Doping with a atoms pentavalentare atom Dopin is constant difference

ons, four of them will Apparatus for studying the spectra of the elements modern physics

Unit 5: modern physics

Because the impurity hasfour5 ofelectrons, Because the impurity atom has 5atom electrons, them will four of them will spectra of the atoms of all elements would have been continuous. This is contrary to all takepartpart inbonding the bonding scheme, sparing valence take in the scheme, sparing oneelements valence extraaone anare excess experimental observations.The spectra of the have discrete nature, andextra electron (excess) electron. The force of attraction onofthe excess electron (excess) The of attraction on the excess electron called lineelectron. spectra, i.e.,force occurring at wavelengths characteristic the element. which is out leftis out weak. Hence, can easily which is left weak.isHence, it can easily beitdetached from be detached from

A bo

Bohr’s Model (1913)

first shell

Fig (15-7b)

Resistors

Diodes and transistors

second shell

Fig ( 13-5a)

Bohr

Bohr’s Model

Fig (15-7c) Inductors

Bohr studied the difficulties faced by Rutherford’s model,Fig (15-7d) free electron level continuous levels

and proposed a model for the hydrogen atom building on Capacitors Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus .

Transformers

shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

Fig (15-7e) 2) Negatively charged electrons move around the nucleus in Fig (15-7f) Switches

in each shell (Fig 13 – 5).

3) The atom is electrically neutral, since Fig the (15-7g) number of of components electrons around the nucleusA different equalssetthe number and of devices positive charges in the nucleus.

(Can you recognize some?)

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

Fig (15-7a)

Fig (13-5b)

Energy levels

359 353 311

Unit 5:

modern physics

Chapter 15: Modern Electronics

Chapter 13: Atomic Spectra

Unit 5: modern physics

whereofni the is the electron hole concentration purecontinuous. silicon, This is contrary to all spectra atoms of allorelements would haveinbeen experimental observations.The spectra of theThis elements have i.e., if n increases, p decreases and vice versa. is called lawa discrete nature, and are spectra, i.e.,approximation, occurring at wavelengths of the element. ofcalled massline action. As an we may say characteristic : in case of n-type +

(15 - 4)

n = ND

p = ni2/ND

(15 - 5)

Fig (15-6b)

potential

In case difference of p-type, -

(15 - 6)

p = NA

n = ni2/NA gas

Doping with a trivalent atom provides an extra hole. A trivalent atom has a core (+3e) and 3 electrons

(15 - 7) slit

prism

Fig (13 â&#x20AC;&#x201C; 4a )

Electronic Components Devices Apparatus for and studying the spectra of the elements Electronic components and devices are the building blocks for all electronic systems (Fig 15 â&#x20AC;&#x201C; 7). Some of these components are simple, e.g., resistor (R), inductor (L), capacitor (C). Some are more complex, such as pn junction (diode), transistor. There are also other specialized devices, such as optoelectronic and control devices. Semiconductors from which most of these devices are made are known to be sensitive to environmental conditions, such as light, heat, pressure, radiation and chemical pollution. That is why they are used as sensors or means for measuring external stimulii. Using these sensors, we can measure the intensity of incident light, temperature, pressure, humidity, pollution, radiation,etc. Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

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-region

transition region -region -region

-region

second shell

Fig ( 13-5a)

Bohr

external potential difference

Bohr’svoltage Model barriar

Fig (15-10b) Fig (15-10a) Motion of electrons and holes Bohr studied free electron level Forwardthe Biosdifficulties faced by Rutherford’s model, and proposed a model for the hydrogen atom building on Rutherford’s findings -region -region:

due to forward bias continuous levels

transition region p-region -region

1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy

Fig (15-11b)

Energy

2) Negatively charged electrons move around the nucleus in

Motion of electrons and holes

current

value. Electrons do not emit radiation as long as they remain due to reverse bias Fig (15-11a) Diode in reverse bias– 5). in each shell (Fig 13 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of Fig (15-12) positive charges in the nucleus. reverse voltage

I - V characteristic in a pn diode

forward voltage

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

and the n-type region to the negative terminal of the battery, the field due to the battery is Bohr’s Model (1913) first shell opposite to the internal field.in the transition region, and therefore, it weakens it. If we reverse the battery, then the two fields will aid each other. In the first case (forward bias), a net current will flow, and in the second case (reverse bias) current is almost zero (Fig 15-12). The action of the pn junction is like a switch, which is closed in the forward direction (conducting) and open (non conducting) in the reverse direction (Fig 15-13). We can make sure that the pn diode is functioning by using an ohmmeter, since the diode should have a small resistance in the forward direction and a large resistance in the reverse

Fig (13-5b)

Energy levels

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pn junction: spectra of the atoms of all elements would have been continuous. This is contrary to all A pn junction (Fig 15 – 8) consists experimental observations.The spectra of an the n–type elements have a discrete nature, and are region p-type i.e., region. The name pn stands for pcalled and line aspectra, occurring at wavelengths characteristic of the element. region and n-region not positive and negative. Also p,n regions are not two regions glued together but an n-material is converted in part to p-material or vice

Fig (15-8) A pn Junction

versa. Holes in the p–type region have high concentration, while holes in the n–type region have low concentration. Therefore, some holes diffuse from the p-type region to the n–type potential difference

region. Also, some electrons diffuse from the n–type region (high concentration for electrons) to the p–type region (low concentration for electrons). Since each region is neutral (the sum of positive charge equals the sum of negative charge), the gas

slit

prism

transfer of some electrons from the n–type region (13 –ions, 4a ) and uncovers an equal number of positive Fig donor

screen

Apparatus studying the spectra the transfer of some holes for from the p–type regionof the elements

uncovers an equal number of negative acceptor ions. This results in a middle region composed of positive ions on one side, and negative ions on the other, while no

Fig (15-9a)

Electrons diffuse from n to p and holes from p to n

electrons or holes exist in this region. This region is called transition (depletion) region. In such a region, an electric field is set up, directed from the positive ions to the negative ions. This electric field causes a drift current to flow in a direction opposite to the diffusion current. At equilibrium, the forward current is balanced

Transition†(depletion) re-

with a reverse current, so that the net current is zero (Fig 15 – 9). (13the – 4bp-type ) If we apply an external voltage suchFigthat

Fig (15-9b)

Transition region depleted from Spectra of some region is connected to the positive terminal of the elements battery electrons and holes, only ions exist

Learn at Leisure

How to convert AC to DC To convert AC to DC several steps may be followed. First a diode may be used as a half wave rectifier (HWR) (Fig 15 - 14a), using a resistor second shell

and a diode (Fig 15-14b). Four diodes may be used in a bridge (Fig 15- 14 c,d) for a full Fig ( 13-5a)

Bohr Output voltage

diode Model Bohr’s

Bohr studied the difficulties faced by Rutherford’s model,

free electron level continuous levels

and proposed a model for the hydrogen atom building on

Rutherford’s findings : 1) At the centerFig of (15-14a) the atom there is a positively chargedFig (15-14b) Waveform of a rectified half wave

nucleus .

A simple half wave rectifier

shells. Each shell (loosely often called orbit) has an energy value. ElectronsVdo not emit radiation as long as they remainV

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of Figin(15-14c) positive charges the nucleus. A full wave rectifier in the positive half cycle

Fig (15-14d) Fig (13-5b)

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

direction. is in (1913) contrast with a linear resistor, where the magnitude of the current is the Bohr’s This Model first shell same, whether or not the voltage polarity is reversed (symmetrical characteristic). A pn diode is important in rectification. It is used in charging car batteries, and mobile batteries, where AC is converted to DC (Fig 15- 14).

