Biomedical Engineering Reference
In-Depth Information
find
T
in terms of
λ
, we write from Eq. (4.8) at time
t
=
T
,
1
2
= e
-λ
T
.
(4.9)
Taking the natural logarithm of both sides gives
-
λ
T
=
ln
2
= -
ln
2,
(4.10)
and therefore
ln
2
λ
0.693
λ
(4.11)
T
=
=
.
Written in terms of the half-life, the exponential decay laws (4.7) and (4.8) become
N
N
0
=
A
A
0
=
e
-0.693
t
/
T
.
(4.12)
The decay law (4.12) can be derived simply on the basis of the half-life. If, for
example, the activity decreases to a fraction
A
/
A
0
of its original value after passage
of time
t
/
T
half-lives, then we can write
1
2
t
/
T
A
A
0
=
.
(4.13)
Taking the logarithm of both sides of Eq. (4.13) gives
ln
A
-
t
-
0.693
t
T
A
0
=
T
ln
2
=
,
(4.14)
from which Eq. (4.12) follows.
Example
Calculate the activity of a 30-MBq source of
2
11
Na after 2.5 d. What is the decay con-
stant of this radionuclide?
Solution
The problem can be worked in several ways. We first find
from Eq. (4.11) and
λ
then the activity from Eq. (4.8). The half-life
T
=
15.0 h of the nuclide is given in
Appendix D. From (4.11),
0.693
T
0.693
15.0 h
= 0.0462 h
-1
.
λ
=
=
(4.15)
24 hd
-1
With
A
0
=
30 MBq and
t
=
2.5 d
×
=
60.0 h,
30 e
-(0.0462
h
-1
×60
h
)
A
=
=
1.88 MBq.
(4.16)
Note that the time units employed for
and
t
must be the same in order that the
λ
exponential be dimensionless.
