PHY-1203

(Physics-II)

4th Chapter: Wave-Particle Duality

 Black Body
 The term black body was introduced by Kirchhoff in 1860. A perfectly
black body is one which absorbs all the electromagnetic radiations incident
on it. As it neither reflects nor transmits any radiation, it appears black. The
main characteristics of such a body is that when heated to a suitable high
temperature, it would emit radiations of all frequencies. As it is a perfect
absorber, it is also a perfect radiant. When black bodies are in equilibrium,
there is a relationship between their temperature and their emission
properties.
 In practice, a perfectly black body in not available. Lamp black and
Platinum black are the nearest approach to a black body. However, a body
showing close approximation to a perfectly black body can be constructed.
 Black Body
• A black body is a perfect emitter and absorber of radiation. In a black body, the
radiation emission depends on its temperature.
• Because of the temperature-emission-frequency relationship, the emission peak
gives us the body’s temperature.
• Stars, as emitting bodies, can be modelled as black bodies. Because of this, the
colour of the star can tell us about its temperature. Red stars will be colder and
blue stars will be hotter.
• Black holes are perfect black bodies, able to absorb all radiation. Several
mechanisms have been predicted and theorized by which they can emit radiation.
One of them is the Hawking radiation.

Astronomical objects can be modelled as black bodies. This is why

black body radiation is very useful in physics, astrophysics, and
astronomy.
 Black Body Radiation
 Black-body radiation is the thermal electromagnetic radiation within, or
surrounding, a body in thermodynamic equilibrium with its environment, emitted
by a black body. When a black body is in equilibrium, the amount of energy
entering a region of the object equals the energy leaving the same region. The
radiation emitted by a black body is known as black body radiation, which has a
continuous spectrum known as the black body spectrum, whose pattern is very
well known.
A perfectly insulated enclosure which is in thermal equilibrium internally contains
black-body radiation, and will emit it through a hole made in its wall, provided the
hole is small enough to have a negligible effect upon the equilibrium.
 Characteristics of the black body spectrum
• The most intensely emitted radiation by an object has a wavelength λ that is
inversely proportional to the temperature. Higher temperatures will emit
intense radiation at shorter wavelengths and the colour will shift to the blue
spectrum. The spectrum of radiation has the following characteristics:

• The spectrum is continuous.

• The spectrum peak moves to shorter
wavelengths as the temperature
increases.
• The spectrum is 0 at 0 – this is the
result of the quantization of energy.
• The spectrum values at larger
wavelengths are smaller.
 Characteristics of the black body spectrum
• The most intensely emitted radiation by an object has a wavelength λ that is
inversely proportional to the temperature. Higher temperatures will emit
intense radiation at shorter wavelengths and the colour will shift to the blue
spectrum. The spectrum of radiation has the following characteristics:

• The spectrum is continuous.

• The spectrum peak moves to shorter
wavelengths as the temperature
increases.
• The spectrum is 0 at 0 – this is the
result of the quantization of energy.
• The spectrum values at larger
wavelengths are smaller.
 Fery’s Black Body
 A double walled hollow copper sphere is taken and is coated with lamp black on its inner
surface (Fig. 3.1). A fine hole O acts as a very narrow opening and a pointed projection P is
made just in front of the hole. When the radiations enter through the narrow opening, they
suffer multiple reflections after falling on the pointed projection. After suffering multiple
reflection, the radiation are completely absorbed. This type of black body shown in Fig. 3.1
was designed by Fery. This body acts as a black body absorber. When this body is placed in a
bath at a fixed temperature, the heat radiations come out of the hole (Fig.3.2). The hole acts as
a black body radiator. It should be remembered that only the hole and not the walls of the
body acts as the black body radiator. Lamp black absorbs about 96% of visible light, while
platinum black absorbs about 98%.

