Biomedical Engineering Reference
In-Depth Information
˙
posed to a uniform, broad beam of monoenergetic neutrons with a fluence rate
,
then the production rate of daughter atoms from neutron absorption is
˙
σN
T
.If
the number of daughter atoms in the sample is
N
and the decay constant is
λ
,then
the rate of loss of daughter atoms from the sample is
λ
N
. Thus, the rate of change
d
N
/d
t
in the number of daughter atoms present at any time while the sample is
being bombarded is given by
d
N
d
t
=
˙
σN
T
-
λ
N
.
(9.33)
To solve this equation, we assume that the fluence rate is constant and that the
original number of target atoms is not significantly depleted, so that
N
T
is also
constant. Without the term
˙
σ
N
T
, which is then constant, Eq. (9.33) would be the
same as the linear homogeneous Eq. (4.2). Therefore, we try a solution to (9.33)
in the same form as (4.7) for the homogeneous equation plus an added constant.
Substituting
N
a
+
be
-
λ
t
=
into Eq. (9.33) gives
e
-
λ
t
=
˙
σN
T
-
a
λ
e
-
λ
t
.
-
b
λ
-
b
λ
(9.34)
=
˙
σ
The exponential terms on both sides cancel, and one finds that
a
N
T
/
λ
.Thus,
the general solution is
˙
σ
N
T
λ
+
b
e
-
λ
t
.
N
=
(9.35)
The constant
b
depends on the initial conditions. Specifying that no daughter
atoms are present when the neutron irradiation begins (i.e.,
N
=
0
when
t
=
0
),
-
˙
σN
T
/
we find from (9.35) that
b
=
λ
. Thus we obtain the final expression
=
˙
σ
N
T
(1 - e
-
λ
t
).
λ
N
(9.36)
The left-hand side expresses the activity of the daughter as a function of the time
t
.
The quantity
˙
σN
T
is called the saturation activity because it represents the max-
imum activity obtainable when the sample is irradiated for a long time (
t
→∞
).
A sketch of the function (9.36) is shown in Fig. 9.10.
When the neutrons are not monoenergetic, the terms in Eq. (9.33) can be treated
in separate energy groups. Alternatively, Eq. (9.36) can be used as is, provided the
proper average cross section is used for
σ
.
Example
A 3-g sample of
32
S is irradiated with fast neutrons having a constant fluence rate
of 155 cm
-2
s
-1
. The cross section for the reaction
32
S(n, p)
32
P is 0.200 barn, and the
half-life of
32
Pis
T
= 14.3 d. What is the maximum
32
P activity that can be induced?
How many days are needed for the level of the activity to reach three quarters of the
maximum?
Solution
The total number of target atoms is
N
T
=
3
32
10
23
10
22
.Themax-
×
6.02
×
=
5.64
×
˙
σ
(155 cm
-2
s
-1
)(0.2
10
-24
cm
2
)(5.64
10
22
)
imum (saturation) activity is
N
T
=
×
×
=