A full wave rectifier in Energy the negative half cycle levels

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reverse bias

anode

Fig (15-13b)

Ideal I-V characteristic diode

gas

slit

prism

Fig (13 â&#x20AC;&#x201C; 4a )

screen

Apparatus for studying the spectra of the elements

Fig (15-13c)

In forward bias the diode is like a closed switch

diode

FigFig(13(15-13d) â&#x20AC;&#x201C; 4b )

In reverse bias of thesome diode iselements like an open switch Spectra

310 362 356

forward bias

Fig (15-13a)

potential Diode symbol difference

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Chapter 13: Atomic Spectra

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cathode spectra of the atoms of all elements would have been continuous. This is contrary to all experimental observations.The spectra of the elements have a discrete nature, and are called line spectra, i.e., occurring at wavelengths characteristic of the element.

Learn atModel Leisure(1913) Bohr’s

first shell

To tune up a TV or radio onto a certain station, we need to adjust the value of a capacitor to set the frequency of the

forward bias

receiver to the frequency of the selected broadcast station. This condition is called resonance. In modern receivers, the capacitor is replaced by a reverse biased pn diode . The width of the transition region increases with increasing reverse bias (Fig 15 -

second shell

15). The increase of the width of the transition region entails an increase of the fixed ionic charge on both sides of the transition

Fig ( reverse 13-5a)bias

Bohr

region with reverse voltage. This is tantamount to capacitor

Bohr’s Model

action. Thus, we can change the value of the capacitor by controlling the revese voltage. This is called electronic tuning Bohr studied the difficulties faced by Rutherford’s model, (and the device is called a varactor). and proposed a model for the hydrogen atom building on

Fig (15-15)

The width of the transition region increases with increasing reverse bias

free electron level continuous levels

Transistor: Rutherford’s findings : transistor by Bardeen, 1)The At the center ofwas the cenceived atom there inis a1955 positively charged

Schockley nucleusand . Brattain. There are many types of transistors, (loosely often called has anfollowed energy pnpshells. or npn.Each Suchshell a transistor consists of aorbit) p-region value. Electrons not emit (pnp), radiationor asanlong as theyfollowed remain by an n-region thendo a p-region n-region each shellthen (Figan13n-region – 5). (npn) (Fig 15- 16). The three by ina p-region 3) The are atomcalled is electrically since number(C). of regions emitter (E)neutral, -base (B) andthecollector electrons around the The nucleus of Consider an npn transistor. first equals junctionthe (np)number is forward

Energy

but we focus here on bipolar transistor 2) Negatively charged electronsjunction move around the (BJT), nucleusi.e., in

Bardeen, Schochley and Brattain

positive charges(pn) in the nucleus. biased. The second junction is reverse biased. In this case, electronsFigare(13-5b) emitted from the

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

Electronic tuning

Energy levels

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wave rectifier (FWR) (Fig 15- 14e). Also, we may obtain a nearly constant current (Fig spectra of the atoms of all elements would have been continuous. This is contrary to all 15-14f) by usingobservations.The a capacitor inputspectra filter (Fig 15- elements 14g). have a discrete nature, and are experimental of the called line spectra, i.e., occurring at wavelengths characteristic of the element.

potential difference

Fig (15-14e)

Waveform of a rectified full wave slit prism gas

Fig (13 â&#x20AC;&#x201C; 4a )

Apparatus for studying the spectra of the elements

Fig (15-14f)

Waveform of a capacitor input filter

input

Output

Fig (13 â&#x20AC;&#x201C; 4b ) Fig of (15-14g) Spectra some elements Capacitor input filter

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screen

Bohr’s Model (1913)

first shell

Unit 5:

Fig ( 13-5a)

Bohr

Bohr’s Model

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

second shell

Fig (13-5b)

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negative emitter (n) diffusing to the base (p), where they wander around in the base until spectra of the atoms of all elements would have been continuous. This is contrary to all picked up by the positive collector (n).A portion of electrons gets recombined with holes. If experimental observations.The spectra of the elements have a discrete nature, and are the emitted electorn current is IE and the portion that reaches the collector Ic is Ic = αe IE, called line spectra, i.e., occurring at wavelengths characteristic of the element. then the protion lost in the base by recombination with holes is IB = (1-αe)IE. This must be the base current supplying holes to the base to make up for the losses due to the recombination process. Therefore, the ratio of the collector current to the base current is: βe =

potential difference

gas

αI α IC = eE = e I B (1- α e )I E 1 − α e

slit

prism

screen

Fig (13 – 4a )

Apparatus for studying the spectra of the elements Fig (15-16a)

A pnp transistor

transistor symbol pnp

Fig (15-16b)

A pnp transistor

Fig (15-16c)

An npn transistor

Fig (13 – 4b )

Spectra of some elements

310 366 360

(15 - 8)

npn Transistor symbol

Fig (15-16d)

An npn transistor

DigitalModel Electronics: Bohr’s (1913)

first shell

converts an image to electrical signals. In TV, the image (video) and sound (audio) are transformed into electrical signals, then into electromagnetic waves. All this occurs at the transmitter. At the receiver, the em signal is transformed back into electrical (video and audio) signals. The electronics which deals with natural quantities is called analog electronics. A new branch of electronics has developed, namely, digital electronics. In this second shell

case, the electrical signal is not transmitted continuously (all values are allowed), but is coded, such that the signal is in terms of one of two possible values representing Fig ( 13-5a) two states Bohr

0 or 1. So, if we want to represent 3, it can be written as 112, where Bohr’s subscript 2 denotes the Model binary system (not eleven). = 1x20+1x21 model, Bohr studied the difficulties faced by3Rutherford’s we may express 17 for in decimal system asatom building on andasproposed a model the hydrogen

free electron level continuous levels

17 = 7x100+1x101 Rutherford’s findings : in the of binary we use weights ofcharged 20, 21, 22 … instead of 100, 101, 102, 1) similarly At the center the system, atom there is athepositively …nucleus .Thus, each . numeral, symbol and alphabet is coded with a binary code. Analog quantities Energy

be encodedcharged by an analog – digital (ADC). At the 2)may Negatively electrons move converter around the nucleus in reciever, digital quantities are decoded usingcalled a digital to analog shells. into Eachanalog shell quantities (loosely often orbit) has an converter energy (DAC). Why do all this? In nature, are unwanted spurious signals, called electrical value.there Electrons do not emit radiation as long as they remainnoise. Noise is caused by the random of electrons. in eachmotion shell (Fig 13 – 5). Electrons are charged particles. As they move randomly, they varying currents. currents ofinterfere with and disturb the 3)cause Theminute atom israndomly electrically neutral, sinceThese the number information bearing signals. We notice that the in weak radio of stations, noise appears as a hiss, electrons - around the nucleus equals number in the (or nucleus. andpositive in weakcharges TV stations with a bad antenna an aerial) noise appears as spots (salt and Fig (13-5b)

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

All electronic systems deal with natural quantities and convert them to electrical signals. As an example, a microphone converts sound to an electrical signal. A video camera

pepper). Electrical noise marrs the useful signals, and is difficult to getEnergy rid of. Inlevels case of digital 369 363 311