Fig. 3.1: Black body absorber Fig. 3.2: Black body emitter
 Plank’s Radiation Law
 According to the classical theory of radiation, energy changes of radiators take
place continuously. The classical theory failed to explain the experimentally
observed distribution of energy in the spectrum of a black body. Max Plank in
1900 deduced an empirical relation to explain the distribution of energy in the
spectrum of a black body observed experimentally. The law is derived with the
following basic postulates:.
1) A black radiation chamber is filled up not only with radiation, but also with
simple harmonic oscillators or resonators known as Plank’s oscillators capable
of vibration with all possible frequencies. The oscillators are of molecular
dimensions.
2) The oscillators or resonators cannot radiate or absorb energy continuously, but
energy is emitted or absorbed in the form of packets or quanta called photon.
Plank assumed that each photon has an energy h where h is the Plank’s
constant and  is the frequency of radiation. As the energy of a photon is h,
the energy emitted (or absorbed) is equal to 0, h, 2h, 3h.....nh.These are
the only available energy states.
 n 

 n e
The average energy per oscillator is given by  n / k  T

 n 0
n 

e
 n / k  T

n 0
n  n 

 / k T
 n x n
  nx n

Taking xe , we get

 n 0
 n 0
........(i )
n  n 

 x
n 0
n
 x
n 0
n

n 

But  nx
n 0
n
 x  2 x 2
 3 x 3
 4 x 4
 ....

 x(1  2 x  3x  4 x  ....)
2 3

x

(1  x ) 2
 n 
and
 x
n 0
n
 1  x  x 2
 x 3
 x 4
 ....
1

(1  x)
Substituting these values in equation (i), we have
 x /(1  x) 2
x
 
1 /(1  x) 1 x
e  / k T

  / k  T
  / k T
1 e (e  1)
But according to Plank’s hypothesis,  =h; therefore average energy of an oscillator
h
 h / k  T
.......(ii)
(e  1)
The number of oscillators per unit volume in the frequency range  and +d is given by

8 2
N  2 d
c
Hence, energy density of radiation between frequencies  and +d is
= (average energy of a Plank’s oscillator )(number of oscillators per unit volume.

8 2 h
Therefore, E d  2 d [ h / k  T ]
c (e  1)

8 h 3 1
 E  [ h / k T ]......(iii )
 1)
2
c (e

This expression represents Plank’s radiation law and this law explains
all experimental results.
 Photoelectric Effect
Photoelectric effect was first discovered by Heinrich Hertz in 1887.

 “When an electromagnetic radiation of sufficiently high frequency

such as ultra-violet light and X-rays is incident on a clean metal
surface, then electrons are emitted from it. The electrons ejected out
of the metal under the action of light are known as photoelectric
effect”. The electrons emitted are known as the photoelectrons.

The well-known photosensitive metals are Na, K, Cs etc.

 Experimental Study of Photoelectric Effect
 Photoelectric phenomenon can be studied with the help of a simple apparatus shown in
Fig 1. It consists of two photosensitive surfaces E and C enclosed in an evacuated
quartz bulb. In the absence of light, there is no flow of current in the circuit and the
ammeter A reads zero. When plate E is exposed to some source of monochromatic
light, current starts flowing.

The explanation of the above

behaviour lies in the fact that when E is
irradiated with light, the incident photons
eject electrons by collision with its atoms.
These photoelectrons are immediately
attracted by the collector plate C thereby
starting current flow as indicated by the
ammeter.

Fig. 3.3
 Einstein’s Photoelectric Equation
 Einstein proposed an explanation of photoelectric effect as early as 1905. He followed
Plank’s idea of quantum theory of light that light consists of photons, According to
this explanation when a single photon is incident on a metal surface, it is completely
absorbed and imparts its energy hf to a single electron. The photon energy is utilized
for two purposes:
(i) partly for getting the electron free from the atom and away from the metal
surface. This energy is known as the photoelectric work function of the
metal and is represented by W0.
(ii) the rest of the photon energy (hf W0) is utilized in imparting kinetic
energy ½ mv2 to the emitted electron.

