IPhO 1983
4.
Theoretical Question IV
Atomics - Problem IV (7 points)
Compton scattering
A photon of wavelength i is scattered by a moving, free electron. As a result the electron stops
and the resulting photon of wavelength 0 scattered at an angle   60 with respect to the
direction of the incident photon, is again scattered by a second free electron at rest. In this
second scattering process a photon with wavelength of  f  1,25  10 10 m emerges at an angle
  60 with respect to the direction of the photon of wavelength  . Find the de Broglie
0
wavelength for the first electron before the interaction. The following constants are known:
h  6,6  10 3 4 J  s - Planck’s constant
m  9,11031 kg - mass oh the electron
c  3,0  108 m / s - speed of light in vacuum
Problem III - Solution
The purpose of the problem is to calculate the values of the speed, momentum and wavelength
of the first electron.
To characterize the photons the following notation are used:
Table 4.1
initial
photon –
final
photon after the
photon
first scattering



momentum p i
p0
pf
energy
E0
Ei
Ef
wavelength
i
i
f
To characterize the electrons one uses
Table 4.2
first electron
first electron second electron Second electron
before collision after collision before collision after collision


momentum p1e
0
0
p2e
energy
E 0e
E 0e
E 2e
E 1e


speed
0
0
v 1e
v 2e
The image in figure 4.1 presents the situation before the first scattering of photon.
Atomics – Problem IV - Solution
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IPhO 1983
Figure 4.1
Figure 4.3
Theoretical Question IV
Figure 4.2
Figure 4.4

To characterize the initial photon we will use his momentum p i and his energy E i
  h h  fi
Pi  
i
c

E  h  f
i
 i
fi 
c
i
is the frequency of initial photon.

For initial, free electron in motion the momentum p oe and the energy E oe are


m 0  v 1e

Poe  m  v 1e 
1  2


2
E  m  c 2  m 0  c
 oe
1  2

( 4.1)
( 4.2)
( 4.3)
where m 0 is the rest mass of electron and m is the mass of moving electron. As usual,
v
  1e . De Broglie wavelength of the first electron is
c
Atomics – Problem IV - Solution
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IPhO 1983
Theoretical Question IV
h
h

1  2
p0e m 0  v 1e
The situation after the scattering of photon is described in the figure 4.2.

To characterize the scattered photon we will use his momentum p 0 and his energy E 0
oe 
h h  fo


Po 
o
c

E  h  f
o
 o
( 4.4).
where
fo 
c
( 4.5)
0
is the frequency of scattered photon.
The magnitude of momentum of the electron ( that remains in rest) after the scattering is zero;
his energy is E 1e . The mass of electron after collision is m 0 - the rest mass of electron at rest.
So,
E1e  m0  c 2
To determine the moment of the first moving electron, one can write the principles of
conservation of moments and energy. That is
 

Pi  poe  p0
( 4.6)
and
( 4.7)
E i  E 0e  E 0  E1e
The conservation of moment on Ox direction is written as
h  f0
h  fi
 m  v 1e  cos 
cos
c
c
( 4.8)
and the conservation of moment on Oy is
m  v 1e  sin 
h  f0
sin
c
( 4.9)
To eliminate  , the last two equation must be written again as

h2 
2


f0  cos  f i 2
m

v

cos


1e

2
c


2
m  v  sin 2   h  f 0 sin 
1e

 c

( 4.10)
and then added.
The result is
m 2  v 12e 
h2  2 2
f 0  f1  2f 0  f i  cos
c2