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Transistor as a of switch: spectra of the atoms all elements would have been continuous. This is contrary to all experimental spectra of the elements have a discrete nature, and are Consideringobservations.The the collector circuit, we have called line at wavelengths characteristic of the element. (15-9) V spectra, = V +i.e., I Roccurring CC

C C

where VCC is the collector battery, and vCE is the voltage difference between the collecttor and the emitter, IC is the collector current and RC is the collector resistance. As IC increases, VCE

Fig (15-18a)

potential difference decreases, until it reaches a value as low as 0.2V

Transistor as a switch (ON condition)

for a high base current. Considering the base as the input ,the collector as the output and the emitter as common (ground), we note that as the slit gas

prism

input increases, (or positive) the transistor is ON, Fig (13 â&#x20AC;&#x201C; 4a ) and the output decreases and vice versa. The

Apparatus for studying the spectra of the elements

circuit behaves as an inverter, for positive voltage in the base (high), current flows in the

collector, and the output voltage is very small (low). If the base voltage is small (or negative) or

Fig (15-18b)

Transistor as a switch (OFF condition)

(low). The transistor is OFF and the current in the collector ceases, and the output voltage on the collector increases (high). The transistor as such operates as a switch (Fig 15-18). Fig (13 â&#x20AC;&#x201C; 4b )

Spectra of some elements

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screen

Fig (15-18c)

Inverter characteristic

operations, such as inversion (NOT), simultaneity or coincidence (AND) and optionality (OR) as

Bohr’s Model (1913)

first shell

1) Inverter (NOT Gate) has one input and one output, and has the following truth table: input output 1 0 0 1 2) AND Gate: has two inputs or more and one output and has the following truth table: input output 00 0 second shell 01 0 10 0 Fig ( 13-5a) 11 1 Bohr Bohr’s Model

input A

output

input B

Bohr studied the difficulties faced by Rutherford’s model, Fig (15-20a) and proposed a model for the hydrogen atom building on

free electron level continuous levels

AND gate symbol

Rutherford’s findings :

1) At the center of the atom there is a positively charged 2) Negatively charged electrons move around the nucleus in (15-20b) shells. Each shell (loosely often calledFig orbit) has an energy States of an AND gate

value. Electrons do not emit radiation as long as they remain

Energy

nucleus .

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, sincelamp the number of electrons around the nucleus equals the number of Fig (15-20c) positive charges in the nucleus.

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

follows :

Fig (13-5b)

An equivalent drawing for an AND gate. The lamp Energy levels does not glow until both switches are closed

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electronics, the information does not lie in the absolute value of the signal (which might be spectra of the atoms of all elements would have been continuous. This is contrary to all contaminated noise), but lies inspectra the code of 0 orhave 1. It adoes not matter experimental byobservations.The of intheterms elements discrete nature,if the andvalue are corresponding to 0 ori.e., to 1occurring has someatnoise superimposed on it. Whatofmatters is the state (0 or 1). called line spectra, wavelengths characteristic the element. This is the main advantage of digital electronics. For this reason, it has permeated our modern life extensively, as in cellular (mobile) telephony, digital satellite TV, and CDs. What has increased the importance of digital electronics is the advent of the computer. Everything that is entered into the computer-whether numbers or letters-must be transformed into a potential binary code. Even images are divided into small elements, each called a pixel (picture difference

element). These too must be encoded. The computer performs all arithmetic and logic

operations using binary (Boolean) algebra. It also stores information in the binary code temporarily in the RAM (Random Access Memory) or permanently in the hard disk, by prism gas for 0 and inslitthe opposite magnetizing in one direction direction for 1.screen

Fig (13 â&#x20AC;&#x201C; 4a )

Logic Gates:

Apparatus for studying the spectra of the elements

Modern applications of electronics, such as computer circuits and modern communication systems depend on digital circuits, called logic gates. These are the circuits that perform logic

input

output

Fig (15-19a)

Not gate symbol

Fig (13 â&#x20AC;&#x201C; 4b )

Fig (15-19a)

An equivalent drawing for a NOT States of a NOT gateSpectra of some elements gate. When the switch is closed (ON) the lamp is (OFF) and vice versa

Fig (15-19b)

310 370 364

current.

pits

Plastic

first shell disk

Collimating coil

lens

All operations performed by the computer are based on these gates and others. These gates can be implemented by transistors. In this case, the transistor may not be looked

sensitive light detector

prism

upon as an amplifier but as a switch . second shell

Thus, we can use the transistor as an inverter (NOT gate).

Bohr A transistor with more than one emitter may

laser

Fig (15-22a) Fig ( 13-5a) A CD - drive

Bohr’s Model

be used as an AND gate, so that the transistor

Energy

will not pass current unless each emitter has positive voltage (1). Bohr studied the difficulties faced by Rutherford’s model, electron level Also, we may envision the transistor as an OR gate in the formfree of a pair of parallel continuous levels and proposed a model for the hydrogen atom building on transistors. If (1) exists at either one of the inputs, one of the transistors conducts and (1) Rutherford’s findings : appears at the output . 1) At the center of the atom there is a positively charged Transistors are also used in memory circuits, where data ( 0 or 1 ) is retained temporarily nucleus . in the RAM and permanently in the hard disk or CD. In a CD, a laser beam engraves a bit in 2) Negatively charged electrons move around the nucleus in a plastic disk for 1 and no bit for 0. This is the Write process. In a CD drive, a laser beam is shells. Each shell (loosely often called orbit) has an energy used for the Read process (Fig 15 – 22a). A DVD is a modified version of a CD with higher value. Electrons do not emit radiation as long as they remain storage capacity. There are also digital cameras, which convert images to electrical signals in each shell (Fig 13 – 5). pixel by pixel, and store them on a magnetic tape, or download them onto a PC (Fig 15 – 3) The atom is electrically neutral, since the number of 23). These cameras use a new technique for handling and transferring electrical charges, electrons around the nucleus equals the number of namely, charge coupled devices (CCD). This makes the cameras light weighted (portable) positive charges in the nucleus. and inexpensive. This is the basis for the camcorder, Fax machinesFigand(13-5b) mobile camera.

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

This canModel be represented Bohr’s (1913) by two switches in parallel, one of them only need be closed to pass

Energy levels

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Thus, there is no output (1) unless both inputs are (1) each, i.e., two conditions or more spectra of the atoms of all elements would have been continuous. This is contrary to all are met to satisfy an output (1). It can be represented by two switches in series. They both experimental observations.The spectra of the elements have a discrete nature, and are have be closed the same time atforwavelengths current to flow and the lamp to glow. calledtoline spectra,at i.e., occurring characteristic of the element. 3) OR Gate has two inputs or more and one output (Fig 15–21). One condition (1) may suffice to have an output (1) . input Output 00 0 01 1 potential difference 10 1 11 1 input A input B gas

output slit

prism

Fig (15-21a) Fig (13 – 4a ) OR gate symbol

Apparatus for studying the spectra of the elements

Fig (15-21b) States of OR gate

Lamp

(15-21c) Fig (13 – 4b )

AnSpectra equivalentofdrawing an OR gate. someforelements One switch need be closed for the lamp to glow

310 372 366

screen

electronic bulletins,(1913) watches and measuring equipment. If Bohr’s Model

first shell

concentrated by laser action, we may have a junction (solid state) laser (Fig 15 – 23 b), which is used in surgery, communication through fiber optics and in modern warfare,

Fig (15-23b)

A junction laser

such as missile guidance and radar (laser radar is called LADAR).

