Biomedical Engineering Reference
In-Depth Information
Example
A solution contains 0.10
Ci of
198
Au and 0.04
Ci of
131
Iattime
t
0. What is the
total beta activity in the solution at
t
= 21 d? At what time will the total activity decay
to one-half its original value?
µ
µ
=
Solution
Both isotopes decay to stable daughters, and so the total beta activity is due to these
isotopes alone. (A small fraction of
131
I decays into
131m
Xe, which does not contribute
to the beta activity.) From Appendix D, the half-lives of
198
Au and
131
I are, respectively,
2.70 days and 8.05 days. At the end of 21 days, the activities
A
Au
and
A
I
of the nuclides
are, from Eq. (4.12),
0.10e
-0.693×21/2.70
10
-4
A
Au
=
=
4.56
×
µ
Ci
(4.17)
and
0.04e
-0.693×21/8.05
10
-3
A
I
=
=
6.56
×
µ
Ci.
(4.18)
The total activity at
t
Ci. To
find the time
t
in days at which the activity has decayed to one-half its original value
of 0.10 + 0.04
=
21 days is the sum of these two activities, 7.02
×
10
-3
µ
=
0.14 Ci, we write
0.07 = 0.1e
-0.693
t
/2.70
+ 0.04e
-0.693
t
/8.05
.
(4.19)
This is a transcendental equation, which cannot be solved in closed form for
t
.The
solution can be found either graphically or by trial and error, focusing in between
two values of
t
that make the right-hand side of (4.19) >0.07 and <0.07. We present
a combination of both methods. The decay constants of the two nuclides are, for Au,
0.693/2.70
0.257 d
-1
and, for I, 0.693/8.05
0.0861 d
-1
. The activities in
=
=
µ
Ci, as
functions of time
t
,are
0.10e
-0.257
t
A
Au
(
t
)
=
(4.20)
and
0.04e
-0.0861
t
.
A
I
(
t
)
=
(4.21)
Figure 4.2 shows a plot of these two activities and the total activity,
A
(
t
)
A
Au
+
A
I
,
calculated as functions of
t
from these two equations. Plotted to scale, the total ac-
tivity
A
(
t
) is found to reach the value 0.07
=
3.50 d. We can improve on
this approximate graphical solution. Direct calculation from Eqs. (4.20) and (4.21)
shows that
A
(3.50)
µ
Ci near
t
=
=
0.0703 and
A
(3.60)
=
0.0689. Linear interpolation suggests the
solution
t
=
3.52 d; indeed, one can verify that
A
(3.52)
=
0.0700
µ
Ci.
The average, or mean, life
τ
of a radionuclide is defined as the average of all of
the individual lifetimes that the atoms in a sample of the radionuclide experience.
It is equal to the mean value of
t
under the exponential curve in Fig. 4.3. Therefore,