Thus, if the mass of an electron is m and its velocity is v, then

1 2 …………………….(i)
hf  W0  mv
2
It is known as Einstein’s photoelectric equation.
 If the photon energy is just sufficient to liberate the electron only, then no
energy would be available for imparting kinetic energy to the electron. Hence, the
above equation reduces to

hf 0  W0
where f0 is called the threshold frequency. Threshold frequency is defined as the
minimum frequency which can eject an electron from metal surface without any velocity.

For frequencies lower than f0, there would be no emission of electrons whereas
for frequencies greater than f0, electrons would be ejected with a certain definite
velocity (and hence kinetic energy)
Substituting this value of w0 in equ. (i), the Einstein’s photoelectric equation becomes

1 2
hf  hf 0  mv …………………….(ii)
2
Equation (ii) suggests that the energy of the emitted photoelectrons is independent
of the intensity of the incident radiation but increases with the frequency.
 Laws of Photoelectric Emission
Photoelectric emission was found to be governed by the following laws:
(i) Photoelectric current (i.e. number of electrons emitted per second)
is directly proportional to the intensity of the incident light.

This can be verified by increasing the

intensity of light and measuring the
corresponding photoelectric current while
holding the frequency of the incident light
constant. Increase in intensity means more
photons and hence ejection of electrons

Fig. 3.4
(ii)For each photosensitive surface, there is a minimum frequency of
radiation (called threshold frequency) at which emission begins.
This fact can be verified by keeping the light intensity constant while varying the
frequency. The current is found to increase with the frequency of the incident light.
Moreover, it is seen that there is a limiting or critical frequency below which no
photoelectrons are emitted. It is called threshold frequency and its value depends
on the nature of the material because for each material there is a certain minimum
energy necessary to liberate an electron. This energy is known as photoelectric
work function or threshold energy W0 where W0 = hf0.

Fig. 3.5
 (iii) The maximum velocity of electron emission (and hence kinetic
energy) varies linearly with the frequency of the incident light but is
independent of its intensity.

Einstein photoelectric equation is given by

1 2
mv max  h( f  f 0 )
2
Or, Emax  f

Hence, increase in the frequency of the

incident light increases the velocity with
which photoelectrons are ejected.

Fig. 3.6
(iv) Photoelectric emission is an instantaneous process. The time lag
between the incident of radiation and emission of first electrons is
less than 10-8 second.

(v) For a given metal surface, stopping potential V0 is directly

proportional to frequency but is independent of the intensity of the
incident light.
Stopping potential is the minimum negative potential given to the collector plate at
which is just sufficient to halt the most energetic photoelectrons emitted.

Now if vmax is the maximum velocity of emission of a photoelectron, e is

the charge of electron and V0 the stopping potential, then the work done by the
retarding potential in stopping the electron = eV0, which gives

1 2
mvmax  eV0
2
Or, Emax  eV0 Joules, or Emax  V0 electron-volt
 If the frequency varies it is found that the stopping potential varies linearly with
frequency. Below threshold frequency, no electrons are emitted, hence stopping potential
is zero for that reason. But as frequency is increased above f0, the stopping potential varies
linearly with the frequency of the incident light..
Einstein photoelectric equation may be expressed in terms of stopping potential as given
below:
1 2
hf  W0  mv max
2
Now, W0  hf 0 and 1 mvmax
2
 eV0
2

 hf  hf 0  eV0
h( f  f 0 )
Or,
V0 
e Fig. 3.7
 Types of Photoelectric Cells
Following three main types are considering:

(i) Photoemissive cell: It depends on the emission of electrons from a metal

cathode when it is exposed to light or other radiation.

(ii)Photovoltaic cell: Here sensitive element is a semiconductor which generates

voltage in proportion to the light or any radiant energy incident on it.