Eelctronic Circuits: Any analog or digital electronic system is composed of electronic components connected together in a closed path called Bohr

second shell

Fig ( 13-5a)

circuit. The components may be passive as resistors, inductors, Bohr’s Model capacitors, or diodes (Fig 15 – 24). Active components include transistors in all types. Bohr studied the difficulties faced by Rutherford’s model, and proposed a model the hydrogen atom building on Circuits formed fromforseparate components and soldered Rutherford’s findings :

free electron level continuous levels Fig (15-24a) Resistors

1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Fig (15-24b)

Different versions of transistors and diodes

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

light from a forward biased heavily doped pn junction is

Fig (13-5b)

Energy levels

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Fig (15-22c)

A CCD element

Fig (15-22b)

potential Digital camera difference

gas

slit

Fig (15-22d)

prism

screen

Fig (13 – 4a )

Storing date onApparatus a magnetic tape the digitalthe spectra of the elements for instudying camera

Fig (15-22e)

Images may be transferred via the internet using a new

Downloading onto a computer

feature called Bluetooth.

axis

Learn at Leisure

transparent dome

LED When current passes through a pn junction, electrons and holes transporting from one side to the other, are annihilted in a recombination process. This process is accompanied by the emission of light in the form of photons. A solid state Fig (13 This – 4b is) called lamp can, thus, be made out of a pn junction. Spectra of some elements

light emitting diode (LED) (Fig 15 – 23 a). It is used in

310 374 368

light

terminal

Fig (15-23a) LED

repeatedly made on(1913) a thin wafer of silicon thousands of Bohr’s Model

first shell

thousands of slices, each called a chip, all carrying exactly the same layout and same specifications. This technique yields low cost electronic circuits due to the mass production involved. The burden is really the initial investment of setting up the foundries, with all the sophisticated equipment including robots, testing systems,

second shell

Fig (15-25d)

and in the design, artwork i.e., the brainwork involved in

Pentium IC

the programming, particularly when such circuits are

Fig ( 13-5a)

Bohr

Bohr’s Model

custom made. The commonplace ICs are, however, inexpensive, since millions are made at a time for the same andtheartwork. Thisfaced is what has made ICs Bohrdesign studied difficulties by Rutherford’s model, popular in botha analog electronic systems. In on and proposed model and for digital the hydrogen atom building fact, instead of designing Rutherford’s findings : highly complicated and costly

free electron level continuous levels

Fig (15-26)

Selective diffusion

ICs or available (off the are sought 1) Atexisting the center of the atom thereshelf) is a ICs positively charged 2) Negatively electrons around the lowest nucleus in namely puttingcharged the right things move together at the shells.cost Eachandshell (loosely oftenefficiency. called orbit) has an energy possible highest possible Electrons not emit radiation as long remain Avalue. collection of do components including ICs asarethey often

Energy

first.nucleus Thus, .design has shifted toward system engineering,

in eachonshell (Fig called 13 – 5).a printed circuit board (PCB). mounted a board 3) The atomis istheelectrically the –number An example motherboardneutral, of a PCsince (Fig 15 27). It of electrons theRAM, nucleus equals the number includes the around processor, Arithmetic Logic Unit of positive charges in the nucleus. (ALU), control circuits etc.

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

times simultaneously. Thus, the wafer is cut up to

Fig (13-5b) Fig (15-27)

Energy levels Motherboard

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Fig (15-24c)

Tansistors as pairs and quadruples potential together are called discrete circuits (Fig 15 – 24). A new era of difference

integrated circuits(ICs) started in the 1960’s at the peak of space research. The goal then was to develop electronic circuits with a new technology ,which would put light weight, compactness, slit gas at prime interest. effectiveness and reliability Theprism answer was

ICs or microchips (Fig 15– 25). The basic Fig idea (13 –is4ato)cram all the

IC’s in different forms

for studying the spectra the elements needed componentsApparatus onto a silicon wafer, where differentofregions

are assigned to needed functions without treating them as separate components. If we want to make a diode, for example, then starting with an n-type wafer, we allow p atoms to diffuse in defined regions in the wafer. This is called selective (planar) diffusion (Fig 15–26). The way this is done is a complicated

Fig (15-25b) IC uncovered

chemical process in which a mask is made and light (recently laser) is used. The process is similar to film developing in photography, and is called photolithography, which means carving on stone. If we now want to make an npn transistor, we ) open up a window in the p-region, Fig and(13 let –n4batoms diffuse Spectra of some elements

selectively there. Interestingly, all these operations are

310 376 370

screenFig (15-25a)

Fig (15-25c)

Microprocessor in comparison with a match head

Alternatively, the two states may be one direction of electron spin, and the other state in the opposite direction. This is called quantum computer. This is the trend of the near future, which is about to materialize. It is in harmony with future direction trends of science in search for minute details of time and space. This has led to the advent of new technologies such as Nano and Femto technologies. Learn at Leisure second shell

Selective Diffusion

How can we make phosphorus atoms for example diffuse in a very small area and not Fig ( 13-5a) Bohr? This happens in several steps. First, we cover the silicon wafer through the rest of the chip Bohr’s Model

with a layer of oxide (Si O2), then we use a mask prepared in a way similar to photographic development. We cover the oxide layer with a photoresist, which is a light sensitive Bohr studied faced by Rutherford’s model, level material. We thentheputdifficulties a mask with opaque and transparent regions onfree topelectron of the photoresist. continuous levels

andthen proposed model for totheultraviolet hydrogen(uv) atomrays. building We exposea the surface When on the photoresist is exposed to uv, : in the region where it is exposed, and remains liquid in the itRutherford’s polymerizesfindings (solidifies)

Energy

1) At the center of the theremask is a and positively unexposed areas. We thenatom lift the use HClcharged acid, which interacts with SiO2 (a nucleus . etching) in the areas where the photoresist is in liquid form (where SiO is process called 2 2) Negatively electrons move around nucleus uncovered withcharged polymerized photoresist). Thetheacid, thus, inopens up a hole in the oxide.

shells. Each shell (loosely often called orbit) through has an energy Then, phosphorus atoms are allowed to diffuse the opening in the oxide, while the value. Electrons not emit areas radiation as theyThus, remainminiaturization depends on the oxide isolates the do remaining fromas long diffusion. in eachofshell (Fig 13Therefore, – 5). accuracy the mask. laser is used in mask making. For more miniaturization, electron are used for shorter λ, and of hence smaller dimensions (why?) 3) The beam atomand is molecular electricallybeam neutral, since the number where they directly carvetheon nucleus the chip. equals But this the cannot be usedofon a large scale, but is used for electrons around number special ICs only. positive charges in the nucleus.

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

that we are Model headed to(1913) reach the size of the atom itself, i.e., 0 and 1 may be stored in the form of Bohr’s first shell an electron being in either one of two states in the atom, ground state or an excited state.

