Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 15.6
Derive the equation for
T
1/2
.
Solution
Start with
N
¼
N
0
e
l
t
. After a time interval equal to one half-life,
t
¼
T
1/2
, the number of radio-
active nuclei remaining is
N
¼
N
0
/2. Therefore,
N
0
=
¼
N
0
e
l
T
1=2
2
Dividing both sides by
N
0
gives
¼
e
l
T
1=2
Taking the natural logarithm of both sides of this equation eliminates the exponential factor on
the right, since ln
1
=
2
e
¼
1:
ln 1
=
2
¼l
T
1=2
Since ln 1/2
¼
0.693 (use your calculator to check this), we have
0
:
693
¼l
T
1=2
0
:
693
l
T
1=2
¼
Let us assume that the number of radioactive nuclei present at
t
¼
0is
N
0
. The number left after
one half-life,
N
0
/2. After the second half-life, the number remaining is again
reduced by one-half. Hence, after a time 2
T
1/2
, passes is
T
1/2
, the number remaining is
N
0
/4, and so forth.
Half-lives range from about 10
22
sto10
21
years.
EXAMPLE PROBLEM 15.7
The half-life of the radioactive nucleus
226
10
3
years. If a sample contains 3
10
16
88
Ra is 1.6
such nuclei, determine the activity.
Solution
First, calculate the decay constant,
l
, using the fact that
10
3
years
10
3
years
10
7
s
10
10
s
T
1=2
¼
1
:
6
¼ð
1
:
6
Þð
3
:
15
=
year
Þ¼
5
:
0
Therefore,
0
693
T
1=2
¼
:
0
:
693
10
11
s
1
l ¼
10
10
s
¼
1
:
4
5
:
0
The activity, or decay rate, of the sample at
t
¼
0 is calculated using the form -d
N
0
¼
d
t
,
R
0
is the
10
16
,
decay rate at
t
¼
0, and
N
0
is the number of radioactive nuclei present at
t
¼
0. Since
N
0
¼
3
we have
10
11
s
1
10
16
10
5
decays
R
0
¼ l
N
0
¼ð
1
:
4
Þð
3
Þ¼
4
:
2
=
s
10
10
decays/s, the activity, or decay rate is
Since 1 Ci
¼
3.7
R
0
¼
11.3
m
Ci.
