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ENGINEERING
PHYSICS
[SUBJECT CODE: PHY1001]
COMMON COURSE MATERIAL FOR FIRST YEAR BTech STUDENTS
DEPARTMENT OF PHYSICS
MANIPAL INSTITUTE OF TECHNOLOGY
MANIPAL UNIVERSITY
SYLLABUS
PHY 1001: ENGINEERING PHYSICS [2 1 0 3]
Optics: Two source interference, Double slit interference, Coherence, Intensity in double slit
interference using phasor method, Interference from thin films, Newton’s rings, Diffraction
and wave theory of light, Single-slit diffraction, Intensity in single-slit diffraction using phasor
method, Diffraction at a circular aperture, Double-slit interference and diffraction combinedIntensity in double-slit diffraction (Qualitative approach), qualitative description of multiple slits
and diffraction grating
[9]
Applied Optics: Spontaneous and stimulated transitions, He-Ne and Ruby laser, Applications of
lasers, Optical fiber, Total internal reflection, angle of acceptance and numerical aperture, types
of optical fiber, types of attenuation, applications of optical fibers.
[5]
Quantum Physics: Black body radiation and Planck’s hypothesis, Stefan’s Law, Wein’s
displacement law, Photoelectric effect, Compton effect, Photons and electromagnetic waves,
Wave properties of particles, de-Broglie hypothesis, Quantum particle (wave packet, phase
speed, group speed). The double-slit experiment revisited, the uncertainty principle
[8]
Quantum Mechanics: An interpretation of quantum mechanics, Wave function and its
significance, particle in a box (infinite potential well), Schrodinger equation, Particle in a well of
finite height, Tunnelling through a potential barrier and its applications, The quantum model
of the hydrogen atom, The wave functions for hydrogen
[8]
Solid State Physics: Free electron theory of metals, Band theory of solids, Electrical conduction
in metals, Insulators and Semiconductors, Superconductivity-Properties and Applications [6]
Reference books
1. Halliday, Resnick, Krane; PHYSICS: Volume 2, (5e), John Wiley & Sons, Inc, 2009
2. Serway & Jewett; PHYSICS for Scientists and Engineers with Modern Physics,
Thomson, 2013
(9e),
COURSE OBJECTIVES
To understand the wave properties of light and thereby applications of interference and
diffraction of light.
To study the working principles of optical fibers and lasers.
To understand the basic principles of quantum physics.
To understand the mechanism of bonding and electrical conductivity in solids.
COURSE OUTCOMES
At the end of this course students will be able to:
Explain the principles of optical methods of testing and measuring of various physical
parameters.
Describe the construction and working of optical fibers and lasers.
Discuss the principles of dual nature of particles and light.
Describe quantum mechanical properties of micro particles such as energy quantization,
tunneling, and quantum mechanical model of hydrogen atom.
Explain electrical conduction properties of materials.
TEACHING PLAN
L/T No.
TOPICS TO BE COVERED
L0
Introduction : course contents, assessments, AMS, availability etc.
L01
Light as an EM (electro-magnetic) wave. Interference of light waves. Coherence. Double-slit
interference
L02
Intensity in double-slit interference.
L03
Interference from thin films. Newton rings.
T04
Tutorial: Problems on topics in L-01 to L-03.
L05
Diffraction and wave theory of light. Single-slit diffraction. Intensity in single-slit diffraction.
T06
Tutorial: Problems on topics in L-05.
L07
Diffraction at a circular aperture. Double-slit interference and diffraction combined.
L08
Multiple slits. Diffraction gratings.
T09
Tutorial: Problems on topics in L-07 & L-08.
L10
Spontaneous and stimulated transitions. Metastable state. Population inversion. Ruby Laser
L11
He-Ne laser. Applications of lasers. Tutorial: Problems on topics in L-10 & L-11.
L12
Optical fibers. Total internal reflection. Angle of acceptance and numerical aperture.
L13
Types of optical fiber. Attenuation in optical fibers. Applications of optical fibers.
T14
Tutorial: Problems on topics in L-12 & L-13.
L15
Black body radiation and Planck’s hypothesis. Stefan’s law. Wien’s displacement law. Rayleigh-Jeans
law.
L16
The photoelectric effect.
T17
Tutorial: Problems on topics in L-15 & L-16.
L18
The Compton effect.
L19
Photons and electromagnetic waves. The wave properties of particles. de Broglie hypothesis.
T20
Tutorial: Problems on topics in L-18 & L-19.
L21
The quantum particle. The double-slit experiment revisited. The uncertainty principle.
T22
Tutorial: Problems on topics in L-21.
L23
An interpretation of quantum mechanics. Wave function and its significance.
L24
The Schrodinger equation. The particle in a “well” of infinite height.
L25
A particle in a “well” of finite height.
L26
Tunneling through a potential energy barrier. Applications.
T27
Tutorial: Problems on topics in L-24 to L-26.
L28
The quantum model of the hydrogen atom.
L29
The wave functions for hydrogen atom.
T30
Tutorial: Problems on topics in L-28 & L-29.
L31
Free-electron theory of metals.
T32
Band theory of solids
L33
Tutorial: Problems on topics in L-31& L-32.
L34
Electrical conduction in metals, insulators and semiconductors.
L35
Superconductivity – properties and applications
T36
Tutorial: Problems on topics in L-34 & L-35.
CONTENTS
Chapter 1
Optics
p01
Chapter 2
Applied Optics
p29
Chapter 3
Quantum Physics
p44
Chapter 4
Quantum Mechanics
p62
Chapter 5
Solid State Physics
p76
EVALUATION SCHEME
Internal assessment
5 Quizzes of 4 marks each
20 Marks
2 Sessional of 15 marks each
30 Marks
End Semester Examination
Total
50 Marks
50 Marks
100 Marks
PHY 1001: ENGINEERING PHYSICS
CHAPTER 1
OPTICS
OBJECTIVES
•
•
•
•
•
To understand the principles of interference and diffraction.
To explain the intensity distribution in interference and diffraction under various
conditions.
To understand the phasor method of adding wave disturbances.
To explain the interference from thin films
To explain the diffraction of light waves at single, multiple slits and circular apertures.
Light is a transverse electro-magnetic wave in which electric (E) and magnetic fields (B)
oscillate in phase, perpendicular to each other and both are perpendicular to the direction of
propagation. The visual sensation of light is due to its E- field and as such in discussing
interference phenomena, one considers superposition of electric fields.
1.1 INTERFERENCE OF LIGHT WAVES
When two or more waves superpose in a region of space, the resultant amplitude of E-field at
any point is the vector sum of the individual amplitudes of the waves and the intensity at that
point is proportional to the square of this E-field amplitude. When waves from two
independent sources superpose, the resultant intensity at any point is the sum of the
intensities due to individual sources and is same throughout the region of superposition. On
the other hand, when two similar waves traveling almost in the same direction superpose,
intensity variation take place in the region of superposition. This redistribution of light
intensity when two or more similar waves superpose is called interference and such similar
waves are called coherent waves.
1.1.1 COHERENT WAVES:
•
Two waves are said to be coherent when they maintain a constant phase difference
between them. For this is to be true the waves must have same wavelength (and hence
same frequency, since velocity of all em waves are same and is equal to 3× 108m/s)
and travel almost in the same direction. Coherence is a necessary condition for
producing stable interference pattern. Suppose phase difference between two waves
keeps changing, the positions of maximum and minimum amplitudes vary with time.
As a result rapid intensity fluctuations will occur which cannot be followed by the eye(
normal eye can resolve fluctuations ~ 16).
•
Coherence depends on the length of the wave trains. Longer the wave train, degree
of coherence is more. Common light sources emit light wave trains of finite length (few
millimeters) accordingly the degree of coherence is less.
Dept. of Physics, MIT Manipal
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Fig. 1.1 Section of infinite wave train and a wave train of finite length
•
Laser light is highly coherent (coherent length of few centimeters to meters).
•
Degree of coherence is very important in telecommunications and holography.
How to produce coherent waves?
Waves from two independent sources cannot be coherent. Because in these sources the
fundamental light emission processes occur in individual atoms, and these atoms do not act
together in a co-operative way (that is, incoherent).
There are two methods of producing coherent waves:
i) Division of wave front: For example, Young double slit experiment (Figure 1.2).Here, two
different portions of a same wave front is made to pass through two narrow slits separated by
large distance d (d>>). Huygen wavelets from the two slits are perfectly coherent.
Fig. 1.2 Production of coherent waves by division of wave front
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ii) Division of amplitude: For example, in Newton rings experiment this mechanism is used.
Light waves reflected from front and rear surfaces of a thin film are perfectly coherent (Figure
1.3).
Fig. 1.3 Production of coherent waves by division of amplitude
When the two coherent waves overlap (i.e., when they travel almost in the same direction)
they produce interference pattern on the screen placed on their path. The fringe pattern
consists of alternating series of bright and dark bands known as interference fringes. For good
contrast of these fringes, the amplitude of the two coherent waves must be comparable.
The interference is constructive when the net intensity is greater than the individual intensities
(Figure 1.4a). The interference is destructive when the net intensity is less than individual
intensities (Figure 1.4b).
Fig. 1.4 (a) Constructive interference of two waves that are in phase (b) Destructive
interference of two waves that are 1800 out of phase
Maximal constructive interference of two waves occurs when their phase difference is 0, 2,
4, .… (the waves are in-phase). During the period of one oscillation ( phase change of 2π or
360˚), the wave disturbance travels a distance of λ, and hence the path and phase difference
are related as
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Phase difference Path difference
2
Thus, the phase differences of 0, 2, 4, .……. is equivalent to path differences of 0, λ,
2λ…... Condition for constructive interference is therefore-path difference = m λ
where m = 0, 1, 2,….
Complete destructive interference of two waves occur when their phase differences are
3, 5 , … (the waves are 180o out of phase) or path differences of λ/2, 2λ/2, 3λ/2 .…
Thus, condition for destructive interference is, path difference = (m+1/2) λ ,
0, 1, 2 …
,
where m =
1.1.2 DOUBLE-SLIT INTERFERENCE
Fig. 1.5 Double slit arrangement and interference pattern
A train of plane light waves is incident on two narrow parallel slits of widths a (<<) separated
by distance d (>>). Each slits emit Huygen wavelets and behave like two independent
coherent sources. The interference pattern on the screen at a distance D consists of bright and
dark fringes.
For D>>d, we can approximate rays r1 and r2 as being parallel. Path difference between two
waves from S1 & S2 on reaching a point P on a screen is S1b = d sin .
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Fig. 1.6 (a) Schematic of double slit arrangement, (b) Showing the path difference between
two rays r1 and r2
For maximum at point P
S1b = m
which can be written as,
d sin = m,
m = 0, 1, 2, . . .m = 0 is the central maximum.
For minimum at point P
d sin ( m 21 )
m = 0, 1, 2, . . .
For small value of , we can make following approximation
sin tan
y
D
In such a case, the path difference
d sin S1b
yd
D
For mth maximum located at ym , we can write
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ym d
m
D
or
ym
m
D
d
where m = 0, 1, 2, . . .
Separation between adjacent maxima (for small ) known as fringe width/band width is,
y ym1 ym
D
( m 1)
y
D
d
m
D
d
d
is independent of m.
The spacing between the adjacent minima is same as the spacing between adjacent maxima.
1.1.3 YOUNG’S DOUBLE SLIT EXPERIMENT
Double slit experiment was first performed by Thomas Young in 1801. So double slit
experiment is known as Young’s Experiment. He used sun light as source for the experiment.
In his experiment, he allowed sun light to pass through narrow opening (S0) and then through
two openings (S1and S2).
Fig. 1.7 Young’s interference experimental set up
1.1.4 INTENSITY IN DOUBLE SLIT INTERFERENCE – PHASOR METHOD
Phasor is a rotating vector.
Electric field components at P due to S1 and S2 are (see figure 1.8)
E1= E0 sin ωt andE2= E0 sin (ωt + ) respectively, where is the phase difference between
them.
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Resultant field E = E1 + E2
Fig. 1.8 Schematic of double slit arrangement
Fig. 1.9 (a) Phasor representation of two waves, (b) Another way of drawing (a)
From phasor diagram (Figure 1.9b),
E=E1+E2
= E sin(t + )
= 2E0cos sin(t + )
But = /2. So above equation can be written as,
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E = 2 E0cos(/2) sin(t+/2)
So, intensity at an arbitrary point P on the screen due to interference of two waves having
phase difference;
4 E 02 cos 2
2
4 0 cos2
2
where 0 E 02 is intensity due to single source.
Since 2dsin/ ,
2 d sin
4 0 cos
From above equation,
At maxima : 2 m
or
At minima : ( 2 m 1)
d sin
or
m
d sin
(m 1 )
2
where m 0, 1, 2, . . .
Fig. 1.10 Intensity variations as a function of phase differences
1.1.5 INTERFERENCE FROM THIN FILMS
A film is said to be thin when its thickness is comparable with the wavelength of the light, i.e,
of the order of a micron. Greater thickness spoils the coherence of the light. In Figure 1.11b,
the region ac looks bright or dark for an observer depending on the path difference between
the rays r1 and r2.
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Fig. 1.11 Thin film interference (a) A soapy water film, (b) Ray diagram
Phase change on Reflection: It has been observed that if the medium beyond the interface
has a higher index of refraction, the reflected wave undergoes a phase change of (=180o). If
the medium beyond the interface has a lower index of refraction, there is no phase change of
the reflected wave. No phase change occur for transmitted light.
Fig. 1.12 Phase changes on reflection at a junction between two strings of different linear
mass densities (a) The incident pulse is in the heavier string, (b) The incident pulse is in the
lighter string
When light pass from one medium to another, its velocity changes and accordingly its
wavelength changes. The type of interference in thin films is determined by the wavelength
n in the film and not the wavelength in air. If n is wavelength in the film of refractive
index n and is the wavelength in vacuum then n = / n
It is therefore optical path length difference that is of interest and not the geometrical path
length difference in discussing interference from thin films.
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Optical path length: Optical path length for a light wave is the vacuum path equivalent of its
geometrical path in the medium.
Distance traveled by light in a medium in the time interval of ‘t’ is d = vt
Refractive index n = c/v
Hence, ct = nd
nd is the optical path corresponding to geometrical path d and is the distance traveled by light
in vacuum in the time ‘t’ that it takes to travel path d.
Equations for thin film interference: Normal incidence (i = 0)
Path difference = 2 d + (½) n (?) + (½) n (?). The terms with question marks are to be used if
there are phase changes at front and rear surfaces respectively.
Assuming air on either side of the film (Figure 1.11b), conditions for Constructive interference:2 d + (½) n = m n
m = 1, 2, 3, . . .
(maxima)
Destructive interference:2 d + (½) n = (m+½) n
m = 0, 1, 2, . . .
(minima)
It can be noted that, it is possible to suppress the unwanted reflections from glass at a chosen
wavelength by coating the glass with a film of proper thickness and in such a case the film is
known as antireflection coating. Moreover, the film may reflect or transmit preferentially a
particular wavelength and in such a case the film is called a monochromator.
Wedge shaped film: When light falls on wedge shaped thin film, it undergoes partial
reflections from upper and lower part of the film thereby producing coherent waves.
Fig. 1.13 Interference in a wedge shaped film
Since the film is thin, the reflected waves are close by and are in a position to interfere.
Constructive interference occurs in certain part of the film [2 d + (½) n= m n] and destructive
interference in others[2 d + (½)n = (m+½)n]. Then bands of maximum and minimum intensity
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appear, called fringes of constant thickness. The locus of the points having the same film
thickness is a straight line and hence straight(linear) fringes are formed.
Newton’s rings: When a plano-convex lens is kept on an optically flat glass plate, a thin film of
air is formed between the two. Monochromatic light falling on this system partly reflects from
upper and lower surfaces of the film (Figure 1.14a). These two coherent waves interfere
constructively or destructively depending on the thickness of the air film. The locus of the
points having the same thickness is a circle and hence alternate bright and dark concentric
circular fringes are formed (Figure 1.15).
Fig.1.14 (a) Newton’s ring set up, (b) the geometry of the set up.
For constructive interference 2d = (m - ½) (assuming normal incidence and air film n = 1)
d R R2 r 2
1
r 2 2
R R 1
R
r R 1 using binomialexpansion
1 r 2
d R R 1 . . .
2 R
r2
2R
Substituting d , in 2d = (m - ½) we get,
r
m 12 R
m 1, 2, . . . for maxima and is known as the order of the rings.