Fig (13-5b)

Energy levels

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Chapter 13: Atomic Spectra

Unit 5: modern physics

ICs have permeated even medical equipment including instrumentation, diagnosis, and spectra of the atoms of all elements would have been continuous. This is contrary to all prognosis. Oneobservations.The day pacemakers spectra and insulin control circuits using microcapsules experimental of the elements have a discrete nature, involving and are called line spectra, at wavelengths characteristic of thewithin. element. microprocessors mayi.e., beoccurring injected into the body to do their work from

Miniaturization, where to ? When the first computer was built in the 1950’s, its capabilities were very limited by today’s standards. potential It was bulky, about the size of an apartment. It was built from vacuum difference

tube (valves). Then, transistors were used. ICs have led to the development of PCs which made computers available to the public. Since the 1970’s PCs are continually being enhanced. Their capacity and capability to do complicated calculations are on the increase, while calculation time is getting shorter, and size and weight are getting smaller. Also, cost gas

slit

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is on the decline. These improvements sound contradictory, but they are happening and at a Fig best (13 –application 4a ) high rate, thanks to the understanding and of the basic concepts of modern Apparatus for studying the spectra of the elements

physics, materials science, chemistry, laser and to the rapid advancement of the technology. There is a common law called Moore’s law, which states that capacity and speed double every 18 months. If a chip (the size of a pin head) contains 100 transistors, this called small scale integration (SSI). If it contains 1000 transistors, it is called medium scale integration (MSI). If it contains 10000 transistors, it is called large scale integration (LSI). If it contains 100000 transistors, it is called very large scale integration (VLSI). If it exceeds that, it is called ultra large scale integration (ULSI). Can you imagine 1 million transistors in a pin head area? What then if you know that the figure in 2005 has reached 300 millions with prospect of even more? If the miniaturization keeps going at that rate what next ? we shall soon be limited by the Fig (13 – 4b ) diffraction of light as the physical dimension will soon approach λ of the used light. It seems Spectra of some elements

310 378 372

screen

(forwardModel connection or forward bias) current flows. If the battery is reversed no current Bohr’s (1913) first shell This is why a diode is used in rectification. • A transistor may be pnp or npn, and can be used as an amplifier, since the ratio of the collector current to the base current βe is large. Therefore, any small change in the base current leads to an amplified change in the collector current. • A transistor may also be used as a switch. It is used in logic gates, such as an inverter (NOT), AND, OR gates.

second shell

• Digital electronics is superceding analog electronics for its ability to overcome electrical noise . Its basic concept is to code information in binary form (0 , 1).Fig ( 13-5a) Bohr Bohr’s Model

• ICs have the advantages of small size and weight, increased speeds and capacity, and yet low cost. This is the reason for the proliferation of PCs. Bohr studied the difficulties faced by Rutherford’s model,

free electron level continuous levels

and proposed a model for the hydrogen atom building on Rutherford’s findings : 1) At the center of the atom there is a positively charged nucleus . shells. Each shell (loosely often called orbit) has an energy value. Electrons do not emit radiation as long as they remain

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Unit modern physics Chapter 13: Atomic Spectra Unit 5: 5: modern physics Chapter 15: Modern Electronics

flows.

Fig (13-5b)

Energy levels

381 375 311

Unit 5:

modern physics

Chapter 15: Modern Electronics

Chapter 13: Atomic Spectra

Unit 5: modern physics

a Nutshell spectra of the atoms of all elements In would have been continuous. This is contrary to all observations.The spectra ions of theandelements discrete nature, are •experimental A metallic crystal consists of positive a cloud have of freea electrons roamingand around called line spectra, i.e., occurring at wavelengths characteristic of the element. the crystal in random motion. There is a force of attraction between the ions and the electron cloud. But the resultant of all forces of attraction on a single free electron is zero . If an electron tries to escape from the metal, a net force of attraction due to the atom layer at the surface pulls it in. • A pure potential silicon (semiconductor) crystal consists of atoms covalently bonded. At low difference

temperatures, there are no free electrons. If temperature increases, some bonds are broken, electrons become free, leaving behind holes. Both electrons and holes move randomly.

gas

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screen

• The number of broken bonds increases with temperature. It may increase also by an Fig (13 4a )photon energy is sufficient to break the external stimulus, such as light, provided that– the bond.

Apparatus for studying the spectra of the elements

• The number of free electrons and holes increases by adding impurities (doping). Thus, the material becomes n-type or p-type. • The conductivity of a semiconductor depends on the conduction of free electrons and holes. Thus, a semiconductor has two current carriers: electrons and holes, while in a metal there is only one current carrier (the electron). Electron concentration in a metal is constant and does not depend on temperature. • Semiconductors are environment-sensitive. They can be used as sensors to light, heat, pressure humidity, chemical pollution, radiation etc. • A diode (pn junction) consists of a Fig p–type (13 –region 4b ) and an n-type region. If the p-side is

Spectra of some elements connected to the positive terminal of the battery and the n-side to the negative terminal

310 380 374

Bohr’s Model (1913)

first shell

Unit 5:

Fig ( 13-5a)

Bohr

Bohr’s Model

Energy

2) Negatively charged electrons move around the nucleus in

in each shell (Fig 13 – 5). 3) The atom is electrically neutral, since the number of electrons around the nucleus equals the number of positive charges in the nucleus.

Chapter 13: Atomic Spectra

free electron level continuous levels

modern physics

second shell

Fig (13-5b)

Energy levels

377 311

Questions and Drills

Unit 5:

modern physics

Chapter 15: Modern Electronics

Chapter 13: Atomic Spectra

Unit 5: modern physics

spectra of the atoms of all elements would have been continuous. This is contrary to all I)experimental Drills: observations.The spectra of the elements have a discrete nature, and are 3, if the density of silicon is 2.33 g/cm3 1)called Calculate the number of siliconatatoms in 1 cmcharacteristic line spectra, i.e., occurring wavelengths of the element. and its atomic mass is 28 (0.5x1023cm-3) 2) If electron or hole concentration in pure silicon is 1x1010cm-3, phosphorus is added at a concentration of 1012cm3, calculate the concentrations of electrons and holes in this case. Is this

potential difference silicon n-type

(n=1012cm-3 p=108cm-3 )

or p-type?

(n - type) 3) Calculate the concentration of aluminum to be added so that silicon returns pure . gas

slit

prism

4) A transistor has αe = 0.99 . Calculate βe. Then calculate the collector current if the base Fig (13 – 4a ) current is 100 µAApparatus for studying the spectra of the elements (βe=99 , Ic = 99x10-4A) 5) The electrical signal in the base of a transistor is 200 µA . The collector current is to be 10 mA. Calculate αe and βe.

(αe = 0.98 , βe = 50)

6) A diode can be represented by a forward resistance 100Ω ,while it is infinity in the reverse direction. We apply +5 V ,and then reverse it to – 5 V. Calculate the current in both cases.

(50 mA, O)

7) If 1 mm2 contains 1 million transistors, calculate the area assigned to each transistor. (10-12m2) II) Essay questions: 1) Discuss the importance of digital electronics and mention 5 applications. Fig followed (13 – 4b )by an inverter. 2) Deduce the truth table for an AND gate Spectra some elements 3) Deduce the truth table for an OR gate of followed by an inverter.

