Final m4 march 2019

Module 4 : Quantum Mechanics & LASER
CONTENTS
I. Quantum Mechanics
1. Introduction to Quantum Mechanics
 Introduction
 Planck’s Law for Energy Density
2. Wave Nature of Particles
 de Broglie’s Hypothesis (De Broglie’s Concept of Matter Wave)
 de Broglie’s Wavelength
 de Broglie’s Wavelength - Extended to Accelerated Electron
3. Heisenberg’s Uncertainty Principle and Applications
 Statement with three equations
 Application of Heisenberg’s Uncertainty Principle
(Non-confinement of the Electrons in the Nucleus)
4. Schrodinger’s Time Independent Wave Equation
5. Applications of Schrodinger’s Wave Equation
 Particle in an One Dimensional Potential Well of Infinite Height
 Eigen Values, Eigen Function & Probability Density for Ground State & First Two
Excited States
6. Wave Function
 Statement
 Physical Significance (Probability Density-Max Born Interpretation & Normalization)
 Properties of Wave Function
7. Numerical Problems
II. LASER
1. Review of Spontaneous & Stimulated Processes
2. Energy Density Using Einstein’s Coefficients
3. Requisites of a Laser System
4. Condition for Laser Action
5. Principle, Construction & Working of CO2 Laser
6. Principle, Construction & Working of Semiconductor Laser
7. Applications of Lasers
 Laser Range Finder (Defense)
 Data Storage (Engineering)
III. Question Bank

MODULE - 4 QUANTUM MECHANICS & LASER
Dr. DIVAKARA S, PROF. & HEAD, DEPT. OF PHYSICS, VVCE, MYSURU Page 2
Books Referred:
1. Concepts of Modern Physics By Arthur Beiser
2. Modern Physics By Kenneth S Krane
3. Quantum Mechanics By Robert Eisberg & Robert Resnick
4. Basic Concepts of Quantum Mechanics By Tarasov L
5. Engineering Physics By S P Basavaraju
6. Quantum Mechanics By Saxena
7. Engineering Physics By M N Avadhanulu & P G Kshirsagar
8. Engineering Physics By Gaur & Gupta
9. Laser & Non-linear Optics By B B Laud
10. Laser Fundamental By Silfvast & William T
11. Lasers & Optical Instruments By Nagabhushan S & Sathyanarayana N
12. Atom, Laser & Spectroscopy By S N Thakar & D K Pai
13. Lasers & its Applications By L Tasasov
14. Optoelectronics & Photonics By Safa O Kasap
15. IBM Red Book – Optical Fiber Communications

Quantum Mechanics
1. Introduction to Quantum Mechanics
In the early years of 20th
century, Max Planck, Albert Einstein, Louis de Broglie, Neils
Bohr, Werner Heisenberg, Erwin Schrodinger, Max Born, Pauli Dirac and others created
the theory now known as quantum mechanics. The History is as follows,
 Planck‟s Black Body Theory (1900)
 Einstein‟s Light Quanta (1905)
 Bohr‟s Model of the Hydrogen Atom (1913)
 de Broglie‟s Hypothesis (1924)
 Schrodinger‟s Wave Equation (1926)
 Heisenberg‟s Uncertainty Principle (1927)
a) Black Body Radiation
 A perfect black body is one which absorbs the radiation of all wavelengths
incident on it.
 The nature of the black body radiation depends only on the temperature.
 Wien studied the spectrum of the radiation issued by a black body by heating
it to higher and higher temperature. He found that the spectrum did include
wide range of wavelengths.
 When a graph of intensity versus wavelength was plotted for the radiation
emitted by the black body at different temperature.
 The black body
radiates mainly in
visible and infrared
region and beyond
ultra-violet region the
energy radiated is very
less.

b) Wien’s Law of Black Body Radiation
 The Wein‟s Law suits only for the shorter wavelength region and high
temperature value of the source.
 Also energy emitted by a black body tends to zero at very high temperature
which is contradiction with the experimental results
c) Rayleigh Jeans Law of Black Body Radiation
 The Rayleigh Jeans Law suits only for longer wavelength region.
 Rayleigh Jeans law predicts that, black body radiates enormous amount of
energy in the shorter wavelength region so that no energy is available for
emission in the longer wavelength region.
 At short wavelengths, there was a major disagreement between the Rayleigh-
Jeans law and experiment. This mismatch became known as the ultraviolet
catastrophe

d) Planck’s Law for Energy Density (Assumptions of Quantum Theory of
Radiation)
A black body consists of a very large number of electrical oscillations, with each
oscillator vibrating with a frequency of its own.
 The value of energy possessed by an oscillator, which is an integral multiple
of hν.
E = nhν
Where „h‟ is Planck‟s constant
„ν‟ is the frequency of vibration and
n=0, 1, 2, …….. etc
 An oscillator may lose or gain energy by absorbing or emitting the a radiation
of frequency „ν‟ whose value is given by,
ν =
Based on the above ideas, he gave a Planck‟s Radiation Law for Energy Density ,
[ ]
Where is the energy per unit volume for wavelengths in the range λ and
λ + dλ.
 Planck‟s Law verifies both shorter (Wein‟s law) and Longer wavelength
(Rayleigh Jean‟s Law) regions.

2. Wave Nature of Particles
 Light exhibits the phenomenon of interference, diffraction, polarization, photoelectric
effect, Raman effect and Compton effect.
 The phenomenon of interference, diffraction and polarization could only be explained
on the basis of „Wave theory of light‟. Thus, this phenomenon indicate that light
possess “Wave Nature”.
 There are certain other phenomenon like photoelectric effect, Raman effect and
Compton effect which could not be explained by wave theory of light and could be
explained through the quantum theory of light. According to which a beam of light
consisting of small packet each having energy „hν‟. These packets are called photon‟s
which behave like particle. Thus, this phenomenon possesses „Particle Nature‟.
 Thus we can conclude that, light exhibits dual nature. It behaves like a wave under
some circumstances and particle under other circumstances. Hence it can be noted
that, light cannot exhibit both the nature simultaneously.
I. de Broglie’s Hypothesis
a) Statement : Every moving material entity having momentum (p) has associated with
it a wave whose wavelength is inversely proportional to the momentum.
i.e.,
b) de Broglie’s Wavelength
From the experiment of interference, diffraction and polarization, it is shown that like
is a wave form. The energy of radiation,
E = hν (1)
Where „ν‟ is the frequency of radiation
The phenomenon of photoelectric effect, Raman effect and Compton effect could be
explained by considering a photon to be a particle of mass „m‟ and its energy
E = mc2
(2)
From equ. (1) & (2),
( )

In general,
The above equation is the de Broglie’s Matter Wave Equation.
c) de Broglie’s Wavelength - Extended to Accelerated Electron
Consider an electron of mass „m‟ under a potential difference of „V‟. Its kinetic
energy is,
(1)
Energy of an electron accelerated under a potential difference of V is,
(2)
From equ. (1) & (2),
√
From De Brogie‟s Hypothesis,
√
√
√
λ =
𝑚𝑣
𝝀
𝟏 𝟐𝟐𝟔𝒏𝒎
√𝑽

3. Heisenberg’s Uncertainty Principle
 A moving particle as a wave group implies that, there are fundamental limits to
the accuracy with which we can measure such particle properties as position and
momentum.
 The particle corresponding to wave group may be located anywhere within the
group at a given time.
 Figure (a) shows a narrow de Broglie‟s wave group; The particle position can be
precisely determined but the wavelength (momentum „p = h/λ‟) cannot be
established because there are not enough waves to measure accurately.
 Figure (b) shows a wide wave group; Now the wavelength can be precisely
determined (more waves) but not the position of the particle.
Thus, we have Heisenberg‟s Uncertainty Principle.
a) Statement
 It is not possible to make simultaneous determination of the position and the
momentum of the particle with unlimited precision (accuracy).
 The product of uncertainty in the position of an object at some instant (∆x) and
the uncertainty in its momentum at same instant(∆p) is equal or greater than h/4 .
i.e. ∆x ∆p ≥ h/4
 The uncertainty involved in the measurement of energy (∆E) and time (∆t) is,
i.e. ∆E ∆t ≥ h/4
 Similarly for angular momentum (∆L) and angle (∆ ) is,
i.e. ∆L ∆ ≥ h/4

b) Physical Significance
 Any instruments cannot measure the quantities more accurately than predicted by
Heisenberg‟s Uncertainty Principles.
 Planck‟s constant is so small that the limitation imposed by the uncertainty
principle is great significant only in the realm of atom.
 It signifies that, the electron exist within the nuclei of an atom.
c) Applications of Heisenberg’s Uncertainty Principle
 Non Confinement of an electron in the nucleus
From Heisenberg‟s Uncertainty Principle,
(1)
We know that, diameter of nucleus is of the order of 10-14
m. Thus, if an electron
is to exist inside the nucleus, then the uncertainty in its position must not exceed
( ) this value, i.e.
Substitute value in equ. (1),
Since the momentum of the electron must at least be equal to the Uncertainty in
the momentum (p=∆p).
An electron may exist within the nucleus but its energy must be,
 This means to say that, in order that an electron may exist inside the nucleus, its
energy must be greater than or equal to 85MeV.
 But the experimental investigation on beta decay emission reveals that, the
kinetic energy of the beta particle is of the order 3 to 4 MeV.
 This clearly indicates that, the emitted electron cannot be the ones which
existed within the nucleus.