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Note that r>0 for m=1. i.e, the first bright ring is at r>0 , and consequently the center must be
dark. This observation can be taken as an experimental evidence for the 180° phase change
on reflection.
Fig. 1.15 Circular interference fringes observed in Newton’s ring set up
EXERCISE
QUESTIONS
1.
2.
3.
4.
What is interference of light waves?
What is coherence? Mention its importance.
How coherent waves are produced? Illustrate with figure.
Write the necessary condition for the constructive and destructive
interference of two light waves in terms of path/phase difference.
5. With necessary diagram, obtain an expression for the fringe-width of its
interference pattern.
6. Obtain an expression for intensity of light in double-slit interference
using phasor diagram.
7. Draw a schematic plot of the intensity of light in double-slit interference
against phase-difference.
8. Explain the following: i) Phase change on reflection ii) Optical path length
9. Write the conditions for constructive and destructive interference of
reflected light from a thin soap film in air, assuming normal incidence.
10. Explain the interference in wedge-shaped thin films.
11. Explain the formation of Newton’s rings and hence obtain an expression
for the radius of mth order bright ring.
PROBLEMS
[2]
[2]
[3]
[2]
[5]
[5]
[2]
[2]
[2]
[2]
[5]
1. Calculate the path difference between two coherent waves in terms of their wavelengths,
for phase differences of – i) 60° ii) 270°.
Ans: i) /6 ii) 3/4
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2. The double slit arrangement is illuminated by light of wavelength 546 nm. The slits are
0.12 mm apart and the screen on which interference pattern appears is 55 cm away.
a) What is the angular position of (i) first minima and (ii) tenth maxima?
b) What is the separation between two adjacent maxima?
Ans: 0.13°, 2.6°, 2.5 mm
3. Monochromatic light illuminates two parallel slits a distance d apart. The first
maximum is observed at an angular position of 15°. By what percentage should
“d” be increased or decreased so that the second maximum will instead be observed
at 15° ?
Ans: 100%
4. A double-slit arrangement produces interference fringes for sodium light (wavelength
= 589 nm) that are 0.23° apart. For what wavelength would the angular separation
be 10% greater ? Assume, the angle is small.
Ans: 650 nm
5. Find graphically the resultant E(t) of the following wave disturbances.
E1 = E0 sin t
E2 = E0 sin (t + 15o)
E3 = E0 sin (t + 30o)
E4 = E0 sin (t + 45o)
Ans: E(t) = 3.83 Eo sin (t + 22.5°)
6. Find the sum of the following quantities (a) graphically ( phasors method) and (b)
algebraically (using trigonometry) :
y1 = 10 sin (t)
y2 = 8.0 sin (t + 30°)
Ans: y = 17.4 sin (t + 13.3°)
7. Source A of long-range radio waves leads source B by 90 degrees. The distance rA to a
detector is greater than the distance rB by 100m. What is the phase difference at the
detector? Both sources have a wavelength of 400m.
Ans: 0°
8. A soap film (n=1.33) in air is 320nm thick. If it is illuminated with white light at normal
incidence, what color will it appear to be in reflected light?
Ans: 567 nm (yellow-green)
9. Lenses are often coated with thin films of transparent substances such as MgF 2 (n=1.38)
to reduce the reflection from the glass surface. How thick a coating is required to produce
a minimum reflection at the center of the visible spectrum? ( wavelength = 550nm)
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Ans: 100 nm
10. If the wavelength of the incident light is λ = 572 nm, rays A and B in Fig 41-24 are
out of phase by 1.50 λ. Find the thickness d of the film.
Ans: 215 nm
11. A broad source of light (wavelength = 680nm) illuminates normally two glass plates 120
mm long that touch at one end and are separated by a wire 0.048mm in diameter at the
other end. How many bright fringes appear over 120 mm distance?
Ans: 141
12. In a Newton’s ring experiment, the radius of curvature R of the lens is 5.0m and its
diameter is 20mm. wavelength= 589nm
How many rings are produced?
How many rings would be seen if the arrangement is immersed in water (n = 1.33)?
Ans: 34, 45
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1.2 DIFFRACTION AND WAVE THEORY OF LIGHT
When light passes through a narrow slit (of width comparable to the wave length of light), the
light not only flare out far beyond the geometrical shadow of the slit; they also gives rise to a
series of alternating light and dark bands. This observation can be explained by assuming that
light must travel as waves and as such bend at the edges of apertures or obstacles on their
path. The phenomenon of bending of light around the edges of obstacles or slits, and hence
its encroachment into the region of geometrical shadow is known as diffraction. As a result of
bending of light waves, the edges of shadows are not very sharp as expected by the rectilinear
propagation of light. Diffraction effects are seen more prominently when the size of the object
causing diffraction have dimensions comparable to the wavelength of light falling on the
object.
Fig. 1.16 Diffraction pattern (a) Poisson spot and (b) razor blade viewed in monochromatic
light
Fig. 1.17 Diffraction at an aperture
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PHY 1001: ENGINEERING PHYSICS
Diffraction occur when coherent wave-fronts of light fall on opaque barrier B, which contains
an aperture of arbitrary shape (Figure 1.17). The diffraction pattern can be seen on screen C.
The pattern formed on the screen depends on the separation of the source and the screen C
from the aperture B. We can consider three cases:
1. Very small separation- when C is very close to B(irrespective of source distance) the waves
travel only a short distance after leaving the aperture and rays diverge very little. The
effects of diffraction are negligible, and the pattern on the screen is the geometrical
shadow of the aperture.
2. Both S and C are at large distance- i.e., both incident and the emerging wave-fronts are
plane (the rays are parallel). One can achieve this condition by using two converging lenses.
This class of diffraction is called Fraunhofer diffraction(Figure 1.18a and b).
3. S and C are at finite distance from the aperture - i.e., incident and emerging wave fronts
are spherical or cylindrical. Diffraction produced in this case is called Fresnel class (Figure
1.18c).
Fig. 1.18 Diffraction (a) and (b) Fraunhofer type, (c) Fresnel type
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1.2.1 SINGLE-SLIT FRAUNHOFER DIFRACTION (QUALITATIVE)
All the diffracted rays arriving at P0travel equal optical paths and hence are in-phase (Figure
1.19). Hence they interfere constructively and produce maximum (central maximum) of
intensity I0 at P0.
Fig. 1.19 Conditions at the central maximum of the diffraction pattern
Consider another point P1 on the screen where the rays leaving the slit at an angle , meet
(Figure 1.20). Ray r1 originates on the top of the slit and ray r2 at its center. If is chosen such
that the path difference between r1 and r2 is (a/2) sin = /2, a condition for destructive
interference of rays r1 and r2.
Fig. 1.20 Conditions at the first minimum of the diffraction pattern
In such a situation, this condition is satisfied for every pair of rays, one from upper half of the
slit and the other corresponding ray from lower half of the slit. Hence each pair of
corresponding rays cancelling each other producing first minima.
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So the condition for first minimum,
a
sin
2
2
or
a sin
This equation shows that, the central maximum becomes wider as the slit is made narrower.
If the slit width is as small as one wavelength ( a= ), the first minimum occurs at = 90° which
implies that the central maxima fills the entire forward hemisphere.
In fig. 1.21, the slit is divided into four equal zones with rays r1, r2 , r3 and r4 leaving the top of
each zone. Let be such that the path difference (a/4) sin, between r1and r2 is
(a/4)
sin = /2.
Fig. 1.21 Conditions at the second minimum of the diffraction pattern
This is satisfied for every pair of rays, separated by a distance a/4. As a result, while the
corresponding rays from first and second quarters of the slit interferes destructively so does
the rays from third and fourth quarters. As a whole, the secondary wavelets from different
parts of the slit interfere destructively resulting in minimum intensity at P 2. Thus, condition for
second minima is, (a/4) sin = /2 or, a sin = 2
In general, the condition for m th minima,
a sin m
m 1, 2, 3, . . .
There is a maximum approximately half way between each adjacent pair of minima.
1.2.2 INTENSITY IN SINGLE SLIT DIFFRACTION (QUANTITATIVE)
Divide the slit of width a into N parallel strips each of width x (this also means that their
separation is also x). The strips are very narrow and can be regarded as radiator of Huygen
wavelets and all the light from a given strip arrives at point P with same phase (Figure 1.22).
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PHY 1001: ENGINEERING PHYSICS
Fig. 1.22 A slit of width a divided into N parallel strips each of width x (Inset shows the
condition at second strip)
The phase difference between waves arriving at point P from two adjacent strips have the
same constant phase difference
2
x sin
The wave disturbance at any point due to each strip can be represented by a vector. To find
the resultant intensity, we have to lay N vectors each of length δEo head to tail, each differing
in direction from the previous one by . The resultant phasor amplitude is found by vector
addition.
Fig. 1.23 Phasor diagram to calculate the intensity in single slit diffraction
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PHY 1001: ENGINEERING PHYSICS
From Figure 1.23,
E 2R sin
Also
2
Em
R
Combining, E
Or , E E m
Em
sin
sin
2
2
where
2
is the phase difference between rays from the top and bottom of the slit. Thus,
2
So,
a sin
2
a
sin
The intensity E
2
sin
E
2
2
m
sin
2
m
where m E m is the max. intensity
From the above eqn., for minima, sin 0
Hence m where m 1 ,2, 3,.....
or, a sin m where m 1, 2, 3,.....
2
Fig. 1.24 Intensity distribution in single slit diffraction
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 1.25The intensity distribution in single-slit diffraction for three different values of a/.
1.2.3 DIFFRACTION AT A CIRCULAR APERTURE
The mathematical analysis of diffraction by a circular aperture shows that the first minimum
occurs at an angle from the central axis given by-
sin 1.22
where d is the diameter of aperture.
d
The equation for first minimum in single slit diffractio n is
sin
a
where a is the slit width
In case of circular aperture, the factor 1.22 arises when we divide the aperture into elementary
Huygens sources and integrate over the aperture.
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 1.26 Diffraction pattern due to a circular aperture
The fact that lens images are diffraction patterns is important when we wish to distinguish two
distant point objects whose angular separation is small. The condition for resolution of such
close objects is known as Rayleigh’s criterion for optical resolution: The images of two closely
spaced sources is said to be just resolved if the angular separation of the two point sources is
such that the central maximum of the diffraction pattern of one source falls on the first
minimum of the diffraction pattern of the other.
R sin 1 1.22
d
since R is very small, it can be appoximated as
R 1.22
d
R is the smallest angular separation for which we can resolve the images of two objects.
Fig. 1.27 Images of two distant point sources formed by a converging lens (a) Well
resolved (b) Just resolved (c) Not resolved
Dept. of Physics, MIT Manipal
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1.2.4 DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
In our analysis of double slit interference we assumed that the slits were arbitrarily narrow
i.e., a<<λ. For such narrow slits, the central part of the screen is uniformly illuminated by the
diffracted waves from each slit. When such waves interfere, they produce interference fringes
of uniform intensity. But, in practice the condition a<<λ is usually not met. For such relatively
wide slits, the intensity of interference fringes is not uniform. Instead, the intensity of the
fringes varies within an envelope due to the diffraction pattern of a single slit as shown in the
Figure 1.28.
Ignoring diffraction effects, the intensity of interference fringes is given byI, INT = Im,INT cos2
Ignoring interference effects, the intensity of diffraction pattern is given by
Im,DIF(sin/)2
The combined effect is the product of the two and is given by Iθ
sin α
Ι m ( cos ) 2
α
I,
DIF
=
2
Fig. 1.28 Double slit interference and diffraction combined
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 1.29 Intensity sketches to illustrate the combined effect of interference and diffraction
1.2.5 MULTIPLE SLITS
In principle, one can use a double slit interference pattern to measure the wavelength, but
fringes being wide there involves an uncertainty in locating their mid points. It has been
observed that increase in the slit number reduces the fringe widths and the precision of
wavelength measurement improves. The second effect of increasing the number of slits is the
appearance of faint secondary maxima, (N-2) in number, as shown in the Figure 1.30.
Fig. 1.30 Intensity pattern for (a) Two-slit diffraction (b) Five-slit diffraction
Following figure shows five slit grating illuminated by monochromatic light of wavelength. A
principal maximum occurs when the path difference between rays from any pair of adjacent
Dept. of Physics, MIT Manipal
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slits is d sin = m , where d is the separation between adjacent slits. Location of principal
maxima is independent of number of slits.
Fig. 1.31 An arrangement of multiple slits (Here N = 5)
Width of the maxima: Consider the Figure 1.32 in which the mth principal maximum occurs at
an angle . We move away from this maximum through an angular displacement to arrive
at the next minimum. This angle is the measure of angular width of the mth maximum.
Fig. 1.32 Width of principal maximum
At minima, the phase difference between adjacent slits is such that,
2
where N is the number of slits, since N phasors form a closed loop
N
Corresponding path difference is,
N
2
L
Dept. of Physics, MIT Manipal
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For the
mth
PHY 1001: ENGINEERING PHYSICS
principal maximum at , we have d sin = m .
For the first minimum at ( + ) after the mth principal maximum is therefore
d sin ( δ ) mλ
λ
N
d sin cos cos sin m
N
1
d sin (d cos ) m N
m (d cos )
N d cos
m N
is the angular half width of mth principal maximum at .
It is seen that, the principal maximum become sharper as number of slits (N) increases as
mentioned earlier. Width of central maximum will be
Nd
.
1.2.6 DIFFRACTION GRATINGS
The diffraction grating, is a useful device for analysing light sources. It consists of a large
number of equally spaced parallel slits. A typical grating might contain N= 10,000 slits
distributed over a width of a few centimetres. They are of two kind: i) Transmission gratings
ii) Reflection gratings. A transmission grating can be made by cutting parallel grooves on a
glass plate with a precision ruling machine. The spaces between the grooves are transparent
to the light and hence act as separate slits. A reflection grating can be made by cutting parallel
grooves on the surface of a reflective material. The reflection of light from the spaces between
the grooves is specular, and the reflection from the grooves is diffuse.
Fig. 1.33 (a) Reflection type grating, (b) Grating spectroscope
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Fig. 1.34 Sample spectra of visible light emitted by a gaseous source
Most gratings used for visible light, whether of the transmission or reflection type, are phase
gratings i.e, there is a periodic change in phase (and a negligible change in amplitude) of the
light as a function of position across the grating. The grating equation is same as that of
multiple slits i.e, d sin = m , d is the slit separation, m 1,2,3,..... and is called order of
the spectra.
EXERCISE
QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10
11
Explain the term diffraction of light. What are the factors that determine
diffraction pattern?
Discuss qualitatively, the Fraunhofer diffraction at a single-slit.
Derive an expression for intensity of diffraction pattern in the case of
single-slit, using phasor diagram.
Draw a schematic plot of the intensity of light in single slit diffraction
against phase difference.
Explain briefly diffraction at a circular aperture.
State and explain Rayleigh’s criterion for optical resolution.
Effect of diffraction is ignored in the case of Young’s double slit
interference. Give reason.
Arrive at the equation for the intensity of double slit diffraction pattern.
Discuss qualitatively, the diffraction due to multiple slits.
Obtain an expression for the half angular width of any principal
maximum in diffraction pattern due to multiple slits.
What is diffraction grating? Write the grating equation.
[3]
[5]
[5]
[2]
[2]
[2]
[2]
[2]
[3]
[5]
[2]
PROBLEMS
1. A slit of width “a“ is illuminated by white light. For what value of “a” does the minimum
for red light ( = 650nm) fall at = 15o?
Ans: 2.51 μm
2. In the above problem what is the wavelength ’ of the light whose first diffraction
maximum (not counting the central maximum) falls at 15o, thus coinciding with the first
minimum of red light?
Ans: 430 nm
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PHY 1001: ENGINEERING PHYSICS
3. Calculate, approximately, the relative (with respect to central maxima) intensities of
the first three maxima in the single-slit diffraction pattern.
Ans: 0.045, 0.016, 0.0083
4. A single slit is illuminated by light whose wavelengths are a and b, so chosen that the
first diffraction minimum of a component coincides with the second minimum of the b
component.
i) What is the relationship between the two wavelengths?
ii) Do any other minima in the two patterns coincide?
Ans: (i) a = 2 b (ii) all minima of a coincide with even numbered minima of b
5. Monochromatic light with wavelength 538 nm falls on a slit with width 25.2m. The
distance from the slit to a screen is 3.48m. Consider a point on the screen 1.13cm from
the central maximum. Calculate (a) (b) (c) ratio of the intensity at this point to the
intensity at the central maximum.