310 382 376

- = 1012cm-3)

screen (NA

General Revision 1) A metallic wire is stretched between two vertical fixed pins .Is the velocity of propagation of a transverse wave through this wire affected by a change in the temperature of the surrounding medium? and why ? 2) Two identical strings, one end of each is fixed to the wall, while the other end is stretched by hand . A transverse pulse is sent through one of the wires, and after a short time another transverse pulse is sent through the other wire. What can be done to make the second pulse catch up with the first pulse? Give reasons. 3) Give reasons: It is easy to see your reflected image on the window glass of a lit room at night when it is dark outside the room. But that is difficult when there is light outside the room. 4) Two rays of light converge on a point on a screen. A parallel glass plate is placed in the path of this light, and the glass plate is parallel to the screen. Will the point of convergence remain on the screen or change position ? Give reasons . 5) Explain and give reasons : When a blue light source is placed at the center of a solid glass cube with a white screen facing each side, a circular spot of light appears on each screen in front of each surface of the cube . When the blue source is replaced by a red color source, the shape of the spot changes from circular to square . 6) In the following figure, a fiber optic has an external layer from glass. Its refractive index is less than that of the glass of the core . If the light beam passes through it as shown in the figure.

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of distance 11 x 10-4 m apart, and the distance between the double slit and the screen was 5m. Find the distance between two successive similar fringes . 12) A rubber hose is connected to a tap, and the water flows through in a steady flow .Explain why the cross sectional area of the flowing water decreases when the end of the rubber hose is directed down, and increases when the end of the rubber hose is directed up. 13) A balloon filled with air is attached to the bottom of a glass tank. Then the tank is filled completely with water ,with this balloon completely immersed in water. Suppose that the tank with its contents were transferred from the Earth to the Moon. Discuss and give reason for all what would happen to the balloon? 14) A hollow copper ball is suspended under the surface of water in a tank. Discuss and give reason for all what would happen to the position of the ball in the tank if it were transferred from the Earth to the Moon? 15) Verify the following statement and correct the mistakes if any:When a person dives in a swimming pool near the bottom , each of the upthrust and pressure exerted over him increases. 16) An ice cube is placed in a glass beaker, then it is filled completely to the rim with water . Discuss in the light of Archimedesâ&#x20AC;&#x2122; principle what changes may happen when the ice melts (fuses) . 17) A glass beaker filled to the rim with water is resting on a scale . A block is placed in water, causing some of it to spill over.The water that is spilled is wiped away ,and the beaker is still filled to the rim . Compare between the initial and final reading on the scale, if the block is made from: a) wood b) iron 378 379 385

a)explain why the direction of the beam does not change at each of S and P. b)explain why there is a total reflection at each of Q and R. c)explain why the double layer fiber optic is preferred to that of a single layer. 7- A teacher gave his students the following figure (A) which expresses a path of light beam from A to B through a triangular prism made of glass which has a critical angle 42Ë&#x161;. He asked the students to draw the path of the beam before reaching A and after leaving B. The figure (B) expresses the attempt of a student. But the teacher made it clear that the angles of X and Y were not correct. Suggest without calculations the changes required to correct the angles X and Y, and give reasons for your suggestion. (Fig A) (Fig B)

8) The tension of a stretched string is changed from 70N to 80N without a change in its length. Calculate the ratio of the fundamental frequencies as a result . 9) A string of length 0.06 m and mass 2.5 x 10-3 kg is stretched by a force of 400N. Find the frequency of the produced tone if it vibrates in three segments. 10) A triangular prism has an angle of 60Ë&#x161; and refractive index of. 2 . Calculate the minimum angle of deviation and the corresponding angle of incidence. 11) A monochromatic light of 66 x 10-8 m wavelength strikes a double slit

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24) Show that Van der Waals’ effect can explain the conversion of gases into liquid state. 25) Give reason : Van der Waals’ effect on gases appears clearly at low temperatures. 26) Gas behavior deviates from that of the ideal gas as its density increases. Disuss this statement . 27) What is the scientific concept on which the Dewar’s flask is designed? 28) Give reason :Liquid helium is preferred as a cryogenic material. 29) Compare between the characteristics of the adiabatic change and the isothermal change . 30) What is meant by the transitional (critical) temperature of a metal? 31) Give reason : We use a superconductive coil in manufactering the levitated train. 32) Give reason : A superconductor is used in making satellite's antenna. 33) Give reason: Meissner effect appears only in superconductive materials. 34) Suppose that the atoms of helium gas have the same average velocity as the atoms of argon gas. Which of them has a higher temperature and why? 35) Calculate the average kinetic energy and root mean square of the velocity of a free electron at 300˚K , where Boltzmann's constant = 1.38 x 10-23 J/˚K , the mass of electron is 9.1 x 10-31kg.

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where the density of water is 1000 kgm-3, that of wood is 550 kgm-3 and that of iron is 7860 kgm-3. 18) State the conditions which make the liquid flow in a steady flow and prove that in a steady flow, the velocity of liquid flow at any point is inversely proportional to the across sectional area of the tube at that point? 19) A major artery of radius 0.5 cm branches out into many capillaries,the radius of each is 0.2 cm . The speed of blood in the main artery is 0.4 m/s , and the speed of blood in each capillary is 0.25 m/s . Find the number of the capillaries ? 20) A cross sectional area of one end of a U- shaped tube is twice the other. When a suitable amount of water is placed in the tube and an amount of oil is poured into the wide end, the surface of water in the tube is lowered by 0.5 cm. Calculate the height of oil in the tube. Knowing that the density of water equals 1000 kg m-3 and that of oil equals 800 kg m-3.

21) The small and large piston cross sectional areas of a hydraulic press are 4 x 10-4m2 and 20 x 10-4m2, respectively, and a force of 200 N is exerted on the small piston. Calculate the mass required to be placed on the large piston to be in the same level with the small piston ,g = 10 m/s2. 22) What is the least area for a floating layer of ice of thickness 5 cm above the water of a river which makes this layer carry a car whose mass is 16 x103kg. The density of water is 1000 kg/m3 and that of ice is 920 kg/ m3. 23) A cork ball has a volume of 5 x10-3 m3,placed in water of density 1000 kgm-3. About 2/5 of its volume was immersed. Calculate the density of the cork and the force needed to immerse all the volume of the ball. 381 380 386

step - up transformer decreases the current. 44) There are three essential factors that must be considered when designing transformers to decrease the loss of the electric energy. What are these factors and how? 45) Give reason: The eddy current is not generated in the metallic blocks unless a magnetic field of variable intensity exists. 46) Compare between an AC generator and a DC generator. 47) Give reason: To increase the power of a motor ,several coils seperated by small angles are used. 48) The following table shows values of resistance of wire of cross sectional area 0.1 m2 and different lengths. Length l m

Resistance R 1

Plot the relation between the length ( l ) on the X axis and Resistance (R) on the Y axis. From the plot find: a) resistance of a part of the wire of length 12 m . b) the resistivity of the material of the wire. c) the conductivity of the material of the wire. 49) A wire 30 cm long and 0.3 cm2 cross sectional area is connected in series with a DC source and an ammeter . The potential difference between the ends of the wire is 382 383 389

