Biomedical Engineering Reference
In-Depth Information
Solution
From Table 12.4, the neutron tissue dose D
n
would be given approximately by the
relation
P
(
E
)
D
n
= 0.149
D
n
= 2 mGy,
(12.17)
or
D
n
=
13.4 mGy.
Tissue-equivalent gases and plastics have been developed for constructing cham-
bers to measure neutron dose directly. These materials are fabricated with the ap-
proximate relative atomic abundances shown in Table 12.3. In accordance with the
proviso mentioned after Eq. (12.11), the wall of a tissue-equivalent neutron cham-
ber must be at least as thick as the range of a proton having the maximum energy
of the neutrons to be monitored.
More often than not, gamma rays are present when neutrons are. In monitoring
mixed gamma-neutron radiation fields one generally needs to know the separate
contributions that each type of radiation makes to the absorbed dose. One needs
this information in order to assign the proper quality factor to the neutron part
to obtain the dose equivalent. To this end, two chambers can be exposed—one
C CO
2
and one tissue equivalent—and doses determined by a difference method.
The response
R
T
of the tissue-equivalent instrument provides the combined dose,
R
T
=
D
γ
+
D
n
. The reading
R
C
of the C CO
2
chamber can be expressed as
R
C
=
D
γ
+
P
(
E
)
D
n
, where
P
(
E
)
is an appropriate average from Table 12.4 for the neutron
field in question. The individual doses
D
γ
and
D
n
can be inferred from
R
T
and
R
C
.
Example
In an unknown gamma-neutron field, a tissue-equivalent ionization chamber regis-
ters 0.082 mGy h
-1
and a C CO
2
chamber, 0.029 mGy h
-1
. What are the gamma and
neutron dose rates?
Solution
The instruments' responses can be written in terms of the dose rates as
R
T
=
D
γ
+
D
n
=
0.082
(12.18)
and
R
C
=
D
γ
+
P
(
E
)
D
n
=
0.029.
(12.19)
Since we are not given any information about the neutron energy spectrum, we must
assume some value of
P
(
E
) in order to go further. We choose
P
(
E
) ∼ 0.15, representa-
tive of neutrons in the lower MeV to keV range in Table 12.4. Subtracting both sides
of Eq. (12.19) from (12.18) gives
D
n
=
0.062 mGy h
-1
.It
(0.082 - 0.029)/(1 - 0.15)
=
D
γ
=
0.020 mGy h
-1
.
follows from (12.18) that
Very often, as the example illustrates, the neutron energy spectrum is not known
and the difference method may not be accurate.