Ans: 0.186°, 0.478 rad (= 27.4°), 0.926
6. A converging lens 32mm in diameter has a focal length f of 24 cm. (a) What angular
separation must two distant point objects have to satisfy Rayleigh’s criterion? Assume
that = 550nm. (b) How far apart are the centers of the diffraction patterns in the focal
plane of the lens?
Ans: 2.1 x 10–5 rad (= 4.3”), 5 μm ( 9λ)
7. In a double slit experiment, the distance D of the screen from the slits is 52cm, the
wavelength is 480nm, slit separation d is 0.12mm and the slit width a is 0.025mm.
i) What is the spacing between adjacent fringes?
ii) What is the distance from the central maximum to the first minimum of the fringe
envelope?
Ans: 2.1 mm, 10 mm
8. What requirements must be met for the central maximum of the envelope of the doubleslit interference pattern to contain exactly 11 fringes?
Ans: slit separation = (11/2) slit width
9. A certain grating has 104 slits with a spacing of d = 2100 nm. It is illuminated with yellow
sodium light ( = 589 nm). Find (a) the angular position of all principal maxima observed
and (b) the angular width of the largest order maximum.
Ans: (a) 16.3°, 34.1°, 57.3° (b) 5.2 x 10–5 rad (= 0.18’)
10. A diffraction grating has 104 rulings uniformly spaced over 25.0mm. It is illuminated at
normal incidence by yellow light from sodium vapor lamp which contains two closely
spaced lines of wavelengths 589.00nm and 589.59nm. (a) At what angle will the first order
maximum occur for the first of these wavelengths? (b) What is the angular separation
between the first order maxima of these lines?
Ans: (a) 13.627° (b) 0.014°
11. Given a grating with 400 rulings/mm, how many orders of the entire visible spectrum
(400-700nm) can be produced?
Ans: 3
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
CHAPTER 2
APPLIED OPTICS
OBJECTIVES:
To explain basic interactions of radiation with matter.
To understand the basic principles and requirements for working of laser.
To identify the various possible applications of laser.
To understand the basic working principle of optical fiber.
To classify different types of optical fibers and explain their merits & demerits.
To analyze different types of losses in optical fibers.
To recognize various applications of optical fibers.
2.1 LASER (Light Amplification by Stimulated Emission of Radiation)
Laser light is highly monochromatic, coherent, directional and can be sharply focused. Each of
these characteristics that are not normally found in ordinary light makes laser a unique and
the most powerful tool. Lasers find a wide variety of applications in the field of scientific
research, engineering and medicine.
2.1.1 INTERACTION OF RADIATION WITH MATTER
There are three possible processes that involve interaction between matter and radiation.
Absorption: Absorption of a photon of frequency f takes place when the energy
difference E2 – E1 of the allowed energy states of the atomic/molecular system equals
the energy hf of the photon. Then the photon disappears and the atomic system moves
to upper energy state E2.
Fig. 2.1 Absorption
Spontaneous Emission: The average life time of the atomic system in the excited state
is of the order of 10–8 s. After the life time of the atomic system in the excited state,
it comes back to the state of lower energy on its own accord by emitting a photon of
energy hf = E2– E1 .
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PHY 1001: ENGINEERING PHYSICS
This is the case with ordinary light sources. The radiations are emitted in different directions
in random manner. Such type of emission of radiation is called spontaneous emission
and the emitted light is not coherent.
Fig. 2.2 Spontaneous Emission
Stimulated Emission: When a photon (called stimulating photon) of suitable frequency
interacts with an excited atomic system, the latter comes down to ground state by
emitting a photon of same energy. Such an emission of radiation is called stimulated
emission. In stimulated emission, both the stimulating photon and the emitted photon(
due to stimulation) are of same frequency, same phase, same state of polarization and
in the same direction. In other words, these two photons are coherent.
Fig. 2.3 Stimulated Emission
All the three processes are taking place simultaneously to varying degrees, in the matter when
it is irradiated by radiation of suitable frequency.
Population inversion: From Boltzmann statistics, the ratio of population of atoms in two
energy states E1 and E2at equilibrium temperature T is,
E E1
nE 2
exp 2
nE 1
k T
[2.1]
where k is Boltzmann constant, n(E1) is the number density of atoms with energy E1 , n(E2)
is the number density of atoms with energy E2 . Under normal condition, where populations
are determined only by the action of thermal agitation, population of the atoms in upper
energy state is less than that in lower energy state (i.e.n(E2)<n(E1), Figure 2.4a).
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 2.4 (a) Normal thermal equilibrium distribution of atomic systems (b) An inverted
population, obtained using special techniques
For the stimulated emission rate to exceed the absorption rate it is necessary to have
higher population of upper energy state than that of lower energy state. This condition
is called population inversion(n(E2)>n(E1), Figure 2.4b). This is a non-equilibrium condition
and is facilitated by the presence of energy states called ‘metastable states’ where the
average life time of the atom is 10-3 s which is much longer than that of the ordinary excited
state ( 10-8s).
Principle of laser: Lasing medium or active medium, resonant cavity and pumping system are
the essential parts of any lasing system (Figure 2.5). Lasing medium has atomic systems
(active centers), with special energy levels which are suitable for laser action. This
medium may be a gas, or a liquid, or a crystal or a semiconductor. The atomic systems
in this may have energy levels including a ground state (E1), an excited state (E3) and
a metastable state (E2). The resonant cavity is a pair of parallel mirrors to reflect the
radiation back into the lasing medium. Pumping is a process of exciting more number
of atoms in the ground state to higher energy states, which is required for attaining
the population inversion.
Fig. 2.5 Block diagram of laser system
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PHY 1001: ENGINEERING PHYSICS
2.1.2 RUBY LASER
In a ruby laser, the lasing medium is a ruby crystal taken in the form of a rod. Pure ruby is
Al2O3. For using it in laser, it is doped with Cr2O3. Cr3+ ions are the active centers, which
has approximately similar energy level structure as shown in Figure 2.7. The ends of the
rod are cut exactly parallel and polished. One end face is fully silvered and the other partially.
The rod along with the silvered end faces serve as a resonant cavity. A helical flash lamp
surrounds the rod. When a current pulse is made to pass through lamp, it flashes an intense
pulse of light. Cr3+ ions in their ground level E1 absorb these photons and are excited to level
E3. The atoms in the state E3 may come down to state E1 by spontaneous emission or
they may come down to meta-stable state (E2) by collision. The atoms in the state E2
come down to state E1 by stimulated emission. When population inversion takes place
at E2, a stray photon of right energy stimulates chain reaction, accumulates more photons,
all coherent. The reflecting ends turn the coherent beam back into active region so
that the regenerative process continues and part of the light beam comes out from the
partially silvered mirror as a laser pulse. The output is an intense pulse of coherent light of
wavelength 693.3nm.
Fig. 2.6 Construction of Ruby laser
Fig. 2.7 The basic three-level scheme for laser operation
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PHY 1001: ENGINEERING PHYSICS
2.1.3 He-Ne LASER
He-Ne Laser has a glass discharge tube filled with He (80%) and Ne (20%) at low
pressure. Helium gas is the “pumping” medium and Neon gas is the “lasing” medium
(Figure 2.8). The simplified energy level diagram (Figure 2.9) shows four levels: Eo, E1,
E2and E3. Electrons and ions in the electrical gas discharge occasionally collide with Heatoms, raising them to level E3 (a meta-stable state). During collisions between He- and
Ne- atoms, the excitation energy of He-atom is transferred to Ne-atom (level E2),
selectively populating E2 due to resonant energy transfer. Thus, population inversion occurs
between levels E2and E1. This population inversion is maintained because (i) the
metastability of level E3 ensures a ready supply of Ne-atoms in level E2 and (ii) level E1
decays rapidly to Eo. Stimulated emission from level E2 to level E1 predominates, and
red laser light of wavelength 632.8nm is generated. The mirror M1 is fully reflective and
the mirror M2 is partially reflective to allow the laser beam to come out. The Brewster’s
windows W ‘s are at polarizing angles to the mirrors, to reduce the reflection loss.
Fig. 2.8 Basic elements of He-Ne gas laser
Fig. 2.9 Atomic levels involved in the operation of He-Ne gas laser
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PHY 1001: ENGINEERING PHYSICS
2.1.4 APPLICATIONS OF LASER
Laser is used in various scientific, engineering and medical applications. It is used in
investigating the basic laws of interaction of atoms and molecules with electromagnetic wave
of high intensity. Laser is widely used in engineering applications like optical communication,
micro-welding and sealing etc. In medical field, laser is used in bloodless and painless surgery
especially in treating the retinal detachment. Also used as a tool in treating dental decay,
tooth extraction, cosmetic surgery.
2.2 OPTICAL FIBERS
Optical fibers are thin, flexible strands of transparent dielectric material such as glass or
plastic. They are basically used to guide infrared & visible light waves through curved paths.
Structure of optical fiber : It consists of a central cylindrical core made of pure glass or plastic
of refractive index n1 surrounded by a cladding made of similar material but of lower refractive
index n2 (n2< n1). But there is a material continuity from core to cladding. The cladding is
enclosed in a polyurethane jacket that protects the fiber from external damaging factors such
as abrasion, crushing & chemical reactions (Figure 2.10). Many such protected fibers are
grouped to form a cable. The diameter of the core varies between 10 to 200 m that of the
cladding varies between 50 to 250 m.
Fig. 2.10 Optical fiber – sectional view
Principle of working : Optical fibers work on the principle of total internal reflection of light.
When a beam of light traveling in an optically denser medium falls on interface separating
denser medium from relatively less dense medium, if the angle of incidence is greater than
particular angle called critical angle(C) for the pair of media, the light undergoes total internal
reflection(Figure 2.11). Total internal reflection is the most superior type of reflection.
Reflection is total in the sense that almost the entire energy is returned to the first medium
through reflection without any loss of energy. Due to this the optical fibers are able to sustain
light signal transmission over very long distances despite large number of reflections.
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 2.11 Total internal reflection
2.2.1 ANGLE OF ACCEPTANCE AND NUMERICAL APERTURE
Consider an optical fiber with refractive index of the material of the core n1and cladding n2
placed in a surrounding medium of refractive index n0. Let a ray AO of light enter the core of
the fiber at an angle 0. Let this ray after refraction through an angle 1at O strikes the
interface between the core and the cladding at the critical angle such that the refracted ray
grazes the interface.
Fig. 2.11 Wave propagation through fiber
Applying Snell’s law of refraction at O, we have,
sin 0 n1
sin 1 n0
sin 0
n1
sin 1
n0
[2.2]
Similarly, applying Snell’s law at B,
sin ( 90 1 )
n
n
2
or cos 1 2
sin 90
n1
n1
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PHY 1001: ENGINEERING PHYSICS
sin 1
1
2
2
2
1
n
n
[2.3]
Substituting Eq. (2.3) in Eq. (2.2) and simplifying,
sin 0
1
n0
n12 n22
[2.4]
0is called the acceptance angle or half angle of the acceptance cone. The acceptance angle
is generally about 5 for a single mode fiber & 10 to 15 for multi mode fibers. The term sin0is
called numerical aperture (NA), which indicates the light gathering power of the optical fiber.
It is evident that any ray that enters the fiber at an angle less than 0, strikes the core-cladding
interface at angle greater than the critical angle and undergoes total internal reflection each
time it strikes the interface. The optical fiber sustains the light signal transmission over a long
distance.
Fractional refractive index change (): It is the ratio of the difference in the refractive indices
(n1n2) between the core & the cladding to the refractive index n1 of the core.
( n1 n2 )
n1
[2.5]
Since n1> n2, is always positive.
Relation between NA & :
From Eq. 2.4, assuming n0 = 1, NA =
n12 n22 = ( n1 n2 )( n1 n2 )
From Eq. 2.5, (n1n2) = n1 and since n1 n2, we can approximate (n1+n2) 2 n1.
Therefore, NA =
( n1 ) ( 2n1 ) n1 2
[2.6]
The light accepting capacity of a fiber can be increased by making large. But there are
practical limitations to achieve this. Also a very large may cause signal distortion.
Skip distance (Ls): Skip distance is the distance between two successive reflections of the ray
of light which propagates through the optical fiber. Consider a portion of the optical fiber
through which a light signal is transmitted.
Fig. 2.12 Skip distance
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
From the Figure 2.12,
Ls d cot 1 d cos ec 2 1 1
n12
Ls d
1
n02 sin 2
( sin 1
n0
sin )
n1
[2.7]
2.2.2 TYPES OF OPTICAL FIBERS
Based on their refractive index profile and geometry, optical fibers may be broadly classified
into
a) single mode step index optical fibers
b) multi-mode step index optical fibers
c) multi-mode graded index optical fibers
Number of modes of transmission through an optical fiber: Depending on the launch angle
into the fiber, there can be hundreds of ray paths or modes by which energy can propagate
down the core. The ray paths corresponding to a given wave front is called a mode. An optical
fiber permits a discrete number of modes to propagate through it. Not all the rays that enter
the acceptance cone sustain propagation. Only those modes that satisfy the coherent phase
condition are successfully propagated. The rays belonging to the same propagating wavefront must remain in step despite the phase changes that occur on reflection and traversing
different optical paths.
The measure of number of modes that are supported (and thereby the fractional power that
can be transmitted) for propagation through an optical fiber is determined by a parameter
called V- number (V), given by
d
n0 n12 n22 where d is the diameter of the core and is the wavelength of the light
V2
propagated. If V >>1, then the number of successfully propagated modes are
.
2
V
Single mode step index optical fiber: A single mode optical fiber consists of a core having a
uniform refractive index n1 that abruptly decreases at the core-cladding interface to a lower
value n2,the refractive index of the cladding. The diameter of the core is narrow (5-10m)
generally a few times the wavelength of the light propagating through it. Only rays nearly
parallel to the fiber axis will travel through. It supports a single mode propagation because of
its narrow core.
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PHY 1001: ENGINEERING PHYSICS
Fig. 2.13 Single mode step-index fiber
Step-Index Multimode fiber: In this case also the refractive index profile is similar to step index
fiber i.e., fiber consists of a core having a uniform refractive index n 1 that abruptly decreases
at the core-cladding interface to a lower value n2 the refractive index of the cladding. But the
diameter of the core is large (50-200m). The comparatively large central core makes it
rugged and easily infused with light, as well as easily terminated and coupled. It also supports
a large number of modes for propagation.
Fig. 2.14 Multi mode step-index fiber
Graded-Index Multimode fiber (GRIN): It consists of a core whose refractive index decreases
gradually from its axis radially outward and becomes equal to the refractive index of the
cladding at the core-cladding interface. The refractive index of the cladding remains uniform.
Dimensions of the core and cladding are similar to that of step index multimode fibers. It
supports a large number of modes for propagation because of its large core diameter.
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 2.15 Multi mode graded index fiber
2.2.3 TYPES OF ATTENUATION
Attenuation is the loss of power of the light signal that occurs during its propagation through
the optical fiber. The main sources of attenuation are
1. absorption
2. scattering
3. other losses
Absorption: Absorption of light during propagation occurs due to the impurities present in the
fiber material and also due to the intrinsic nature of the material itself.
The impurities generally present are
a) Transition metals such as iron, chromium, cobalt, copper etc present in the starting
materials.
b) The hydroxy ions (OH-) that enter into the fiber material at the time of fabrication.
The photons absorbed by the impurities may be lost as heat or may be reemitted as light
energy of different wavelengths and different phase from the one that is propagated. Hence
it results in a loss. Intrinsic absorption occurs by the pure material itself even if the material is
free from impurities and in-homogeneities. Intrinsic absorption though quite less compared
to the loss due to the impurities, it cannot be eliminated.
Scattering: Glass is a heterogeneous mixture of oxides of silicon, phosphorus, germanium etc.
Structural in-homogeneities in the core index will set in the fiber material during solidification
of glass from its molten state. It will also result in a fluctuation of the molecular density. These
in-homogeneities act as scattering centers. Since their dimensions are smaller than the
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
wavelength of the light propagated through the fiber, the energy loss that occurs due to such
1
scattering resemble Rayleigh scattering that is proportional to 4 . The losses due to these
scattering cannot be eliminated by any process. There are other structural in-homogeneities
& defects that set in during fabrication of the fiber that contribute to the loss due to scattering.
