Biomedical Engineering Reference
In-Depth Information
7.6/0.581 = 13 cm. A shield of this thickness can be interposed anywhere between
the source and the point of exposure. Usually, shielding is placed close to a source to
realize the greatest solid-angle protection.
Until now we have discussed monoenergetic photons. When photons of different
energies are present, separate calculations at each energy are usually needed, since
the attenuation coefficients and buildup factors are different.
Example
A 10-Ci point source of
24
Na is to be stored at the bottom of a pool of water.
The radionuclide emits two photons per disintegration with energies 2.75 MeV and
1.37 MeV in decaying by
-
emission to stable
24
Mg. How deep must the water be if
the exposure rate at a point 6 m directly above the source is not to exceed 20 mR h
-1
?
What is the exposure rate at the surface of the water right above the source?
β
Solution
The mass attenuation coefficients for the two photon energies can be obtained from
Fig. 8.9. Since we are dealing with water, they are numerically equal to the linear
attenuation coefficients. Thus,
µ
1
= 0.043 cm
-1
and
µ
2
= 0.061 cm
-1
, respectively,
for the 2.75-MeV and 1.37-MeV photons. The approach we use is to consider the
harder photons first and find a depth of water that will reduce their exposure rate to a
level somewhat below 20 mR h
-1
, and then see what additional exposure rate results
from the softer photons. The final depth can be adjusted to make the total 20 mR h
-1
.
The exposure rate from 2.75-MeV photons at a distance
d
=
6 m with no shielding is
0.5
CE
d
2
0.5
×
10
×
2.75
X
2.75
=
0.382 R h
-1
.
=
=
(15.6)
6
2
To reduce this level to 20 mR h
-1
under conditions of good geometry requires a water
depth
µ
1
x
given by
382e
-
µ
1
x
,
20
=
(15.7)
or
2.95 relaxation lengths. From Fig. 15.2, one can see that the buildup factor
B
1
for 2.75-MeV photons for a water shield of this thickness is about 3.4. The number
of relaxation lengths that compensate for this amount of buildup is
y
µ
1
x
=
1.22.
Adding this amount to the preceding gives an estimate
µ
1
x
= 2.95 + 1.22 = 4.17. At
this depth, the buildup factor has increased to about 4.4, for which the compensating
depth added to the first estimate is
y
=
ln
4.4 = 1.5. Therefore, we try an estimated
relaxation length of 2.95 + 1.5
=
ln
3.4
=
=
4.5. Thus, we obtain for the shielded exposure rate
for the more energetic photons,
X
2.75
= 4.4 × 382e
-4.5
= 19 mR h
-1
.
(15.8)
For the 1.37-MeV photons, the thickness of this shield in relaxation lengths is larger
by the ratio of the attenuation coefficients; that is,
6.4.
From Fig. 15.2, the buildup factor for the 1.37-MeV photons is estimated to be, ap-
µ
2
x
=
4.5
×
(0.061/0.043)
=