4. One Dimensional, Time Independent Schrodinger’s Wave Equation
Consider a wave equation for a de Broglie‟s wave travelling along X-axis direction can
be written in complex notation as,
(1)
Where „ψ‟ is the wave function, „A‟ is constant, „ω‟ is angular frequency and „k‟ is wave
number of the wave.
Differentiate the equation (1) twice with respect to „t‟,
(2)
Consider a travelling wave equation travelling in X-axis direction,
Where „y‟ is the displacement and „v‟ is the velocity of the wave.
Similarly, for de Broglie‟s wave
(3)
Substituting equ. (2) in equ. (3),
(4)
We know from de Broglie‟s Hypothesis,

( ) (5)
We know that, the total energy (E) is the sum of Potential energy (V) and Kinetic energy
(1/2 mv2
).
(6)
(7)
* +
Since equation (8) is independent of time than the above equation can be written as,
This is the time independent Schrodinger‟s wave equation using complex notation for a
particle in one dimension.
𝒅 𝟐
𝝍
𝒅𝒙 𝟐 +
𝟖𝒎𝝅 𝟐
𝒉 𝟐 𝑬 𝑽 𝝍 𝟎

5. Application of Schrodinger’s Wave Equation
Particle in an One Dimensional Potential Well of Infinite Height
or
(Particle in an one dimensional box)
Diagram :
To ∞ To ∞
V = ∞
Ψ = 0
a
V = 0
Ψ 0
●
Particle
V = ∞
Ψ = 0
X = 0 X = a
Explanation :
 Consider a particle moving inside a one dimensional box of length 'a‟.
 The particle is free to move between the walls of the box X = 0 and X = a.
 Potential energy of the particle (V)
V = 0 for 0 < X < a
V = ∞ for X 0
V = ∞ for X a
 Wave function (Ψ)
Ψ 0 for 0 < X < a
Ψ = 0 for X 0
Ψ = 0 for X a
[Ψ = 0 for X 0 & X a as the particle is always present inside the box]

Derivation :
Consider a one dimensional, time independent Schrodinger‟s wave equation,
+
V = 0 inside the box,
+
+ (1)
Where (2)
The general solution for equation (1),
Ψ = A Sinkx + B Coskx (3)
Where A and B are constants. The value of these constant can be evaluated by
applying boundary conditions.
(a) Ψ = 0 at X = 0
From equ. (3),
0 = A Sink(0) + B Cosk(0)
B = 0 (4)
(b) Ψ = 0 at X = a and B = 0
From equ. (3),
0 = A Sinka + 0 Coska
0 = A Sin(ka)
ka = Sin-1
0
ka = n
k = (5)
Substituting equ. (4) & equ. (5) in equ. (3),
(6)
For every value of n = 1, 2, 3, 4, …………., there will be a corresponding value of
Wave functions are called Eigen function.
𝜓 𝒏 𝑨 𝑺𝒊𝒏 (
𝒏𝝅
𝒂
) 𝒙

(7)
For every value of n = 1, 2, 3, 4, …………., there will be a corresponding value of
Energy values are called Eigen value.
Normalisation of wave function
From normalisation condition,
∫ | |
∫ | n (
n
) |
∫ ( )
∫ [ ( ) ]
[
( )
( )
]
[ ]
[ ]
√
Normalized wave function of the particle,
𝐸 𝒏
𝒏 𝟐
𝒉 𝟐
𝟖𝒎𝒂 𝟐
𝜓 𝒏
𝟐
𝒂
𝑺𝒊𝒏 (
𝒏𝝅
𝒂
) 𝒙

Eigen Functions, Probability Density and Energy Values for a Particle in an Infinite Potential Well
Case 1: n = 1 ground State, least energy state, first quantum state, zero point energy
state
(a) Eigen Function √ ( )
Ψ1 = 0 for x = 0 and x =a
Ψ2 = Maximum for x = a/2
(b) Probability Density
| |2
= 0 at x = 0, a
| |2
= Maximum at x = a/2
(c) Energy of Particle (E1)
Case 2: n = 2 1st
Excited State, second quantum state, next to ground state
(a) Eigen Function √ ( )
Ψ2 = 0 for x = 0, a/2, a
Ψ2 = Maximum for x = a/4, 3a/4

| |2
= 0 at x = 0, a/2, a
| |2
= Maximum at x = a/4, 3a/4
Case 3: n = 3 2nd
Excited State, third quantum state
(b) Eigen Function √ ( )
Ψ3 = 0 for x = 0, a/3, 2a/3, a
Ψ3 = Maximum for x = a/6, 3a/6, 5a/6
| |2
= 0 at x = 0, a/3, 2a/3, a
| |2
= Maximum at x = a/6, 3a/6, 5a/6

6. Wave Function
a) Statement
 The quantity whose variation makes up matter wave is called wave function (ψ).
 For Example : 1. In water waves, the quantity varies periodically is the height of
the Water.
2. In sound waves, it is pressure.
 The value of wave function associated with a moving body at a particular point X,
Y, Z in space at time „t‟ is related to the likelihood of finding the body at the time.
 Wave function is denoted by ψ (X, Y, Z, t)
b) Physical Significance of wave function
 Ψ is a complex quantity and has no direct physical significance by itself but if we
consider Ψ Ψ*
we will get a real quantity.
 The wave function Ψ has no direct physical significance by itself. There is a
reason, the probability something be in a certain place at a given time must lie
between 0 (the object is not there) and 1 (there).
 An intermediate probability, say 0.2 means that there is 20% chance of finding the
object.
(i) Probability Density | | (Max Born Interpretations)
Wave function (Ψ) is a complex quantity, +
Where a & b are the real function of the variable (X, Y, Z, t) & i=√
The complex conjugate of Ψ is Ψ*,
Then, +
| | | + |
The probability of finding the particle per unit volume is called probability
density (| | ).
| |
| |

(ii) Normalization of Wave Function
The probability per unit length of finding particle at position „x‟ and time
„t‟ is,
| |
Probability of finding the particle at (x,t)
| |
Along X-axis, -∞ < x < ∞
∫ | |
If the particle does not exist in space then P = 0
If the particle exist somewhere at all-time P = 1
∫ | |
For three dimensions,
∫ | |
The above equation is called condition for normalization and any wave
function satisfies the above condition then such wave function is called
normalized wave function.

c) Properties of wave function
(i) Ψ is single valued everywhere
(ii) Ψ is finite everywhere
(iii) Ψ and its first derivatives with respect to its variable are continuous
everywhere
(iv) For bound states, ψ must vanish at infinite. If ψ is a complex function then
ψψ* must vanish at infinity.
(v) If the wave function f(x) satisfies the above properties then it is called as
Eigen function.
(vi) Ψ must be normalized, which means that ψ must go to zero as
inorder that ∫| | over all space be a finite constant.
From graph; At x=P, f(x) has 3 values (f1, f2,
f3). i.e. if f(x) were to be a wave function,
then the probabilities of finding the particle
has different values at same location hence
the wave function is not acceptable.
From graph; At x=R, f(x) = ∞. Thus f(x) were
to be a wave function, then it signifies a large
probabilities of finding the particle at x = R
which violate the uncertainty principle and
wave function becomes unacceptable.
From graph; At x=Q, f(x) is truncated at
A and restarts at B. Between A & B, it is
not defined. If f(x) were to be a wave
function, then the state of the system at
x= Q cannot be discontinued. Hence the
wave function is not acceptable.

