Biomedical Engineering Reference
In-Depth Information
Since 84.0 MeV s
-1
is absorbed by the slab, 1000 - 84 = 916 MeV s
-1
is transmitted.
The fraction of the total transmitted energy carried by the uncollided photons is there-
fore 818/916 = 0.893.
(e) With
0.0270 cm
2
g
-1
0.0271 cm
2
g
-1
,wehave
µ
tr
=
0.0270/0.0271 = 0.9963. Equation (8.58) implies that the fraction of the secondary-
electron initial kinetic energy that is emitted as bremsstrahlung is
µ
en
/
ρ
=
and
µ
tr
/
ρ
=
µ
en
/
1-
µ
en
g
=
µ
tr
=
1 - 0.9963
=
0.0037.
(8.65)
Example
A
137
Cs source is stored in a laboratory. The photon fluence rate in air at a point in
the neighborhood of the source is 5.14
10
7
m
-2
s
-1
. Calculate the rate of energy
absorption per unit mass (dose rate) in the air at that point.
×
Solution
The desired quantity is given by Eq. (8.61). The mass energy-absorption coefficient of
air for the photons emitted by
137
Cs (
h
0.662 MeV, Appendix D) is, from Fig. 8.12,
µ
en
/
ρ
= 0.030 cm
2
g
-1
. The incident fluence rate is
ν =
˙
= 5.14 × 10
7
m
-2
s
-1
,andso
the energy fluence rate is
˙
=
˙
10
7
MeVm
-2
s
-1
10
3
MeV cm
-2
s
-1
.
h
ν =
3.40
×
=
3.40
×
(8.66)
Thus, Eq. (8.61) gives
µ
en
ρ
D
=
˙
102 MeV g
-1
s
-1
.
=
(8.67)
Expressed in SI units,
102 MeV
gs
J
MeV
×
10
3
g
kg
D
10
-13
=
×
1.60
×
(8.68)
10
-8
Jkg
-1
s
-1
0.0587 mGy h
-1
.
=
1.63
×
=
(8.69)
The unit, J kg
-1
, is called the gray (Gy). Note that the dose rate is independent of the
temperature and pressure of the air. An exact balance occurs between the amount of
energy absorbed and the amount of mass present, regardless of the other conditions.
For a given material, the mass interaction coefficients (Figs. 8.8, 8.9, 8.11, and 8.12)
are independent of the density. On the other hand, the value of the linear coefficients
does depend on density. For example, Fig. 8.13 applies to water in the liquid phase
(unit density), whereas the values given in Figs. 8.9 and 8.11 are the same for both
liquid water and water vapor.
