Biomedical Engineering Reference
In-Depth Information
Example
(a) What is the effective half-life of
198
Au in the stomach? (b) What is the effective
decay constant there?
Solution
(a) The effective half-life is given by Eq. (16.10). From Appendix D, the radiological
half-life is
T
R
=
2.70 d. From Fig. 16.8, we find that the metabolic half-time is
T
ST
=
0.693/(24 d
-1
)
0.693/
λ
ST
=
=
0.0289 d. Equation (16.10) gives
T
R
T
ST
T
R
+
T
ST
=
2.70
×
0.0289
2.70 + 0.0289
=
T
Eff
=
0.0286 d.
(16.23)
(b) The effective decay constant is
0.693
T
Eff
=
0.693
0.0286 d
=
24.2 d
-1
.
λ
Eff
=
(16.24)
One can also calculate
λ
Eff
as the sum of the radiological and metabolic constants.
One has
λ
R
= 0.693/(2.70 d) = 0.257 d
-1
and, from Fig. 16.8,
λ
ST
= 24 d
-1
, giving
λ
Eff
=
λ
R
+
24.3 d
-1
.
λ
ST
=
Example
For a single ingestion of
198
Au at time
t
=
0, calculate the fractions of the initial
activity in the stomach, small intestine, and upper large intestine as functions of
time. In Eq. (16.7),
f
1
=
0.1.
Solution
We evaluate the rate constants for Eqs. (16.11), (16.17), and (16.22). It is convenient
to express time in days. From the last example, the radioactive decay constant is
λ
R
=
(0.257 d
-1
) [(1/24) dh
-1
]
0.0107 h
-1
. With the help of Fig. 16.8 and the given value,
=
f
1
=
0.1, we find
(6/24) h
-1
1-0.1
f
1
λ
SI
1-
f
1
=
0.1
×
= 0.0278 h
-1
.
λ
B
=
(16.25)
Thus, with time in hours,
0.0107 +
24
1.01 h
-1
.
λ
1
=
λ
R
+
λ
ST
=
24
=
(16.26)
6
24
+ 0.0278 = 0.289 h
-1
.
λ
2
=
λ
R
+
λ
SI
+
λ
B
= 0.0107 +
(16.27)
0.0107 +
1.8
0.0857 h
-1
.
λ
3
=
λ
R
+
λ
ULI
=
24
=
(16.28)
The fraction of the ingested activity in the stomach after
t
hours is, from Eq. (16.11),
q
ST
(
t
)
A
0
e
-λ
1
t
e
-1.01
t
.
=
=
(16.29)