Formulas at a glance
 E = hν =
 [ ]

√ √

√
 ∆x ∆p ≥ h/4 ( ) = -
 ∆E ∆t ≥ h/4
 ∆L ∆ ≥ h/4

 √

 √ n ( )
 | |
 ∫ | |
 ∫ | |

01 Calculate the energy density per unit wavelength in a blackbody cavity at 1500K at a
wavelength of 6000Å.
02 Compute the de Broglie’s wavelength for a neutron moving with one tenth part of the velocity
of light.
03 Calculate the de Broglie’s wavelength associated with 400g cricket ball with a speed of
90Km/hr.
v
04 Compare the de Broglie’s wavelength associated with (a) 10g bullet travelling at 500m/s. (b)
An electron with kinetic energy 100Mev.
05 Calculate the wavelength associated with the electron carrying an energy 2000ev and also
calculate electron momentum.
𝑇 𝐾
𝑈𝜆 ?
Sol :
𝜆 Å
𝑈𝜆
𝜋 𝑐
𝜆
[
𝑒
𝜈
𝑘𝑇
]
𝑈𝜆
𝜋 𝑋 𝑋 𝑋 𝑋
𝑋
𝑒
𝑋 𝑋 𝑋 8
𝑋 𝑋 𝑋 𝑋
𝑈𝜆 𝟕 𝟐𝟑 𝑱/𝒎 𝟒
𝑣 𝐶 𝑋 7
𝑚/𝑠
𝜆 ?
Sol :
𝜆
𝑝 𝑚𝑣
𝑋
𝑋 𝑋 𝑋
𝟏 𝟑𝟐 𝑿 𝟏𝟎 𝟏𝟒
𝒎
𝑚 𝑔 𝐾𝑔
𝑣
𝐾𝑚
𝑟
𝑣
𝑋
𝑋
𝑚/𝑠
𝜆 ?
Sol :
𝜆
𝑝 𝑚𝑣
𝑋
𝑋
𝟔 𝟔𝟐𝟓 𝑿 𝟏𝟎 𝟑𝟓
𝒎
𝑎 𝑚 𝑔 𝑣 𝑚/𝑠
𝑏 𝐸 𝑀𝑒𝑉
𝜆 𝐵/𝜆 𝑒 ?
Sol :
𝐸 𝑋 𝑋 𝑋 J
= 𝑋 𝑋 J
𝜆 𝐵
𝑝 𝑚𝑣
𝑋
𝑋
𝟏 𝟑𝟐𝟓 𝑿 𝟏𝟎 𝟑𝟒
𝒎
𝜆 𝑒
√ 𝑚𝐸
𝑋
√ 𝑋 𝑋 𝑋 𝑋
𝟏 𝟐𝟐𝟕 𝑿 𝟏𝟎 𝟏𝟑
𝒎
𝜆 𝐵
𝜆 𝑒
𝟏 𝟎𝟖 𝑿 𝟏𝟎 𝟐𝟏
𝝀 𝑩 < 𝝀 𝒆
𝐸 𝑒𝑉
𝜆 ?
Sol :
𝐸 𝑋 𝑋 J
p=?
𝜆
√ 𝑚𝐸
𝑋
√ 𝑋 𝑋 𝑋 𝑋 𝑋
𝟎 𝟐𝟕𝟒Å
𝐸
𝑝
𝑚
𝒑 √𝟐𝒎𝑬 √ 𝑋 𝑋 𝑋 𝑋 𝑋 𝟐 𝟒𝟏𝟓 𝑿 𝟏𝟎 𝟐𝟑
𝑵𝒔

06 Estimate the potential difference through which a proton is needed to be accelerated so that its
de Broglie’s wavelength becomes equal to 1Å.
07 Calculate the de Broglie’s wavelength of Helium nucleus (Alpha particle) that is accelerated
through 500V.
08 A particle of mass 0.5Mev/c2 has a kinetic energy 100ev. Find its de Broglie’s wavelength.
09 Calculate the de Broglie’s wavelength of thermal neutron at 300K.
10 Compare the energy of photon with that of a neutron when both are associated with a
wavelength of 1Å.
11 Calculate the de Broglie’s wavelength of proton whose kinetic energy is equal to the rest mass
energy of electron and mass of proton is equal to 1836 times the mass of electron.
𝜆 Å
Sol :
V=?
𝜆
√ 𝑚𝑒𝑉
𝑉
𝑚𝑒 𝜆
𝑋
𝑋 𝑋 𝑋 𝑋
𝟎 𝟎𝟖𝟐𝑽
𝜆 ?
𝑀𝑎𝑠𝑠 𝑚
𝑚 𝑝 + 𝑚 𝑛
𝑚 ≈ 𝑚 𝑛
Sol :
V=500v
Charge=2e
𝜆
√ 𝑚𝑒𝑉
𝑋
√ 𝑋 𝑋 𝑋 𝑋 𝑋 𝑋 𝑋
𝜆 𝟒 𝟓𝟑 𝑿 𝟏𝟎 𝟏𝟑
𝑽
𝜆 ?
𝑚
𝑋 𝑋 𝑋
𝑋
𝑚 𝑋 𝐾𝑔
𝐸 𝑒𝑉 𝑋 7
𝐽
Sol :
m=0.5MeV/c2
𝜆
√ 𝑚𝐸
𝑋
𝜆 𝟏 𝟐𝟒Å
𝜆 ?
𝑇 𝐾
Sol :
𝜆
√ 𝑚𝑘𝑇
𝑋
𝟏 𝟒𝟓𝟑Å
𝜆 Å
𝐸 𝑃 𝑜𝑡𝑜𝑛
𝐸 𝑁𝑒𝑢𝑡𝑟𝑜𝑛
?
Sol : 𝐸 𝑝 𝑜𝑡𝑜𝑛
𝑐
𝜆
𝑋 𝑋 𝑋
𝑋
𝟏 𝟗𝟖𝟗 𝑿 𝟏𝟎 𝟏𝟓
𝑱
𝜆
√ 𝑚𝐸
𝐸 𝑛𝑒𝑢𝑡𝑟𝑜𝑛
𝑚 𝜆
𝑋
𝑋 𝑋 𝑋 𝑋
𝟏 𝟑𝟏𝟐 𝑿 𝟏𝟎 𝟐𝟎
𝑱
𝐸 𝑃 𝑜𝑡𝑜𝑛
𝐸 𝑁𝑒𝑢𝑡𝑟𝑜𝑛
𝟏 𝟓𝟏𝟓 X 𝟏𝟎 𝟓
𝜆 ?
𝑚 𝑝 𝑚 𝑒 𝑋 7
𝐾𝑔
𝐸 𝑝 𝑚 𝐶 𝑋 𝐽
Sol :
𝜆
√ 𝑚𝐸
𝑋
𝟒 𝑿 𝟏𝟎 𝟏𝟒
𝒎

12 Compare the momentum, the total energy and the kinetic energy of an electron with a
wavelength of 1Å with that of a photon with same wavelength.
13 For a particle moving in a free space, prove that √ where id the de Broglie’s
wavelength and γ is the frequency associated with the quantum energy carried by the electron.
14 A particle of mass 0.65Mev/c2
has a kinetic energy 80ev. Find the de Broglie’s wavelength and
particle velocity of the wave.
15 Find the de Broglie’s wavelength of an electron accelerated through a potential difference of
182V and object of mass 1Kg moving with a speed of 1m/s. compare the result and comment.
𝜆 Å
𝑃𝑒
𝑃 𝑝 𝑜
?
𝐸 𝑒
𝐸 𝑝 𝑜
?
𝐾𝐸 𝑜𝑓 𝐸 𝑒
𝐾𝐸 𝑜𝑓 𝐸 𝑝 𝑜
?
Sol :
𝑎 𝜆
𝑝
𝑝
𝜆
𝑃 𝑒 𝑃 𝑝 𝑜 𝑋 𝑁𝑠
𝑷 𝒆
𝑷 𝒑𝒉𝒐
𝟏
𝑏 𝐸 𝑝 𝑜
𝐶
𝜆
𝑋 𝑋 𝑋
𝑋
𝟏 𝟗𝟖𝟗 𝑿 𝟏𝟎 𝟏𝟓
𝑱
𝐸 𝑒 KE + Rest M ss Energy m𝑣 + 𝑚 𝐶
𝑃
𝑚
+ 𝑚 𝐶
𝐸𝑒
𝑋
𝑋 𝑋
+ 𝑋 𝑋 𝑋 𝑋 7
+ 𝑋
Ignore the KE by comparing with rest mass energy, 𝐸 𝑒 𝟖 𝟏𝟗𝑿𝟏𝟎 𝟏𝟒
J
𝐸 𝑒
𝐸 𝑝 𝑜
𝑐
𝐾𝐸 𝑜𝑓 𝐸 𝑒
𝐾𝐸 𝑜𝑓 𝐸 𝑝 𝑜
𝑋
𝑋
𝟎 𝟎𝟏𝟐
Sol :
𝜆
𝑝 𝑚𝑣 𝑚 𝜗𝜆
𝜆 𝜗
𝑚
𝝀√ 𝝑
𝑚
𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝜆 ?
𝑚
𝑋 𝑋 𝑋
𝑋
𝑚 𝑋 𝐾𝑔
𝐸 𝑒𝑉
Sol :
m=0.65MeV/c2
𝜆
√ 𝑚𝐸
𝑋
𝟏 𝟐𝟐Å
𝜆
𝑝 𝑚𝑣
𝑣
𝑚 𝜆
𝑋
𝑋 𝑋 𝑋
𝟒 𝟕 𝑿 𝟏𝟎 𝟔
𝒎/𝒔
𝑎 𝑉 𝑉
𝑏 𝑚 𝐾𝑔
𝑣 𝑚/𝑠
𝜆 𝑒/𝜆 𝑜 ?
Sol :
𝜆 𝑒
√ 𝑚𝑒𝑉
𝑋
√𝑉
𝑋
√
𝟎 𝟗𝟏Å
𝜆 𝑜
𝑝 𝑚𝑣
𝑋
𝑋
𝟔 𝟔𝟐𝟓 𝑿 𝟏𝟎 𝟑𝟒
𝒎
𝜆 𝑒
𝜆 𝑜
𝟏 𝟑𝟕 𝑿 𝟏𝟎 𝟐𝟑
𝝀 𝒆 > 𝝀 𝒐