Their sources are trapped gas bubbles, un-reacted starting materials etc. However these can
be reduced to a great extent by improved methods of manufacturing.
Other losses: a) Due to dimensional irregularities and imperfections in the fibers (that are
called microscopic bends) the light may not sustain total internal reflection. The energy will
escape from the core.
Fig. 2.16 Signal attenuation in optical fiber due to microscopic bending
b) Macroscopic bends occur during wrapping the fiber on a spool or negotiating a curve during
cable laying. Fibers can withstand bends of curvature up to about 10cm without significant
loss. For higher curvature (smaller radius of curvature) than this, the loss increases
exponentially.
Amplification is therefore needed in communication applications at regular intervals in order
to compensate for the losses that occur despite all precautions. An optical repeater is used to
boost the signal.
2.2.4 APPLICATIONS
Optical communication: In optical communication, fibers are used to carry information. An
optical communication system uses a transmitter, which encodes a message into an optical
signal, a channel (optical fiber), which carries the signal to its destination, and a receiver, which
reproduces the message from the received optical signal. Optical communication has several
advantages – can carry large data in digital form, interference and noise free.
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
Fig. 2.17 Simplified block diagram of optical communication system
Optical fibers are also used in sensors, flexible fiberscope (endoscope) and other industrial
applications.
EXERCISE
QUESTIONS
2.1
1
2
3
4
5
6
2.2
7
8
LASER
Mention the characteristics of a laser beam.
Explain the following terms with reference to lasers:
(a) spontaneous emission
(b) stimulated emission
(c) metastable state
(d) population inversion
(e) pumping
(f) active medium
(g) resonant cavity.
Explain the principle of a laser.
Explain construction and working of ruby laser with necessary
diagrams.
Explain construction and working of He-Ne laser with necessary
diagrams.
Mention any four applications of laser.
OPTICAL FIBERS
What is total internal reflection?
Explain the following terms with reference to optical fibres:
(a) acceptance cone half-angle,
(b) numerical aperture,
(c) modes of propagation in an optical fibre.
[2]
[1 EACH]
[2 ]
[5]
[5]
[2]
[1]
[1 EACH]
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PHY 1001: ENGINEERING PHYSICS
9
Obtain an expression for numerical aperture in terms of refractive
index of core and cladding and then arrive at the condition for ray
propagation in an optical fibre.
Define fractional refractive index change and obtain a relation
between & numerical aperture NA.
What is skip distance? With neat diagram, derive an expression for
it.
Briefly explain the different types of optical fibers with necessary
diagrams.
Explain the different types of attenuations possible in optical
fibers? Explain them.
Mention any four applications of optical fiber.
10
11
12
13
14
[5]
[3]
[3]
[5]
[4]
[2]
PROBLEMS
2.1
1.
LASER
A three level laser of the type shown in figure,
emits laser light at a wavelength 550 nm, near
the centre of the visible band. If the optical
mechanism is shut off, what will be the ratio
of the population of the upper level E2 to that
of the lower level E1 at 300 K ? At what
temperature for the condition of (a) would the
ratio of populations be half ?
Ans: (a) N2/N1= 1.77 x 10-38
(b) T=37800k
2.
A ruby laser emits light at a wavelength of 694.4 nm. If a laser pulse is
emitted for 12.0 ps and the energy release per pulse is 150 mJ, (a) what is
the length of the pulse, and (b) how many photons are there in each pulse
? Ans: (a) 3.6 x 10-3 m
(b) 5.25 x 1017
3.
It is entirely possible that techniques for modulating the frequency or
amplitude of a laser beam will be developed so that such a beam can serve
as a carrier for television signals, much as microwave beams do now. Assume
also that laser systems will be available whose wavelengths can be precisely
tuned to anywhere in the visible range (400 nm to 700 nm). If a television
channel occupies a bandwidth of 10 MHz, how many channels could be
accomodated with this laser technology ?
Comment on the intrinsic
superiority of visible light to microwaves as carriers of information.
Ans: The number of signals that can be sent in this range is = 3.21 x 107
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PHY 1001: ENGINEERING PHYSICS
4.
A He-Ne laser emits light at a wavelength of 632.8 nm and has an output
power of 2.3 mW. How many photons are emitted each minute by this
laser when operating ? Ans: n=4.4 x 1017
5.
An atom has two energy levels with a transition wavelength of 582 nm. At
300 K, 4.0 x 1020 atoms are in the lower state. (a) How many occupy the
upper state under conditions of thermal equilibrium ? (b) Suppose, instead,
that 7.0 x 1020 atoms are pumped into upper state, with 4.0 x 1020 atoms in
the lower state. How much energy could be released in a single laser pulse
? Ans: (a) N2 = 6.6 x 10 -16 i i.e. N2 = 0 (b) 240J
2.2
OPTICAL FIBER
6.
A step index optical fibre 63.5 m in core-diameter has a core of refractive index
1.53 and a cladding of index 1.39. Determine (a) the numerical aperture for the
fibre, (b) the critical angle for core-cladding interface, (c) the acceptance cone
half-angle (the maximum entrance angle) (d) the number of reflections in 1.0 m
length of the fibre for a ray at the maximum entrance angle, (e) the number of
reflections in 1.0 m length of the fibre for a ray at half the maximum entrance
angle.
(a) Ans: (a) NA=0.64 (b) ӨC =65.3° (c)ӨO = 39.75°
(d) Ls=138µm Number of reflections per unit length=1/Ls=7250
(e)Ls =278.63µm Number of reflections per unit length=1/Ls=3588
7.
A glass optical fibre of refractive index 1.450 is to be clad with another to ensure
total internal reflection that will contain light traveling within 5 of the fibre-axis.
What maximum index of refraction is allowed for the cladding?
Ans: The maximum index of refraction allowed for cladding should be less than
1.444
8.
The numerical aperture of an optical fibre is 0.2 when surrounded by air.
Determine the refractive index of its core. The refractive index of the cladding is
1.59. Also find the acceptance cone half-angle when the fibre is in water.
Refractive index of water is 1.33.
Ans: R.I of core =1.6 and Acceptance angle Ө0=8.650
9.
The angle of acceptance of an optical fibre is 30 when kept in air. Find the angle
of acceptance when it is in a medium of refractive index 1.33. Ans: Angle of
acceptance=220
10.
Calculate the V-number for a fiber of core diameter 40µm and with refractive
indices of 1.55 and 1.50 respectively for core and cladding when the wavelength of
the propagating wave is 1400nm. Ans: V=35
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CHAPTER 3
QUANTUM PHYSICS
OBJECTIVES:
To learn certain experimental results that can be understood only by particle
theory of electromagnetic waves.
To learn the particle properties of waves and the wave properties of the particles.
To understand the uncertainty principle.
3.1 BLACKBODY RADIATION AND PLANCK’S HYPOTHESIS
A black body is an ideal system that absorbs all radiation incident on it and also it is a perfect
emitter, emitting all radiations which is the function of temperature. A small hole cut into a
cavity is the most popular and realistic example. None of the incident radiation escapes but
absorbed in the walls of the cavity. This causes a heating of the cavity walls. The oscillators in
the cavity walls vibrate and re-radiate at wavelengths corresponding to the temperature of
the cavity, thereby producing standing waves. Some of the energy from these standing waves
can leave through the opening. The electromagnetic radiation emitted by the black body is
called black-body radiation.
Fig. 3.1 A physical model of a blackbody
•
•
The black body is an ideal absorber of incident radiation.
A black-body reaches thermal equilibrium with the surroundings when the incident
radiation power is balanced by the power re-radiated.
• The emitted "thermal" radiation from a black body characterizes the equilibrium
temperature of the black-body.
• Emitted radiation from a blackbody does not depend on the material of which the walls
are made.
Basic laws of radiation
(1) All objects emit radiant energy.
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(2) Hotter objects emit more energy (per unit area) than colder objects. The total power of the
emitted radiation is proportional to the fourth power of temperature. This is called Stefan’s
Law and is given by P = A e T4
where P is the power radiated from the surface of the object (W), T is equilibrium surface
temperature (K),σ is Stefan-Boltzmann constant (= 5.670 x 10−8 W/m2K4), A is surface area of
the object (m2) and e is emissivity of the surface (e=1 for a perfect blackbody).
(3) The peak of the wavelength distribution shifts to shorter wavelengths as the black body
temperature increases. This is Wien’s Displacement Law and is given by
λm T = constant = 2.898 × 10−3 m.K , or λm T−1
where λm is the wavelength corresponding to peak intensity and T is equilibrium temperature
of the blackbody.
Fig. 3.2 Intensity of blackbody radiation versus wavelength at two temperatures
(4) Rayleigh-Jeans Law: The intensity or power per unit area I (,T)d, emitted in the
wavelength interval to +d from a black body is given by,
2 c kB T
I( ,T )
4
kB is Boltzmann constant, c is speed of light in vacuum, T is equilibrium blackbody
temperature. It agrees with experimental measurements only for long wavelengths.It predicts
an energy output that diverges towards infinity as wavelengths become smaller and is known
as the ultraviolet catastrophe.
(5) Planck‘s Law: Max Planck developed a theory of blackbody radiation to explain the
experimental observations, that lead to an equation for I (,T). To derive the law, Planck made
two assumptions concerning the nature of the oscillators in the cavity walls:
(i) The energy of an oscillator is quantized hence it can have only certain discrete values:
En = n h f
where n is a positive integer called a quantum number, f is the frequency of cavity oscillators,
and h is a constant called Planck’s constant. Each discrete energy value corresponds to a
different quantum state, represented by the quantum number n.
(ii) The oscillators emit or absorb energy only when making a transition from one quantum
state to another. Difference in energy will be integral multiples of hf.
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PHY 1001: ENGINEERING PHYSICS
Fig. 3.3 Comparison of experimental results and the curve predicted by the Rayleigh–Jeans
law for the distribution of blackbody radiation
Fig. 3.4 Allowed energy levels for an oscillator with frequency f
In Planck’s model, the average energy associated with a given wavelength is the product of
the energy of transition and a factor related to the probability of the transition occurring. As
the energy levels move farther apart at shorter wavelengths, the probability of excitation
decreases, as does the probability of a transition from the excited state. This “turns the curve
over” and brings down to zero at shorter wavelengths overcoming the ultra-violet catastrophe
problem. Using this approach, Planck could derive an equation for the distribution of energy
from a black as,
I( ,T )
2 h c 2
1
5
e
hc
λkB T
1
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PHY 1001: ENGINEERING PHYSICS
where I (,T) d is the intensity or power per unit area emitted in the wavelength interval d
from a black body, h - Planck’s constant, kB - Boltzmann's constant, c - speed of light in
vacuum and T - equilibrium temperature of blackbody .
The Planck‘s Law gives a distribution that peaks at a certain wavelength, the peak shifts to
shorter wavelengths for higher temperatures, and the area under the curve grows rapidly with
increasing temperature.This law is in agreement with the experimental data.
The results of Planck's law:
The denominator [exp(hc/λkT)] tends to infinity faster than the numerator (λ–5), thus
resolving the ultraviolet catastrophe and hence arriving at experimental observation:
I (λ, T) 0 as λ 0.
hc
For very large λ,
exp( hckT ) 1 k T I( ,T ) 2 c 4 k T
i.e. I (λ, T) 0 as λ.
From a fit between Planck's law and experimental data, Planck’s constant was derived to be h
= 6.626 × 10–34 J-s.
3.2 PHOTOELECTRIC EFFECT
Ejection of electrons from the surface of certain metals when it is irradiated by an
electromagnetic radiation of suitable frequency is known as photoelectric effect. Figure 3.5
shows an experimental set up for studying Photoelectric Effect. When the tube is kept in dark,
the ammeter reads zero indicating no current. However when plate E is illuminated by light
having an appropriate frequency, a current is detected by the ammeter. By applying a proper
negative potential to the plate C, current in the ammeter can be reduced to zero. In such a
case the potential difference is called stopping potential Vs.
Fig. 3.5Apparatus for studying Photoelectric Effect (T – Evacuated glass/ quartz tube, E –
Emitter Plate / Photosensitive material / Cathode, C – Collector Plate / Anode, V – Voltmeter,
A - Ammeter)
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Experimental Observations:
Fig. 3.6 Photoelectric current versus applied potential difference for two light intensities
1. No electrons are emitted if the incident light frequency falls below a cutoff frequency.
2. Photo current increases with the light intensity but is independent of its frequency.
3. Kinetic energy of the most energetic photoelectrons is independent of light intensity
but depends on its frequency.
4. For small voltages, photo current increases with applied potential but for large values
of V, current gets saturated.
5. Electrons are emitted from the surface of the emitter almost instantaneously
Classical Predictions
1. If light is really a wave, it was thought that if one shines a light of any fixed
frequency/wavelength, at sufficient intensity electrons should absorb energy continuously
from the em waves and electrons should be ejected.
2. As the intensity of light is increased (made it brighter and hence classically, a more
energetic wave), kinetic energy of the emitted electrons should increase i.e., kinetic energy
must depend on intensity and not on frequency.
3. Measurable/ larger time interval between incidence of light and ejection of
photoelectrons.
A successful explanation of the observed facts was given by Einstein:
Einstein’s Interpretation of electromagnetic radiation
1. Electromagnetic waves carry discrete energy packets (light quanta called photons now).
2. The energy E, per packet depends on frequency f as E = hf.
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PHY 1001: ENGINEERING PHYSICS
3. More intense light corresponds to more photons, not high energy photons.
4. Each photon of energy E moves in vacuum at the speed of light: c = 3 x 10 8 m/s and each
photon carries a momentum p = E/c.
Einstein’s theory of photoelectric effect
A photon of the incident light gives all its energy hf to a single electron (absorption of energy
by the electrons is not a continuous process as envisioned in the wave model). This energy is
used for two purposes : i) work is to be done to remove the electron out of the material, ( and
is known as work function of the metal) ii) to impart kinetic energy to the emitted electron .
Thus according to law of conservation of energy hf = + Kmax , is called the work function
of the metal. It is the minimum energy with which an electron is bound in the metal. This
equation is known as Einstein’s photoelectric equation.
All the observed features of photoelectric effect could be explained by Einstein’s photoelectric
equation:
1. Equation shows that Kmax depends only on frequency of the incident light.
2. Almost instantaneous emission of photoelectrons due to one -to –one interaction between
photons and electrons.
3. Ejection of electrons depends on light frequency since photons should have energy greater
than the work function in order to eject an electron.
4. The cutoff frequency fc is related to by fc = /h. If the incident frequency f is less
than fc , there is no emission of photoelectrons.
The graph of kinetic energy of the most energetic photoelectron Kmax vs frequency f is a
straight line( since Kmax = hf - ). Slope of the graph gives Planck constant h, Y- intercept gives
the work function and X-intercept gives the critical frequency fc.
Fig. 3.7 A representative plot of Kmax versus frequency of incident light for three different
metals
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3.3 COMPTON EFFECT
When X-rays are scattered by free/nearly free electrons, they suffer a change in their
wavelength which depends on the scattering angle. This scattering phenomenon is known as
Compton Effect.
Classical Predictions: Oscillating electromagnetic waves(classically,X-rays are em waves)
incident on electrons should have two effects: i) oscillating electromagnetic field causes
oscillations in electrons. Each electron first absorbs radiation as a moving particle and then reradiates in all directions as a moving particle and thereby exhibiting two Doppler shifts in the
frequency of radiation. ii) radiation pressure should cause the electrons to accelerate in the
direction of propagation of the waves.
Because different electrons will move at different speeds after the interaction, depending on
the amount of energy absorbed from electromagnetic waves, the scattered waves at a given
angle will have all frequencies(Doppler- shifted values).
Compton’s experiment and observation: Compton measured the intensity of scattered X-rays
from a solid target (graphite) as a function of wavelength for different angles. The experimental
setup is shown in Figure 3.8. Contrary to the classical prediction, only one frequency for
scattered radiation was seen at a given angle. This is shown in the Figure 3.9.
The graphs for three nonzero angles show two peaks, one at o and the other at’ >o . The
shifted peak at ’ is caused by the scattering of X-rays from free electrons. Shift in wavelength
was predicted by Compton to depend on scattering angle as
λ′ − λ =
h
(1−cos θ)
mc
,
where m is the mass of the electron, c velocity of light, h Planck’s constant.
ℎ
This is known as Compton shift equation, and the factor
is called the Compton
𝑚𝑐
ℎ
wavelength and 𝑚 𝑐 = 2.43 pm.
