1
QUANTUM MECHANICS
1.1 INTRODUCTION
The failureness of classical mechanics to explain about observed spectrum of black body radiation was
a main cause to develop a new mechanics, called Quantum Mechanics. Although the stability of atoms,
the observed variation of the specific heat of solids or gases, the origin of discrete spectra of atoms, the
detailed description of the motion of atomic particles and photoelectric effect, compton effect, raman
effect etc. were some other facts which could not be explained by the classical mechanics.
The basic idea of quantum theory was introduced by Max Planck in 1900 to explain the black
body radiations, but the subsequent developments and interpretations were made by a number of physicists
including Einstein, Bohr, deBroglie, Heisenberg, Schrödinger, Born and Dirac.
1.2 BLACK BODY RADIATION
A black body is an object that absorbs all the radiations that fall
on it, i.e. no radiation passes through it and none is reflected.
This body looks black under outside illumination. A hollow object
with a small hole, acts like a black body. If the black body is
heated to a high temperature, it emits exactly same wavelengths
and intensities which would be obsorbed by the body. The
radiation emitted by a black body is called black body radiation.
T1  T2
u  
T2
Wavelength (  )
(a)
Fig. 1.1 Black body model
T1
T1
Spectral energy density
Spectral energy density
In 1899, Lummer and Pringsheim studied the distribution
of energy amongst the different wavelengths or frequencies of a
thermal spectrum of a black body radiation. The experimental
results of this study are illustrated in Figs. 1.2 ( a) and (b).
T1  T2
u
T2
Frequency ()
(b)
Fig. 1.2 (a) and (b) Black body radiation spectrum
1.2
Engineering Physics-II
Wien and Rayleigh-Jeans studied the black body radiations and derived a formula to explain these
results. But the Wien’s formula* holds good only in the region of shorter wavelengths (higher frequencies)
and the Rayleigh-Jean’s fromula** holds good in the region of longer wavelengths (lower frequencies).
Hence Wien’s formula and Rayleigh-Jean’s formula do not precisely agree with the experimental results
given by Lummer and Pringsheim.
1.2.1 Planck’s Quantum Theory
In order to explain black body radiation, Max Planck, in the
year 1900, proposed the quantum theory. According to this
theory, matter is composed of a large number of oscillating
particles which vibrate with different frequencies. In quantum
theory, the energy of oscillating particle is quantized. The
oscillating particle cannot have any energy but only those
energies which are given by
E = nh
where h is Planck’s constant whose value is 6.625  10 34
Joule-second,  is the oscillator frequency and n is an integer
(n = 1, 2, 3. . . .). In this theory, it is also assumed that the
vibrating particle does not radiate energy continuosuly but
discretely, i.e., in the form of some minimum quantity of energy
known as quanta or photon.
While studing black body radiations, Max Planck in 1901
concluded that the emission and absorption of thermal energy
is not a continuous process but it takes place in discrete amount
i.e. an integral multiple of a certain energy unit ‘ h’. According
to this postulate, the exchange of energy between light and
matter is not continuous, but it is in the form of small bundles
or packets of definite energy proportional to frequency of light.
These small packets of energy are called photons or quanta.
Karl Ernst Ludwig Max Planck (April
23, 1858 – October 4, 1947) was a
German physicist. He was considered the founder of quantum theory.
According to planck the energy could
only be a multiple of an elementary unit
E = h  . where h is Planck’s constant
and  is the frequency of the radiation. It was in recognition of his monumental accomplishment that Planck
was awarded the Nobel Prize in Physics in 1918.
Max Planck derived a theoretical expression for the energy distribution on the basis of quantum
theory of heat radiation and Maxwell-Boltzmann distribution Law. He deduced following formula
u d  
8h3
d
3
h
c e kT  1