16 An electron has a de Broglie’s wavelength of 3nm and rest mass of 0.511eV/c2
. Determine its
particle (group) velocity.
17 An electron has a speed of 4.8 x 105
m/s accurate to 0.012% with what accuracy can be located
the position of the electron.
18 A spectral line of wavelength 5461Å has a width of 10-4
Å. Evaluate the minimum time spent by
the electrons in its upper energy state between the excitation and de-excitation process.
19 A nucleon is confined to a nucleus of diameter 4 x 10-8
m with an accuracy of 0.005%.
Calculate the minimum uncertainty in the momentum of the nucleon. Also calculate the
minimum kinetic energy of the nucleon in eV.
20 The position and momentum of 1KeV electron are simultaneously determined and if its
position is located within 1Å. What is the percentage of uncertainty in its momentum.
𝜆 nm
𝑚
𝑋 𝑋
𝑋
𝑚 𝑋 𝐾𝑔
𝑣 ?
Sol :
m=0.511eV/c2
𝜆
𝑝 𝑚𝑣
𝑣
𝑚 𝜆
𝑋
𝑋 𝑋 𝑋
𝟐 𝟒𝟑 𝑿 𝟏𝟎 𝟏𝟎
𝒎/𝒔
𝑣 X 𝑋
𝑥 ?
Sol :
= 57.6m/s
𝑥
𝜋𝑚 𝑣
𝑋
𝜋 𝑋 𝑋 𝑋
𝟏 𝟎𝟎𝟓𝜇𝒎
𝑥 p
𝜋
𝜆
𝑚
𝑡 ?
Sol :
λ=5461Å
𝑡
λ
𝜋𝑐 λ
𝑋
𝜋𝑋 𝑋 𝑋
𝟖𝒏𝒔
𝐸 𝑡
𝜋
𝑡
𝜋 𝐸 𝜋
𝑐
λ
𝜋 𝑐(
λ
) λ
(Ignore –ve sign)
𝑥 X 𝑋
𝑝 ?
Sol :
= 2 X 10-12
m
E=?
𝐸 𝑚𝑣
𝑝
𝑚
𝑋
𝑋 𝑋
𝟐 𝟎𝟕𝟗 𝑿 𝟏𝟎 𝟏𝟗
𝑱 𝟏 𝟐𝟗𝒆𝑽
𝑥 p
𝜋
𝑝
𝜋 𝑥
𝑋
𝜋 𝑋 𝑋
𝟐 𝟔𝟑 𝑿 𝟏𝟎 𝟐𝟑
𝑵𝒔 ≈ 𝒑
𝑥 Å
𝑜𝑓 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 𝑖𝑛 𝑝 ?
Sol :
E=1KeV=1X 103
X1.602X10-19
J
E=1.602X10-16
J
𝐸 𝑚𝑣 𝑝 √ 𝑚𝐸 √ 𝑋 𝑋 𝑋 𝑋 𝟏 𝟕𝟏 𝑿 𝟏𝟎 𝟐𝟑
𝑵𝒔
𝑜𝑓 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 𝑖𝑛 𝑝
𝑝
𝑝
𝑋 𝟑 𝟏
𝑥 p
𝜋
𝑝 𝜋 𝑥
𝑋
𝜋 𝑋 𝑋
𝟎 𝟓𝟑 𝑿 𝟏𝟎 𝟐𝟒
𝑵𝒔

21 On the basis of uncertainty principal estimate the minimum energy of an electron that is that
is confined in a cubic cavity of volume 50 x 50 x 50Å3
.
22 Write down the uncertainty relation connecting position and momentum variables. Show that
this relation can be written as ΔX Δ ≥ 2
/ 4
23 Show that Δt Δ ≥ 2/ c for photon.
24 An electron is confined to a box of length 10-9
m, calculate the minimum uncertainty in its
velocity.
25 An electron is bound in an one dimensional potential well of width 1Å but of infinite wall
height. Find its energy value in ground state, and also in the first two excited state. Jan 2019
𝑋 𝑌 𝑍 Å
𝑋 𝑋
𝑌 𝑌
𝑍 𝑍
Sol :
𝐸 𝐸 𝑋 + 𝐸 𝑌 + 𝐸 𝑍
𝑝 𝑥
𝑚
+
𝑝 𝑌
𝑚
+
𝑝 𝑍
𝑚
𝐸
𝑚
[(
𝜋 𝑋
) + (
𝜋 𝑌
) + (
𝜋 𝑍
) ]
𝑚𝜋
[
𝑋
]
𝐸
𝑋 𝑋
𝑋 𝑋 𝑋𝜋 𝑋 𝑋
𝟏 𝟖𝟑 𝑿 𝟏𝟎 𝟐𝟐
𝑱
Sol :
𝑡 λ
λ
𝜋
𝑋 𝑃
𝜋
𝑋
𝜋 𝑝 𝜋 λ
𝜋 (
λ
) λ
(Ignore –ve sign)
𝑡 λ
λ
𝜋𝑐
𝐸 𝑡
𝜋
𝑡
𝜋 𝐸 𝜋
𝑐
λ
𝜋 𝑐(
λ
) λ
(Ignore –ve sign)
Sol :
Sol :
∆X=1Å
∆v=?
𝑋 𝑃
𝜋
𝑋𝑚 𝑣
𝜋
𝑣
𝜋𝑚 𝑋
𝑋
𝜋 𝑋 𝑋 𝑋 𝑋
𝟓𝟕 𝟗𝒎/𝒔
𝐸 𝐸 𝐸 ?
Sol :
a=1Å 𝑬 𝒏
𝒏 𝟐
𝒉 𝟐
𝟖𝒎𝒂 𝟐
𝐹𝑜𝑟 𝑛 𝐸
𝑚𝑎
( 𝑋 )
𝑋 𝑋 𝑋 𝑋
𝟔 𝟎𝟑𝟐 𝑿 𝟏𝟎 𝟏𝟖
𝑱 𝟑𝟕 𝟔𝒆𝑽
𝑚𝑎
𝐸 𝟏𝟓𝟎 𝟓𝒆𝑽
𝑚𝑎
𝐸 𝟑𝟑𝟖 𝟕𝒆𝑽

26 A particle is moving in a one dimensional potential box of infinite height of width 2.8nm.
Calculate the probability of finding the particle within an interval of 0.8nm at the center of the
box. When it is in its state of least energy.
27 A quantum particle confined to one dimensional box of width ’a’ is known to be in its first
excited state. What is the probability of finding the particle over an interval of a/2 marked
symmetrically at the center of the box or What is the probability of finding the particle in
central half.
28 Show that probability of locating an electron in an infinite potential well between 0 and a/2 is
independent of its quantum state.
29 Calculate the probability of locating an electron in an infinite potential well between 0 and a/n.
where n is quantum state.
𝑃 ?
Sol :
a=2.8nm
∆X=0.8nm
X=a/2
n=1
𝑃 |𝜓| 𝑋
𝑎
𝑠𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎
𝑆𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑃
𝑎
𝑆𝑖𝑛 (
𝑋 𝜋
𝑎
)
𝑎
𝑋
𝑎
𝑆𝑖𝑛 (
𝜋
) 𝑋
𝑋
𝑎
𝑋 𝑋
𝑋
𝑃 𝟎 𝟓𝟕 𝒐𝒓 𝟓𝟕
𝑃 ?
Sol :
Width=a
n=2
𝑃 ∫ | 𝜓| 𝑋
𝑎
𝑎
∫
𝑎
𝑠𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎
𝑎 𝑎
∫ 𝑆𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎
𝑎
𝑃
𝑎
𝑋 ∫ [ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋] 𝑋
𝑎
𝑎 𝑎
[ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋] 𝑎
𝑎
𝑃
𝑎
[(
𝑎 𝑎
) ] 𝟎 𝟓𝟎 𝒐𝒓 𝟓𝟎
Sol :
𝑃 ∫ | 𝜓| 𝑋
𝑎/
∫
𝑎
𝑠𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎/
𝑎
∫ 𝑆𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎/
𝑃
𝑎
𝑋 ∫ [ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋] 𝑋
𝑎/
𝑎
[ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋]
𝑎/
𝑃
𝑎
*(
𝑎
) + 𝟎 𝟓𝟎 𝒐𝒓 𝟓𝟎
Sol :
𝑃 ∫ | 𝜓| 𝑋
𝑎/𝑛
∫
𝑎
𝑠𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎/𝑛
𝑎
∫ 𝑆𝑖𝑛 (
𝑛𝜋
𝑎
) 𝑋 𝑋
𝑎/𝑛
𝑃
𝑎
𝑋 ∫ [ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋] 𝑋
𝑎/𝑛
𝑎
[ 𝐶𝑜𝑠 (
𝑛𝜋
𝑎
) 𝑋]
𝑎/𝑛
𝑃
𝑎
*(
𝑎
𝑛
) +
𝟏
𝒏