Fig. 3.8 Schematic diagram of Compton’s apparatus. The wavelength is measured with a
rotating crystal spectrometer for various scattering angles. (In the figure 90° scattering is
shown).
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PHY 1001: ENGINEERING PHYSICS
Fig. 3.9 Scattered x-ray intensity versus wavelength for Compton scattering at = 0°, 45°,
90°, and 135° showing single frequency at a given angle
Derivation of the Compton shift equation: Compton could explain the experimental result by
treating the X-rays not as waves but rather as point like particles (photons) having energy E =
hfo = hc/o, momentum p = hf/c = h/and zero rest energy. Photons collide elastically with
free electrons initially at rest and moving relativistically after collision.
Fig. 3.10 Quantum model for X-ray scattering from an electron
Let o , po = h/o and Eo = hc/o be the wavelength, momentum and energy of the incident
photon respectively.’, p’ = h/’ and E’ = hc/’ be the corresponding quantities for the
scattered photon.
We know that, for the electron, the total relativistic energy 𝐸 = √p2 c 2 + m2 c 4
Kinetic energy K = E − m c2
1
And momentum p = mv.
where
2
1 vc 2
v and m are the speed and mass of the electron respectively.
In the scattering process, the total energy and total linear momentum of the system must
be conserved.
For conservation of energy we must have,Eo = E’ + K
ie,
Eo = E’ + (E − m c2)
Or Eo − E’ + m c2=𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 + 𝑚2 𝑐 4 = 𝑝2 𝑐 2 + 𝑚2 𝑐 4
Squaring both the sides,
For conservation of momentum,
x-component: 𝑝𝑜 = 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝 𝑐𝑜𝑠 𝜙
y-component: 0 = 𝑝′ 𝑠𝑖𝑛 𝜃 − 𝑝 𝑠𝑖𝑛 𝜙
Rewriting these two equations
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PHY 1001: ENGINEERING PHYSICS
′
𝑝𝑜 − 𝑝 𝑐𝑜𝑠 𝜃 = 𝑝 𝑐𝑜𝑠 𝜙
𝑝′ 𝑠𝑖𝑛 𝜃 = 𝑝 𝑠𝑖𝑛 𝜙
Squaring both the sides and adding,
𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 = 𝑝2
Substituting this 𝑝2 in the equation :
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = 𝑝2 𝑐 2 , one gets
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = (𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 )𝑐 2
Substituting photon energies and photon momenta one gets
(
ℎ𝑐
𝜆𝑜
−
ℎ𝑐
𝜆
2
) + 2(
′
ℎ𝑐
𝜆𝑜
−
ℎ𝑐
ℎ𝑐
𝜆
𝜆𝑜
) 𝑚𝑐 2 = (
′
2
) − 2(
ℎ𝑐
𝜆𝑜
ℎ𝑐
ℎ𝑐
2
) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( ′ )
𝜆
Simplifying one gets
2
2
( ℎ𝑐
) − 2 ( ℎ𝑐
) (ℎ𝑐
) + (ℎ𝑐
) + 2 ℎ𝑐 (
𝜆
𝜆
𝜆′
𝜆′
𝑜
i.e.,
OR,
𝑜
−
ℎ𝑐
+ ( 𝜆1 −
𝜆𝑜 𝜆′
𝑜
1
𝜆′
) 𝑚𝑐 2 = −
′
(𝜆𝜆−𝜆𝜆𝑜′ ) 𝑚𝑐 2 =
𝑜
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆𝑜
−
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆′
2
2
) 𝑚𝑐 2 = ( ℎ𝑐
) − 2 ( ℎ𝑐
) (ℎ𝑐
) 𝑐𝑜𝑠 𝜃 + (ℎ𝑐
)
𝜆′
𝜆
𝜆
𝜆′
𝑜
𝑜
𝑐𝑜𝑠 𝜃
(1 − 𝑐𝑜𝑠 𝜃)
Compton shift:
𝝀′ − 𝝀𝒐 =
𝒉
𝒎𝒄
(𝟏 − 𝒄𝒐𝒔 𝜽)
3.4PHOTONS AND ELECTROMAGNETIC WAVES [DUAL NATURE OF LIGHT]
•
•
•
Light exhibits diffraction and interference phenomena that are only explicable in terms of
wave properties.
Photoelectric effect and Compton Effect can only be explained taking light as photons /
particle.
This means true nature of light is not describable in terms of any single picture, instead
both wave and particle nature have to be considered. In short, the particle model and the
wave model of light complement each other.
3.5 de BROGLIE HYPOTHESIS- WAVE PROPERTIES OF PARTICLES
We have seen that light comes in discrete units (photons) with particle properties (energy E
and momentum p) that are related to the wave-like properties of frequency and wavelength.
Louis de Broglie postulated that because photons have both wave and particle characteristics,
perhaps all forms of matter have wave-like properties, with the wavelength λ related to
momentum p in the same way as for light.
deBroglie wavelength: 𝜆 =
ℎ
𝑝
=
ℎ
𝑚𝑣
where h is Planck’s constant and p momentum , m mass
and v is speed of the particle. The electron accelerated through a potential difference of V,
1
has a non-relativistic kinetic energy
𝑚 𝑣 2 = 𝑒 ∆𝑉where e is electron charge.
2
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Hence, the momentum (p) of an electron accelerated through a potential difference of V is
𝑝 = 𝑚 𝑣 = √2 𝑚 𝑒 ∆𝑉
ℎ
𝜆 =
𝑝
𝐸
Frequency of the matter wave associated with the particle isℎ , where E is total relativistic
energy of the particle
Davisson-Germer experiment and G P Thomson’s electron diffraction experiment confirmed
de Broglie relationship p = h /. Subsequently it was found that atomic beams, and beams of
neutrons, also exhibit diffraction when reflected from regular crystals. Thus de Broglie's
formula seems to apply to any kind of matter. Now the dual nature of matter and radiation is
an accepted fact and it is stated in the principle of complementarity, which states that wave
and particle properties of either matter or radiation are complement to each other.
3.6 THE QUANTUM PARTICLE
Quantum particle is a model by which particles having dual nature are represented. We must
choose one appropriate behavior for the quantum particle (particle or wave) in order to
understand a particular behavior.
To represent a quantum wave, we have to combine the essential features of both an ideal
particle and an ideal wave. An essential feature of a particle is that it is localized in space. But
an ideal wave is infinitely long (non-localized) as shown in Figure 3.11.
Fig. 3.11 Section of an ideal wave of single frequency
Now to build a localized entity from an infinitely long wave, waves of same amplitude, but
slightly different frequencies are superposed (Figure 3.12).
Fig. 3.12 Superposition of two waves Wave1 and Wave2
If we add up large number of waves in a similar way, the small localized region of space where
constructive interference takes place is called a wavepacket, which represents a quantum
particle (Figure 3.13).
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PHY 1001: ENGINEERING PHYSICS
Fig. 3.13 Wave packet
Mathematical representation of a wave packet: Superposition of two waves of equal
amplitude, but with slightly different frequencies, f1 and f2, traveling in the same direction are
considered. The waves are written as
𝑦1 = 𝐴 𝑐𝑜𝑠(𝑘1 𝑥 − 𝜔1 𝑡) and 𝑦2 = 𝐴 𝑐𝑜𝑠(𝑘2 𝑥 − 𝜔2 𝑡)
where 𝑘 = 2𝜋/𝜆 ,
𝜔 = 2𝜋𝑓
The resultant wave y = y1 + y2
𝛥𝑘
𝑦 = 2𝐴 [𝑐𝑜𝑠 ( 2 𝑥 −
where k = k1 – k2 and = 1 – 2.
𝛥𝜔
𝑘 +𝑘
𝑡) 𝑐𝑜𝑠 ( 1 2 2 𝑥
2
−
𝜔1 +𝜔2
𝑡)]
2
Fig. 3.14 Beat pattern due to superposition of wave trains y1 and y2
The resulting wave oscillates with the average frequency, and its amplitude envelope (in
square brackets, shown by the blue dotted curve in Figure 3.14) varies according to the
difference frequency. A realistic wave (one of finite extent in space) is characterized by two
different speeds. The phase speed, the speed with which wave crest of individual wave moves,
is given by
𝜔
𝑣𝑝 = 𝑓 𝜆
or
𝑣𝑝 = 𝑘
The envelope of group of waves can travel through space with a different speed than the
individual waves. This speed is called the group speed or the speed of the wave packet which
is given by
𝑣𝑔 =
(𝛥𝜔
)
2
(𝛥𝑘
)
2
=
𝛥𝜔
𝛥𝑘
For a superposition of large number of waves to form a wave packet, this ratio is
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
In general these two speeds are not the same.
Relation between group speed (vg) and phase speed (vp):
𝜔
𝑣𝑃 = 𝑘 = 𝑓 𝜆
𝜔 = 𝑘 𝑣𝑃
But
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
=
𝑑(𝑘𝑣𝑃 )
𝑑𝑘
= 𝑘
𝑑𝑣𝑃
𝑑𝑘
+ 𝑣𝑃
Substituting for k in terms of λ, we get
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PHY 1001: ENGINEERING PHYSICS
𝑣𝑔 = 𝑣𝑃 − 𝜆 (
𝑑𝑣𝑃
𝑑𝜆
)
Relation between group speed (vg) and particle speed (u):
𝐸
𝜔 = 2𝜋𝑓 = 2𝜋
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
and
ℎ
2𝜋
𝑑𝐸
ℎ
2𝜋
𝑑𝑝
ℎ
=
=
2𝜋
𝑘 =
𝜆
2𝜋
=
ℎ⁄𝑝
=
2𝜋𝑝
ℎ
𝑑𝐸
𝑑𝑝
For a classical particle moving with speed u, the kinetic energy E is given by
𝐸 =
1
2
𝑚 𝑢2 =
𝑝2
2𝑚
and
𝑑𝐸 =
2 𝑝 𝑑𝑝
2𝑚
or
𝑑𝐸
𝑑𝑝
=
𝑝
𝑚
= 𝑢
𝑑𝜔
𝑑𝐸
=
= 𝑢
𝑑𝑘
𝑑𝑝
i.e., we should identify the group speed with the particle speed, speed with which the energy
moves. To represent a realistic wave packet, confined to a finite region in space, we need the
superposition of large number of harmonic waves with a range of k-values.
𝑣𝑔 =
3.7 DOUBLE–SLIT EXPERIMENT REVISITED
One way to confirm our ideas about the electron’s wave–particle duality is through an
experiment in which electrons are fired at a double slit. Consider a parallel beam of monoenergetic electrons incident on a double slit as in Figure 3.15. Let’s assume the slit widths are
small compared with the electron wavelength so that diffraction effects are negligible. An
electron detector is positioned far from the slits at a distance much greater than d, the
separation distance of the slits. The detector is movable along the y direction in the drawing
and so can detect electrons diffracted at different values of .
If the detector collects electrons for a long enough time, we find a typical wave interference
pattern for the counts per minute, or probability of arrival of electrons. Such an interference
pattern would not be expected if the electrons behaved as classical particles, giving clear
evidence that electrons are interfering, a distinct wave-like behavior.
In the interference pattern the minimum occurs when 𝑑 𝑠𝑖𝑛 𝜃 = 𝜆/2
The electron wavelength is given by
𝜆 = ℎ/𝑝
For small angle ,
𝑠𝑖𝑛 𝜃 ≅
𝜃
=
ℎ
2𝑝𝑑
Fig. 3.15 (a) Schematic of eelectron beam interference experiment, (b) Photograph of a
double-slitinterference pattern produced by electrons
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PHY 1001: ENGINEERING PHYSICS
This experiment proves the dual nature of electrons : The electrons are detected as particles
at a localized spot at some instant of time, but the probability of arrival at that spot is
determined by finding the intensity of two interfering waves. If slit 2 is blocked half the time,
keeping slit 1 open, and slit 1 blocked for remaining half the time, keeping 2 open, the
accumulated pattern of counts/ min is shown by blue curve in Figure 3.16. That is interference
pattern is lost and the result is simply the sum of the individual results.
Fig. 3.16 Results of the two-slit electron diffraction experiment with each slit closed half the
time (blue) the result with both slits open (interference pattern is shown in brown)
The observed interference pattern when both the slits are open, suggests that each particle
goes through both slits at the same time. We are forced to conclude that an electron interacts
with both the slits simultaneously shedding its localized behaviour. If we try to find out which
slit the particle goes through, the interference pattern vanishes. Means, if we know which path
the particle takes, we lose the fringes. We can only say that the electron passes through both
the slits.
3.7 UNCERTAINTY PRINCIPLE
It is fundamentally impossible to make simultaneous measurements of a particle’s position
and momentum with infinite accuracy. This is known as Heisenberg uncertainty principle.
The uncertainties arise from the quantum structure of matter.
For a particle represented by a single wavelength wave existing throughout space, is
precisely known, and according to de Broglie hypothesis, its p is also known accurately. But
the position of the particle in this case becomes completely uncertain.
This means = 0, p =0; but x =
In contrast, if a particle whose momentum is uncertain (combination of waves / a range of
wavelengths are taken to form a wave packet), so that x is small, but is large. If x is
made zero, and thereby p will become .
In short
( x ) ( px) ≥ h / 4
where x is uncertainty in the measurement of position x of the particle and px is uncertainty
in the measurement of momentum px of the particle.
One more relation expressing uncertainty principle is related to energy and time which is given
by
( E ) ( t ) ≥ h / 4
where E is uncertainty in the measurement of energy E of the system when the
measurement is done over the time interval t.
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EXERCISE
QUESTIONS
3.1 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS
1 Explain (a) Stefan’s law (b) Wien’s displacement law (c) Rayleigh-Jeans
law (d) Planck’s hypothesis (e) Black body (f) Ultraviolet catastrophe
2 Sketch schematically the graph of wavelength vs intensity of radiation
from a blackbody.
3 Explain Planck’s radiation law.
4 Write the assumptions made in Planck’s hypothesis of blackbody
radiation.
3.2. THE PHOTOELECTRIC EFFECT
5 (a) Explain photoelectric effect.
(b) What is work function of metal?
6 What are the observations in the experiment on photoelectric effect?
7 What are the classical predictions about the photoelectric effect?
8 Explain Einstein’s photoelectric equation.
9 Which are the features of photoelectric effect-experiment explained by
Einstein’s photoelectric equation?
10 Sketch schematically the following graphs with reference to the
photoelectric effect: (a) photoelectric current vs applied voltage (b)
kinetic energy of most-energetic electron vs frequency of incident light.
3.3 THE COMPTON EFFECT
11 Explain Compton effect.
12 Explain the experiment on Compton effect.
13 Derive the Compton shift equation.
3.4 PHOTONS AND ELECTROMAGNETIC WAVES
14 Explain the wave properties of the particles.
3.5 THE QUANTUM PARTICLE
15 Explain a wave packet and represent it schematically.
16 Explain (a) group speed (b) phase speed
17 Show that the group speed of a wave packet is equal to the particle speed.
3.6 THE DOUBLE–SLIT EXPERIMENT REVISITED
18 (a) Name any two phenomena which confirm the particle nature of light.
(b) Name any two phenomena which confirm the wave nature of light.
3.7 THE UNCERTAINTY PRINCIPLE
19 Explain Heisenberg uncertainty principle.
20 Write the equations for uncertainty in (a) position and momentum (b)
energy and time.
[1 EACH]
[1]
[2]
[2]
[1 EACH]
[5]
[3]
[2]
[2]
[1EACH]
[2]
[5]
[5]
[2]
[2]
[1+1]
[3]
[1]
[1]
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21 Mention two phenomena which can be well explained by the uncertainty
relation.
[1]
PROBLEMS
3.1 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS
1 THERMAL RADIATION FROM DIFFERENT OBJECTS
Find the peak wavelength of the blackbody radiation emitted by each of the following.
A. The human body when the skin temperature is 35°C
B. The tungsten filament of a light bulb, which operates at 2000 K
C. The Sun, which has a surface temperature of about 5800 K.
Ans: 9.4 μm, 1.4 μm, 0.50 μm
2 THE QUANTIZED OSCILLATOR
A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m. The
spring is stretched 0.40 m from its equilibrium position and released.
A. Find the total energy of the system and the frequency of oscillation according to
classical calculations.
B. Assuming that the energy is quantized, find the quantum number n for the system
oscillating with this amplitude.