(iii)Photoconductive cell: It uses a semiconductor material whose resistance

changes in accordance with the radiant energy received.
 Problems
(1) The wavelength of light falling on the surface of a metal of work function 2.3 eV is 4300 Ǻ. With
what velocity will the electron be emitted? [v=4.55×105 m/s]
(2) The stopping potential for the electrons emitted from a metal due to photoelectric effect is found to
be 1 V for light of 2500 Ǻ. Calculate the work function of the metal in eV. [W=3.968eV]
(3) A surface having work function 1.51 eV is illuminated by light of wavelength 4000 Ǻ. Calculate
(i) the maximum kinetic energy of the ejected electrons and (ii) the stopping potential.
[K.E=2.552×10-19 J; 1.595 V]
(4) If light of  = 6000 Ǻ falls on a metal surface and emits photoelectrons with a velocity of 4×105
m/s, what is photoelectric threshold wavelength? [Ans: 7,631 Ǻ]
(5) A photon of wavelength 3310 Ǻ falling on a photo cathode ejects an electron of energy 3×10-19 J
and one of wavelength 5000 Ǻ ejects an electron of energy 0.972×10-19J. Calculate the value of
Plank’s constant and the threshold wavelength for the photo cathode.[ h=6.62×10-31 J-s;
0=6620×10-10m]
(6) The stopping potential is 4.6 V for light of frequency 2×1015 Hz. When light of frequency 4×1015
Hz is used, the stopping potential is 12.9 V. Calculate the value of Plank’s constant. [h=6.44×10-34
J-s]
(7) The energy required to remove an electron from sodium is 2.3 eV. Does sodium show
photoelectric effect for orange light with wavelength 680 nm?
 Compton Effect
Definition:
“When a photon strikes a stationary electron, wavelength of photon is
increased after collision. This phenomenon is referred to as Compton effect."
OR

“It is the phenomenon in which a photon of frequency  is scattered by an

electron and the scattered photon has a frequency less than
that of the incident photon is called Compton effect.“
As a result of the Compton effect, the photons transfer some of their energy
to the electrons. It is mainly through the Compton effect that matter absorbs radiant
energy. Compton effect was first observed by Arthur Compton in 1923 and this
discovery led to his award of the 1927 Nobel Prize in Physics.
 Derivation of Compton Shift
 Let a beam of monochromatic X-ray photon strikes an electron at rest and is scattered
away from its original direction of motion while the electron receives an impulse and
begin to move. The photon loses an amount of energy that is the same as the kinetic
energy gained by the electron.

Fig. 3.8
If the initial photon has the frequency  associated with it , the scattered photonhas
the lower frequency ’, where
Loss in photon energy = gain in electron energy

 h  h   KE
Suppose the photon is scattered through an angle  and the electron moves in a
direction  (Fig. a). Let P be the momentum of the recoil electron and m0 its rest mass.
Energy and momentum are conserved in this process.
From the principle of conservation of energy, we have
total initial energy = total final energy


h  m0 c  h  P c  m0 c
2 2 2 2 4

or , P 2 c 2  m02 c 4  h(  )  m0 c 2
Squaring this equation, we get
2 2 2
0
4 2
 2

P c  m c  h (  )  2h(  )m0 c  m0 c 2 2 4

2 2 2
Pc 2m0 c
or , 2
 (   ) 2
 (   ) ……………..(i)
h h
 Momentum is a vector quantity and in elastic collision between two bodies it is
conserved in each of two mutually perpendicular directions. In the present case,
resolving the momenta along and at right angles to the direction of the incident photon.
We get
(i) in the original photon direction
initial momentum = final momentum
h h 
0 cos   P cos
c c
Pc
or , cos     cos  ……….(ii)
h Fig. 3.9
(ii) in the direction at right angles to the original direction

initial momentum = final momentum

h 
0 sin   P sin 
c
Pc
Or, sin     sin  ……….(iii)
h
Squaring equs. (ii) and (iii) and then adding, we have