( 4.11)
or
Atomics – Problem IV - Solution
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IPhO 1983
m 02  c 2
v 
1   1e 
 c 
2
Theoretical Question IV

 v 12e  h 2  f 02  f12  2f 0  f i  cos

( 4.12)
The conservation of energy (4.7) can be written again as
m  c 2  h  f1  m 0  c 2  h  f 0
( 4.13)
or
m0  c 2
v 
1   1e 
 c 
2
 m 0  c 2  h  f 0  f1 
( 4.14)
Squaring the last relation results
m 02  c 4
v 
1   1e 
 c 
 m 02  c 4  h 2  f 0  f1   m 0  h  c 2  f 0  f1 
2
2
( 4.15)
Subtracting (4.12) from (4.15) the result is
2m 0  c 2  h  f 0  f1   2h 2  f1  f 0  cos  2h 2  f1  f  0
( 4.16)
or
h
m0  c
1  cos   c 
f1
c
f0
( 4.17)
Using

h
m0  c
( 4.18)
the relation (4.17) becomes
  1  cos   i  0
( 4.19)
The wavelength of scattered photon is
0  i    1  cos 
( 4.20)
shorter than the wavelength of initial photon and consequently the energy of scattered photon is
greater that the energy of initial photon.
i  0

E i  E 0
( 4.21)
Let’s analyze now the second collision process that occurs in point N . To study that, let’s
consider a new referential having Ox
direction on the direction of the photon scattered after
the first collision.
Atomics – Problem IV - Solution
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IPhO 1983
Theoretical Question IV
The figure 4.3 presents the situation before the second collision and the figure 4.4 presents the
situation after this scattering process. The conservation principle for moment in the scattering
process gives
h
h
    cos  m  v 2e  cos 
 0
f
( 4.22)

h
 sin  m  v  sin   0
2e

 f
To eliminate the unknown angle  must square and then add the equations (4.22)
That is
2
 h

h
  cos   m  v 2e  cos  2
 0 f


2
 h

2


 sin   m  v 2e  sin  

 f
( 4.23)
or
h

 f
2
2
  h 
2h2
2
    
cos  m  v 2e 
  0  0  f
( 4.24)
The conservation principle of energy in the second scattering process gives
h c
0
 m0  c 2 
h c
f
 m c2
( 4.25)
(4.24) and (4.25) gives
h2 c2
2f

h2 c2
20

2h2 c2
cos  m 2  c 2  v 22e
0  f
( 4.26)
and
2
 1
 1
1
1
h  c      m 02  c 4  2h  c 3  m 0      m 2  c 4
 f 0 
 f 0 
2
2
( 4.27)
Subtracting (4.26) from (1.27), one obtain
 h
 1  cos   f  0

 m 0 c
      1  cos 
0
 f
( 4.28)
That is
f  0

E f  E 0
Atomics – Problem IV - Solution
( 4.29)
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IPhO 1983
Theoretical Question IV
Because the value of f is know and  can be calculate as
f  1,25  10 10 m


6,6  10 34


m  2,41 10 12 m  0,02  10 10 m

31
8
9,1 10  3  10

( 4.30)
the value of wavelength of photon before the second scattering is
0  1,23  10 10 m
( 4.31)
Comparing (4.28) written as:
f  0    1  cos 
( 4.32)
and (4.20) written as
i  0    1  cos 
( 4.33)
clearly results
i  f
( 4.34)
The energy of the double scattered photon is the same as the energy of initial photon. The
direction of “final photon” is the same as the direction of “initial” photon. Concluding, the final
photon is identical with the initial photon. The result is expected because of the symmetry of the
processes.
Extending the symmetry analyze on electrons, the first moving electron that collides the initial
photon and after that remains at rest, must have the same momentum and energy as the second
electron after the collision – because this second electron is at rest before the collision.
That is


p1e  p2e
( 4.35)

E1e  E 2e
Taking into account (4.24), the moment of final electron is
p2 e  h
1

2
f

1
2  cos

2
f  1  cos  f  f  1  cos 
( 4.36)
The de Broglie wavelength of second electron after scattering (and of first electron before
scattering) is
 1

1
2  cos



2
2
 f   1  cos  f  f  1  cos  
f


1e  2e  1 
( 4.37)
Numerical value of this wavelength is
1e  2e  1,24  10 10 m
( 4.38)
Professor Delia DAVIDESCU, National Department of Evaluation and Examination–Ministry of
Education and Research- Bucharest, Romania
Professor Adrian S.DAFINEI,PhD, Faculty of Physics – University of Bucharest, Romania
Atomics – Problem IV - Solution
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