C. Suppose the oscillator makes a transition from the n = 5.4 x 1033 state to the state
corresponding to n = 5.4 x 1033 – 1. By how much does the energy of the
oscillator change in this one-quantum change.
Ans: 2.0 J, 0.56 Hz, 5.4 x 1033, 3.7 x 10–34 J
3 The human eye is most sensitive to 560 nm light. What is the temperature of a black body
that would radiate most intensely at this wavelength?
Ans: 5180 K
4 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into an oven.
Find the number of photons per second escaping the hole and having wavelengths
between 500 nm and 501 nm.
Ans: 1.30 x 1015/s
5 The radius of our Sun is 6.96 x 108 m, and its total power output s 3.77 x 1026 W. (a)
Assuming that the Sun’s surface emits as a black body, calculate its surface temperature.
(b) Using the result, find max for the Sun.
Ans: 5750 K, 504 nm
6 Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz, (b)
3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons
and state the classification of each on the electromagnetic spectrum.
Ans: 2.57 eV, 1.28 x 10–5 eV, 1.91 x 10–7 eV, 484 nm, 9.68 cm, 6.52 m
7 An FM radio transmitter has a power output of 150 kW and operates at a frequency of
99.7 MHz. How many photons per second does the transmitter emit?
Ans: 2.27 x 1030 photons/s
3.2 THE PHOTOELECTRIC EFFECT
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PHY 1001: ENGINEERING PHYSICS
8 THE PHOTOELECTRIC EFFECT FOR SODIUM: A sodium surface is illuminated with light
having a wavelength of 300 nm. The work function for sodium metal is 2.46 eV. Find
A. The maximum kinetic energy of the ejected photoelectrons and
B. The cutoff wavelength for sodium.
Ans: 1.67 eV, 504 nm
9 Molybdenum has a work function of 4.2eV. (a) Find the cut off wavelength and cut off
frequency for the photoelectric effect. (b) What is the stopping potential if the incident
light has wavelength of 180 nm?
Ans: 296 nm, 1.01 x 1015 Hz, 2.71 V
10 Electrons are ejected from a metallic surface with speeds up to 4.60 x 105 m/s when light
with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b)
What is the cut-off frequency for this surface?
Ans: 1.38 eV, 3.34 x 1014 Hz
11 The stopping potential for photoelectrons released from a metal is 1.48 V larger
compared to that in another metal. If the threshold frequency for the first metal is 40.0
% smaller than for the second metal, determine the work function for each metal.
Ans: 3.70 eV, 2.22 eV
12 Two light sources are used in a photoelectric experiment to determine the work function
for a metal surface. When green light from a mercury lamp ( = 546.1 nm) is used, a
stopping potential of 0.376 V reduces the photocurrent to zero. (a) Based on this what is
the work function of this metal? (b) What stopping potential would be observed when
using the yellow light from a helium discharge tube ( = 587.5 nm)?
Ans: 1.90 eV, 0.215 V
3.3 THE COMPTON EFFECT
13 COMPTON SCATTERING AT 45°: X-rays of wavelength o = 0.20 nm are scattered from
a block of material. The scattered X-rays are observed at an angle of 45° to the incident
beam. Calculate their wavelength.
What if we move the detector so that scattered X-rays are detected at an angle larger
than 45°? Does the wavelength of the scattered X-rays increase or decrease as the angle
increase?
Ans: 0.200710 nm, INCREASES
14 Calculate the energy and momentum of a photon of wavelength 700 nm.
Ans: 1.78 eV, 9.47 x 10–28kg.m/s
15 A 0.00160 nm photon scatters from a free electron. For what photon scattering angle
does the recoiling electron have kinetic energy equal to the energy of the scattered
photon?
Ans: 70°
16 A 0.880 MeV photon is scattered by a free electron initially at rest such that the scattering
angle of the scattered electron is equal to that of the scattered photon ( = ).
(a)
Determine the angles & . (b) Determine the energy and momentum of the scattered
electron and photon.
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10–22
3.4
17
18
19
Ans: 43°, 43°, 0.602 MeV, 3.21 x
kg.m/s, 0.278 MeV, 3.21 x 10–22 kg.m/s
PHOTONS AND ELECTROMAGNETIC WAVES
THE WAVELENGTH OF AN ELECTRON: Calculate the de- Broglie wavelength for an
electron moving at 1.0 x 107 m/s.
Ans: 7.28 x 10–11 m
THE WAVELENGTH OF A ROCK: A rock of mass 50 g is thrown with a speed of 40 m/s.
What is its de Broglie wavelength?
Ans: 3.3 x 10–34 m
AN ACCELERATED CHARGED PARTICLE: A particle of charge q and mass m has been
accelerated from rest to a nonrelativistic speed through a potential difference of V. Find
an expression for its de Broglie wavelength.
Ans: λ =
h
√2 m q Δv
20 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find the
wavelength of a photon having the same energy.
Ans: 7.09 x 10–10 m, 4.14 x 10–7 m
21 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a nickel
lattice. If the first maximum in the diffraction pattern was observed at = 50.0°, what was
the lattice spacing a between the vertical rows of atoms in the figure?
Ans: 2.18 x 10–10 m
3.5 THE QUANTUM PARTICLE
22 Consider a freely moving quantum particle with mass m and speed u. Its energy is E= K=
mu2/2. Determine the phase speed of the quantum wave representing the particle and
show that it is different from the speed at which the particle transports mass and energy.
Ans: vGROUP = u ≠ vPHASE
3.6 THE DOUBLE–SLIT EXPERIMENT REVISITED
23 Electrons are incident on a pair of narrow slits 0.060 m apart. The ‘bright bands’ in the
interference pattern are separated by 0.40 mm on a ‘screen’ 20.0 cm from the slits.
Determine the potential difference through which the electrons were accelerated to give
this pattern.
Ans: 105 V
3.7 THE UNCERTAINTY PRINCIPLE
24 LOCATING AN ELECTRON: The speed of an electron is measured to be 5.00 x 103 m/s to
an accuracy of 0.0030%. Find the minimum uncertainty in determining the position of
this electron.
Ans: 0.383 mm
25 THE LINE WIDTH OF ATOMIC EMISSIONS: The lifetime of an excited atom is given as 1.0
x 10-8 s. Using the uncertainty principle, compute the line width f produced by this finite
lifetime?
Ans: 8.0 x 106 Hz
26 Use the uncertainty principle to show that if an electron were confined inside an atomic
nucleus of diameter 2 x 10–15 m, it would have to be moving relativistically, while a proton
confined to the same nucleus can be moving nonrelativistically.
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Ans: vELECTRON 0.99996 c, vPROTON 1.8 x 107 m/s
27 Find the minimum kinetic energy of a proton confined within a nucleus having a diameter
of 1.0 x 10–15 m.
Ans: 5.2 MeV
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CHAPTER 4
QUANTUM MECHANICS
OBJECTIVES:
To learn the application of Schrödinger equation to a bound particle and to learn
the quantized nature of the bound particle, its expectation values and physical
significance.
To understand the tunneling behavior of a particle incident on a potential barrier.
To learn the quantum model of H-atom and its wave functions.
4.1 AN INTERPRETATION OF QUANTUMMECHANICS
Experimental evidences proved that both matter and electromagnetic radiation exhibit wave
and particle nature depending on the phenomenon being observed. Making a conceptual
connection between particles and waves, for an electromagnetic radiation of amplitude E, the
probability per unit volume of finding a photon in a given region of space at an instant of time
as
PROBABILITY
𝑉
∝ 𝐸2
Fig. 4.1 Wave packet
Taking the analogy between electromagnetic radiation and matter-the probability per unit
volume of finding the particle is proportional to the square of the amplitude of a wave
representing the particle, even if the amplitude of the de Broglie wave associated with a
particle is generally not a measureable quantity. The amplitude of the de Broglie wave
associated with a particle is called probability amplitude, or the wave function, and is denoted
by .
In general, the complete wave function for a system depends on the positions of all the
particles in the system and on time. This can be written as
(r1,r2,…rj,…,t) = (rj) e–it
where rj is the position vector of the jthparticle in the system.
For any system in which the potential energy is time-independent and depends only on the
position of particles within the system, the important information about the system is
contained within the space part of the wave function. The wave function contains within it
all the information that can be known about the particle.||2is always real and positive, and
is proportional to the probability per unit volume, of finding the particle at a given point at
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PHY 1001: ENGINEERING PHYSICS
some instant. If represents a single particle, then
-called the probability density- is
the relative probability per unit volume that the particle will be found at any given point in the
volume.
One-dimensional wave functions and expectation values: Let be the wave function for a
particle moving along the x axis. Then P(x) dx = ||2dx is the probability to find the particle
in the infinitesimal interval dx around the point x. The probability of finding the particle in the
arbitrary interval a ≤ x ≤ b is
𝑏
𝑃𝑎𝑏 = ∫𝑎 |𝜓|2 𝑑𝑥 .
The probability of a particle being in the interval a ≤ x ≤ b is the area under the probability
density curve from a to b. The total probability of finding the particle is one. Forcing this
condition on the wave function is called normalization.
+∞
∫−∞ |𝜓|2 𝑑𝑥 = 1 .
Fig. 4.2 An arbitrary probability density curve for a particle
All the measureable quantities of a particle, such as its position, momentum and energy can
be derived from the knowledge of . eg, the average position at which one expects to find the
particle after many measurements is called the expectation value of x and is defined by the
equation
+∞
〈𝑥〉 ≡ ∫−∞ 𝜓 ∗ 𝑥 𝜓 𝑑𝑥 .
The important mathematical features of a physically reasonable wave function (x) for a
system are
(x) may be a complex function or a real function, depending on the system.
(x) must be finite, continuous and single valued everywhere.
The space derivatives of, must be finite, continuous and single valued
everywhere.
must be normalizable.
4.2 THE SCHRÖDINGER EQUATION
The appropriate wave equation for matter waves was developed by Schrödinger. Schrödinger
equation as it applies to a particle of mass m confined to move along x axis and interacting
with its environment through a potential energy function U(x) is
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PHY 1001: ENGINEERING PHYSICS
2
2
ℏ 𝑑 𝜓
+𝑈𝜓 = 𝐸𝜓
2 𝑚 𝑑𝑥 2
where E is a constant equal to the total energy of the system (the particle and its
environment) and ħ = h/2.This equation is referred to as the one dimensional, timeindependent Schrödinger equation.
Application of Schrödinger equation:
−
1. Particle in an infinite potential well (particle in a box)
2. Particle in a finite potential well
3. Tunneling
4.3PARTICLE IN AN INFINITE POTENTIAL WELL(PARTICLEINA“BOX”)
Fig. 4.3 (a) Particle in a potential well of infinite height, (b) Sketch of potential well
Consider a particle of mass m and velocity v, confined to bounce between two impenetrable
walls separated by a distance L as shown in Figure (a). Figure (b) shows the potential energy
function for the system.
U(x) = 0,
for 0 <x<L,
U (x) = ,
for x≤ 0, x≥L
Since U (x)= , for x< 0, x>L , (x) = 0 in these regions. Also (0) =0 and (L) =0. Only those
wave functions that satisfy these boundary conditions are allowed. In the region 0 <x<L, where
U = 0, the Schrödinger equation takes the form
𝑑2 𝜓
2𝑚
+
𝐸 𝜓 = 0
𝑑𝑥 2
ℏ2
Or
𝑑2 𝜓
𝑑𝑥 2
= − 𝑘2 𝜓 ,
where 𝑘 2 =
2𝑚𝐸
ℏ2
or
𝑘 =
√2𝑚𝐸
ℏ
The most general form of the solution to the above equation is
(x) = Asin(kx) + B cos(kx)
where A and B are constants determined by the boundary and normalization conditions.
Applying the first boundary condition,
i.e., at x = 0, = 0 leads to
0 = A sin 0 + B cos 0
or B = 0 ,
And at x = L , = 0 ,
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PHY 1001: ENGINEERING PHYSICS
0 = A sin(kL) + B cos(kL) = A sin(kL) + 0 ,
SinceA 0 , sin(kL) = 0 .
k L = n π ; ( n = 1, 2, 3, ……….. )
Now the wave function reduces to
𝜓𝑛 (𝑥) = 𝐴 𝑠𝑖𝑛 (
𝑛𝜋𝑥
𝐿
)
To find the constant A, apply normalization condition
+∞
∫−∞ |ψ|2 dx = 1
𝐿1
𝐴2 ∫0
2
[1 − 𝑐𝑜𝑠(
2𝑛𝜋𝑥
)] 𝑑𝑥
𝐿
𝐿
𝑛𝜋𝑥
∫0 𝐴2 [𝑠𝑖𝑛 (
𝐿
2
)] 𝑑𝑥 = 1 .
= 1
2
𝐴 = √𝐿
Solving we get
2
Thus
𝜓𝑛 (𝑥) = √𝐿 𝑠𝑖𝑛 (
Since
𝑘 =
We get,
√2𝑚𝐸
ℏ
∴
or
𝑛𝜋𝑥
𝐿
)
√2𝑚𝐸
and
ℏ
𝐿 =
is the wave function for particle in a box.
kL = nπ
𝑛𝜋.
ℎ2
𝐸𝑛 = ( 8 𝑚 𝐿2) 𝑛2 ,
n = 1, 2, 3,
. . . . .
Each value of the integer n corresponds to a quantized energy value, E n .
The lowest allowed energy (n = 1),
𝐸1 =
ℎ2
8 𝑚 𝐿2
.
This is the ground state energy for the particle in a box.
Excited states correspond to n = 2, 3, 4,…which have energies given by 4E1 , 9E1 , 16E1….
respectively.
Energy level diagram, wave function and probability density sketches are shown below.
Fig. 4.4 Energy level diagram for a particle in potential well of infinite height
Since ground state energy E1 ≠0, the particle can never be at rest.
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PHY 1001: ENGINEERING PHYSICS
Fig. 4.5 Sketch of (a) wave function, (b) Probability density for a particle in potential well of
infinite height
4.4 A PARTICLE IN A POTENTIAL WELL OF FINITE HEIGHT
Fig. 4.6Potential well of finite height U and length L
Consider a particle with the total energy E, trapped in a finite potential well of height U such
that
U(x) = 0 , 0 <x<L,
U(x) = U , x≤ 0, x≥L
Classically, for energy E<U, the particle is permanently bound in the potential well. However,
according to quantum mechanics, a finite probability exists that the particle can be found
outside the well even if E<U. That is, the wave function is generally nonzero in the regions I
and III. In region II, where U = 0, the allowed wave functions are again sinusoidal. But the
boundary conditions no longer require that the wave function must be zero at the ends of the
well.
Schrödinger equation outside the finite well in regions I & III
𝑑2 𝜓
𝑑𝑥 2
=
2𝑚
ℏ2
(𝑈 − 𝐸) 𝜓 ,
or
𝑑2 𝜓
𝑑𝑥 2
= 𝐶2 𝜓
where
𝐶2 =
2𝑚
ℏ2
(𝑈 −
𝐸)
General solution of the above equation is
(x) = AeCx + B e−Cx
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PHY 1001: ENGINEERING PHYSICS
where A and B are constants.
A must be zero in Region III and B must be zero in Region I, otherwise, the probabilities
would be infinite in those regions. For solution to be finite,
I = AeCx
for x≤ 0
-Cx
III = Be for x≥L
This shows that the wave function outside the potential well decay exponentially with
distance.
Schrodinger equation inside the square well potential in region II, where U = 0
𝑑2 𝜓𝐼𝐼
𝑑𝑥 2
+ (
2𝑚
ℏ2
2𝑚𝐸
𝐸) 𝜓𝐼𝐼 = 0 ,
ℏ2
= 𝑘2
General solution of the above equation
𝜓𝐼𝐼 = 𝐹 𝑠𝑖𝑛[𝑘𝑥] + 𝐺 𝑐𝑜𝑠[𝑘𝑥]
To determine the constants A, B, F, G and the allowed values of energy E, apply the four
boundary conditions and the normalization condition:
𝑑𝜓
At x = 0 , I(0) = II(0) and [ 𝑑𝑥𝐼]
At x = L , II(L) = III(L)
𝑥=0
=
[
and
[
𝑑𝜓𝐼𝐼
]
𝑑𝑥 𝑥=0
𝑑𝜓𝐼𝐼
]
𝑑𝑥 𝑥=𝐿
=
[
𝑑𝜓𝐼𝐼𝐼
]
𝑑𝑥 𝑥=𝐿
+∞
∫
|𝜓|2 𝑑𝑥 = 1
−∞
Figure 4.5 shows the plots of wave functions and their respective probability densities.