30 A particle moving in one dimensional box is described by the wave function Ψ = X√ for
0<x<1 and Ψ = 0 elsewhere. Find the probability of finding the particle within the interval (0,
½).
31 An excited atom has an average time of 10-8
s. During this period, it emits a photon and returns
to the ground state. What is the minimum uncertainty in the frequency of photon.
32 The first excited state energy of an electron in an infinite potential well is 240eV. What will be
its ground state energy when the width of the potential well is doubled.
33 An electron is confined to move between two rigid walls separated by 20Å. Find the de
Broglie’s wavelength representing the first 3 allowed energy states of the electron and the
corresponding energies
𝐸 𝐸 𝐸 ?
𝜆 𝜆 𝜆 ?
Sol :
𝑎 Å
Sol :
∆t=10-8
s
∆ν =?
𝐸 𝑡
𝜋
𝜗 𝑡
𝜋
𝜗 𝑡
𝜋
𝜗
𝜋 𝑡 𝜋 𝑋
𝟖 𝑿 𝟏𝟎 𝟔
𝑯𝒛
𝜓 𝑋√
Sol :
P=?
0<X<1/2
𝑃 ∫ |𝜓| 𝑑𝑥 ∫ 𝑋√ 𝑑𝑥
//
𝑃 ∫ 𝑋 𝑑𝑥
/
𝑃 [
𝑋
]
/
[ ]
𝟏
𝟖
𝒐𝒓 𝟏𝟐 𝟓
Sol :
n=2, E2=240eV
Width=2a, E1=?
𝑬 𝒏
𝒏 𝟐
𝒉 𝟐
𝟖𝒎𝒂 𝟐
𝝀
𝒉
√𝟐𝒎𝑬
𝑚𝑎
( 𝑋 )
𝑋 𝑋 𝑋( 𝑋 )
𝟏 𝟓𝟏 𝑿 𝟏𝟎 𝟐𝟎
𝑱 𝒂𝒏𝒅 𝝀 𝟏
𝒉
√ 𝟐𝒎𝑬 𝟏
𝟒𝒏𝒎
𝑚𝑎
𝐸 𝟔 𝟎𝟑 𝑿 𝟏𝟎 𝟐𝟎
𝑱 𝒂𝒏𝒅 𝝀 𝟐
𝒉
√ 𝟐𝒎𝑬 𝟐
𝟐𝒏𝒎
𝑚𝑎
𝐸 𝟏𝟑 𝟔 𝑿 𝟏𝟎 𝟐𝟎
𝑱 𝒂𝒏𝒅 𝝀 𝟑
𝒉
√ 𝟐𝒎𝑬 𝟑
𝟏 𝟑𝟑𝒏𝒎
𝑬 𝒏
𝒏 𝟐
𝒉 𝟐
𝟖𝒎𝒂 𝟐
𝑚𝑎
𝑎
√ 𝑚𝐸
𝑋
𝑋 𝑋 𝑋 𝑋 𝑋
𝑚𝑎
( 𝑋 )
𝑋 𝑋 𝑋 𝑋
𝟑𝟗𝟕 𝑿 𝟏𝟎 𝟏𝟖
𝑱 𝟏𝟓𝒆𝑽
𝑎 𝟑 𝟏𝟕 𝑿 𝟏𝟎 𝟐𝟎
𝒎 𝟐𝒂 𝟔 𝟑𝟒 𝑿 𝟏𝟎 𝟐𝟎
𝒎

34 Find the value of B for which the wave function
√
is normalized in the
region > > .
35 An electron is traped in a 1-D potential well of infinite height and of width of 0.1nm. Calculate
the energy required to excite it from its ground state to fifth excited state.
Sol : ∫ |𝜓| 𝑑𝑥 ∫ |𝜓| 𝑑𝑥
∫
𝐵
𝑋 + 𝑎
𝑑𝑥
𝐵
𝑎
∫
𝑎
𝑋 + 𝑎
𝑑𝑥
𝐵
𝑎
*𝑡𝑎𝑛
𝑥
𝑎
+
𝐵
𝑎
*
𝜋
+
𝐵 √
𝑎
𝜋
Sol :
a=0.1nm
E6-E1=? 𝑬 𝒏
𝒏 𝟐
𝒉 𝟐
𝟖𝒎𝒂 𝟐
𝑛
𝑚𝑎
𝑋 ( 𝑋 )
𝑋 𝑋 𝑋 ( 𝑋 )
𝑋 7
𝐽
𝑛
𝑚𝑎
𝑛 𝐸 𝑋 𝑋 7
𝐽
𝑬 𝟔 𝑬 𝟏 84.31 X 𝟏𝟎 𝟏𝟕
J

LASER
Introduction
 The Term ”LASER” is an acronym for Light Amplification by Stimulated Emission
of Radiation. Lasers are devices that produce intense beams of light which are,
(a) Monochromatic: The wavelength (color) of laser light is extremely pure
(monochromatic) when compared to other sources of light.
(b) Coherent: All of the photons (energy) that make up the laser beam have a
fixed phase relationship (coherence) with respect to one another or The light
from a laser is said to be coherent, which means that the wavelengths of
the laser light are in phase in space and time.
(c) Highly Collimated: Light from a laser typically has very low divergence. It can
travel over great distances or can be focused to a very small spot with a
brightness which exceeds that of the sun.
(c) Directional: Lasers emit light that is highly directional, that is, laser light is
emitted as a relatively narrow beam in a specific direction.
Because of these properties, lasers are used in a wide variety of applications in all walks
of life.
 History on LASER
(a) In 1954, Charles Townes and Arthur Schawlow invented the Maser (microwave
amplification by stimulated emission of radiation), using ammonia gas and
microwave radiation - the maser was invented before the (optical) laser.
(b) The basic operating principles of the laser were put forth by Charles Townes and
Arthur Schalow from the Bell Telephone Laboratories in 1958.
(c) In 1960, Theodore Maiman invented the Ruby Laser considered to be the first
successful optical or light laser.
(d) The first gas laser (helium neon) was invented by Ali Javan in 1960. The gas laser
was the first continuous-light laser and the first to operate "on the principle of
converting electrical energy to a laser light output.
(e) In 1962, Robert Hall created a revolutionary type of laser that is still used in many of
the electronic appliances and communications systems that we use every day
(Semiconductor injection Laser).

1. Review of Spontaneous and Stimulated Emission
(Interaction of radiation with matter)
a) Induced Absorption
 Induced absorption is the absorption of an incident photon by a system is elevated
from a lower energy state to a higher energy state are called induced absorption.
 Consider an atom present in the ground state (E1) absorbs the energy from an
incident photon and elevated to excited state (E2).
+
b) Spontaneous Emission
 Spontaneous emission is the emission of a photon, when a system transits from a
higher energy state to a lower energy state without the aid of any external agency.
 Consider an atom in the excited state (E2) comes to ground state (E1) by emitting
a photon of energy ∆E = E2 - E1 without being aided by any external energy.
+
E2
E1

c) Stimulated Emission
 Stimulated emission is the emission of a photon by a system when a system,
under the influence of a passing photon of right energy, due to which the system
transits from a higher energy state to a lower energy state. The photon thus
emitted is called the stimulated photon and will have same phase, energy and
direction of movement as that of passing photon (incident photon) called the
stimulated emission.
 Consider an atom in the excited state (E2) comes to ground state (E1) by emitting
a two photon (Stimulated & incident photon) travelling in same direction and with
exactly the same energy (∆E = E2 - E1) with the aid of any external energy
(incident photon having the energy precisely equal to (∆E = E2 - E1).
+ + +
 Comparison

