Modern Physics

Compton effect
X-rays are incident on a suitable material (graphite), scattered radiation consists of two
wavelengths. One having same wavelength as that of incident radiation while the other
having the wavelength greater than that of incident radiations. This scattering phenomenon
is known as compton effect.

Let an X-ray of the photon of energy E = hν colloide with an electron initially at rest.
The photon scattered through an angle φ with its energy E = hν 0 . The electron itself recoils
with a velocity v in a direction making an angle θ with the direction of the incident photon.
The change in wavelength which is also called compton shift is given by,

h
∆λ = λ0 − λ = (1 − cos θ)
m0 c
This relation is known as compton equation. Here, λ is the wavelength of incident X-
rays, λ0 is the wavelength of scattered X-rays, θ is the scattering angle amd m is the mass
h
of the electron. The quantity mc is known as the compton wavelength and it is a universal
constant.
The compton shift depends only on the scattering angle θ but is independent of energy
or wavelength of the incident photon. Also, λ0 is always greater than λ, i.e., the scattered
photon has a longer wavelength than the incident photon. The change in wavelength ranges
from 0◦ to 180◦ . i.e.,
Maximum wavelength shift for θ = 180◦ =⇒ ∆λ = h
m0 c

Minimum wavelength shift for θ = 0◦ =⇒ ∆λ = 0

Physical significance
Compton effect explained on the basis that the X-ray photon colloide with electrons and
during collision, exchange of energy takeplace through particle-particle collision. Thus,
compton effect signifies the particle nature of light.

1
de-Broglie hypothesis
In 1924, Louis de Broglie of France suggested that matter, like radiation, has dual nature,
i.e., matter which is made of discrete particles such as atoms, protons, electrons, etc., might
exhibit wave like properties under appropriate conditions. The existence of de Broglie waves
was experimentally demonstrated in 1927 by Davisson and Germer.
Matter waves: The waves associated with material particles like electrons, protons,
neutrons, atoms or molecules are called Matter Waves or de Broglie Waves.
Louis de Broglie hypothetically suggested that, when a particle has a momentum p, its
motion is associated with a wavelength λ, called de Broglie wavelength given by,

h
λ=
p
The de Broglies prediction is concerned with wave-particle duality, i.e., the particle
nature of waves and wave nature of particles.

de-Broglie wavelength of electrons (Extension to the case of electron particle)

Consider an electron accelerated by a potential difference of V volts. If m is the mass and
v is the velocity of the electron, then the energy acquired by the electron is given by,
1
eV = mv 2
2
Multiplying by m both sides and rearranging we get,

2meV = m2 v 2
If p is the momentum of the electron, then,

2meV = p2
or
√
p= 2meV ......(1)
We know that, de-Broglie wavelength is given by,

h
λ= ......(2)
p
From equation (1) and (2),

h
λ= √
2meV
Since m, h and e are universal physical constants, then the above equation becomes,
1.226
λ = √ nm
V
From the above equation it is clear that
1
λ= √
V

2
Phase velocity
The velocity with which a wave travels is called phase velocity and is also called wave
velocity. If a point is marked on the wave representing the phase of the particle, then the
velocity with which the phase propogates from one point to another is called phase velocity.

Consider a simple harmonic wave originatng at the origin O and travelling with a velocity
along X-direction. At t = 0, let the particle at the origin O be disturbed, it executes simple
harmonic motion. Its displacement from the mean position at any instant of time is given
by the equation,

y = A sin ωt
Where, A is the amplitude of the vibration and ω is the angular frequency.
Consider a particle of the medium at the point p at a distance x from O. As the wave
travels with a constant speed in a given medium, the time taken by disturbance to reach
p is x/v. Therefore, displacement of the particle p at a time t will be same as that of the
particle at O at a time (t − x/v) is given by,
 x
y = A sin ω t −
v
This equation represents the displacement of the particle at p at time t. Then the above
equation becomes,
 ωx 
y = A sin ωt −
v

y = A sin(ωt − kx)
Where, k = ωv is called the propogation constant or the wave number. The term,
(ωt − kx) gives the phase of the particle and it will be same for all uniphase points in a
periodic wave. Therefore, for uniphase points,

d
(ωt − kx) = 0
dt
dx
ω−k =0
dt
dx ω
=
dt k

3
 If x is the distance travelled, then the term ( dx
dt ) is known as phase velocity. Therefore,

ω
vp =
k

Group velocity
Matter waves can be considered as a resultant wave due to the superposition of many
component waves where the velocities are slightly differ. Thus, a wave group or wave
packet is formed. The velocity with which the wave group travels is called group velocity,
which is same as the particle velocity.

Consider a wave group formed by the superposition of two waves, which slightly differ
in their velocities with amplitude A travelling in the same direction and are represented by,

y1 = A sin[ωt − kx] − − − − − −(1)

y2 = A sin[(ω + ∆ω)t − (k + ∆k)x] − − − − − −(2)

Where, y1 and y2 are instantaneous displacements, ω and (ω + ∆ω) are the angular
velocities, k and (k + ∆k) are the wave numbers and x is the displacement at instant of
time t.
The resultant displacement y is given by,

y = y2 + y1

y = A sin[(ω + ∆ω)t − (k + ∆k)x] + A sin[ωt − kx]

We know that,
   
A−B A+B
sin A + sin A = 2 cos sin
2 2
Therefore,

   
(ω + ∆ω)t (k + ∆k)x ωt − kx (ω + ∆ω)t (k + ∆k)x ωt − kx
y = 2A cos − − sin − +
2 2 2 2 2 2
   
(∆ω)t (∆k)x (2ω + ∆ω)t (2k + ∆k)x
y = 2A cos − sin −
2 2 2 2

4
 Since, ∆ω and ∆k are very small compared with ω and k, respectively, then, (2ω+∆ω) ≈
2ω and (k + ∆k) ≈ 2k.
Therefore,
 
(∆ω)t (∆k)x
y = 2A cos − sin(ωt − kx) − − − − − −(3)
2 2
From equations (1) and (3), we get,
 
(∆ω)t (∆k)x
R = 2A cos −
2 2
At t = 0 and t = 0, then the maximum amplitude is given by,

Rmax = 2A
Therefore, the group velocity is given by,
 
x (∆ω)/2
vg = =
t (∆k)/2
∆ω dω
vg = lim∆k→0 =
∆k dk

Relation b/w phase velocity and group velocity

Consider a particle of mass m having vp and vg are the phase velocity and group velocity,
respectively. If λ is the de-Broglie wavelength, then we know that,
Phase velocity,
ω
vp = − − − −(1)
k
Group velocity,

dω
vg = − − − −(2)
dk
where ω is the angular frequency of the wave and k is the wave number.
From equation (1), we have

ω = vp k
Therefore,

d(vp k)
vg =
dk
dvp
vg = vp + k
dk
or

dvp dλ
vg = vp + k − − − −(3)
dλ dk
We know that, propogation constant,

5
 2π
k=
λ
or
2π
λ=
k
Differentiating we get,
 
dλ d 2π
=
dk dk k
dλ 2π
= − 2 − − − −(4)
dk k
Then equation (3) becomes,
 
dvp 2π
vg = vp + k − 2
dλ k
 
dvp 2π
vg = vp −
dλ k
Therefore,

dvp
vg = vp − λ
dλ
This is the relation b/w group velocity and phase velocity.