Biomedical Engineering Reference
In-Depth Information
The preceding conditions are met for a set of
N
identical radioactive atoms, ob-
served for a time
t
. Therefore, if many such sets of
N
atoms are prepared and
observed for time
t
, the numbers of atoms that decay from each set are expected
to be represented by the binomial distribution. After the next example, illustrating
radioactive decay as a Bernoulli process, we derive this distribution and show how
it and its spread depend upon
N
,
n
,
p
,
q
, and the observation time
t
.
Example
A sample of
N
10 atoms of
42
K (half-life
=
=
12.4 h) is prepared and observed for a
time
t
=
3h.
(a) What is the probability that atoms number 1, 3, and 8 will decay during this
time?
(b) What is the probability that atoms 1, 3, and 8 decay, while none of the others
decay?
(c) What is the probability that exactly three atoms (any three) decay during the 3
hours?
(d) What is the probability that exactly six atoms will decay in the 3 hours?
(e) What is the chance that no atoms will decay in 3 hours?
(f ) What is the general formula for the probability that exactly
n
atoms will decay,
where 0
10?
(g) What is the sum of all possible probabilities from (f )?
(h) If the original sample consisted of
N
≤
n
≤
=
100 atoms, what would be the chance
that no atoms decay in 3 hours?
Solution
(a) The decay constant for
42
Kis
λ
= 0.693/(12.4 h) = 0.0559 h
-1
. The probability that
a given atom survives the time
t
=
3 h without decaying is, by Eq. (11.1),
e
-
λ
t
e
-0.0559×3
e
-0.168
q
=
=
=
=
0.846.
(11.3)
The probability that a given atom will decay is
p
=
1-
q
=
0.154.
(11.4)
The probability that atoms 1, 3, and 8 decay in this time is
p
3
= (0.154)
3
= 0.00365.
(11.5)
(b) The answer to (a) is independent of the fate of the other seven atoms. The
probability that none of the others decay in the 3 hours is
q
7
(0.846)
7
0.310. The
probability that only atoms 1, 3, and 8 decay while the others survive is therefore
=
=
p
3
q
7
=
(0.00365)(0.310)
=
0.00113.
(11.6)
(c) The last answer,
p
3
q
7
, gives the probability that a particular, designated three
atoms decay—and only those three—in the specified time. The probability that exactly
three atoms (any three) decay is
p
3
q
7
times the number of ways that a group of three
can be chosen from among the
N
=
10 atoms. To make such a group, there are 10