2. Energy Density Using Einstein’s Coefficients
 Consider two energy level system E1 & E2 (E2>E1) and there be N1 atoms with energy E1
and N2 atoms with the energy E2 per unit volume. N1 and N2 are called number density
and be the energy density per unit volume of the system of frequency range ν and
ν+dν and be the energy density.
 Consider the absorption and also the two emission process
a) Induced Absorption
b) Spontaneous emission
c) Stimulated Emission
 At the thermal equilibrium condition, the rate of absorption is equal to the sum of
rate of spontaneous emission and stimulated emission.
+
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝛼 𝑁 𝑈𝜈
The number of absorption per unit time per unit volume is
called rate of absorption.
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝐵 𝑁 𝑈𝜈 (1)
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝛼 𝑁
The number of emission per unit time per unit volume is
called rate of spontaneous emission.
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝐴 𝑁 (2)
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝛼 𝑁 𝑈𝜈
The number of emission per unit time per unit volume is
called rate of stimulated emission.
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝐵 𝑁 𝑈𝜈 (3)
N2
N1
N2
N1
N2
N1

+
[
( )
]
[
( )( )
]
From Boltzmann‟s Law,
[
( )( )
]
From Planck‟s law,
[ ]
Comparing equ. (6) & equ. (7),
Equ. (7) becomes,
This is the equation for energy density at thermal equilibrium using Einsteins‟ Coefficients.
𝑈𝜈
𝐴
𝐵
[
𝑒
𝜈
𝑘𝑇
]

3. Condition for LASER action
a) Population Inversion
 Consider three energy level E1, E2 and E3 of a system (E3>E2>E1). Let E2 be a
metastable state of a system.
 Let the atom be excited from E1 to E3 by the supply of appropriate energy from an
external source.
 From the E3, the atoms undergo spontaneous download transitions rapidly to E2 and
E1 states. But E2 is a metastable state, those atoms which get into that state stay over
long duration (10-3
s) because of which the population of E2 state increases steadily.
Since the atoms are being excited continuously from E1 level its population goes on
decreasing. The population of E2 state overtakes that of E1 which condition is
known as population inversion.
 Once the population of E2 exceeds that of E1, the stimulated emissions outnumber
the spontaneous emissions and soon stimulated photon, all identical in respect of
phase, wavelength and direction, grow to a very large number which build up the
laser light. Hence the condition for laser action is achieved by means of population
inversion.
 Population Inversion is the state of a system, at which the population of a particular
higher energy state is more than that of a specified lower energy state „N2>N1‟.
b) Pumping: The method of providing population inversion is known as pumping.
c) Lasing: The process which leads to emission of stimulated photon due to population
inversion is often referred to as lasing.
10
-3
s
10
-8
s Radiationless transition due to lattice vibration
Ground State, E1, N1
Metastable State, E2, N2
Excited State, E3, N3

4. Requisites of a LASER system
The requisites of a laser system are,
 An excitation source for pumping action
The excitation source providing energy in an appropriate form for pumping the
atoms to higher energy levels. The energy input may be in the form of light energy.
This kind of pumping is called optical pumping (Ruby laser). If the pumping is
achieved by electrical energy input then it is called electrical pumping (He-Ne
Laser).
 An active medium which supports population inversion
 A part of the input energy is absorbed by the active medium in which
population occurs at a certain stage
 The heart of the laser ia s certain medium called an active medium. It
contains atoms, ions or molecules capable of decaying from their high energy
state by emitting electromagnetic waves.
 A laser cavity or resonance cavity
 A resonant cavity consists of an active medium bound between two mirrors,
one fully silvered and the other partially silvered.
 The separation between the mirrors allows only selected wavelength of
radiation to bounce back and forth. The direction of the travel of the photon
is parallel to the axis of resonant cavity and the repeated bouncing of light
back and forth increases the chances of more stimulated emissions.
 Two types of waves exists in an active medium.
(i) The two waves interfere constructively if there is no phase difference
between the two.
(ii) But their interference became destructive if the phase difference is
Active Medium

Constructive Interference
Destructive Interference
 For construction interference, the distance between the two mirror should be
equal to integral multiple of λ/2.
( )

5. Principle, Construction and Working of Carbon Dioxide Laser
a) Introduction – Vibration energy level of a CO2 Laser
A carbon dioxide molecule has two oxygen atoms between which there is a carbon
atom. It has 3 different modes of vibration.
(i) Symmetric Stretching Mode
----------●----------------●----------------●------------
 In this mode, oxygen atoms oscillate along the molecule axis either
approaching towards carbon atom or departing from the carbon atom. The
carbon atom remains stationary.
 CO2 molecules will have intermediate energy in this state.
 This state is denoted by (100) state
(ii) Asymmetric Stretching Mode
---------●→------------←●-----------------●→--------
 In this mode, all the 3 atoms oscillate along the molecular axis but the two
oxygen atoms move in one direction while the carbon atom moves in
opposite direction.
 The molecule possesses highest energy in this state.
 This is the (001) mode.
(iii) Bending Mode
--------●----------------------●---------------------●---------------
 In this mode, all the three atoms oscillate normal to the molecular axis.
While vibrating the two oxygen atoms pull together in one direction as the
carbon atom is in opposite direction.
 The energy of molecule in this state is least among the three modes of
vibration.
 This is (010) state
O C O
⇄ ⇆
O C O
↓ C ↓
O ↑ O
Molecular Axis
Molecular Axis
Molecular Axis

Carbon Dioxide LASER
a) Principle
 For N2 state, the vibrational levels are metastable.
 There is a close coincidence in energy between its first excited state (ν=1)
and the state for asymmetric stretch mode (001) of the CO2 molecule.
 This helps in causing population inversion in CO2 gas laser by means of
resonance transfer of energy.
b) Construction of CO2 laser
 A CO2 laser consists of a discharge tube of 2.5cm in diameter and of 5m length.
 The tube is water cooled and is filled with a mixture of CO2, N2 and He gas in the
ratio 1:2:3 (sometimes, traces of H2 or H2O is added because during discharge, some
of the CO2 molecules breaks into CO and O. The H2 or H2O vapours help to reoxidise
CO to CO2).
 The pressure inside the tube is 6 – 17 torr (1 torr =133.3pa).
(The actual size, pressure and proportion of gases vary with the particular application
of the laser and also the applied field- AC/DC)
 Two optically plane mirrors are fixed on either side of the tube normal to its axis. One
of the mirrors is fully silvered (100% reflection) whereas the other is partially
silvered (1% transmission).
 Brewster‟s windows are used to get plane polarized beam of laser.

c) Working of CO2 laser
 When a suitable voltage is applied across the two electrodes, a glow discharge of gases is
initiated in the tube. During discharge, many electrons are free from the gas atoms.
 These free electrons accelerate towards the positive electrode at which time they begin to
collide with the N2 and CO2 molecules in the path.
a) The N2 molecule are raised to the first vibrational level (Metastable, ν=1)
+ +
b) Same way, many of CO2 molecules also be raised to (001) state (Asymmetric stretch
energy)
Ground State Ground State
Nitrogen Carbon Dioxide

 The N2 molecule remains in metastable state for long time, for increase of population.
Now, there is a close coincidence of energy of CO2 gas (001) with (ν=1). Therefore
N2 molecule in metastable state collides with CO2 molecule in the ground state. Because
of matching energy, resonant transfer of energy takes place from N2 to CO2 molecules,
the CO2 moves to (001) state and N2 molecules returns to ground state.
+ +
 Thus, the population of the (001) state of CO2 increases rapidly which leads to population
inversion.
 The ground state, (010) state, (020) state, (100) state and (001) state are the E1, E2, E3, E4
and E5 levels respectively.
 Population inversion takes place between E5 level to E3 & E4 levels and two laser
transitions takes place,
a) Transition from E5 to E4 level which gives rise to radiation of wavelength 10.6μm
(Infrared region)
b) Transition from E5 to E3 level which gives rise to radiation of wavelength 9.6μm
(Infrared region)
 The CO2 molecules in the E3 and E4 levels undergo collision with those in ground state
and also by resonance energy transfer down to E2 level.
 Since ≈ ≈ contributes to the population of E2 level.
 By absorbing the transferred energy and the surroundings thermal energy, CO2 molecules
in the E1 level are also excited to E2 level, thus increasing the population of E2 level.
 The CO2 molecules in the E2 level undergone collision with He and H2O vapour and
come down to ground state.
 Helium gas transfers the heat of discharge to tube well. Thus bringing down the thermal
excitation of CO2 from ground state to E2 level which reducing population inversion.

6. Principle, Construction and Working of Homo Junction Semiconductor LASER
[or Gallium Arsenide(GaAs) Laser]
a) Principle
In a semiconductor laser, the population of energy levels near to bottom of the
conduction band is increased by driving a current. When the current suppressed a
threshold value, population inversion takes place and the stimulated emission
overtakes the spontaneous emission thus leading to lasing.
b) Construction of GaAs semiconductor laser
 Gallium Arsenide laser diode is a single crystal of GaAs and consists of heavily
doped n and p type sections. .The n section is formed by doping with tellurium
and p section is obtained by doping with Zinc (The doping concentration is very
high and is of order 1017
to 1019
dopant atom per cm3
).
 A junction is connected to DC supply in a forward bias and junction acts like the
active medium (width of the pn junction varying from 1μm to 100 ).
 Resonance cavity required for energy amplification is obtained by cleaving the
front and back of the semiconductor material. The cleaved surface must be
perfectly flat and parallel. The back face is made fully reflecting and front face is
made partially reflecting for laser beam exit.

