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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 combined-

Intensity 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, (9e),

Thomson, 2013

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.


L15 Black body radiation and Planck’s hypothesis. Stefan’s law. Wien’s displacement law. Rayleigh-Jeans

law.

L16 The photoelectric effect.


L18 The Compton effect.

L19 Photons and electromagnetic waves. The wave properties of particles. de Broglie hypothesis.


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.


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


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 50 Marks

5 Quizzes of 4 marks each 20 Marks

2 Sessional of 15 marks each 30 Marks

End Semester Examination 50 Marks

Total 100 Marks

PHY 1001: ENGINEERING PHYSICS

Dept. of Physics, MIT Manipal 1

CHAPTER 1

OPTICS

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.

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.



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



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



differencePathdifferencePhase

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 , where m =

0, 1, 2 …

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 .



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

)m(sind 21

m = 0, 1, 2, . . .

For small value of , we can make following approximation

D

ytansin

In such a case, the path difference

D

dybSsind 1

For mth maximum located at ym , we can write



d

Dmy

or

mD

dy

m

m

where m = 0, 1, 2, . . .

Separation between adjacent maxima (for small ) known as fringe width/band width is,

d

Dy

d

Dm

d

D)1m(

yyy m1m

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.



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,



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;

2cosE4 22

0

source. single todueintensity is E where

2

cos4

2

00

2

0

...2,1,0, m where

(m sind :minima At

m sind :maxima At

equation, above From

/dsin2 Since

)21or)1m2(

orm2

sindcos4

,

20

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.



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.



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



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)

2

12

22

R

r1RR

rRRd

R2

r...

R

r

2

11RRd

1Rr

22

expansionbinomialusing

Substituting d , in 2d = (m - ½) we get,

rings. theoforder theasknown is and maximafor ...,2,1

21

m

Rmr



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. What is interference of light waves? [2]

2. What is coherence? Mention its importance. [2]

3. How coherent waves are produced? Illustrate with figure. [3]

4. Write the necessary condition for the constructive and destructive

interference of two light waves in terms of path/phase difference.

[2]

5. With necessary diagram, obtain an expression for the fringe-width of its

interference pattern.

[5]

6. Obtain an expression for intensity of light in double-slit interference

using phasor diagram.

[5]

7. Draw a schematic plot of the intensity of light in double-slit interference

against phase-difference.

[2]

8. Explain the following: i) Phase change on reflection ii) Optical path length [2]

9. Write the conditions for constructive and destructive interference of

reflected light from a thin soap film in air, assuming normal incidence.

[2]

10. Explain the interference in wedge-shaped thin films. [2]

11. Explain the formation of Newton’s rings and hence obtain an expression

for the radius of mth order bright ring.

[5]

PROBLEMS

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



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 MgF2 (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)



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



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



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



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.



sinaor2

sin2

a

minimum, first for condition the So

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 P2. Thus, condition for

second minima is, (a/4) sin = /2 or, a sin = 2

...3,2,1, m

minima, m for condition the general, In th

msina

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).



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

sinx

2

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



2 where E

E Combining,

R

E Also

1.23, Figure From

m

sinE,Or

2sin

2

E

2sinR2E

m

m

is the phase difference between rays from the top and bottom of the slit. Thus,

sina

sina

2 So,

2

3,..... 2, 1, where sin or,

3,..... 2,, 1 where Hence

0sin minima,for eqn., above theFrom

intensity max. theis wheresin

sinE intensity The

2

m

2

m

2

2

m

2

mma

mm

E

E

m

Fig. 1.24 Intensity distribution in single slit diffraction



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-

aperture.of diameter the is d where d

22.1sin

slit width theis a wheresin

isn diffractioslit singlein minimumfirst for equation The

a

In case of circular aperture, the factor 1.22 arises when we divide the aperture into elementary

Huygens sources and integrate over the aperture.



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.

d22.1

is

d22.1sin 1

R

R

R as dappoximate be can it small, very since

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



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 I, DIF =

Im,DIF(sin/)2

The combined effect is the product of the two and is given by2

2)(

α

αsin cos mθ ΙI

Fig. 1.28 Double slit interference and diffraction combined



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



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,

loop closed a form phasors since slits, ofnumber theis where2

NNN

Corresponding path difference is,

N2L



For the mth principal maximum at , we have d sin = m .