P 2c 2
2
(sin 2
  cos 2
 )  (    cos  ) 2
   2
sin 2

h
P 2c 2
Or,
2
  2
 2  cos     2
cos 2
    2
sin 2

h
 2 2
Pc
2
   2  cos   cos    sin 
2 2 2 2 2

h
2
 
   2 cos  (cos   sin  )
2 2 2

   2  cos  
2 2

 (  ) 2  2   2  cos


 (  )  2 (1  cos )
2 ……….(iv)

The left hand side of equs. (i) and (iv) are the same, therefore, we have
2
2m0 c
(  )  2
(  )  (  )  2 (1  cos )
2

h
2
2m0 c
Or, (  )  2 (1  cos )
h
    h
Or,  (1  cos )
  m0 c 2

1 1 h
Or,   (1  cos ) ……….(v)
   m0 c 2

To express this equ. In terms of the wave-lengths  and ’, we substitute

c c
 and  where, c is the velocity of light in free space
 
  h
Then we get,   2
(1  cos )
c c m0 c
h
     (1  cos ) ……….(vi)
m0 c
This is the expression for the Compton shift in the wave-length of the X-
rays scattered by electrons in a light element.
 h
The quantity c  is the Compton wave-length of the electron. The numerical
m0 c
value of c is
h 6.626 10 34
c    0.02426  10 10
m
m0 c 9.1110  2.998 10
31 8

Case -1 :
If  = 0, then

h h
    (1  cos 0)  (1  1)  0
m0 c m0 c
   
Case -II : If  = 90°, then
h h
    (1  cos 90)  (1  0)
m0 c m0 c
h 0
  0.02426  which is equal to Compton wave-length
m0 c
Case -III : If  = 180°, then
h h
    (1  cos180)  (1  1)
m0 c m0 c
2h 0
  0.0484 
m0 c
which is the maximum change of wave-length. Hence the maximum change of wave-
length is equal to 0.0484Ǻ and in this situation the photon will be reflected by the
electron.
Problems
(1) An X-ray beam of wave-length 0.300 Ǻ undergoes a 60° Compton scattering. Find
the wavelength of the scattered photon and the energy of electron after scattering.
[Ans: 0.31212 Ǻ ; 2.571210-16 J]
(2) Find the maximum wave-length of Compton scattering X-rays when 1 Ǻ X-rays are
incident on a sample.
(3) An X-ray photon of wave-length 0.1 Ǻ is reflected at angle 9o° with its original
direction after collision with an electron at rest. Find the energy it will loose on
collision.
(4) X-rays of wavelength 10.0 pm are scattered from a target. (a) Find the wavelength
of the X-rays scattered through 45°. (b) Find the maximum wavelength present in
the scattered X-rays. (c) Find the maximum kinetic energy of the recoil electrons.
[Ans:10.7 pm; 14.9 pm;6.54×10-15 J]
(5) Photon of energy 1.02 MeV undergoes Compton scattering through 180°. Calculate
the energy of the scattered photon. [Ans: 0.204 Mev]
 Pair Production
 “The conversion of a photon into an electron and a positron, when the photon
traverse the strong electric field surrounding a nucleus, is called pair production.

In this process, the photon disappears and is

converted to an electron-positron pair. This process
can take place only when the photon energy
exceeds 2m0c2. The pair production process can not
occur in free space and usually takes place in the
presence of a nuclear field. The rest mass energy
m0c2 of an electron or positron is 0.51MeV, hence
pair production requires a photon energy of at least
1.02 MeV. Any additional photon energy becomes
kinetic energy of the electron and positron.
Fig. 3.10 : In the process of pair production, a
photon of sufficient energy materializes into an
electron and a positron
 Pair Annihilation
 The inverse of pair production occurs when a positron is near an electron and the
two come together under the influence of their opposite electric charges. Both
particles vanish simultaneously, with the lost mass becoming energy in the form of
two gamma ray photons. This phenomenon is known as pair annihilation.

