Biomedical Engineering Reference
In-Depth Information
Therefore, the net count rate is
r
n
= 142.6 - 28.5 = 114 cpm. The standard deviation
can be found from either of the expressions in (11.51). Using the first (which does
not depend on the calculated values,
r
g
and
r
b
), we find
1426
(10 min)
2
+
2561
(90 min)
2
=
3.82 min
-1
σ
nr
=
=
3.82 cpm.
(11.52)
(b) Since the counter efficiency is
=
0.28, the inferred activity of the sample is
(114 min
-1
)/0.28
A
6.78 Bq. The standard deviation of the
activity is
σ
nr
/
= (3.82 min
-1
)/0.28 = 13.6 dpm = 0.227 Bq.
(c) A 5% uncertainty in the net count rate is 0.05
r
n
=
=
r
n
/
=
=
407 dpm
=
5.70 cpm. For
the true net count rate to be within this range of the mean at the 95% confidence
level means that 5.70 cpm
0.05
×
114
=
2.91 cpm. Using the
second expression in (11.51) with the background rate as before (since we do not yet
know the new value of
n
g
), we write
=
1.96
σ
nr
(Table 11.2), or that
σ
nr
=
142.6 min
-1
t
g
+
28.5 min
-1
90 min
2.91 min
-1
σ
nr
=
=
.
(11.53)
Solving, we find that
t
g
=
17.5 min.
(d) Yes. The relative uncertainties remain the same and scale according to the effi-
ciency. If the efficiency were larger and the counting times remained the same, then
a larger number of counts and less statistical uncertainty would result.
Optimum Counting Times
If the total time
T
t
g
+
t
b
for making the gross and background counts is fixed,
one can partition the individual times
t
g
and
t
b
in a certain way to minimize the
standard deviation of the net count rate. To find this partitioning, we can write the
second equality in (11. 51) as a function of either time variable alone and minimize
it by differentiation.
4)
Substituting
T
-
t
g
for
t
b
and minimizing the mathematical
expression for the
variance
, rather than the standard deviation (simpler than deal-
ing with the square root), we write
=
r
g
t
g
+
d
d
t
g
σ
d
d
t
g
r
b
T
-
t
g
2
nr
=
=
0.
(11.54)
It follows that
-
r
g
r
b
(
T
-
t
g
)
2
=
(11.55)
t
g
+
0
or
-
r
g
t
g
+
r
b
t
b
=
0,
(11.56)
4 Alternatively, the function
f
(
t
g
,
t
b
)
solving the two equations,
∂
f
/
∂
t
g
=
0and
(
t
g
+
t
b
), involving the
Lagrange multiplier
λ
, can be minimized by
≡
σ
nr
-
λ
∂
f
/
∂
t
b
=
0. Equation (11.57) results.