For the first minimum at ( + ) after the mth principal maximum is therefore

N

λmλδd )(sin

Nmsincossind

cos

1

Nm)cos d(sin d

Nmdm )cos (

cos d 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 dN

.

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



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, 3,.....2,1,m and is called order of

the spectra.

EXERCISE

QUESTIONS

1. Explain the term diffraction of light. What are the factors that determine

diffraction pattern? [3]

2. Discuss qualitatively, the Fraunhofer diffraction at a single-slit. [5]

3. Derive an expression for intensity of diffraction pattern in the case of

single-slit, using phasor diagram. [5]

4. Draw a schematic plot of the intensity of light in single slit diffraction

against phase difference.

[2]

5. Explain briefly diffraction at a circular aperture. [2]

6. State and explain Rayleigh’s criterion for optical resolution. [2]

7. Effect of diffraction is ignored in the case of Young’s double slit

interference. Give reason.

[2]

8. Arrive at the equation for the intensity of double slit diffraction pattern. [2]

9. Discuss qualitatively, the diffraction due to multiple slits. [3]

10 Obtain an expression for the half angular width of any principal

maximum in diffraction pattern due to multiple slits.

[5]

11 What is diffraction grating? Write the grating equation. [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



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 double-

slit 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



CHAPTER 2

APPLIED OPTICS

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 .

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.



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,

Tk

EEexp

En

En 12

1

2

[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).



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



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



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 He-

atoms, 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



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.



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,

0

1

1

0

n

n

sin

sin

1

0

10 sin

n

nsin [2.2]

Similarly, applying Snell’s law at B,

1

21

n

n

90sin

)90(sin

or

1

21

n

ncos



21

22

1n

n1sin [2.3]

Substituting Eq. (2.3) in Eq. (2.2) and simplifying,

22

21

0

0 nnn

1sin [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.

1

21

n

)nn( [2.5]

Since n1> n2, is always positive.

Relation between NA & :

From Eq. 2.4, assuming n0 = 1, NA = 22

21 nn = )nn)(nn( 2121

From Eq. 2.5, (n1n2) = n1 and since n1 n2, we can approximate (n1+n2) 2 n1.

Therefore, NA = 2n)n2()n( 111 [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



From the Figure 2.12,

1eccosdcotdL 12

1s

1sinn

ndL

220

21

s

( sinn

nsin

1

01 ) [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 wave-

front 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

22

210 nnn

dV

where d is the diameter of the core and is the wavelength of the light

propagated. If V >>1, then the number of successfully propagated modes are 2

V 2

.

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.



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 n1 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.



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



wavelength of the light propagated through the fiber, the energy loss that occurs due to such

scattering resemble Rayleigh scattering that is proportional to 4

1

. 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.



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 LASER

1 Mention the characteristics of a laser beam. [2]

2 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. [1 EACH]

3 Explain the principle of a laser. [2 ]

4 Explain construction and working of ruby laser with necessary

diagrams. [5]

5 Explain construction and working of He-Ne laser with necessary

diagrams. [5]

6 Mention any four applications of laser. [2]

2.2 OPTICAL FIBERS

7 What is total internal reflection? [1]

8 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. [1 EACH]



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. [5]

10 Define fractional refractive index change and obtain a relation

between & numerical aperture NA. [3]

11 What is skip distance? With neat diagram, derive an expression for

it. [3]

12 Briefly explain the different types of optical fibers with necessary

diagrams. [5]

13 Explain the different types of attenuations possible in optical

fibers? Explain them. [4]

14 Mention any four applications of optical fiber. [2]

PROBLEMS

2.1 LASER

1. 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



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



CHAPTER 3

QUANTUM PHYSICS

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.

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.



(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,

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.

4

B Tkc2)T,(

I



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,

1

1ch2)T,(

e5

2

Tλk

hc

B

I



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.

For very large λ,

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)

Tkc2)T,(1exp 4

Tk

ch)(kT

hc

I



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.



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 108 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



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 re-

radiates 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

m c(1−cosθ),

where m is the mass of the electron, c velocity of light, h Planck’s constant.

This is known as Compton shift equation, and the factor ℎ

𝑚 𝑐 is called the Compton

wavelength and ℎ

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).