e e  
 

The total mass of the positron and electron is equivalent of 1.02 MeV, and each photon has an
energy h of 0.51 MeV plus half the kinetic energy of the particles relative to their centre of
mass.
 Wave-Particle Duality of Matter
 The concept of wave nature of matter arose from the dual character of radiation which
sometimes behaves as a wave and at other times as a particle. Radiations including visible
light, infrared, ultraviolet and x-rays etc behave as waves in propagation experiments based
of interference and diffraction. These experiments conclusively prove the wave nature of
these radiations because they require the presence of two waves at the same position at the
same time. However, radiation behaves as a particle in interaction experiments which
includes black-body radiations, photoelectric effect and Compton effect. Here radiation
interact with matter in the form of photons. The quantum theory explains these experiments
which is clear evidence that wave consists of discrete particle like packets of energy, called
photons. Thus radiation (light) has dual nature.
Wave-particle duality is the concept that all matter exhibits the properties of both
waves and particles. In 1924, Lewis de-Broglie proposed that matter has dual characteristic
just like radiation. It means when the matter is moving it shows the wave properties (like
interference, diffraction etc.) are associated with it and when it is in the state of rest then it
shows particle properties. Thus, the matter has dual nature. The waves associated with
moving particles are matter waves or de-Broglie waves.
 De Broglie Wave-length
 In 1924, French theoretical Physicist Louis de-Broglie made a very bold and novel suggestion
that like radiation matter has also dual (i.e. particle-like and wave-like) characteristics. De-
Broglie proposed that any particle which is in motion, is associated with a wave. According to
this hypothesis, all matter particles like electrons, protons, neutrons, atoms or molecules have
an associated wave with them. This wave of matter is called matter wave or de-Broglie
wave or pilot wave.

According to the Plank’s quantum theory, the total energy of the photon is given by

E  h ……………..(i)
If m is the mass of the photon, then from Einstein mass-energy relation, we have

E  mc 2 ……………..(ii)
From equations (i) and (ii), we have
h
h  mc 2 or , m  2 ……………..(iii)
c
Now the momentum of the photon is given by

h
p  mass  velocity  2  c
c
h
p
c
This can be expressed in terms of  as

h h
p or ,   ……………..(iv)
 p
This is the wave-length of a photon in terms of its momentum.

The momentum of a particle of mass m and velocity v is p = mv and therefore its de-Broglie
wave-length is accordingly
h
 ……………..(v)
mv
where, m is the relativistic mass.
The relativistic mass of a particle is given by

m0
m
1 v c
2 2

h 1 v c h
2 2

  1 v c
2 2 ……..(vi)
m0 v m0 v
Equation (vi) is the expression for the wave-length of the matter waves.
 Group Velocity and Phase Velocity
 Consider two waves that have the same amplitude A but differ by an amount  in
angular frequency and an amount k in wave number. They can represented by the
equations

y1  A cos(t  kx)
y2  A cos[(  )t  (k  k ) x]
The superposition of the two waves will yield a single wave packet or wave group. Let us find
the velocity vg with which the wave group travels.

The resultant displacement y at any time t and any position x is the sum of y1 and y2

y  y1  y2
 A cos(t  kx)  A cos[(   )t  (k  k ) x]
1 1
 2 A cos [( 2   )t  (2k  k ) x] cos ( t  k x)
2 2
 and k are small compared with  and k respectively. Therefore,

2    2
2k  k  2k

Fig. 1

 k ……..(i)
 y  2 A cos[( t x) cos( t  kx)
2 2
This is the analytical expression for resultant wave (wave packet) due to superposition of
the two waves. The second cosine function is the original wave. The coefficient of this cosine can
be considered to be an amplitude that varies with x and t. This variation of amplitude is called
modulation of the wave.
Hence, equ (i) represents wave of angular frequency  and wave number k that has
1
superimposed upon it a modulation of angular frequency  and of wave number 1 k .
2 2
The effect of the modulation is thus to produce successive wave groups, as in Fig. 1.
The phase velocity v is the rate at which the phase of the wave propagates in space