The other two remaining sides perpendicular to the junction are roughened to
suppress reflections of the photons.

c) Working of GaAs semiconductor laser
 The figure shows the energy level diagram of GaAs semiconductor diode. EFn
and EFp are the Fermi levels in the n-type region and p-type region respectively.
They continue into the junction as shown in the figure. As there are electrons in
EFn and holes in EFp. The population condition is established at the junction.
 The energy band diagram shows a pn-junction with valence band (V),
conduction band (C) and separated by a energy gap (Eg). At T = 0K, the
conduction band is completely empty and the valence band is completely filled
energy state.
C
Energy Gap (Eg)
V
 Initially, the concentration of electrons in the energy levels at the bottom of the
conduction band will still be lesser than that in the energy levels at the top of the
valence band and recombination results in only spontaneous emissions.
Conduction Band
Valence Band

 When the pn junction is heavily forward bias with a large current. The electrons
from the valence band raises to conduction band but this is an unstable state and
with a short time (10-13
s), electrons in the conduction band drops to the lower
level in that band. The lowest level of conduction band is filled with electrons
and the top of valence band is filled is full of holes.
 There is a significant increase in the concentration of electron in the conduction
band near the junction on the n side and also the concentration of holes in the
valence band on the p side.
i.e. In conduction band, more number of electrons moves from n-type to p-type
and In valence band, more number of holes moves from p-type to n-type. As a
result, large number of electrons confined in pn junction of conduction band and
also large number of holes confined in pn junction of valence band.
 This indicates the population inversion at the pn junction. Now one of the
electron from the conduction band drops to valence band to recombine with a
hole and energy is associated with this recombination is emitted as a photon of
light. This energy is in the form of electromagnetic radiation.
← Electrons
← Holes
p-Type pn Junction n-type

At this stage, a photon released by a spontaneous emission may trigger
stimulated emissions over a large number of recombination. i.e. photons emitted
from the recombination interns give its energy to the electrons in conduction
band for more recombination of electron and holes to takes place. As a result
large numbers of photons are released and these photons are in phase with each
other and have same wavelength, thus travel together and reflected at the end
face. As a result energy amplification will takes place which is leading to buildup
of laser radiation of high power.
 As the current is being passed continuously, more electron get excited, rise to
conduction band, new holes generated in the valence band. This maintains the
population inversion. While the recombination of the electron-hole pairs
continues with generation of laser beam.
 The energy gap of GaAs is 1.4 eV, the wavelength of the emitted light is,
Note : (a) When PN Junction is unbiased
(b) When PN junction is biased
n-type pn junction p type
n-type pn junction p type

7. Application of laser
a) Laser Rangefinder in Defense
 The principle of laser rangefinder is the same as that of conventional RADAR.
 a) It is the best equipment to find the distance of range of enemy‟s target.
b) In Military, all battle tanks are fitted with laser rangefinder which is interfaced
with computer to provide information in a digital readout form within 1% of
actual distance.
c) It is also used for continuous tracking and ranging of missiles & aircrafts from
ground or from air.
Working
 A high powered pulsed laser beam from a solid state laser device (Nd-YAG laser)
(Neodymium-doped Yttrium Aluminum Garnet)is directed towards the enemy
target from a transmitter.
 The pulses are narrow with high peak power. Upon incidence, the beam bounces
from the surface of the target as a reflection.
 A part of reflected beam called „echo‟ is received as a signal by a receiver. Inside
the receiver there will be an interface filter. It is an narrow band optical filter
tuned to the frequency of the laser light so that all the background noise entering
the receiver is wiped off.
 Then the signal is amplified by using a photomultiplier.
 The range finder high speed clock measures the exact time from the instant the
pulses left the unit and until they returned and then convert it into distance.

b) Application of laser in use of compact disc (Data Storage)
 A compact disc is a thin circular disc of about 12cm diameter and its shining side
is made up of metal & plastic consisting of three layers. At the bottom is a layer
of polycarbonate which is tough but it is brittle plastic. Above that is a layer of
aluminum coated with plastic and lacquer (Metal protective).
 The information is created in digital form in the CD by using a laser beam. The
laser beam burns and etches (carve/stamp) bumps on its surface at certain
specific intervals on a track. These bumps are called pits. Presence of a bump in
a fixed length in a track indicates a zero.
 An unburnt space in a specific length of the track remains flat on the length is
called „land‟ and represents the number one.
 Thus the laser beam can store information by burning some length (for zeroes)
and leaving some length unburned (for ones) in the binary language.
 While reading the CD, a laser beam scans the tracks. As it is bounced, it follows
the patterns of pits and lands. A photocell converts these into electric pulses.
 In turn, an electric circuit generates zeroes and ones. A decoder converts the
binary number into a changing pattern of electric currents in the analog form.
 CD – 700MB, DVD-4.7GB, BD(Blue-ray Disc-5 times the DVDs), BDXL
(Blue-ray Disc EXtra Large – 50Gb to 128GB)
 In a compact disc (CD), digital information is stored as a sequence of raised
surfaces called “pits” and recessed surfaces called “lands”. Both pits and lands
are highly reflective and are embedded in a thick plastic material with refractive
index of 1.5.

Formulas at a glance
 =
 [ ]




 ( )
01 A laser medium at thermal equilibrium temperature 300K has two energy level with a
wavelength separation of 1µm. Find the ratio of population densities of the upper and lower
levels.
02 The ratio of population of two energy levels is 1.059 x 10-30
. Find the wavelength of light emitted
at 300K.
𝑇 𝐾
𝑁
𝑁
?
Sol :
λ = 10-6
m
𝑁
𝑁
𝟏 𝟑𝟔 𝑿 𝟏𝟎 𝟐𝟏
𝑁
𝑁
𝑒
𝑐
𝜆𝑘𝑇
𝑁
𝑁
𝑒
6 6 𝑋 𝑋 𝑋 8
6 𝑋 8 𝑋 𝑋
𝑒 8
𝑁
𝑁
𝑋
𝑇 𝐾
Sol :
λ = ?
𝑙𝑛
𝑁
𝑁
𝑙𝑛 𝑒
𝑐
𝜆𝑘𝑇
𝑙𝑛
𝑁
𝑁
𝑐
𝜆𝑘𝑇
𝑙𝑛 𝑒
𝜆
𝑐
𝑘𝑇𝑙𝑛
𝑁
𝑁
𝑋 𝑋 𝑋
𝑋 𝑋 𝑋 ln 𝑋
𝟔𝟗𝟔𝒏𝒎
𝑁
𝑁
𝑒
𝑐
𝜆𝑘𝑇

03 The ratio of population between the upper and lower levels of an atomic system is 10-6
. If the
radiation emitted by a transition between these levels is 1500nm then what must be the
temperature of an atoms.
04 Atomic transitions between two specific levels give rise to light of wavelength 550nm. What is the
ratio of Einstein’s coefficients for spontaneous and stimulated emissions.
05 Show that the ratio of rate of spontaneous emission and absorption is given by [1 – e-hγ/KT
].
06 Show that the ratio of rate of spontaneous emission and stimulated emission is given by [ehγ/KT
-1].
𝑁
𝑁
𝑇 ?
Sol :
λ = 1500nm
𝑙𝑛
𝑁
𝑁
𝑙𝑛 𝑒
𝑐
𝜆𝑘𝑇
𝑙𝑛
𝑁
𝑁
𝑐
𝜆𝑘𝑇
𝑙𝑛 𝑒
𝑇
𝑐
𝑘𝜆𝑙𝑛
𝑁
𝑁
𝑋 𝑋 𝑋
𝑋 𝑋 𝑋 ln
𝟔𝟗𝟓𝑲
𝑁
𝑁
𝑒
𝑐
𝜆𝑘𝑇
𝐴
𝐵
?
Sol :
λ = 550nm
𝐴
𝐵
𝟏 𝟎𝟎𝟎𝟖 𝑿 𝟏𝟎 𝟏𝟑
𝐴
𝐵
𝜋 𝜗
𝐶
𝜋
𝐶
(
𝐶
𝜆
)
𝜋
𝜆
𝑋 𝜋 𝑋 𝑋
𝑋
Sol : 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑝𝑜𝑛𝑡𝑒𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑏𝑡𝑖𝑜𝑛
𝐴 𝑁
𝐵 𝑁 𝑈 𝜈
𝐴
𝐵
𝑒
𝛄
𝑘𝑇
𝐴
𝐵
[
𝑒
𝛄
𝑘𝑇
]
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑝𝑜𝑛𝑡𝑒𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑏𝑡𝑖𝑜𝑛
𝑒
𝛄
𝑘𝑇 [ 𝑒
𝛄
𝑘𝑇 ] [ 𝑒
𝛄
𝑘𝑇]
Sol :
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡𝑖𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
𝐴 𝑁
𝐵 𝑁 𝑈 𝜈
𝐴
𝐵
𝐴
𝐵
[
𝑒
𝛄
𝑘𝑇
]
[ 𝑒
𝛄
𝑘𝑇 ]