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 = √p2c2 + m2c4

Kinetic energy K = E − m c2

And momentum p = mv. where

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 = √𝑝2𝑐2 + 𝑚2𝑐4

Squaring both the sides, (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐2 + 𝑚2𝑐4 = 𝑝2𝑐2 + 𝑚2𝑐4

For conservation of momentum, x-component: 𝑝𝑜 = 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝 𝑐𝑜𝑠 𝜙

y-component: 0 = 𝑝′ 𝑠𝑖𝑛 𝜃 − 𝑝 𝑠𝑖𝑛𝜙

Rewriting these two equations

2

2

cv1

1



𝑝𝑜 − 𝑝′ 𝑐𝑜𝑠 𝜃 = 𝑝 𝑐𝑜𝑠 𝜙

𝑝′ 𝑠𝑖𝑛 𝜃 = 𝑝 𝑠𝑖𝑛 𝜙

Squaring both the sides and adding,

𝑝𝑜2 − 2𝑝𝑜𝑝

′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 = 𝑝2

Substituting this 𝑝2 in the equation :

(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐2 = 𝑝2𝑐2, one gets

(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐2 = (𝑝𝑜2 − 2𝑝𝑜𝑝

′ 𝑐𝑜𝑠 𝜃 + 𝑝′2)𝑐2

Substituting photon energies and photon momenta one gets

(ℎ𝑐

𝜆𝑜−

ℎ𝑐

𝜆′)2

+ 2(ℎ𝑐

𝜆𝑜−

ℎ𝑐

𝜆′) 𝑚𝑐2 = (

ℎ𝑐

𝜆𝑜)2

− 2(ℎ𝑐

𝜆𝑜) (

ℎ𝑐

𝜆′) 𝑐𝑜𝑠 𝜃 + (ℎ𝑐

𝜆′)2

Simplifying one gets

(ℎ𝑐

𝜆𝑜)2

− 2 (ℎ𝑐

𝜆𝑜) (ℎ𝑐

𝜆′) + (ℎ𝑐

𝜆′)2

+ 2 ℎ𝑐 ( 1

𝜆𝑜− 1

𝜆′) 𝑚𝑐2 = ( ℎ𝑐

𝜆𝑜)2

− 2( ℎ𝑐

𝜆𝑜) (ℎ𝑐

𝜆′) 𝑐𝑜𝑠 𝜃 + (ℎ𝑐

𝜆′)2

i.e., −ℎ𝑐

𝜆𝑜𝜆′ + ( 1

𝜆𝑜− 1

𝜆′) 𝑚𝑐2 = −ℎ𝑐

𝜆𝑜𝜆′ 𝑐𝑜𝑠 𝜃

OR, (𝜆′−𝜆𝑜 𝜆𝑜𝜆′) 𝑚𝑐2 =

ℎ𝑐

𝜆𝑜𝜆′(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,

has a non-relativistic kinetic energy 1

2𝑚 𝑣2 = 𝑒 ∆𝑉where e is electron charge.

Compton shift:

𝝀′ − 𝝀𝒐 = 𝒉

𝒎𝒄(𝟏 − 𝒄𝒐𝒔𝜽)



Hence, the momentum (p) of an electron accelerated through a potential difference of V is

𝑝 = 𝑚 𝑣 = √2 𝑚 𝑒 ∆𝑉

𝜆 = ℎ

𝑝

Frequency of the matter wave associated with the particle is𝐸

ℎ , 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).



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(𝑘1𝑥 − 𝜔1𝑡) and 𝑦2 = 𝐴 𝑐𝑜𝑠(𝑘2𝑥 − 𝜔2𝑡)

where 𝑘 = 2𝜋/𝜆 , 𝜔 = 2𝜋𝑓 The resultant wave y = y1 + y2

𝑦 = 2𝐴 [𝑐𝑜𝑠 (𝛥𝑘

2𝑥 −

𝛥𝜔

2𝑡) 𝑐𝑜𝑠 (

𝑘1+𝑘2

2𝑥 −

𝜔1+𝜔2

2𝑡)]

where k = k1 – k2 and = 1 – 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

𝑣𝑝 = 𝑓 𝜆 or 𝑣𝑝 = 𝜔

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)

(𝛥𝑘2

) =

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):

𝑣𝑃 = 𝜔

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



𝑣𝑔 = 𝑣𝑃 − 𝜆 ( 𝑑𝑣𝑃

𝑑𝜆)