 ……..(ii)
v 
k
The velocity vg of the wave group (group velocity) is the rate at which the envelope of the wave
packet propagates


vg 
k
When  and k have continuous spreads, the group velocity is given by

d
vg  ……..(iii)
dk
 Phase Velocity
 The phase velocity or wave velocity of a monochromatic wave is the velocity with
which a definite phase (crest, trough etc) of the wave travels through the medium.
In other words, it is the velocity with which a plane progressive wave-front travels
towards. It is denoted by v i.e.,

 2 
v     [where,  is the wave’s angular frequency (rad./sec)
2 k k and k is the angular wave number (rad./m)]

Group Velocity
 The group of waves is called a wave packet and with time the packet moves
forwards in the medium with a velocity, called the group velocity. It is defined by
the equation
d
vg 
dk
 Relation Between Group Velocity and
Particle Velocity
 Let us consider a particle moving with a velocity v. According to de-Broglie
hypothesis, let it consists of a wave group or wave packet. The velocity of the
whole group or point of maximum amplitude is known as the group velocity,
which is
d
vg  .............(i )
dk
where,  is the angular frequency and k is the wave number or propagation constant and
both are function of v.

We know the total energy of the particle is given by

2
m0 c
E  mc  2
................(ii ) [ where, m is the relativistic mass and

1 v c
2 2 m0 is the rest mass of the body]
and the momentum is
m0 v
p  mv  .................(iii )
1  v2 c2
Now, the angular frequency

  2  2
E  E  h
h
2 m0 c 2
 ........................(iv )
h 1 v c 2 2

The propagation constant of the associated de-Broglie wave is

2 p
k  2  
h
 h
p
2 m0 v
 ............( v )
h 1 v c2 2
Now from equation (i), we have
d d dv
vg   ..............( vi)
dk dk dv
Differentiating equ.(iv) with respect to v, we have

d 2m0 c 2 1 2v
 [ (1  v c ) .( 2 )
2 2 3 2

dv h 2 c
2m0 v
 ...........( vii)
h (1  v c )
2 2 32

Now, differentiating equation (v) with respect to v, we get

dk 2 m0 1 1 2v
 [  v ( )(1  v c ) ( 2 )]
2 2 3 2

dv  1 v c
2 2
2 c
2 m0 1 v2 1
 [1  2 ]
h 1 v c
2 2 c (1  v c )
2 2
 dk 2 m0 1 v2 c2
or ,  [1  2 2 2 ]
dv h 1 v2 c2 c (c  v )
2 m0 1 c2 2 m0 1 1
 
2 (c  v )
2 2 2 (1  v 2 c 2 )
h 1 v c
2 h 1 v c
2

2 m0 1
 ............( viii )
h (1  v c )
2 2 32

Putting the values of d/dv and dk/dv in equ.(vi), we have

d dv 2m0 v h
vg    (1  v c )
2 2 32

dk dv h (1  v c )
2 2 32
2m0

 vg  v
So, the de-Broglie wave group associated with a moving particle
travels with the same velocity as the particle.
 Heisenberg's Uncertainty Principle
 The limits of accuracy with which the position and momentum of a particle can be
obtained by the uncertainty principle according to which

x p  
where, x and p are the respective uncertainties in the determination of the position
and momentum of the particle. That is, the product of the uncertainties in
determining the position and momentum of a particle is equal to or greater than
Plank's constant. Thus the uncertainty principle states:

“It is impossible to know both the exact position and exact momentum of an object at the
same time.”

As indicated by this principle, if position is measured accurately, then measurement

of momentum becomes correspondingly inaccurate and vice versa. The principle
only gives probability of fining a particle in a given space in place of certainty.
 Fig. 2 (a) Fig. 2(b)

 Figure 1(a) shows a narrow de-Broglie wave group. The position of the particle
can be precisely determined, but the wave-length (and hence the particle’s
momentum) can not be established because there are not enough waves to measure
accurately.
Figure 2(b) shows a wide wave group. Now the wave-length can be precisely
determined ut not the position of the particle.
 Proof of Uncertainty Principle
 In the figure, a particle is moving with velocity v which is located within a wave
packet that moves with the velocity vg = v. The particle is somewhere within the
region x of the wave packet.

