Biomedical Engineering Reference
In-Depth Information
A number of approximate formulas for neutron shielding, based on removal
cross sections, have been developed for reactor cores having various shapes and
other characteristics. We will not attempt to cover them here. A simple, useful for-
mula is available, however, for radioactive neutron sources.
4)
Because of the small
intensities compared with fission sources, relatively thin shields are needed. There-
fore, the scattered neutrons contribute significantly to the dose outside the shield,
and their effect can be represented by a buildup factor
B
. The dose-equivalent rate
H
outsideashieldofthickness
T
at a distance
R
cm from a point source of strength
S
neutrons s
-1
is given by
BSqe
-
r
T
4
πR
2
H
=
,
(15.31)
where
r
is the removal cross section and
q
is the dose-equivalent rate per unit
neutron fluence rate (e.g., Sv h
-1
per neutron cm
-2
s
-1
) for neutrons of the source
energy. The factor
q
can be obtained from Table 12.5;
H
and
q
will have the same
units for dose equivalent. For Po-Be and Po-B sources with a water or paraffin
shield at least 20 cm thick,
B
=
5
.
Example
Calculate the dose-equivalent rate 1.6 m from an unshielded 3.0
10
10
Bq
210
Po-
Be source, which emits 2.05 × 10
6
neutrons s
-1
. By what factor is the rate reduced by
a 25-cm water shield? What is the dose-equivalent rate behind 50 cm of water?
×
Solution
In Eq. (15.31) the presence of the shield introduces the factors
B
exp
(-
r
T
). For the
unshielded source,
H
0
=
Sq
/4
πR
2
and
S
= 2.05 × 10
6
neutrons s
-1
. Table 9.2 shows
the average energy to be 4.2 MeV, for which Table 12.5 indicates that about 16 neu-
trons cm
-2
s
-1
give a dose-equivalent rate of 0.025 mSv h
-1
. Therefore we have
0.025 mSv h
-1
16 cm
-2
s
-1
0.00156 mSv h
-1
cm
2
s;
q
=
=
(15.32)
and so
(2.05
×
10
6
s
-1
)(0.00156 mSv h
-1
cm
2
s)
4
H
0
=
0.0099 mSv h
-1
=
(15.33)
π
(160 cm)
2
0.103 cm
-1
for the unshielded source. With
r
=
from Table 15.5,
B
=
5, and
T
=
25 cm, the dose-equivalent rate is reduced by the factor
B
e
-
r
T
5e
-0.103×25
=
=
0.38.
(15.34)
The rate with 50 cm of water interposed is
10
6
5
×
2.05
×
×
0.00156
H
=
= 2.9 × 10
-4
mSv h
-1
.
e
-0.103×50
(15.35)
4
π
(160)
2
4 Protection Against Neutron Radiation Up to
30 Million Electron Volts,
Handbook 63
,
National Bureau of Standards, Washington,
DC (1957).