Relation between group speed (vg) and particle speed (u):

𝜔 = 2 𝜋 𝑓 = 2 𝜋 𝐸

ℎ and 𝑘 =

2 𝜋

𝜆 =

2 𝜋

ℎ 𝑝⁄ =

2 𝜋 𝑝

ℎ

𝑣𝑔 = 𝑑𝜔

𝑑𝑘 =

2 𝜋

ℎ 𝑑𝐸

2 𝜋 ℎ

𝑑𝑝 =

𝑑 𝐸

𝑑 𝑝

For a classical particle moving with speed u, the kinetic energy E is given by

𝐸 = 1

2 𝑚 𝑢2 =

𝑝2

2 𝑚 and 𝑑𝐸 =

2 𝑝 𝑑𝑝

2 𝑚 or

𝑑 𝐸

𝑑 𝑝 =

𝑝

𝑚 = 𝑢

𝑣𝑔 = 𝑑𝜔

𝑑𝑘 =

𝑑 𝐸

𝑑 𝑝 = 𝑢

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.

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 mono-

energetic 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

2 𝑝 𝑑

Fig. 3.15 (a) Schematic of eelectron beam interference experiment, (b) Photograph of a

double-slitinterference pattern produced by electrons



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.



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 [1 EACH]

2 Sketch schematically the graph of wavelength vs intensity of radiation

from a blackbody. [1]

3 Explain Planck’s radiation law. [2]

4 Write the assumptions made in Planck’s hypothesis of blackbody

radiation. [2]

3.2. THE PHOTOELECTRIC EFFECT

5 (a) Explain photoelectric effect.

(b) What is work function of metal? [1 EACH]

6 What are the observations in the experiment on photoelectric effect? [5]

7 What are the classical predictions about the photoelectric effect? [3]

8 Explain Einstein’s photoelectric equation. [2]

9 Which are the features of photoelectric effect-experiment explained by

Einstein’s photoelectric equation? [2]

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. [1EACH]

3.3 THE COMPTON EFFECT

11 Explain Compton effect. [2]

12 Explain the experiment on Compton effect. [5]

13 Derive the Compton shift equation. [5]

3.4 PHOTONS AND ELECTROMAGNETIC WAVES

14 Explain the wave properties of the particles. [2]


15 Explain a wave packet and represent it schematically. [2]

16 Explain (a) group speed (b) phase speed [1+1]

17 Show that the group speed of a wave packet is equal to the particle speed. [3]

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. [1]

20 Write the equations for uncertainty in (a) position and momentum (b)

energy and time. [1]



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



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.



Ans: 43°, 43°, 0.602 MeV, 3.21 x 10–22 kg.m/s, 0.278 MeV, 3.21 x 10–22 kg.m/s

3.4 PHOTONS AND ELECTROMAGNETIC WAVES

17 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

18 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

19 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


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.



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



CHAPTER 4

QUANTUM MECHANICS

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

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.



some instant. If represents a single particle, then ||2 -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

𝑃𝑎𝑏 = ∫ |𝜓|2 𝑑𝑥𝑏

𝑎 .

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



−ℏ2

2 𝑚

𝑑2𝜓

𝑑𝑥2+ 𝑈 𝜓 = 𝐸 𝜓

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, time-

independent 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

𝑑2𝜓

𝑑𝑥2 +

2 𝑚

ℏ2 𝐸 𝜓 = 0

Or 𝑑2𝜓

𝑑𝑥2 = − 𝑘2 𝜓 , where 𝑘2 = 2 𝑚 𝐸

ℏ2 or 𝑘 = √ 2 𝑚 𝐸

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 ,



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 ( 𝑛 𝜋 𝑥

𝐿)

To find the constant A, apply normalization condition

∫ |ψ|2 dx+∞

−∞ = 1 or ∫ 𝐴2 [𝑠𝑖𝑛 (

𝑛 𝜋 𝑥

𝐿)]

2

𝑑𝑥𝐿

0 = 1 .

𝐴2 ∫ 12[1 − 𝑐𝑜𝑠(

2 𝑛 𝜋 𝑥

𝐿)] 𝑑𝑥

𝐿

0 = 1

Solving we get 𝐴 = √ 2

𝐿

Thus 𝜓𝑛(𝑥) = √2

𝐿𝑠𝑖𝑛 (

𝑛 𝜋 𝑥

𝐿) is the wave function for particle in a box.