The group velocity vg is expressed as

d d (2 )
vg  
dk d (2  )
x
d 2 d ……..(i)
or, v g   
d (1  ) d
From de-Broglie wave-length, we know

h h dp ……..(ii)
 d   2
p p Fig. 3
Putting the value of d from equation (ii) in equation (i)
d 2 d
v g   2
 p
2

 (h dp p )
2
h dp
2 d d ……..(iii) h
h h   or , 2 p 2  h 2
h dp dp p
For a wave packet traveling along the x-direction and of length x which takes a time
t to pass a reference position (in figure),

x ……..(iv)
vg 
t
Therefore from equation (iii) and (iv), we get

x d 
h h
t dp p x
 x px  h  t ……..(v)
 Now, the minimum time required to measure the frequency of the wave must be
the time for the passage of one complete wave-length to pass a reference point.
This time of passage of one complete cycle is related to the frequency by

1
t 

 t   1 ……..(vi)
Substituting this expression in equation (v), we get

x px  h
A more sophisticated approach shows that

h
x p x  or, x px  
2
 The corresponding energy uncertainty is

E  h
Hence, we have
1
h  t 
E  
t

 E t  h
This is the energy-time uncertainty relation
 Whether electron can reside inside the nucleus?
(i) Nuclear size:
Typical nuclei are less than 10-14 m in radius. If an electron exists inside nucleus, the
uncertainty in its position x may not exceed 10-14 m. According to Heisenberg’s
uncertainty principle, uncertainty in the electron’s momentum is
 6.33 10 34 J  sec
p    1.0  10  20
J  sec/m
x 2 10 m 14

If this is the uncertainty in the electron’s momentum, the momentum itself must be
at least comparable in magnitude. Therefore, approximate momentum of the
electron = p = 1.010-20 kg-ms-1. We know

E  p 2 c 2  m02 c 4
 (1.0 10 20 ) 2  (3 108 ) 2  (9.11031 ) 2  (3 108 ) 4
 3 1012 Joule  2 107 eV
 E  20 MeV
  E  20 MeV
This shows that if an electron exists in the nucleus, the kinetic energy of
the electron must be more than 20 MeV. Electrons of such large energy are never
found to be emitted during -decay. The maximum energy of a -particle emitted is
only 2 to 4 MeV. Hence, we conclude that electrons cannot be present within the
nuclei.

(ii) Magnetic Moment:

Protons and electrons are endowed with magnetic properties. The magnetic
moment of an electron is about one thousand times that of a proton. If electrons exist
inside nucleus, the magnetic moment of electrons will have a dominating influence
and so nuclear magnetic moments ought to be of the same order of magnitude as that
of the electron. However, the observed magnetic moments of nuclei are comparable
with that of the proton. This experimental fact goes against the electrons existing
inside the nucleus.

Due to these reasons, it is concluded that electrons cannot exist in the nucleus.
Problems
1. A particle of mass 0.51 MeV/c2 has kinetic energy 100eV. Find its de-Broglie wave-
length. [=1.23410-10m]

2. A proton is confined to a nucleus of radius 510-15 m. Calculate the minimum

uncertainty of momentum and kinetic energy of the proton (mp= 1.6710-27 kg). [

3. An electron is confined to a box of length 10-10m. Calculate the minimum

uncertainty in the measurement of its velocity. [v=1.58 106 ms-1]

4. An electron has a speed of 600 m/s with an accuracy of 0.005%. Calculate the
certainty with which we can locate the position of the electron? [x=0.003846m]

5. The position and momentum of a 1 KeV electron are simultaneously determined. If

its position is located to within 1Å, what is the percentage of uncertainty in its
momentum? [p/p  100= 6.173%]