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Unit I: Quantum Physics
Lectures on Engineering Physics
BE - 201
1.3 Compton Effect
A. H. Compton observed that “when a monochromatic beam of high frequency (lower
wavelength) radiation (e.g., X-rays and γ-ray) is scattered by a substance, the scattered
radiation contains two type of wavelengths one having same wavelength as that of incident
radiation while the other having the wavelength greater (or lower frequency) than that of
incident radiations. This effect is known as Compton Effect.
Quantum Explanation: The explanation was given by Compton which was based on quantum
theory of light. According to quantum theory when photon of energy hυ strikes with the
substance some of the energy of photon is transferred to the electrons, therefore the energy (or
frequency) of photon reduces and wavelength increases.
Various assumptions were made for explaining the effect these were:
(i)
(ii)
(iii)
Compton Effect is the result of interaction of an individual particle and free electron
of target.
The collision is relativistic and elastic.
The laws of conservation of energy and momentum hold good.
Figure 1.6 : Compton Scattering
1.3.1 Derivation of Compton Shift
Before collision
The energy of incident photon
The energy of electron
The momentum of the photon
The momentum of electron
 h
 mo c 2
h

c
=0
After collision
 h 
 mc 2
h 

c
= mv
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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Unit I: Quantum Physics
Lectures on Engineering Physics
BE - 201
The energy of the system before collision = h  mo c 2
The energy of the system after collision = h   mc 2
According to the principle of conservation of energy
h  mo c 2  h   mc 2
i.e.
mc 2  h  h   mo c 2
…….. (1.32)
According to the principle of conservation of linear momentum along and perpendicular to the
direction of incident photon (i.e., along x and y axis), we have
h
h 
0 
cos   mv cos 
c
c
i.e.
mvc cos   h  h  cos 
And
0
i.e.
mvc sin   h  sin 
…….. (1.33)
h 
sin   mv sin 
c
……..(1.34)
Squaring (1.33) and (1.34) and then adding, we get
m 2 v 2 c 2   h  h  cos     h  sin  
2
2
Or
m2v 2c 2   h    h   cos 2   2  h  h   cos    h   sin 2 
Or
m2v 2c 2   h    h    2  h  h   cos 
2
2
2
2
2
……..(1.35)
Squaring equation (1.32), we get
m2c 4  mo 2c 4   h    h    2  h  h    2mo c 2  h  h 
2
2
……..(1.36)
Subtracting (1.35) from (1.36), we get
m2c4  m2v2c2  mo 2c4  2  h  h  cos  1  2moc2 h  h 
……..(1.37)
According to the theory of relativity
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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Unit I: Quantum Physics
m
mo
1
v2
c2
, m2 
Lectures on Engineering Physics
BE - 201
mo 2
 v2 
1  2 
 c 
 v2 
m2 1  2   mo 2
 c 
Or
m 2 c 2  m 2 v 2  mo 2 c 2
Multiplying both sides by c2, we get
m2 c 4  m2 v 2 c 2  mo 2c 4
……..(1.38)
Using equation (1.38) equation (1.37) becomes
0  2  h  h  cos  1  2moc2 h  h 
2  h  h  1  cos    2moc2 h  h 
Or
  
h

1  cos  
 
mo c 2
1 1
h
 
1  cos  
   mo c 2
……..(1.39)
To find the relation in term of wavelength, let us substitute    c   and  c  , we thus have
      
Compton shift  
h
1  cos  
mo c
h
1  cos  
mo c
……..(1.40)
From above equations (1.39) and (1.40) following conclusions can be drawn
1.
2.
3.
4.
5.
The wavelength of the scattered photon λ’ is greater than the wavelength of incident
photon λ.
∆ λ is independent of the incident wavelength.
∆ λ have the same value for all substance containing free electron
∆ λ only depend on the scattering angle .
when  = 0; cos  = 1
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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Unit I: Quantum Physics
Lectures on Engineering Physics
BE - 201
∆ λ = λ’ – λ = 0
6.
 = λ’ = λ, the scattered wavelength is same as the incident wavelength in the
direction of incidence.
when = ; cos  =0
∆ λ = λ’ – λ =
 
h
mo c
h
 0.02426 Ao = λc ;
mo c
……..(1.41)
Where λc is called the Compton wavelength of the electron.
7.
when = ; cos  = -1
max 
2h
 0.04652 Ao
mo c
……..(1.42)
Why Compton Effect is not observed in visible spectrum
The maximum change in wavelength max is 0.04652 Ao or roughly 0.05 Ao. This small
therefore can not be observed for wavelength longer than few angstrom units. For exampleIncident
Radiation
X-ray
Visible
Incident
max
Percentage of incident
radiation
Wavelength
1 Ao
0.05 Ao
5% (detectable)
o
5000 A
o
0.05 A
0.001% (undetectable)
1.3.2 Direction of Recoil electron
Dividing equation (1.36) by (1.35) direction of recoil electron is given by
tan  
h  sin 
h  h  cos 
tan  
c  sin 
c   c   cos 
tan  
 sin 
    cos 
……..(1.43)
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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Unit I: Quantum Physics
Lectures on Engineering Physics
BE - 201
1.3.3 Kinetic Energy of Recoil Electron
The kinetic energy gained by electron is equal to the energy loss by the scattered photon
 m  mo  c2  h  h 
   

   
 m  mo  c 2  hc 
    
E  hc 

   
Where     
h
1  cos  
mo c
……..(1.44)
1.3.4 Verification of Compton Effect:
A beam of monochromatic X-ray of known wavelength λ is made incident on a graphite
scatterer. These were observed at different angles with a Bragg spectrometer.
Collimating
Slits
Scattered X ray
Monochromatic
X ray
X ray
Spectrometer
Scatterer
Figure 1.7 : Experimental verification of Compton Effect
The intensity distribution verses wavelength for various angles is plotted as shown in figure. It is
obvious that the curves have two peaks, one corresponding to modified radiations and the other
to unmodified radiations.
The difference between the two peaks on the wavelength axis gives the Compton shift ( λ). The
curve shows the greater is the scattering angle; greater is the Compton shift in accordance with
      
h
1  cos  
mo c
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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Unit I: Quantum Physics
Lectures on Engineering Physics
BE - 201
At θ = 90o, λ = 0.02426Å
Unmodified radiation
 = 0o
 = 45o
∆λ
Modified radiation
λ’
 = 90o
∆λ
λ'
 = 135o
∆λ
 = 180o
∆λmax
λ
λ'
λ
Thus Compton Effect is experimentally verified.
Dr. Amita Maurya,, Asstt. Prof., Physics, People’s College of Research & Technology, Bhopal
For more lecture notes visit http://sites.google.com/site/dramitamaurya
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