Biomedical Engineering Reference
In-Depth Information
k
α
t
g
k
α
k
2
α
t
g
+4
r
b
t
g
+
t
b
t
g
t
b
+
r
b
t
g
+
t
b
t
g
t
b
1/2
2
t
g
+
1
+
k
β
.
(11.78)
2
This general result gives the net rate that corresponds to the minimum detectable
true activity for a given background rate
r
b
and arbitrary choices of
α
,
β
,andthe
counting times. When the latter are equal
(
t
g
=
t
b
=
t
)
, Eq. (11.78) gives for the
number of net counts with the minimum detectable true activity
√
8
n
b
+
1+
k
2
=
2
n
b
k
α
k
α
α
8
n
b
2
=
r
2
t
1+
k
2
1+
k
2
1/2
.
k
α
√
2
n
b
α
4
n
b
+
α
8
n
b
+
k
β
(11.79)
With the help of Eq. (11.70), we can also write
2
=
1
+
k
β
2
n
b
1+
k
2
1+
k
2
1/2
k
α
√
2
n
b
α
4
n
b
+
α
8
n
b
.
(11.80)
The minimum detectable true activity is given by
A
II
=
2
,
(11.81)
t
where
is the counter effi
ci
ency. As with Eqs. (11.70) and (11.72), we obtain a
simple formula when
k
α
/
√
n
b
1
:
(
k
α
+
k
β
)
2
n
b
.
2
=
(11.82)
When the background count
B
is accurately known, we h
av
e seen by Eq. (11.73)
that the minimum significant count difference is
1
=
k
α
√
B
. If a sample has ex-
actly the minimum detectable true activity, then the expected number of net counts
2
is just
k
β
st
andard d
eviations greater than
1
. The standard deviation of the net
count rate is
(
B
+
2
)
.Thus,
2
=
k
α
√
B
+
k
β
B
+
2
.
(11.83)
Solving for
2
, we find
2
=
√
B
k
α
+
1+
k
√
B
+
k
2
k
2
β
2
√
B
+
k
β
β
4
B
(11.84)
(Background accurately known).
This expression for
2
is then to
be
used in Eq. (
11
.81) to obtain the minimum
detectable true activity. When
k
α
/
√
B
1
and
k
β
/
√
B
1
, one has, approximately,
in place of Eq. (11.82),
(
k
α
+
k
β
)
√
B
2
=
(Background accurately known)
.
(11.85)