Since 𝑘 = √ 2 𝑚 𝐸

ℏ and k L = n π

We get, √ 2 𝑚 𝐸

ℏ 𝐿 = 𝑛 𝜋 .

∴ 𝐸𝑛 = (ℎ2

8 𝑚 𝐿2) 𝑛2 , n = 1, 2, 3, . . . . .

Each value of the integer n corresponds to a quantized energy value, En .

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.



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

𝑑2𝜓

𝑑𝑥2 =

2 𝑚

ℏ2 (𝑈 − 𝐸) 𝜓 , or

𝑑2𝜓

𝑑𝑥2 = 𝐶2 𝜓 where 𝐶2 =

2 𝑚

ℏ2 (𝑈 −

𝐸)

General solution of the above equation is

(x) = AeCx + B e−Cx



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

III = Be-Cxfor 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

𝑑2𝜓𝐼𝐼

𝑑𝑥2 + ( 2 𝑚

ℏ2 𝐸)𝜓𝐼𝐼 = 0 , 2 𝑚 𝐸

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:

At x = 0 , I(0) = II(0) and [𝑑𝜓𝐼

𝑑𝑥]𝑥=0

= [𝑑𝜓𝐼𝐼

𝑑𝑥]𝑥=0

At x = L , II(L) = III(L) and [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.



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

𝑈(𝑟) = − 𝑘𝑒𝑒

2

𝑟

where ke = 1/40= 8.99 x 109 N.m2/C2 Coulomb constant and r is radial distance of

electron from H-nucleus.



Fig. 4.9 Spherical polar coordinate system

The time-independent Schrödinger equation in 3-dimensional space is

−ℏ2

2 𝑚 (𝜕2𝜓

𝜕𝑥2+

𝜕2𝜓

𝜕𝑦2+

𝜕2𝜓

𝜕𝑧2) + 𝑈 𝜓 = 𝐸 𝜓

Since U has spherical symmetry, it is easier to solve the Schrödinger equation in

spherical polar coordinates (r, ,)where 𝑟 = √𝑥2 + 𝑦2 + 𝑧2 ,

is the angle between z-axis and �⃗� .

is the angle between the x-axis and the projection of �⃗� onto the xy-plane.

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,

𝐸𝑛 = − ( 𝑘𝑒𝑒

2

2 𝑎𝑜)

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



n = 5 O shell = 4 g subshell

n = 6 P shell = 5 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 𝑎𝑜3𝑒𝑥𝑝 (− 𝑟

𝑎𝑜) where ao is Bohr radius ( = 0.0529 nm).

|1s|2 is the probability density for H-atom in 1s-state:

|𝜓1𝑠|2 =

1

𝜋 𝑎𝑜3 𝑒𝑥𝑝 (−

2 𝑟

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 (− 2 𝑟

𝑎𝑜)



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(2 −

𝑟

𝑎𝑜) 𝑒𝑥𝑝 (−

𝑟

𝑎𝑜)

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



EXERCISE

QUESTIONS

4.1 AN INTERPRETATION OF QUANTUM MECHANICS

1 What is a wave function ? What is its physical interpretation ? [2]

2 What are the mathematical features of a wave function? [2]

4.2 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. [5]

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]

5 Sketch the lowest three energy states, wave-functions, probability

densities for the particle in a one-dimensional “box”. [3]

6 The wave-function for a particle confined to moving in a one-dimensional

box is

ψ(x) = A sin(nπx

L) . Use the normalization condition on to show that

𝐴 = √2

𝐿 .

[2]

7 The wave-function of an electron is ψ(x) = A sin(nπx

L) . Obtain an

expression for the probability of finding the electron between x = a and

x = b. [3]

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. [5]

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. [3]

4.5 TUNNELING THROUGH A POTENTIAL ENERGY BARRIER

10 Give a brief account of tunneling of a particle through a potential energy

barrier. [4]


11 Give a brief account of quantum model of H-atom. [2]

4.7 THE WAVE FUNCTIONS FOR HYDROGEN



12 The wave function for H-atom in ground state isψ1S(r) =

1

√πao3exp (− r

ao) . Obtain an expression for the radial probability density

of H-atom in ground state.

