Biomedical Engineering Reference
In-Depth Information
Fig. 4.6
Activities as functions of time when
T
2
>
T
1
and
N
20
=
0. No equilibrium conditions occur. Eventually, only the
daughter activity remains.
Therefore, the activity of the 1 mg sample of
90
Sr is
10
-3
g
138 Ci g
-1
A
1
=
×
=
0.138 Ci,
(4.48)
which is also equal to the activity
A
2
of the
90
Y. The latter has a specific activity
1600 y
226
90
SA
2
=
×
(4.49)
1
24
d
h
×
1
365
y
d
64.0 h
×
10
5
Ci g
-1
.
=
5.50
×
(4.50)
Therefore, the mass of
90
Y in secular equilibrium with 1 mg of
90
Sr is
0.138 Ci
10
5
Ci g
-1
= 2.51 × 10
-7
g = 0.251
µ
g.
(4.51)
5.50
×
Example
A sample contains 1 mCi of
191
Os at time
t
= 0. The isotope decays by
β
-
emission
into metastable
191m
Ir, which then decays by
emission into
191
Ir. The decay and
γ
half-lives can be represented by writing
-
--------→
15.4 d
β
γ
--------→
4.94 s
191
76
Os
191m
77
Ir
191
77
Ir.
(4.52)
(a) How many grams of
191
Os are present at
t
=
0?
(b) How many millicuries of
191m
Ir are present at
t
=
25 d?