(1.1)
This is known as Planck’s radiation formula. It solved the problem of ultraviolet catastrophe.
Planck’s radiation formula explained the spectral energy density of black body completely. Hence, the
concept of quantum of energy (quantum theroy) was the foundation of modern physics. Einstein built
on this idea and proposed the quantization of electromagnetic radiation in 1905 to explain the photoelectric
effect.
* Wien’s formula: u    d  
8hc  hc / kT
e
d  , (for small values of T).
5
** Rayleigh-Jean’s formula: u    d  
8kT
d  , (for large values of T).
4
1.3
Quantum Mechanics
1.3 PHOTOELECTRIC EFFECT
The emission of electrons from a metal plate when illuminated by electromagnetic radiation of suitable
frequency or wavelength is called photoelectric effect. The emitted electrons are called photoelectrons.
The phenomenon was discovered by Hertz in 1887 when he allowed ultraviolet light to fall on
zinc plate. The phenomenon was experimentally verified by Hallwachs and Lenard. Afterwards, it was
discovered that alkali metals like lithium, sodium, potassium, rubidium and cesium eject electrons when
visible light falls on them. Millikan investigated the effect with a number of alkali metals over a wide
range of light frequencies and was given Nobel Prize in 1923.
1.3.1 Explanation of Photoelectric Effect
Failure of Classical Theory: The classical theory, which assumes that light travels in the form of
electromagnetic waves, fails to explain the experimental observations on photoelectric effect. The main
drawbacks of the classical theory are:
(i) According to the classical theory, if the
intensity of incident radiation were increased,
Photon
i.e. more energy is given to the electorns,
Electron
the energy of emitted electrons must increase.
1
2
E=hv
K = mv
This is contrary to the experimental
2
observation. On increasing the intensity of
Photo
incident radiation the photoelectrons are Sensitive
ejected with the same kinetic energy.
Material
(ii) The existence of a threshold frequency for a
given material cannot be explained on the
Fig. 1.3
basis of wave theory of radiation.
(iii) The instantaneous ejection of photoelectrons from the metal surface cannot be explained by
classical theory. If the illumination is faint, the classical theory predicts that time of several
minutes would have lapsed before a single electron is released. This is contrary to the
experimental observation.
Einstein’s Explanation of Photoelectric Effect: Einstein applied Planck’s quantum theory to
explain the phenomenon of photoelectric effect in 1905. According to this theory, light travels in the
form of energy packats called photons (or quanta). The velocity of these photons is equal to that of light
( c  3  108 m/s), and their energy is h, where ‘h’ is Planck’s constant and  the frequency of light.
Thus, when light of frequency  is incident on a metallic surface, the energy of the photon is E = h.
This energy is supplied to an electron present in the surface of photosensitive material, where this
energy is utilised in two parts (Fig. 1.3).
(i) A part of its energy is used to release electron from the atom and away from the metal
surface. This energy is known as photoelectric work function of the metal and is denoted
by W.
(ii) The rest of the energy is utilised in imparting kinetic energy to the ejected electron. If the
ejected electron moves with a velocity , its kinetic energy is K =
1 2
m .
2
1.4
Engineering Physics-II
 Energy of incident photon = Energy spent in removing an electron (work-function)
+ Kinetic energy of electron.
i.e.,
E=W+K

1
hv  W  mv 2
2
(1.2)
Further, if the frequency of incident light is equal to the threshold frequency v0 , then the electron
will be just-ejected from the metal surface but its velocity (or kinetic energy ) will be zero.
Hence put  = v0 and v = 0 in Eq. (1.2), we get
(1.3)
W  hv0
Putting this value of W in Eq. (1.2), we get

hv  hv0 
1
mv 2
2
h (v  v0 ) 
1 2
mv
2
(1.4)
This equation is called Einstein’s photoelectric equation. In terms of stopping potential V0 , we
have
eV0 
and hence
1
m 2
2
(1.5)
h (v  v0 )  eV0
V0 

(1.6)
hv
h
v 0
e
e
(1.7)
This is another form of Einstein’s photoelectric equation.
Eq. (1.7) represents a straight line shown in Fig. 1.4. The slope of
this straight is h/e. As h and e are both constant, the slope of all such
straight lines is constant.
Corresponding to the threshold frequency, we define long
wavelength limit (0). It represents the upper limit of the wavelength
for photoelectric effect. Its physical significance is that radiation having
wavelength longer than 0 would not be able to eject electorns from a
given metal surface whereas those having <0 will eject. The value
of 0 is given by
c  0  0

0 
c ch

v0 W

0 
3  108  6.625  10 34
W
Fig. 1.4
1.5
Quantum Mechanics

19.875  10 26
metre
W

19.875  1016
Å
W
(1.8)
Here W is expressed in Joules. If W expressed in eV, then
0 

19.875  10 16
W  1.602  1019
12400
Å
W  eV 
(1.9)
From Eq. (1.9), the value of photoelectric work function is given by
W
12400
eV
 0 (Å)
(1.10)
The Einstein’s photoelectric equation (Eq. 1.2) can explain all experimental results. For a particular
material, work function W is constant, hence
1
K .E  mv 2  h
2
(1.11)
2  
Thus the increase in frequency of incident light causes increase in velocity of photoelectrons
provided its intensity remains cosntant.
An increase in the intensity of incident radiation is equivalent to an increase in the number of
photons falling on the emitting surface. If the frequency of incident radiation is above the threshold
frequency    0  , the number of emitted photoelectrons will increase . In this way intensity of emitted
electrons (photoelectric current ) is directly proportional to the intensity of incident radiation.
Example 1.1: A certain spectral line has wavelength 4000 Å. Calculate the frequency and energy
in eV of the photon associated with it. (Given c  3  108 m/s and h  6.63  1034 J-s)
Solution: Frequency of the photon is given by  
c

