Biomedical Engineering Reference
In-Depth Information
The ideal geometry and other conditions represented by Fig. 8.10 and Eq. (8.59)
are approached in practice only to various degrees of approximation. The nonuni-
formity and finite width of real beams, for example, are two factors that usually
deviate significantly from the ideal. The computation of attenuation and energy
absorption in thick slabs is treated with the help of buildup factors (Charter 15).
Nevertheless,
µ
en
is frequently useful for estimating absorbed energy in a number
of situations, as the following examples illustrate.
Example
A parallel beam of 1-MeV photons is normally incident on a 1.2-cm aluminum
slab (
2.70 g cm
-3
)atarateof10
3
s
-1
. The mass attenuation and mass energy-
absorption coefficients are, respectively, 0.0620 cm
2
g
-1
and 0.0270 cm
2
g
-1
. (a) What
fraction of the photons is transmitted without interacting? (b) What fraction of the
incident photon energy is transmitted by the slab? (c) How much energy is absorbed
per second by the slab? (d) What fraction of the transmitted energy is carried by the
uncollided photons? (e) If the mass energy-transfer coefficient is 0.0271 cm
2
g
-1
, what
fraction of the initial kinetic energy transferred to the electrons in the slab is emitted
as bremsstrahlung?
ρ
=
Solution
(a) The values given for the mass attenuation and mass energy-absorption coefficients
can be checked against Figs. 8.8 and 8.11. The attenuation coefficient is
×
2.70 = 0.167 cm
-1
. The fraction of photons that penetrate the 1.2-cm slab without
interacting is therefore e
-(0.167×1.2)
µ
=
0.0620
0.818.
(b) The energy transmitted by the slab can be inferred from Eq. (8.59), irrespective
of the beam width and uniformity, provided the slab can be regarded as “thin.” This
criterion is satisfied to a good approximation, since 0.818 of the incident photons do
not interact (
=
=
0.0729 cm
-1
. It follows from Eq. (8.59) that the fraction of the incident photon energy
that is transmitted by the slab is
µ
x
=
0.200). The energy absorption coefficient is
µ
en
=
0.0270
×
2.70
E
E
0
= e
-
µ
en
x
= e
-0.0729×1.2
= 0.916,
(8.63)
where
E
and
E
0
are the transmitted and incident rates of energy flow. Whereas the
result (8.63) is approximate, the answer to part (a) is exact.
(c) The rate of energy absorption by the slab is the difference between the incident and
transmitted rates. The result (8.63) implies that the fraction of the incident photon
energy that is absorbed by the slab is 0.084. With the incident rate
E
0
=
(1.0 MeV)
×
(10
3
s
-1
)
10
3
MeV s
-1
, we find that the rate of energy absorption in the slab is
=
1.0
×
0.084
E
0
=
84.0 MeV s
-1
.
(d) From part (a) it follows that the rate of energy transmission through the slab by
the uncollided photons is (exactly)
E
0
e
-
µx
10
3
818 MeV s
-1
.
=
1.0
×
×
0.818
=
(8.64)