07 A pulsed laser emits photons of wavelength 780nm with 20mW average power per pulse.
Calculate the number of photons contains in each pulse. If the pulse duration is 10ns. Jan 2019
08 A pulse from laser with 1mW lasts for 10ns. If the number of photons emitted per second is 3.491
x 107
. Calculate the wavelength of laser.
09 A He-Ne laser is emitting a laser beam with an average power of 45mW. Find the number of
photons emitted per second by the laser. The wavelength of emitted radiation is 6328Å.
10 The transition to the ground state from the upper and lower energy state in a Ruby laser results
in emission of photons of wavelength 6298Å and 6943Å respectively. Estimate the energy values
of the two energy levels in eV and also their ratio of populations.
11 Find the number of modes of standing waves and their frequency separation in the resonant
cavity of 1m of He-Ne operating at a wavelength of 632.8nm.
Sol :
λ = 780nm
P=20mW
N = ?
t = 10ns
𝑁
𝐸
𝐸
𝑃𝑡
(
𝑐
𝜆
)
𝑃𝑡𝜆
𝑐
𝑋 𝑋 𝑋 𝑋 7 𝑋
𝑋 𝑋 𝑋 8
𝑁 7.86 X 𝟏𝟎 𝟖
/𝒑𝒖𝒍𝒔𝒆
Sol :
λ = ?
P=1mW
N =3.491 X 107
t = 10ns
𝑁
𝐸
𝐸
𝑃𝑡
(
𝑐
𝜆
)
𝑃𝑡𝜆
𝑐
𝑋 𝑋 𝑋 𝑋 7 𝑋
𝑋 𝑋 𝑋 8
𝜆
𝑐𝑁
𝑃𝑡
𝑋 𝑋 𝑋 8 𝑋 X
𝑋 𝑋 𝑋
=694.4nm
Sol :
λ = 6328Å
P=45mW
N =?
t = 1s
𝑁
𝐸
𝐸
𝑃𝑡
(
𝑐
𝜆
)
𝑃𝑡𝜆
𝑐
𝑋 𝑋 𝑋 𝑋
𝑋 𝑋 𝑋 8
𝑁 1.43 X1017
/second
𝑁
𝑁
?
Sol :
𝜆 = 6943Å
𝜆 = 6298Å
𝑁
𝑁
𝑒
𝐸
𝑘𝑇 𝑒
𝐸 𝐸
𝑘𝑇 𝑒
7 𝑋 𝑋 ⬚
𝑋 𝑋
𝟎 𝟖𝟓𝟕
𝐸
𝑐
𝜆
𝑋 𝑋 𝑋 8
𝑋
𝟐 𝟖𝟔𝟒 𝑿 𝟏𝟎 𝟏𝟗
𝑱
𝐸
𝑐
𝜆
𝑋 𝑋 𝑋 8
𝑋
𝟐 𝟖𝟕 𝑿 𝟏𝟎 𝟏𝟗
𝑱
𝜗 𝑚 𝜗 𝑚 ?
Sol :
𝜆 = 632.8nm
L=1m
M=?
𝐿 (
𝜆
) 𝑚 𝑚
𝐿
𝜆
𝑋
𝑋
𝟑𝟏𝟔𝟎𝟓𝟓𝟔
𝐿 (
𝜆
) 𝑚 𝜆
𝐿
𝑚
𝐶
𝜗
𝐿
𝑚
𝜗 𝑚
𝑚𝐶
𝐿
𝜗 𝑚
𝑚 𝐶
𝐿
𝜗 𝑚 𝜗 𝑚
𝐶
𝐿
𝑋
𝑋
𝟏 𝟓 𝑿 𝟏𝟎 𝟖
𝑯𝒛

12 A laser operating at 632.8nm emits 3.182 x 1016
photons per second. Calculate the output power
of the laser if the input power is 100W. Also find the percentage power converted into coherent
light energy.
13 A semiconductor laser has a peak emission radiation of wavelength of 1.24nm. What is its band
gap value in eV.
14 Calculate the ration of (i) Einstein’s Coefficients and (ii) stimulated to spontaneous emissions, for
a system in thermal equilibrium at 300K in which radiation of wavelength 1.39µm are emitted.
15 Calculate the wavelength of Laser emitted from an extrinsic semiconductor laser if the band gap
is 0.02eV. To which region of EM spectrum does it belong.
𝑃𝑜𝑢𝑡 ?
𝑃𝑖𝑛 𝑊
𝑜𝑓 𝑃𝑜𝑤𝑒𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 ?
Sol :
𝜆 = 632.8nm
N=3.182 X 1016
/s
𝐸 𝑁 𝑋 𝐸 𝑃𝑡 𝑁 𝑋 𝐸
𝑃 𝑁 𝑋
𝑐
𝜆
𝑋 𝑋 𝑋 𝑋 𝑋
𝑋
𝟎 𝟎𝟏𝑾
𝑜𝑓 𝑃𝑜𝑤𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
𝑋 𝑋 𝟎 𝟎𝟏
𝐸𝑔 ?
Sol :
𝜆 = 1.24nm
𝐸 𝑔 𝜗
𝑐
𝜆
𝑋 𝑋 𝑋
𝑋
𝟏 𝟔 𝑿 𝟏𝟎 𝟏𝟗
𝑱 𝟏𝒆𝑽
𝜆 𝜇𝑚
Sol :
𝐴
𝐵
?
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑝 𝑒𝑚𝑖𝑠𝑠𝑠𝑖𝑜𝑛
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑡 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
?
T=300K 𝐴
𝐵
𝟔 𝟐 𝑿 𝟏𝟎 𝟏𝟓
𝐴
𝐵
𝜋 𝜗
𝐶
𝜋
𝐶
𝐶
𝜆
𝜋
𝜆
( 𝑋 𝜋 𝑋 𝑋 )
( 𝑋 )
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑠𝑝𝑜𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
𝐵 𝑁 𝑈 𝜈
𝐴 𝑁 ⬚
𝐵
𝐴
𝐴
𝐵
[
𝑒
𝛄
𝑘𝑇
]
𝑒
𝛄
𝑘𝑇 𝑒
𝑐
𝜆𝑘𝑇
𝑒
𝑋 𝑋 𝑋
𝑋 𝑋 𝑋 𝑋
= 𝟗 𝟕𝟏 𝑿 𝟏𝟎 𝟏𝟔
𝐸𝑔 𝑒𝑉
Sol :
𝜆 = ?
𝐸 𝑔 𝜗
𝑐
𝜆
𝜆
𝑐
𝐸 𝑔
𝑋 𝑋 𝑋
𝑋 𝑋
𝑛𝑚
This EM Spectrum belongs to IR Region

IV. Question Bank
Module 4: Quantum Mechanics & LASER
Q. No. Question Bank
01
Give a brief account of blackbody radiation and Planck‟s law leading to the quantization
of energy.
Or
Mention the assumptions of quantum theory of radiation (or Planck‟s law).
02 Explain the wave nature of particle.
03
State de Bronglie hypothesis. Derive an expression for de Bronglie wavelength of an
electron accelerated by a potential difference V.
04 State and explain Heisenberg‟s uncertainty principle. Mention its physical significance.
05 Show that, non-confinement of electron in the nucleus of an atom.
06 What are wave functions? Mention their properties & physical significance.
07 Set up one dimensional, time-independent Schrodinger‟s wave equation.
08
Obtain the time-independent Schrodinger‟s wave equation for a particle in an one
dimensional potential well of infinite height and solve to get the Eigen value and Eigen
function.
09
Find the Eigen value, Eigen function and probability density for a particle in an one
dimensional potential well of infinite height for ground state & first two excited state.
10 Explain the term spontaneous emission and stimulated emission.
11
Derive an expression for energy density of radiation under thermal equilibrium condition
in terms of Einstein coefficients.
12 Explain the condition for laser action.
13 Explain the requisites of a laser system.
14
Explain the principle, construction and working of CO2 Laser with the help of suitable
diagrams.
15
Explain the principle, construction and working of semiconductor Laser with the help of
suitable diagrams.
16 Describe how a laser range finder is made use of in defense.
17 Explain how data storage is achieved in a compact disk.
18 Numerical Problem on Quantum Mechanics & LASER

Final m4 march 2019

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Final m4 march 2019