Here c  3  108 m/s and   4  107 m
3  108
 7.5  1014 Hz
4  10 7
Energy of photon is given by


E = h
 6.63  1034  7.5  1014 Joule
Ans.
1.6
Engineering Physics-II
6.63  1034  7.5  1014
eV
1.6  1019
= 3.108 eV

Ans.
Example 1.2: Radiations of wavelength 5400 Å fall on a metal plate whose work function is 1.9 eV.
Find the kinetic energy of the emitted photoelectrons and their stopping potential.
Solution: Kinetic energy of emitted photoelectrons is given by
K
1 2
m  E  W
2
hc

But
E  h 
Here
h  6.62  1034 , c  3  108 m/s ,   5400 Å  5.4  107 m and W = 1.9 eV

6.63  1034  3  108
 2.3 eV
5.4  10 7  1.6  10 19
K = E – W = 2.3 – 1.9 = 0.4 eV
E
Ans.

Stopping potential is given by
V0 
K 0.4  1.6  1019

 0.4 Volt
e
1.6  1019
Ans.
Example 1.3: If the speed of photoelectrons is 106 m/s, what should be the frequency of the
incident radiation on a potassium metal having work function 2.3 eV?
Solution: According to Einstein’s photoelectric equation
hv = W +
1 2
m
2
Here W = 2.3 eV = 2.3  1.610 19 J , m  9.1  10 31 kg , v = 106 m/s
hv  2.3  1.6  10 19 
1
 9.1  10 31  1012
2
 8.3  1019 J
 Frequency  

8.3  10 19
h
8.3  10 19
6.62  10 34
 1.24  1015 Hz
Ans.
Example 1.4: The energy of photoelectrons emitted from a sensitive plate is 1.46 eV. If its
threshold is 4500 Å, calculate the wavelength of incident radiation.
1.7
Quantum Mechanics
Solution: Kinetic energy of emitted electron is given by
1 1 
K  h    0   hc   
  0 
1 K
1


 hc  0
or
Here K= 1.46 eV  1.46  1.6  10 19 J , h  6.63  1034 J-s, c  3  108 m/s, 0  4.5  107 m

1
1.46  1.6  10 19
1


34
8
 6.62  10  3  10
4.5  10 7
 2.222  10 6  1.176 10 6
 3.398  106 m1
 Wavelength of incident radiation

1
3.398  106
 2.943  107 m  2943 Å
Ans.
Example 1.5: In an experiment tungsten cathode which has a threshold 2300 Å is irradiated by
ultraviolet light of wavelength 1800 Å. Calculate (i) the maximum energy of emitted , photoelectrons
and (ii) the work function for tungsten. Give your results in electron-volts. Given 1 eV = 1.6  10 19 J
and velocity of light c = 3  108 m/s.
Solution: From Einstein’s photoelectric equaiton
c c 
   
1 2
m max  h  h0  h     hc  0
2
  0 
  0 
Here h  6.62  1034 J-s,   1800 Å  1.8  107 m , 0  2.3  107 m

1 2
6.62  10 34  3  108  2.3  1.8 
m max 
 2.3  1.8  10 7 
2
1.6  10 19

6.62  3  0.5
= 1.5 eV
1.6  2.3  1.8
Ans.
hc
Work function W  h0  
0

W
6.62  1034  3  108 = 5.39 eV
2.3  107  1.6  1019
Ans.
Example 1.6: A metallic surface, when illumintated with light of wavelength 1 , emits electrons
with energies up to a maximum value E1 and when illuminated with light of wavelength  2   2  1  , it
emits electrons with energies up to a maximum value E2 . Prove that Planck’s constant h and the work
1.8
Engineering Physics-II
function W are given by,
h
 E2  E1  1 2
c  1   2 
and W 
E2  2  E11
1   2
Solution: Maximum kinetic energy E1 is given by
E1 
hc
W
1
(1)
and E2 is given by
hc
W
2
Subtracting (1) from (2), we get
 1
  
1
 E2  E1   hc     hc  1 2 
  2 1 
 1 2 
E2 

h
 E2  E1  1 2
c  1   2 
(2)
Hence proved.
From Eqs. (1) and (2), we have
E11  hc  W 1
(3)
E2  2  hc  W  2
(4)
Subtracting (3) from (4), we get
E2  2  E11  W  1   2 

