Biomedical Engineering Reference
In-Depth Information
Substituting from Eq. (8.12) for
h
ν
gives
sin
θ
ϕ
=
)
.
(8.22)
tan
/
mc
2
)(1 -
cos
θ
(1 +
h
ν
The trigonometric term in
θ
can be conveniently expressed in terms of the half-
2
sin
2
(
angle. Since
sin
θ
=
2
sin
(
θ
/2)
cos
(
θ
/2)
and
1-
cos
θ
=
θ
/2)
, it reduces to
sin
θ
1-
cos
θ
2
sin
(
θ
/2)
cos
(
θ
/2)
2
sin
2
(
cot
2
.
(8.23)
=
=
θ
/2)
Equation (8.22) can thus be written in the compact form
1+
h
mc
2
tan
ϕ
cot
2
=
.
(8.24)
When
θ
is small, cot
/2 is large and
ϕ
is near 90
◦
. In this case, the photon travels in
the forward direction, imparting relatively little energy to the electron, which moves
off nearly at right angles to the direction of the incident photon. As
θ
θ
increases
from 0
◦
to 180
◦
, cot
/2 decreases from ∞ to 0. Therefore,
ϕ
decreases from 90
◦
to 0
◦
. The electron recoil angle
ϕ
in Fig. 8.4 is thus always confined to the forward
direction (
0
θ
90
◦
), whereas the photon can be scattered in any direction.
≤
ϕ
≤
Example
In the previous example a 1.332-MeV photon from
60
Co was scattered by an electron
at an angle of 140
◦
. Calculate the energy acquired by the recoil electron. What is the
recoil angle of the electron? What is the maximum fraction of its energy that this
photon could lose in a single Compton scattering?
Solution
Substitution into Eq. (8.19) gives the electron recoil energy,
1 - (-0.766)
0.511/1.332 + 1 - (-0.766)
=
T
=
1.332
1.094 MeV.
(8.25)
Note from Eq. (8.15) that
T
+
h
ν
= 1.332 MeV =
h
ν, as it should. The angle of recoil
of the electron can be found from Eq. (8.24). We have
cot (140
◦
/2)
1 + 1.332/0.511
=
tan
ϕ
=
0.101,
(8.26)
5.76
◦
. This is a relatively hard collision in which the
photon is backscattered, retaining only the fraction 0.238/1.332
from which it follows that
ϕ
=
0.179 of its energy
and knocking the electron in the forward direction. From Eq. (8.20),
=
1.332
2 + 0.511/1.332
= 1.118 MeV.
2
×
T
max
=
(8.27)
The maximum fractional energy loss is
T
max
/
h
ν =
1.118/1.332
=
0.839.
