E
1
=
hf
1
E
2
=
hf
2
p
1
=
h
/
λ
1
p
2
=
h
/
λ
2
φ
θ
m
p
e
=
E
2
–
E
0
2
1
–
c
Continued
Fig. 321
The scattering of x rays can be treated as a collision of a photon of initial momentum
h
/
l
1
and a free electron. Using conservation of momentum and energy, the momentum of the
scattered photon
/
2
can be related to the initial momentum, the electron mass, and the
scattering angle. The resulting Compton equation for the change in the wavelength of the x ray is
Equation 342.
More
More
Derivation of Compton’s
Equation
Let
1
and
2
be the wavelengths of the incident and scattered x rays, respectively,
as shown in Figure 321. The corresponding momenta are
Eh
f
h
11
p
55
5
1
cc
1
and
2
p
2
c
2
using ƒ
5
c
. Since Compton used the
K
a
line of molybdenum (
5
0.0711 nm; see
Figure 318
b
), the energy of the incident x ray (17.4 keV) is much greater than the
binding energy of the valence electrons in the carbon scattering block (about 11 eV);
therefore, the carbon electron can be considered to be free.
Conservation of momentum gives
p
5
p
1
p
12
e
or
222
p
5
p
1
p
2
2
p
·
p
e
1
2
341
22
5
p
1
p
2
2
pp
cos
u
1
2
where
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 Fall '10
 Guna
 Electron, Compton, P1 P2, p1 p2 cos, p2 e p2

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