Download Compton Effect

Quantum Mechanics
https://www.youtube.com/watch?v=cz27jq4ATL4
Reference:
Concepts of Modern Physics “A. Beiser”
1
Compton Effect
In 1923, Compton while studying the scattering of X-rays by a block of graphite
observed that two types of X- rays were found in scattered rays. One, whose
wavelength is same as that of incident X-rays and other, whose wavelength is
greater than that of incident X-rays. This difference in wavelength called Compton
shift which depends on scattering angle and this effect is known as Compton
Effect.
Compton effect is the outcome of collision between the high energy photon and free e-.
pp sin 
pe  0
pp 
pp cos 
pe cos 
h
c
after collision
before collision
pe sin 
•Loss in photon energy=gain in electron energy  K  h-h  K
•Since rest mass of photon is zero, E=pc
again E=h , therefore, p=h /c.
Now if we consider momentum conservation in the direction
of original photon and in the direction perpendicular to it then
we have
h
h
 0  cos   pcos  ...(i)
c
c
h
0
sin   psin  ...(ii)
c
h
h
 0  cos   pcos  ...(i)
c
c
h
0
sin   psin  ...(ii)
c
pc cos  h  h  cos
pc sin   h  sin 
By squaring and adding we get
p 2 c 2  (h ) 2  2h 2  cos   (h ) 2 .......(iii )
total energy of Electron
E  K  m0 c 2  m02c 4  p 2c 2
 K  m0 c

2 2
 m02 c 4  p 2c 2
or , p 2 c 2  K 2  2m0 c 2 K
Since K  h  h '
We have p c  (h )  2h 2   (h )2  2m0c 2 (h  h ) .... (iv)
2 2
2
Equating equations  iii  and  iv  we obtain
2m0c 2 (h  h )  2h2 (1  cos  ) .......(v)
Dividing equation  v  by 2h 2c 2 we get
m0 c        
(1  cos  )
  
h c c  c c
m c  1 1  (1  cos  )
or , 0    
h    
 
Therefore,
h

     
(1  cos  ) .........(vi )
m0 c
Equation  vi  shows Compton shift.
h
c 
 0.0024nm or 2.4 pm is known as compton wavelength
m0 c
for scattering particle (electron).
Note:
Case I: when  =0o , λ=0 Case II: when  =90o , λ=λ c Case III : when   180o ,   2c  4.8 pm
Realtion between  and 
pc sin   h  sin 
pc cos  h  h  cos 
h  sin 
2sin( / 2)cos( / 2)

h  h  cos 
 /    cos
h
      
(1  cos  )
m0c
1 1
h
 
(1  cos )
2

  m0c
tan  

h

(1  cos )  1 let
2
  m0c

  (1  cos  )  1


2sin( / 2)cos( / 2)
tan  
 (1  cos  )  1  cos 
2sin( / 2)cos( / 2)
tan  
(  1)(1  cos  )
2sin( / 2)cos( / 2)
tan  
(  1)2sin 2 ( / 2)
h

m0c 2
cot  / 2
tan  
h
1
m0 c 2
Compton effect
• (h/m0c) is known as Compton wavelength of
the scattering particle. For an electron it is
0.024 Å.
• This phenomenon gives a very strong evidence
in support of the quantum theory of radiation.
• Observe the difference in wavelength for
various values of .
Quest:
Q1: An x-ray photon is scattered by an electron. The frequency of the
scattered photon relative to that of the incident photon (a) increases, (b)
decreases, or (c) remains the same.
Ans: (b)
Q2. A photon of energy E0 strikes a free electron. The scattered photon with
energy E moves in opposite direction that of the incident photon. In this
Compton effect interaction, the resulting kinetic energy of the electron is (a)
E0 , (b) E , (c) E0  E , (d) E0 + E , (e) none of the above.
Ans: (c)
Q3. Deduce the expression for the maximum kinetic energy of the recoiled
electron.
Q4. Why Compton effect not observed with visible light.
Q5. Explain the presence of unmodified scattered radiation along with
modified radiations for non zero scattering angle.