Fig. 4.7Sketch of (a) wave function, (b) Probability density for a particle in potential well of
finite height
It is seen that wavelengths of the wave functions are longer than those of wave functions of
infinite potential well of same length and hence the quantized energies of the particle in a
finite well are lower than those for a particle in an infinite well.
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4.5TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
Consider a particle of energy E approaching a potential barrier of height U, (E<U). Potential
energy has a constant value ofU in the region of width L and is zero in all other regions. This is
called a square barrier and U is called the barrier height. Since E<U, classically the regions II
and III shown in the figure are forbidden to the particle incident from left. But according to
quantum mechanics, all regions are accessible to the particle, regardless of its energy.
Fig. 4.8 Tunneling through a potential barrier of finite height
By applying the boundary conditions, i.e.and its first derivative must be continuous at
boundaries (at x = 0 and x = L), full solution to the Schrödinger equation can be found which is
shown in figure. The wave function is sinusoidal in regions I and III but exponentially
decaying in region II. The probability of locating the particle beyond the barrier in region III
is nonzero. The movement of the particle to the far side of the barrier is called tunneling or
barrier penetration. The probability of tunneling can be described with a transmission
coefficient T and a reflection coefficient R.
The transmission coefficient represents the probability that the particle penetrates to the
other side of the barrier, and reflection coefficient is the probability that the particle is
reflected by the barrier. Because the particles must be either reflected or transmitted we
have, R + T = 1.
An approximate expression for the transmission coefficient, when T<< 1 is
T ≈ e−2CL , where 𝐶 =
√ 2 𝑚 (𝑈−𝐸)
ℏ
.
4.6 THE QUANTUM MODEL OF THE HYDROGEN ATOM
The potential energy function for the H-atom is
𝑘𝑒 𝑒 2
𝑈(𝑟) = −
𝑟
where ke = 1/40= 8.99 x 109 N.m2/C2 Coulomb constant and r is radial distance of
electron from H-nucleus.
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Fig. 4.9 Spherical polar coordinate system
The time-independent Schrödinger equation in 3-dimensional space is
ℏ2 𝜕 2 𝜓 𝜕 2 𝜓 𝜕 2 𝜓
−
(
+
+
) +𝑈𝜓 = 𝐸𝜓
2 𝑚 𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
Since U has spherical symmetry, it is easier to solve the Schrödinger equation in
spherical polar coordinates (r, ,)where 𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2 ,
⃗ .
is the angle between z-axis and 𝒓
⃗ onto the xy-plane.
is the angle between the x-axis and the projection of 𝒓
It is possible to separate the variables r, θ, φ as follows:
(r, , ) = R(r) f() g()
By solving the three separate ordinary differential equations for R(r), f(), g(), with
conditions that the normalized and its first derivative are continuous and finite
everywhere, one gets three different quantum numbers for each allowed state of the
H-atom. The quantum numbers are integers and correspond to the three independent
degrees of freedom.
The radial function R(r) of is associated with the principal quantum number n. Solving
R(r), we get an expression for energy as,
𝑘 𝑒2
𝐸𝑛 = − ( 2𝑒𝑎 )
𝑜
1
𝑛2
= −
13.606 𝑒𝑉
𝑛2
,
n = 1, 2, 3,
. . .
which is in agreement with Bohr theory.
The polar function f() is associated with the orbital quantum number . The azimuthal
function g() is associated with the orbital magnetic quantum number m.
The
application of boundary conditions on the three parts of leads to important
relationships among the three quantum numbers: n can range from 1 to , can range
from 0 to n–1 ; [n allowed values]. m can range from –to+ ; [(2+1) allowed values].
All states having the same principal quantum number are said to form a shell. All states
having the same values of n and are said to form a subshell:
n = 1 K shell
= 0 s subshell
n = 2 L shell
= 1 p subshell
n = 3 M shell
= 2 d subshell
n = 4 N shell
= 3 f subshell
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n = 5 O shell
n = 6 P shell
.. ..
.. ..
.. ..
.. ..
=4
=5
.. ..
.. ..
g subshell
h subshell
.. ..
.. ..
4.7 WAVE FUNCTIONS FOR HYDROGEN
The potential energy for H-atom depends only on the radial distance r between nucleus
and electron. Therefore some of the allowed states for the H-atom can be represented
by wave functions that depend only on r (spherically symmetric function). The simplest
wave function for H-atom is the 1s-state (ground state) wave function (n = 1, = 0):
𝜓1𝑠 (𝑟) =
1
√𝜋 𝑎𝑜3
𝑒𝑥𝑝 (−𝑎𝑟𝑜 ) where ao is Bohr radius ( = 0.0529 nm).
|1s|2 is the probability density for H-atom in 1s-state:
1
2𝑟
|𝜓1𝑠 |2 =
3 𝑒𝑥𝑝 (− 𝑎𝑜 )
𝜋 𝑎𝑜
The radial probability density P(r) is the probability per unit radial length of finding
the electron in a spherical shell of radius r and thickness dr. P(r)dr is the probability of
finding the electron in this shell.
P(r) dr = ||2 dv = ||2 4r2 dr
P(r) = 4r2 ||2
Fig. 4.10 A spherical shell of radius r and thickness dr has a volume equal to 4 r2dr
Radial probability density for H-atom in its ground state:
𝑃1𝑠 = (
4 𝑟2
𝑎𝑜3
) 𝑒𝑥𝑝 (− 2𝑎𝑜𝑟 )
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Fig. 4.11 (a) The probability of finding the electron as a function of distance from the nucleus
for the hydrogen atom in the 1s (ground)state. (b) The cross section in the xy plane of the
spherical electronic charge distribution for the hydrogen atom in its 1s state
The next simplest wave function for the H-atom is the 2s-state wave function (n = 2,
= 0):
𝜓2𝑆 (𝑟) =
1
√32𝜋𝑎𝑜3
𝑟
𝑟
(2 − 𝑎 ) 𝑒𝑥𝑝 (− 𝑎 )
𝑜
𝑜
2s is spherically symmetric (depends only on r). Energy corresponding to n = 2 (first excited
state is E2= E1/4 = –3.401 eV.
Fig. 4.12 Plot of radial probability density versus r/a0 (normalized radius) for 1s and 2s states
of hydrogen atom
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EXERCISE
QUESTIONS
4.1
1
2
4.2
AN INTERPRETATION OF QUANTUM MECHANICS
What is a wave function ? What is its physical interpretation ?
What are the mathematical features of a wave function?
THE SCHRODINGER EQUATION
4.3 THE PARTICLE IN A “BOX”
3 By solving the Schrödinger equation, obtain the wave-functions for a
particle of mass m in a one-dimensional “box” of length L.
4 Apply the Schrödinger equation to a particle in a one-dimensional “box”
of length L and obtain the energy values of the particle.
5 Sketch the lowest three energy states, wave-functions, probability
densities for the particle in a one-dimensional “box”.
6 The wave-function for a particle confined to moving in a one-dimensional
box is
ψ(x) = A sin(nπx
) . Use the normalization condition on to show that
L
𝐴 = √2𝐿 .
[2]
[2]
[5]
[5]
[3]
[2]
nπx
L
7 The wave-function of an electron is ψ(x) = A sin( ) . Obtain an
expression for the probability of finding the electron between x = a and
x = b.
4.4 A PARTICLE IN A WELL OF FINITE HEIGHT
8 Sketch the potential-well diagram of finite height U and length L, obtain
the general solution of the Schrödinger equation for a particle of mass
m in it.
9 Sketch the wave-functions and the probability densities for the lowest
three energy states of a particle in a potential well of finite height.
4.5 TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
10 Give a brief account of tunneling of a particle through a potential energy
barrier.
4.6 THE QUANTUM MODEL OF THE HYDROGEN ATOM
11 Give a brief account of quantum model of H-atom.
4.7 THE WAVE FUNCTIONS FOR HYDROGEN
[3]
[5]
[3]
[4]
[2]
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12 The
wave
1
√πa3o
function
for
H-atom
in
ground
state
isψ1S (r) =
exp (− aro) . Obtain an expression for the radial probability density
of H-atom in ground state.
Sketch schematically the plot of this vs. radial distance.
13 The wave function for H-atom in 2s state is ψ2S (r) =
1
√32πa3o
(2 − 𝑎𝑟𝑜 ) exp (− aro) .
Write
the
expression
for
the
[4]
radial
probability density of H-atom in 2s state. Sketch schematically the plot
of this vs. radial distance.
14 Sketch schematically the plot of the radial probability density vs. radial
distance for H-atom in 1s-state and 2s-state.
[3]
[3]
PROBLEMS
4.1 AN INTERPRETATION OF QUANTUM MECHANICS
1 A WAVE FUNCTION FOR A PARTICLE
A particle wave function is given by the equation (x) = A e–ax2 .
(A) What is the value of A if this wave function is normalized?
(B) What is the expectation value of x for this particle?
Ans: A = (2a/π)¼ ,
x = 0
2 A free electron has a wave function ψ(x) = A exp[𝑖(5.0 × 1010 )𝑥]
where x is in meters. Find (a) its de Broglie wavelength, (b) its momentum, and (c) its
kinetic energy in electron volts.
Ans: 1.26 x 10–10m, 5.27 x 10–24kg.m/s, 95.5 eV
4.2 THE SCHRODINGER EQUATION
4.3 THE PARTICLE IN A “BOX”
3 A BOUND ELECTRON
An electron is confined between two impenetrable walls 0.20 nm apart. Determine the
energy levels for the states n =1 ,2 , and 3.
Ans: 9.2 eV, 37.7 eV, 84.8 eV
4 ENERGY QUANTIZATION FOR A MACROSCOPIC OBJECT
A 0.50 kg baseball is confined between two rigid walls of a stadium that can be modeled
as a “box” of length 100 m. Calculate the minimum speed of the baseball. If the baseball
is moving with a speed of 150 m/s, what is the quantum number of the state in which
the baseball will be?
Ans: 6.63 x 10–36 m/s, 2.26 x 1037
5 A proton is confined to move in a one-dimensional “box” of length 0.20 nm. (a) Find the
lowest possible energy of the proton. (b) What is the lowest possible energy for an
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electron confined to the same box? (c) Account for the great difference in results for (a)
and (b).
Ans: 5.13 x 10–3 eV, 9.41 eV
6 MODEL OF AN ATOM
(A) Using the simple model of a particle in a box to represent an atom, estimate the
energy (in eV) required to raise an atom from the state n =1 to the state n =2.
Assume the atom has a radius of 0.10 nm and that the moving electron carries the
energy that has been added to the atom.
(B) Atoms may be excited to higher energy states by absorbing photon energy.
Calculate the wavelength of the photon that would cause the transition from the
state n =1 to the state n =2.
Ans: 28.3 eV, 43.8 nm
4.4 A PARTICLE IN A WELL OF FINITE HEIGHT
4.5 TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
7 TRANSMISSION COEFFICIENT FOR AN ELECTRON
A 30-eV electron is incident on a square barrier of height 40 eV. What is the probability
that the electron will tunnel through the barrier if its width is (A) 1.0 nm? (B) 0.10 nm?
Ans: 8.5 x 10–15, 0.039
8 An electron with kinetic energy E = 5.0 eV is incident on a
barrier with thickness L = 0.20 nm and height U = 10.0 eV
as shown in the figure.What is the probability that the
electron (a) will tunnel through the barrier? (b) will be
reflected?
Ans: 0.0103, 0.990
4.6 THE QUANTUM MODEL OF THE HYDROGEN ATOM
9 THE n = 2 LEVEL OF HYDROGEN:
For a H-atom, determine the number of allowed states corresponding to the
principal quantum number n = 2, and calculate the energies of these states.
Ans: 4 states (one 2s-state + three 2p-states),
–3.401 eV
10 A general expression for the energy levels of one-electron atoms and ions is
𝐸𝑛 = −
𝜇 𝑘𝑒2 𝑞12 𝑞22
2 ℏ2 𝑛2
, where ke is the the Coulomb constant, q1 and q2 are the
charges of the electron and the nucleus, and μ is the reduced mass, given byμ =
m1 m2
m1 +m2
.
The wavelength for n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm
(visible red light). What are the wavelengths for this same transition in
(a)
positronium, which consists of an electron and a positron, and
(b)
singly ionized helium ?
Ans: 1310 nm, 164 nm
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4.7 THE WAVE FUNCTIONS FOR HYDROGEN
11 THE GROUND STATE OF H-ATOM:
Calculate the most probable value of r (= distance from nucleus) for an electron
in the ground state of the H-atom. Also calculate the average value r for the
electron in the ground state.
Ans: ao , 3 ao/2
12 PROBABILITIES FOR THE ELECTRON IN H-ATOM:
Calculate the probability that the electron in the ground state of H-atom will be
found outside the Bohr radius.
Ans: 0.677
13 For a spherically symmetric state of a H-atom the schrodinger equation in
spherical coordinates is−
ℏ2
2𝑚
𝜕2 𝜓
( 𝜕𝑟 2 +
2 𝜕𝜓
𝑟
) −
𝜕𝑟
𝑘𝑒 𝑒 2
𝑟
𝜓 = 𝐸 𝜓 . Show that the 1s
wave function for an electron in H-atom 𝜓1𝑆 (𝑟) =
1
√𝜋𝑎𝑜3
𝑒𝑥𝑝 (− 𝑎𝑟𝑜 ) satisfies the
schrodinger equation.
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CHAPTER 5
SOLID STATE PHYSICS
OBJECTIVES:
To comprehend the electrical properties of metals, semiconductors and insulators
To understand the effect of doping on electrical properties of semiconductors
To understand superconductivity and its engineering applications
5.1 FREE-ELECTRON THEORY OF METALS
Quantum based free electron theory of metals is centered on wave nature of electrons. In this
model, one imagines that the outer-shell electrons are free to move through the metal but
are trapped within a three-dimensional box formed by the metal surfaces. Therefore,
each electron is represented as a particle in a box and is restricted to quantized energy
levels. Each energy state can be occupied by only two electrons (one with spin up & the
other with spin down) as a consequence of exclusion principle. In quantum statistics, it is
shown that the probability of a particular energy state E being occupied by an electrons is
given by
f E
1
E EF
exp
1
kT
[5.1]
where f(E) is called the Fermi-Dirac distribution function and EF is called the Fermi energy.
Plot of f(E) versus E is shown in figure 5.1.
Fig. 5.1 Plot of Fermi-Dirac distribution function f(E) versus energy E at (a) T = 0K and (b) T >
0K
At zero kelvin (0 K), all states having energies less than the Fermi energy are occupied, and
all states having energies greater than the Fermi energy are vacant. i.e. Fermi energy is the
highest energy possessed by an electron at 0 K (Figure 5.1a). As temperature increases (T >
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0K), the distribution rounds off slightly due to thermal excitation and probability of Fermi level
being occupied by an electron becomes half (Figure 5.1b). In other words, Fermi energy is that
energy state at which probability of electron occupation is half. The Fermi energy EF also
depends on temperature, but the dependence is weak in metals.
Density of states: From particle in a box problem, for a particle of mass m is confined to move
in a one-dimensional box of length L, the allowed states have quantized energy levels given
by,
h2
2 2 2
2
En
n
n
8 m L2
2 m L2
n = 1, 2, 3 . . .
[5.2]
An electron moving freely in a metal cube of side L, can be modeled as particle in a threedimensional box. It can be shown that the energy for such an electron is
2 2
nx2 ny2 nz2
E
2
2mL
[5.3]
where m is mass of the electron and nx, ny, nz are quantum numbers(positive integers). Each
allowed energy value is characterized by a set of three quantum numbers (nx, ny, nz - one
for each degree of freedom). Imagine a three-dimensional quantum number space whose
axes represent nx, ny, nz. The allowed energy states in this space can be represented as
dots located at positive integral values of the three quantum numbers as shown in the
Figure 5.2.
Fig. 5.2 Representation of the allowed energy states in a quantum number space (dots
represent the allowed states)
Eq. 5.3 can be written as
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E
n2
Eo
n x2 ny2 nz2
where E o
2 2
and n
2 m L2
[5.4]
E
Eo
Eq. 5.4 represents a sphere of radius n. Thus, the number of allowed energy states
having energies between E and E+dE is equal to the number of points in a spherical
shell of radius n and thickness dn. In this quantum number space each point is at the
corners of a unit cube and each corner point is shared by eight unit cubes and as such the
contribution of each point to the cube is 1/8 th. Because a cube has eight corners, the effective
point per unit cube and hence unit volume is one. In other words, number of points is equal
to the volume of the shell. The “volume” of this shell, denoted by G(E)dE.
