Biomedical Engineering Reference
In-Depth Information
Fig. 15.7
Decay scheme of
42
K.
Solution
With no shielding, the exposure rate at
r
=
1 m is given by Eq. (12.27):
X
1.37 R h
-1
.
=
0.5
CE
=
0.5
×
10
×
(0.18
×
1.52)
=
(15.3)
We make an initial estimate of the shielding required to reduce this to 2.5 mR h
-1
on the basis of narrow-beam geometry. The number of relaxation lengths
µx
needed
would then satisfy the relation
2.5
1370
=
e
-
µx
10
-3
,
=
1.82
×
(15.4)
or
µx
= 6.31. The energy of the photons emitted by
42
K is 1.52 MeV. We see from
Fig. 15.1 (point source) that for photons of this energy in lead and a thickness of
6.31 relaxation lengths, the value of the buildup factor is about 3. To keep the re-
quired reduction (15.4) the same when the buildup factor is used, the number of
relaxation lengths in the exponential must be increased. The number
y
of added
relaxation lengths that compensates a buildup factor of 3 is given by e
-
y
=
1/3, or
y
=
ln
3
=
1.10. Added to the initial value, the estimated shield thickness becomes
6.31 + 1.10
7.41 relaxation lengths. Inspection of Fig. 15.1 shows that the buildup
factor has now increased to perhaps 3.5. Thus, a better guess is
y
=
ln
3.5
=
1.25,
with an estimated shield thickness of 6.31 + 1.25 = 7.56, which we round off to 7.6
relaxation lengths. It remains to verify a final solution numerically by trial and error.
For
µx
= 7.6 in Fig. 15.1, we estimate that
B
= 3.6. The reduction factor with buildup
included is then
=
B
e
-
µx
3.6e
-7.6
10
-3
,
=
=
1.8
×
(15.5)
which agrees with (15.4) within our degree of precision. From Fig. 8.8, we obtain for
the mass attenuation coefficient
0.051 cm
2
g
-1
.With
11.4 g cm
-3
for lead,
µ
/
ρ
=
ρ
=
0.581 cm
-1
. The required thickness of lead shielding is
x
we have
µ
=
=
7.6/
µ
=