W
E2  2  E11
1   2
Hence proved.
Example 1.7: A metallic surface, when illuminated with light of wavelength 3333 Å, emits electrons
with energies upto 0.6 eV, and when illuminated with light of wavelength 2400 Å, it emits electrons with
energies up to 2.04 eV. Calculate Planck’s constant and the work function of the metal.
Solution: Utilising the results of the previous example, Planck’s constant is given by
 E  E1  1 2
h 2
c  1   2 
Here
E1  0.6 eV  0.6  1.6  1019 J
E2  2.04 eV  2.04  1.6  1019 J
1  3.333  107 m
2  2.4  107 m

h

(2.04  0.6)  1.6  1019  3.333  2.4  10 14
3  108  (3.333  2.4)  107
1.44  1.6  3.333  2.4  1034
 6.58 1034 J-s
3  0.933
1.9
Quantum Mechanics
Work function is given by
W
 2.04  2.4  0.6  3.333  10
E2  2  E11

c  1   2 
3.33  2.4   10 7
7
 3.1 eV
Ans.
Example 1.8: The work function of a metal surface is 1.2 eV . Calculate the kinetic energy of the
fastest and slowest photoelectrons and the retarding potential when light of frequency (5.5  1014 Hz)
falls on the surface.
Solution: The kinetic energy of fastest photoelectrons is given by
K max  h  W
Here h  6.62  1034 J-s,  5.5  1014 Hz , W = 1.2 eV

6.62  1034  5.5  1014
 1.2
1.6  1019
= 2.27 – 1.2 = 1.07 eV
K max 
Ans.
The kinetic energy of sowest electrons is zero.
The stopping potential is given by
V0 
K max 1.07  1.6  1019

= 1.07 Volt
e
1.6  1019
Ans.
1.4 MATTER WAVE
After the discovery of Planck’s quantum theory of radiation, it
became an established fact that radiation has dual character
behaving either as waves or as particles under suitable
circumstances. Interference and Diffraction phenomena
demonstrate their wave behaviour while Photoelectric Effect,
Compton Effect etc. (i.e., emission and absorption of photons)
demonstrate their particle behaviour.
Towards the middle of the nineteenth century, it became
clear to the physicists that the physical universe is made up of
two great entities viz. the energy and the matter. The dual nature
of radiation stimulated the scientists, as to why the other physical
entity, i.e., matter must also not possess the dual nature like
radiant energy. Louis de Broglie, a French physicist put forward
a daring suggestion in 1924 that just like radiation, matter also
shows dual nature. In other words, particles of matter, e.g.,
electrons also show wave like properties. His duality hypothesis
was based on the following observations.
(i) The whole energy in this universe is in the form of
electromagnetic radiation and matter.
(ii) As nature loves symmetry, hence matter and energy
must be mutually symmetrical. If radiant energy
has dual character, matter must also possess dual
character.
Louis-Victor-Pierre-Raymond, 7th
duc de Broglie (August 15, 1892 –
March 19, 1987) was a French physicist
and a Nobel laureate. In physics the de
Broglie hypothesis is the statement
that all matter (any object) has a wavelike nature (wave-particle duality). The
de Broglie relations show that the
wavelength is inversely proportional to
the momentum of a particle and that the
frequency is directly proportional to the
particle’s kinetic energy. The hypothesis
was advanced by Louis de Broglie in 1924
in his PhD thesis. He was awarded the
Nobel Prize for Physics in 1929 for this
work, which made him the first person to
receive a Nobel Prize on a PhD thesis.
1.10
Engineering Physics-II
The waves which were assumed to be associated with matter on the basis of de Broglie’s hypothesis
are called as Matter Waves or de Broglie Waves. The principle of Complimentarity is also applicable
here. This implies that in no experiment, matter is found to exist both as a particle and as a wave
simultaneously. The two aspects of wave and particle are always complimentary to each other.
1.4.1 The de Broglie Wavelength
According to Planck’s quantum theory of radiation, the energy of a photon of radiation of frequency 
is given by
E = h
( h = Planck’s constant)
(1.12)
If photon is considered as a particle whose effective mass is m, then according to Einstein’s massenergy equivalence principle, the energy E of photon is
E  mc 2
(c = velocity of light)
(1.13)
Since the energy of the photon in the two cases is same, hence
h  mc 2

mc 
h
c
The quantity mc is the momentum p (say) of the photon of effective mass m and travelling with the
speed of light, i.e., c.
h
p  mc 
( c   )




h
p
(1.14)
This equation provides a connection between a wave like property of radiation, the wavelength
() and a particle like property, the momentum ( p).
de Broglie suggested that this same relationship connects the particle like and wave like properties
of matter. That is, there is a sinusoidal wave having a wavelength associated with a free particle moving
with linear momentum ‘p’. Thus the wavelength of the wave associated with a particle having mass ‘ m’
and moving with velocity ‘v’ is given by

h
h

p m
(1.15)
The wavelength of a particle computed according to Eq. (1.15) is called its de Broglie wavelength.
Note that Planck’s constant provides the connecting link between the wave and particle natures of both
matter and energy.
1.4.2 Important Points to Note
h
, m represents the relativistic mass of
m
the particle. If the velocity of the particle is comparable to that of light, then mass of the
particle is given by:
(1) In the expression for de Broglie wavelength  