1
1
G(E) dE = 4 n 2 dn n 2 dn
8
2
G( E ) dE
1
2
1
E E 2
d
E o E o
G( E ) dE
1
2
using the relation n
E 12 1 12
E o 2 E dE
Eo
2 2
G( E ) dE 41
2
2mL
3 2
E
1
2
1
4
3 2
Eo
E
1
2
E
Eo
dE
dE
3
2 m 2 L3 1 2
G( E ) dE
E dE ,
2 2 3
L3 V
Number of states per unit volume per unit energy range, called density of states, g(E) is
given by
g(E) = G(E)/V
3
G( E )
2 m 2 12
g( E ) dE
dE
E dE
V
2 2 3
4 2 m
g( E ) dE
h3
3
2
E
1
2
dE
h
2
Finally, we multiply by 2 for the two possible spin states of each particle.
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g( E ) dE
8 2 m
h3
3
2
E
1
2
dE
[5.5]
g(E) is called the density-of-states function.
Electron density: For a metal in thermal equilibrium, the number of electrons N(E) dE,
per unit volume, that have energy between E and E+dE is equal to the product of the
density of states and the probability that a state is occupied. that is,
N(E)dE = [ g(E)dE ] f(E)
8 2 m
N( E ) dE
h3
3
2
E
1
dE
E EF
1
exp
kT
2
[5.6]
Plots of N(E) versus E for two temperatures are given in figure 5.3.
Fig. 5.3 Plots of N(E) versus E for (a) T = 0K (b) T = 300K
If ne is the total number of electrons per unit volume, we require that
8 2 m
ne N( E ) dE
h3
0
3
2
0
E
1
dE
E EF
1
exp
kT
2
[5.7]
At T = 0K, the Fermi-Dirac distribution function f(E) = 1 for E <EF and f(E) = 0 for E >EF. Therefore,
at T = 0K, Equation 5.7 becomes
8 2 m
ne
h3
3
2
EF
E
0
1
2
8 2 m
dE
h3
3
2
EF
2
3
3
2
16 2 m
3 h3
3
2
3
EF 2
[5.8]
Solving for Fermi energy at 0K, we obtain
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PHY 1001: ENGINEERING PHYSICS
EF 0
h2
2m
3 ne
8
2
3
[5.9]
The average energy of a free electron in a metal at 0K is Eav = (3/5)EF.
5.2 BAND THEORY OF SOLIDS
When a quantum system is represented by wave function, probability density ||2 for that
system is physically significant while the probability amplitude not. Consider an atom such
as sodium that has a single s electron outside of a closed shell. Both the wave functions S ( r )
and S ( r ) are valid for such an atom [ S ( r ) and S ( r ) are symmetric and anti symmetric
wave functions]. As the two sodium atoms are brought closer together, their wave functions
begin to overlap. Figure 5.4 represents two possible combinations : i) symmetric - symmetric
and ii) symmetric – antisymmetric . These two possible combinations of wave functions
represent two possible states of the two-atom system. Thus, the states are split into two
energy levels. The energy difference between these states is relatively small, so the two states
are close together on an energy scale.
Fig. 5.4 The wave functions of two atoms combine to form a composite wave function : a)
symmetric-symmetric b) symmetric-antisymmetric
When two atoms are brought together, each energy level will split into 2 energy levels. (In
general, when N atoms are brought together N split levels will occur which can hold 2N
electrons). The split levels are so close that they may be regarded as a continuous band of
energy levels. Following figure shows the splitting of 1s and 2s levels of sodium atom
when : (a) two sodium atoms are brought together (b)five sodium atoms are brought
together (c) a large number of sodium atoms are assembled to form a solid. The close
energy levels forming a band are seen clearly in (c).
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Fig.5.5 Splitting of 1s and 2s levels of sodium atoms due to interaction between them
Some bands may be wide enough in energy so that there is an overlap between the
adjacent bands. Some other bands are narrow so that a gap may occur between the allowed
bands, and is known as forbidden energy gap. The 1s, 2s, and 2p bands of solid sodium
are filled completely with electrons. The 3s band (2N states) of solid sodium has only
N electrons and is partially full; The 3p band, which is the higher region of the overlapping
bands, is completely empty as shown in Figure 5.6
Fig. 5.6 Energy bands of a sodium crystal
5.3 ELECTRICAL CONDUCTION IN METALS, INSULATORS AND SEMICONDUCTORS
Good electrical conductors contain high density of free charge carriers, and the density
of free charge carriers in insulators is nearly zero. In semiconductors free-charge-carrier
densities are intermediate between those of insulators and those of conductors.
Metals: Metal has a partially filled energy band (Figure 5.7a). At 0K Fermi level is the
highest electron-occupied energy level. If a potential difference is applied to the metal,
electrons having energies near the Fermi energy require only a small amount of
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additional energy to reach nearby empty energy states above the Fermi-level. Therefore,
electrons in a metal experiencing a small force (from a weak applied electric field) are free to
move because many empty levels are available close to the occupied energy levels. The model
of metals based on band theory demonstrates that metals are excellent electrical conductors.
Insulators: Consider the two outermost energy bands of a material in which the lower band is
filled with electrons and the higher band is empty at 0 K (Figure5.7b). The lower, filled band is
called the valence band, and the upper, empty band is the conduction band. The energy
separation between the valence and conduction band, called energy gap Eg, is large for
insulating materials. The Fermi level lies somewhere in the energy gap. Due to larger energy
gap compare to thermal energy kT (26meV) at room temperature, excitation of electrons from
valence band to conduction band is hardly possible. Since the free-electron density is nearly
zero, these materials are bad conductors of electricity.
Semiconductors: Semiconductors have the same type of band structure as an insulator, but
the energy gap is much smaller, of the order of 1 eV. The band structure of a semiconductor
is shown in Figure 5.7c. Because the Fermi level is located near the middle of the gap for a
semiconductor and Eg is small, appreciable numbers of electrons are thermally excited from
the valence band to the conduction band. Because of the many empty levels above the
thermally filled levels in the conduction band, a small applied potential difference can easily
raise the energy of the electrons in the conduction band, resulting in a moderate conduction.
At T = 0 K, all electrons in these materials are in the valence band and no energy is available
to excite them across the energy gap. Therefore, semiconductors are poor conductors at very
low temperatures. Because the thermal excitation of electrons across the narrow gap is more
probable at higher temperatures, the conductivity of semiconductors increases rapidly with
temperature. This is in sharp contrast with the conductivity of metals, where it decreases with
increasing temperature. Charge carriers in a semiconductor can be negative, positive, or both.
When an electron moves from the valence band into the conduction band, it leaves behind a
vacant site, called a hole, in the otherwise filled valence band.
Fig. 5.7 Band structure of (a) Metals (b) Insulators (c) Semiconductors
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In an intrinsic semiconductor (pure semiconductor) there are equal number of conduction
electrons and holes. In the presence of an external electric field, the holes move in
the direction of field and the conduction electrons move opposite to the direction of
the field. Both these motions correspond to the current in the same direction (Figure
5.8).
Fig. 5.8 Movement of electrons and holes in an intrinsic semiconductor
Doped Semiconductors (Extrinsic semiconductors) : Semiconductors in their pure form are
called intrinsic semiconductors while doped semiconductors are called extrinsic
semiconductors. Doping is the process of adding impurities to a semiconductor. By
doping both the band structure of the semiconductor and its resistivity are modified.
If a tetravalent semiconductor (Si or Ge) is doped with a pentavalent impurity atom
(donor atom), four of the electrons form covalent bonds with atoms of the
semiconductor and one is left over (Figure 5.9). At zero K, this extra electron resides in
the donor-levels, that lie in the energy gap, just below the conduction band. Since the
energy Ed between the donor levels and the bottom of the conduction band is small,
at room temperature, the extra electron is thermally excited to the conduction band.
This type of semiconductors are called n-type semiconductors because the majority of
charge carriers are electrons (negatively charged).
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Fig. 5.9 n-type semiconductor – two dimensional representation and band structure
If a tetravalent semiconductor is doped with a trivalent impurity atom (acceptor atom),
the three electrons form covalent bonds with neighboring semiconductor atoms, leaving
an electron deficiency (a hole) at the site of fourth bond (Figure 5.10). At zero K, this
hole resides in the acceptor levels that lie in the energy gap just above the valence
band. Since the energy Ea between the acceptor levels and the top of the valence band
is small, at room temperature, an electron from the valence band is thermally excited
to the acceptor levels leaving behind a hole in the valence band. This type of
semiconductors are called p-type semiconductors because the majority of charge carriers
are holes (positively charged).The doped semiconductors are called extrinsic
semiconductors.
Fig. 5.10 p-type semiconductor – two dimensional representation and band structure
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5.4 SUPERCONDUCTIVITY-PROPERTIES AND APPLICATIONS
Superconductor is a class of metals and compounds whose electrical resistance decreases to
virtually zero below a certain temperature Tc called the critical temperature. The critical
temperature is different for different superconductors. Mercury loses its resistance
completely and turns into a superconductor at 4.2K. Critical temperatures for some important
elements/compounds are listed below.
Element/Compound
Tc (K)
La
6.0
NbNi
10.0
Nb3Ga
23.8
Fig. 5.11 Plot of Resistance Vs Temperature for normal metal and a superconductor
Meissner Effect: In the presence of magnetic field, as the temperature of superconducting
material is lowered below Tc, the field lines are spontaneously expelled from the interior of
the superconductor(B = 0, Figure 5.12). Therefore, a superconductor is more than a perfect
conductor; it is also a perfect diamagnet. The property that B = 0 in the interior of a
superconductor is as fundamental as the property of zero resistance. If the magnitude of the
applied magnetic field exceeds a critical value Bc, defined as the value of B that destroys a
material’s superconducting properties, the field again penetrates the sample. Meissner effect
can be explained in the following way.
A good conductor expels static electric fields by moving charges to its surface. In effect, the
surface charges produce an electric field that exactly cancels the externally applied field inside
the conductor. In a similar manner, a superconductor expels magnetic fields by forming
surface currents. Consider the superconductor shown in Figure 5.12. Let’s assume the sample
is initially at a temperature T>Tc so that the magnetic field penetrates the cylinder. As the
cylinder is cooled to a temperature T<Tc, the field is expelled. Surface currents induced on the
superconductor’s surface produce a magnetic field that exactly cancels the externally applied
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
field inside the superconductor. As expected, the surface currents disappear when the
external magnetic field is removed.
Fig. 5.12 A superconductor in the form of a long cylinder in the presence of an external
magnetic field.
BCS Theory: In 1957. Bardeen, Cooper and Schrieffer gave a successful theory to explain the
phenomenon of superconductivity, which is known as BCS theory. According to this theory, two
electrons can interact via distortions in the array of lattice ions so that there is a net attractive
force between the electrons. As a result, the two electrons are bound into an entity called a
Cooper pair, which behaves like a single particle with integral spin. Particles with integral spin
are called bosons. An important feature of bosons is that they do not obey the Pauli exclusion
principle. Consequently, at very low temperatures, it is possible for all bosons in a collection
of such particles to be in the lowest quantum state and as such the entire collection of Cooper
pairs in the metal is described by a single wave function. There is an energy gap equal to the
binding energy of a Cooper pair between this lowest state and the next higher state.. Under
the action of an applied electric field, the Cooper pairs experience an electric force and move
through the metal. A random scattering event of a Cooper pair from a lattice ion would
represent resistance to the electric current. Such a collision would change the energy of the
Cooper pair because some energy would be transferred to the lattice ion. There are no
available energy levels below that of the Cooper pair (it is already in the lowest state),
however, and none available above because of the energy gap. As a result, collisions do not
occur and there is no resistance to the movement of Cooper pairs.
Applications: Most important and basic application of superconductors is in high field
solenoids which can be used to produce intense magnetic field. Superconducting magnets are
used in magnetic resonance imaging (MRI) technique. Magnetic levitation, based on Meissner
effect, is another important application of superconductors. This principle is used in maglev
vehicles. Detection of a weak magnetic field and lossless power transmission are some other
important applications of superconductors.
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
EXERCISES
QUESTIONS
5.1 FREE-ELECTRON THEORY OF METALS
1 Write the expression for Fermi-Dirac distribution function. Sketch
schematically the plots of this function for zero kelvin and for
temperature above zero kelvin.
2 Derive an expression for density-of-states.
3 Assuming the Fermi-Dirac distribution function , obtain an expression
for the density of free-electrons in a metal with Fermi energy EF, at
zero K and, hence obtain expression for Fermi energy EF in a metal at
zero K. [ Given: density-of-states function
[3]
[5]
3
8 2 m 2 12
g( E ) dE
E dE ]
h3
5.2 BAND THEORY OF SOLIDS
4 Explain the formation of energy bands in solids with necessary diagrams.
5.3 ELECTRICAL CONDUCTION IN METALS, INSULATORS AND
SEMICONDUCTORS
5 Distinguish between conductors, insulators and semiconductors on the
basis of band theory
6 Indicate the position of (a) donor levels (b) acceptor levels, in the
energy band diagram of a semiconductor.
[5]
[5]
[3]
[2]
7 What is the difference between p-type and n-type semiconductors?
Explain with band diagram.
8 With necessary diagrams, explain doping in semiconductors.
8 Why the electrical conductivity of an intrinsic semiconductor increases
with increasing temperature?
5.4 SUPERCONDUCTIVITY-PROPERTIES AND APPLICATIONS
9 What are superconductors? Draw a representative graph of Resistance Vs
Temperature for a superconductor.
10 Explain Meissner effect.
11 Explain briefly the BCS theory of superconductivity in metals.
[3]
[5]
[2]
[2]
[3]
[3]
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
PROBLEMS
5.1 FREE-ELECTRON THEORY OF METALS
1 Each atom of gold (Au) contributes one free-electron to the metal. The
concentration of free-electron in gold is 5.90 x 1028/m3. Compute the Fermi
Energy of gold.
Ans: 5.53 eV
2 Sodium is a monovalent metal having a density of 971 kg/m3 and a molar mass
of 0.023 kg/mol. Use this information to calculate (a) the density of charge
carriers and (b) the Fermi energy.
Ans: 2.54 x 1028/m3, 3.15 eV
3 Calculate the energy of a conduction electron in silver at 800 K, assuming the
probability of finding an electron in that state is 0.950. The Fermi energy is
5.48 eV at this temperature.
Ans: 5.28 eV
4 Show that the average kinetic energy of a conduction electron in a metal at
zero K is (3/5) EF
Suggestion: In general, the average kinetic energy is
1
E AV
E N( E ) dE
ne
where the density of particles
ne N( E ) dE
0
N( E ) dE
8 2 m
h3
3
2
1
2
E dE
E EF
1
exp
kT
5 (a) Consider a system of electrons confined to a three-dimensional box. Calculate
the ratio of the number of allowed energy levels at 8.50 eV to the number at
7.00 eV. (b) Copper has a Fermi energy of 7.0 eV at 300 K. Calculate the ratio
of the number of occupied levels at an energy of 8.50 eV to the number at
Fermi energy. Compare your answer with that obtained in part (a).
Ans: (a) 1.10 (b) 1.46x10-25
5.2. BAND THEORY OF SOLIDS
5.3 ELECTRICAL CONDUCTION IN METALS, INSULATORS AND SEMICONDUCTORS
6 Most solar radiation has a wavelength of 1 μm or less. What energy gap should
the material in solar cell have in order to absorb this radiation ? Is silicon (Eg=
1.14 eV) appropriate ?
Ans: 1.24 eV or less; yes
Dept. of Physics, MIT Manipal
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PHY 1001: ENGINEERING PHYSICS
7 The energy gap for silicon at 300 K is 1.14 eV. (a) Find the lowest-frequency photon
that can promote an electron from the valence band to the conduction band. (b) What
is the wavelength of this photon?
Ans: 2.7x1014 Hz, 1090 nm
8 The longest wavelength of radiation absorbed by a certain semiconductor is 512 nm.
Calculate the energy gap for this semiconductor.
Ans: 2.42 eV
5.4 SUPERCONDUCTIVITY-PROPERTIES AND APPLICATIONS
Dept. of Physics, MIT Manipal
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