Sketch schematically the plot of this vs. radial distance. [4]

13 The wave function for H-atom in 2s state is ψ2S(r) =

1

√32πao3(2 − 𝑟

𝑎𝑜) exp (− r

ao) . Write the expression for the radial

probability density of H-atom in 2s state. Sketch schematically the plot

of this vs. radial distance. [3]

14 Sketch schematically the plot of the radial probability density vs. radial

distance for H-atom in 1s-state and 2s-state. [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



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


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 𝑘𝑒

2𝑞12𝑞2

2

2 ℏ2𝑛2, 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 =m1m2

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



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

2 𝑚 (𝜕2𝜓

𝜕𝑟2+

2

𝑟

𝜕𝜓

𝜕𝑟) −

𝑘𝑒𝑒2

𝑟 𝜓 = 𝐸 𝜓 . Show that the 1s

wave function for an electron in H-atom 𝜓1𝑆(𝑟) = 1

√𝜋𝑎𝑜3𝑒𝑥𝑝 (− 𝑟

𝑎𝑜) satisfies the

schrodinger equation.



CHAPTER 5

SOLID STATE PHYSICS

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

1

kT

EEexp

1Ef

F

[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 >

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



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,

2

2

222

2

2

n nLm2

nLm8

hE

n = 1, 2, 3 . . . [5.2]

An electron moving freely in a metal cube of side L, can be modeled as particle in a three-

dimensional box. It can be shown that the energy for such an electron is

2z

2y

2x2

22

nnnLm2

E

[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



2

o

2z

2y

2x n

E

Ennn [5.4]

and whereo

2

22

oE

En

Lm2E

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.

dnn2

1n4G(E) dE 22 dn

8

1 =

relation the using o

21

oo

21

E

En

E

Ed

E

EdE)E(G

dEEEdEEEE

EdE)E(G 2

12

3

o412

1

212

1

o

o

21

dEEmL2

dE)E(G 21

23

2

22

41

VL,dEELm

2

2dE)E(G 32

1

32

323

Number of states per unit volume per unit energy range, called density of states, g(E) is

given by

g(E) = G(E)/V

dEEm

2

2dE

V

)E(GdE)E(g 2

1

32

23

2

hdEE

h

m24dE)E(g 2

1

3

23

Finally, we multiply by 2 for the two possible spin states of each particle.



dEEh

m28dE)E(g 2

1

3

23

[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)

1Tk

EEexp

dEE

h

m28dE)E(N

F

21

3

23

[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

0 F

21

3

23

0

e

1Tk

EEexp

dEE

h

m28dE)E(Nn

[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

23

F3

23

23

F32

3

23E

0

21

3

23

e Eh3

m216E

h

m28dEE

h

m28n

F [5.8]

Solving for Fermi energy at 0K, we obtain



3

2

e2

F8

n3

m2

h0E

[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 )r(S

and )r(S are valid for such an atom [ )r(S

and )r(S 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).



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



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



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).



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



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



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.



EXERCISES

QUESTIONS


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.

[3]

2 Derive an expression for density-of-states. [5]

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

dEEh

m28dE)E(g 2

1

3

23

]

[5]

5.2 BAND THEORY OF SOLIDS

4 Explain the formation of energy bands in solids with necessary diagrams. [5]

5.3 ELECTRICAL CONDUCTION IN METALS, INSULATORS AND

SEMICONDUCTORS

5 Distinguish between conductors, insulators and semiconductors on the

basis of band theory [3]

6 Indicate the position of (a) donor levels (b) acceptor levels, in the

energy band diagram of a semiconductor. [2]

7 What is the difference between p-type and n-type semiconductors?

Explain with band diagram.

[3]

8 With necessary diagrams, explain doping in semiconductors. [5]

8 Why the electrical conductivity of an intrinsic semiconductor increases

with increasing temperature? [2]


9 What are superconductors? Draw a representative graph of Resistance Vs

Temperature for a superconductor. [2]

10 Explain Meissner effect. [3]

11 Explain briefly the BCS theory of superconductivity in metals. [3]



PROBLEMS


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

dE)E(NEn

1E

e

AV

where the density of particles

0

e dE)E(Nn

1Tk

EEexp

dEE

h

m28dE)E(N

F

2

1

3

2

3

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



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


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