36) An amount of an ideal gas has a mass of 0.8 x 10-3 kg , a volume of 0.285 x 10-3 m-3,at a temperature of 12o C and under pressure of 105N/m2 . Calculate the molecular mass of the gas where the universal gas constant equals 8.31 J/˚K. 37) Calculate the mean kinetic energy of an oxygen molecule at a temperature of 50oC, where Boltzmann's constant is equal to 1.38 x 10-23 J/˚K. 38) If the temperature at the surface of the Sun is 6000oK, find the root mean square speed of hydrogen molecules at the surface of the Sun, knowing that the hydrogen is in its atomic state. Its atomic mass =1, Avogadro's number (NA)=6.02 x 1023, and Boltzmann's constant=1.38 x 10-23 J/˚K . 39) Give reason : Efficiency of the battery increases by the decrease of its internal resistance. 40) In electric circuits connected in parallel, thick wires are used at the ends of the battery,but at the ends of each resistor less thick wires are used. Why? 41) What is meant by: a) the effective value of AC. b) eddy current. c) the sensitivity of a galvanometer. d) the efficiency a transformer. 42) What is the physical concept for the operation of the following devices: galvanometer – transformer – current divider (or shunt)– potential multiplier. 43) Give reason :The step- down transformer increases the current, and the 383 382 388

b) the efficiency of a transformer=90%. c) eddy currents. d) the effective value of an AC current =2A. 55) A step -down transformer of efficiency 100% is to be used to light a lamp of power 24 W at a potential difference 12 V. If the power source applied to the transformer is 240 V, the number of turns of the secondary coil is 480 turn. 1) calculate the current passing through each of the primary and secondary coils 2) the number of the turns of the primary coil. 56) When an electric current is flowing through a perpendicular wire in a uniform magnetic field, the wire is affected by a force. Which of the following instruments is based on this principle: (1) electromagnet. (2) motor. (3) generator. (4) transformer. 57) Calculate the emf of a source if the work done to transfer 5C is 100 J. 58)Three resistors 10 1 , 20 1 , 30 1 are connected to a power supply . If the currents are 0.15 A , 0.2 A , 0.05 A, respectively. Calculate the equivalent resistor for this circuit, and illustrate your answer with a labeled diagram. 59) Two resistors 400 1 and 300 1 are connected in series to a 130V power supply. Compare between the readings of a voltmeter of resistance 200 1 when connected across each resistor seperately ( neglecting the internal resistance of the power supply). 60) A wire has length 2 m and cross sectional area 0.1 m2 is connected to a source with emf 10 V. Calculate the resistivity and conductivity of its material if it carries a 384 385 391

0.8 V, when a current of 2A passes through it. Calculate the conductivity of the wire material. 50) A rectangular coil of N turns and surface area A is placed parallel to the lines of a regular magnetic field of flux density B Tesla. If the coil starts rotation from this position with a regular angular velocity Ď&#x2030;, until it compelets half a revolution. Clarify with a labeled diagram how the value of the emf changes with the rotational angle during this time, and what is the maximum value of the induced emf generated in this coil. 51) A galvanometer has a resistance 40 â&#x201E;Ś and reads up to 20 mA. Calculate the resistance of the shunt required to convert it into an ammeter, reading up to 100 mA. If the coil of the galvanometer is connected to a potential multiplier with resistance 210 â&#x201E;Ś.Calculate the maximum potential difference to be read. 52) Compare between each of a) a step -down transformer and a step- up transformer in terms of function, use, and number of turns of the secondary coil. b) dynamo and motor in terms of function and use. 53) Why does the transmission of the electric power from a generating station require wires under high voltage? Choose the correct answer and give account a) to be able to use the transformers . b) to insure that the current will flow for a long distance. c) to minimize the loss in the electric energy . d) to minimize the resistance of the wires. 54)What is meant by : a) the coefficient of mutual induction between two coils =2H. 385 384 390

386 387

387 386

388 389

389 388

390 391

391 390

392 393

393 392

394 395

395 394

49)

V I 0.8 R = = 0.4 Ω 2

RA 0.4 × 0.3 × 10 −4 ρe = = lL 30 × 10 −2

ρe = 4 × 10 −5 Ω.m

I R

g g 51) Rs = I − I g

200 × 10-3 × 400 100 × 10 −3 −20 × 10 −3 = 10Ω V = Vg + R m I g =

= (20 × 10 −3 ×40) + (210 × 20 × 10 −3 ) V = 5V

55) Is =

Pw 24 = = 2A Vs 12 Vs I p = Vp Is

12 I p = 240 2 24 Ip = = 0.1A 240 Vs Ns = Vp N p 12 480 = 240 N p Np =

480 × 240 = 9600 12

396 397 403

32) A superconductive material is used in making satelliteâ&#x20AC;&#x2122;s antenna, because it has no electrical resistance. This makes it easily affected by the weakest of electromagnetic waves. 33) Meissner effect appears only in superconductive materials, because they offer no electrical resistance to the electrons. They are easily affected by external magnetic fields and retain the kinetic energy acquired due to this field without any loss in the form of thermal energy. Hence, current flows continuously ,leading in turn to a counter magnetic field , so that the net field inside is zero (diamagnetic) .This is why a superconducting magnet is always repulsive to the external magnetic field . 34) The temperature of argon gas is higher than that of helium gas, because the mass of the argon atom is more than that of a helium atom , so the kinetic energy of an argon atom is more than that of a helium atom. 2 35) 12 mv = 32 kT

1 mv2 = 3 x 1.38 x 10-23 x 300 2 2

= 6.21 x 10-21 J

(

)

-21 v = 6.21 x 10 -31x 22 2= 1.17 x 105 m/s 9.1 x 10

39) As the internal resistance of the battery decreases, the lost work done (wasted energy) is reduced during operation. 40) Because the current intensity in parallel circuits is greater at the input and output compared to the current in each branch.

397 396 402

398 381

57) VB =

100 W = =20V Q 5

64)The wasted potential difference is the potential (voltage) drop on the internal resistance of the cell . VB I = R+r = 12 =4.8 2+0.5 Ir = 4.8 X 0.5 =2.4V Percent drop = 2.4 X 100 = 20% 12

399 398 404

serial

quantity

symbol

unit

potential energy

power

W , Js-1 (watt)

impulse

Pw Iimp

temperature

t˚C , t˚F , T˚K

Celsius, Fahrenheit, Kelvin

quantity of matter

mole

pressure

pascal , Nm-2

27 atmospheric pressure

quantity of heat

Qth

specific heat

Cth

J kg-1 ˚K

heat capacity

qth

JK-1

31 latent heat for evaporation

Bth

J kg-1

32 latent heat for fusion

Lth

J kg-1

33 volume expansion coefficient

αV βP

per degree rise

34 pressure expansion coefficient

-1

per degree rise

kg/s

36 volume rate of flow

m3/s

37 viscosity coefficient

ηvs

Ns m-2

efficiency

______

electric charge

Q,q

C (Coulomb)

electron charge

41 potential difference

V (Volt)

mass rate of flow

battery voltage 401 407

Appendix 1

Symbols and Units of Some Physical Quantities serial

quantity

symbol

unit

displacement

x,y,z,d

m (meter)

area

volume

Vol

time

s (second)

