Biomedical Engineering Reference
In-Depth Information
(c) How many atoms of
191m
Ir decay between
t
= 100 s and
t
102 s?
(d) How many atoms of
191m
Ir decay between
t
= 30 d and
t
= 40 d?
=
Solution
As in the last two examples, the parent half-life is large compared with that of the
daughter. Secular equilibrium is reached in about 7
34 s. Thereafter, the ac-
tivities
A
1
and
A
2
of the
191
Os and
191m
Ir remain equal, as they are in secular equilib-
rium. During the periods of time considered in (b) and in (d), however, the osmium
will have decayed appreciably; and so one deals with an example of transient equilib-
rium. The problem can be solved as follows.
(a) The specific activity of
191
Os is, from Eq. (4.27),
×
4.9
=
1600
×
365
226
191
=
10
4
Ci g
-1
.
SA
1
=
×
4.49
×
(4.53)
15.4
The mass of the sample, therefore, is
10
-3
Ci
10
-8
g.
10
4
Ci g
-1
=
2.23
×
(4.54)
4.49
×
(b) At
t
= 25 d,
A
2
=
A
1
= 1 × e
-0.693×25/15.4
= 0.325 mCi.
(4.55)
(c) Between
t
= 100 s and 102 s secular equilibrium exists with the
osmium source essentially still at its original activity. Thus the
191m
Ir decay rate at
t
=
10
7
s
-1
.
During the next 2 s the number of
191m
Ir atoms that decay is
2
100 s is
A
2
=
1mCi
=
3.7
×
10
7
.
(d) This part is like (c), except that the activities
A
1
and
A
2
do not
stay constant during the time between 30 and 40 d. Since
transient equilibrium exists, the numbers of atoms of
191m
Ir
and
191
Os that decay are equal. The number of
191m
Ir atoms that
decay, therefore, is equal to the integral of the
191
Os activity
during the specified time (
t
in days):
3.7 × 10
7
40
30
10
7
×
3.7
×
=
7.4
×
-0.0450
e
-0.0450
t
40
10
7
3.7
×
e
-0.693
t
/15.4
d
t
=
(4.56)
30
10
8
(0.165 - 0.259)
=
-8.22
×
10
7
.
=
7.73
×
(4.57)