periodic time

velocity / speed

m s-1

angle

α,θ,φ

deg , rad

rad s-1

angular velocity

mass

m,M

electron mass

density

acceleration

13 acceleration due to gravity

linear momentum

force

weight

torque

work

energy

kinetic energy

KE 400 406

kg m-3 m s-2 m s-2 kg m s-1 N , kg ms-2 N(Newton) Nm J(Joule) J J

Appendix 2 Fundamental Physical Constants symbol

value

1-Universal gravitation constant

6.677x10-11 N m2 kg-2

2-Boltzmann constant

1.38x10-23 JK-1

3-Avogadro’s number

k NA

4- Universal gas constant

8.31x103 J.kmol-1 K-1

5-Coulomb’s law constant

9x109 Nm2C-2

6-Permeability of free space

4 x10-7 Weber m-1A-1

7-Speed of light in vacuum

3x108 m.s-1

8-Elementary charge

1.6x10-19 C

9- Electron rest mass

9.1x10-31 kg

1.673x10-27 kg

12-Planck’s constant

6.63x10-34 Js

13-Atomic mass unit

1.66x10-27 kg

14-Rydberg constant

1.096x107 m-1

15-Neutron rest mass

1.675x10-27 kg

Physical Constant

e me

10-Specific charge of electron 11-Proton rest mass

16-Molar volume of ideal gas at S.T.P

6.02x1026 Molecule.kmol-1

1.79x1011 C.kg-1

22.4x10-3 m3

17-Standard gravity at the Earth’s surface

9.8066 ms-2

18-Equatorial radius of the Earth

6.374x106 m

19- Mass of the Earth

5.976x1024 kg

20-Mass of the Moon

7.35x1022 kg

21- Mean radius of the Moon’s orbit

3.844x108 m

around the Earth 403 409

serial

quantity

symbol

unit

emf

43 electromotive force (emf) 44

field intensity

Vm-1

electric flux

φe

Gauss

electric current

A (Ampere)

electrical resistor

Ω (Ohm)

resistivity

Ωm

conductivity

ρe σ

transistor gain

αe , βe

______

51 magnetic field intensity

Am-1

52 magnetic flux density

Tesla , Wb m-2

Ω-1 m-1

magnetic flux

φm

Wb (Weber)

self inductance

H (Henry)

mutual inductance

permeability

Weber A-1 m-1

magnetic dipole

Nm Tesla-1

speed of light

ms-1

frequency of wave

Hertz (Hz)

60 frequency of electric current 61

wave length

refractive index

______

dispersive power

n ωα

402 408

______

Appendix 3

Standard Prefixes power of 10

name

10-24

Yocto

10-21

Zepto

10-18

Atto

10-15

Femto

10-12

Pico

10-9

Nano

10-6

Micro

10-3

Milli

10-2

Centi

10-1

Deci

100

___

101

Deka

102

Hecto

103

Kilo

106

Mega

109

Giga

1012

Tera

1015

Peta

1018

Exa

1021

Zetta

1024

Yotta

405 411

Physical Constant

symbol

22-Mass of the Sun

23- Mean radius of the Earth’s orbit around the Sun

res

24-Period of the Earth’s orbit around the Sun 25- Diameter of our galaxy 26- Mass of our galaxy 27- Radius of the Sun 28- Sun’s radiation intensity at the Earth’s surface

404 410

value 1.989x1030 kg 1.496x1011 m

3.156x107 s

7.5x1020 m

2.7x1041 kg

7x108 m

0.134 J cm-2 s-1

Appendix 5 Gallery of Scientists Ibn Malka

A pioneer in medicine and the discoverer

(1072 -1152 )

of the laws of motion

Ibn Unis

A pioneer in astronomy and the inventor

(952 -1009 )

of the simple pendulum.

Al Baironi

A pioneer in geography and astronomy.

(973 - 1048 ) Ibn Al-Haytham

A pioneer in mathematics, astronomy,

(965 - 1040)

medicine and the founder of optics.

Al Kindy

A pioneer in philosophy, physics ,

(800 - 873)

particularly optics.

Edison (Thomas)

The inventor of the phonograph and the

(1847-1931) Arkhimêdês

(287 -212 BC) Avogadro (Amedeo)

electric lamp, and other inventions “1000". The discoverer of the ratio of the radius of a circle to its circumference, buoyancy and the reflecting mirror.

The discoverer of the molcular theory

(1776 - 1856)

607 413

Appendix 4

Greek Alphabet

606 412

Torricelli (Evangelista) (1608 - 1647) Galileo (Galilei) (1564 - 1642)

The inventor of the barometer

The inventor of the telescope and the discoverer of accelration due to gravity.

Galvani (Luigi) (1737 - 1798)

The discoverer of the electric charge in muscles.

Dalton (John) (1766 - 1844)

Rutherford (Ernest) (1871 - 1937) Ruhmkorff (Heinrich) (1803 - 1877)

The discoverer of the law of mixing gases.

The discoverer of radioactivty.

The discoverer of the induction coil.

Rontgen (Wilhelm)

The discoverer of X-rays.

(1845 - 1923) Schrodinger (Erwin) (1887 - 1961)

The discoverer of Quantum Mechanics.

Al-Khazin

A pioneer in hydrostatics.

609 415

Ampére (André - Marie)

He performed studies on electricity,

(1775 - 1836)

telegraph and magnetism.

Oersted (Christian)

The founder of the theory of

(1777 - 1851) Ohm (George)

electromagnetism in 1820 The discoverer of Ohm’s law

(1789 - 1854) Einstein (Albert) (1879 - 1955) Pascal (Blaise)

He was awarded Nobel prize in 1921 for his explanation of the photoelectric effect, the founder of the theory of relativity The discoverer of Pascal’s rule.

(1623 - 1662) Al Joazri

A pioneer in fine mechanics and water clocks.

Bragg (William)

The founder of X-ray diffraction.

(1862 - 1942) Bohr (Neils)

He produced a model for the atom.

(1885 - 1962) Boyle (Robert)

The discoverer of Boyle’s law.

(1627 - 1691)

608 414

Lenz (Heinrich)

The discoverer of Lenzâ&#x20AC;&#x2122;s rule.

(1804 - 1865) Planck (Max)

The discoverer of the photon and the blackbody radiation.

(1858 - 1947) Maxwell (James)

The discoverer of Maxwellâ&#x20AC;&#x2122;s equations.

Newton (Isaac)

The discoverer of the laws of motion, gravity and colors.

(1642 - 1727) Hertz (Heinrich)

The discoverer of the electromagnetic waves

(1857 - 1894) Huygens (Christian)

He proposed the secondary sources in the from of a wave.

(1629 - 1695) Young (Thomas)

The discoverer of interference.

(1773 - 1829)

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Faraday (Michael)

The discoverer of the laws of

(1791 - 1867)

electromagnetics.

Van Der Waals (Johannes)

The discoverer of Van Der Waalsâ&#x20AC;&#x2122;

(1837 - 1923) Fraunhofer (Joseph Von) (1787 - 1826)

effect. He interpreted the atomic spectra and diffraction

Volta (Alessandro)

The inventor of the battery.

(1745 - 1827) Fermi (Enrico) (1901 - 1954) Kamelingh (Onnes) (1853 - 1926) Kepler (Johannes) (1571 - 1630)

He contributed to the atomic bomb.

The discoverer of liquid helium.

The discoverer of the laws of planetary motion.

Copernicus (Nicolas)

He proved that the Earth rotates around

(1473 - 1543)

the Sun.

Kirchhoff (Gustav)

The discoverer of Kirchhoffâ&#x20AC;&#x2122;s law.

(1824 - 1887)

410 416

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Appendix 6 Selected Physics Sites on the Internet

http://www.dke-encyc.com http://imagine.gsfc.nasa.gov http://csep10.phys.utk.edu http://www.howstuffworks.com http://www.colorado.edu/physics/2000/index.pl http://scienceworld.wolfram.com/physics http://www.physlink.com http://www.intuitor.com/moviephysics http://www.newport.com/spectralanding http://www.mathpages.com/home/iphysics.htm http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html http://www.smsec.com

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