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32
t
tt
pd
//
.ln/
A
i
30 60
30 2
0 693 10 30
xxe
./
99
Time (days)
Activity
3 Kinetics ofRadioactive Decay
Problem 3.5 (continued)
Time of decay is 10h, which is 5 times the daughter half-life, suggesting a
transient equilibrium.
Using Eq, (3.17), activity A after 10h is
mC
51 0
.
The time to reach maximum daughter activity is given by the formula:
tt
144
max
A typical example of transient equilibrium is 99Mo (t
(t
=6h). Because 87% of 99Mo decays to
1/2
12 12 12 12
//
pd
tt
//
12 12
p
d
=66h) decaying to
99m
1/2
Tc, and the remaining 13% to the
(3.18)
99m
Tc
ground state, Eqs. (3.15), (3.16), and (3.17) must be multiplied by a factor of 0.87.
Therefore, in the time–activity plot, the
the 99Mo parent activity (Fig.3.5). Also, the
about 23h, that is, 4(t
(Eq. (3.18)).
1/2)d
99m
Tc daughter activity will be lower than
99m
Tc activity reaches maximum in
100
50
10
Fig. 3.5 Plot of logarithm of 99Mo and
The activity of the daughter
decays to
would be higher than the parent activity after reaching equilibrium, as recognized from Eq. (3.17),
99m
Tc. If 100% of the parent were to decay to the daughter, then the daughter activity
and Fig.3.4
5
2
99m
Tc is less than that of the parent 99Mo, because only 87% of 99Mo
Mo (66 hrs)
m99
Tc (6 hrs)
2143
99m
Tc activities versus time showing transient equilibrium.
56
AA
t
21
Time (hours)
RADIO ACTIVITY
3.5 Successive Decay Equations
33
3.5.3 Secular Equilibrium
When λd >> λp, that is, when the parent half-life is much longer than that of the
daughter nuclide, in Eq. (3.16), we can neglect λp compared to λd. Then Eq.
(3.16) becomes
d
t
p
(3.19)
Equation (3.19) represents the secular equilibrium. This equilibrium holds when
the half-life of the parent is much longer than that of the daughter nuclide by more
than a factor of 100 or so. In secular equilibrium, both parent and daughter activities
are equal, and both decay with the half-life of the parent nuclide. A semilogarithmic
plot of activity versus time representing secular equilibrium is shown in Fig.3.6.
Typical examples of secular equilibrium are
(t
=100min), and 68Ge (t
1/2
Fig. 3.6 Plot of activity
versus time illustrating
secular equilibrium. In
equilibrium, the daughter
activity becomes equal to
that of the parent
=280days) decaying to 68Ga (t
1/2
100
113
Sn (t
Parent
Daughter
=117days) decaying to
1/2
=68min).
1/2
113m
In
10
1
0481
6
34
3.6 Questions
3 Kinetics ofRadioactive Decay
1. Calculate (a) the total number of atoms and (b) the total mass of
a 30-mCi (1.11-GBq)
131
I sample (t
=8.0days).
1/2
131
I present in
2. Calculate (a) the disintegration rate per minute and (b) the activity in curies and
becquerels present in 1mg of
201
Tl (t
=73h).
1/2
3. A radiopharmaceutical has a biologic half-life of 10h in humans and a physical
half-life of 23h. What is the effective half-life of the radiopharmaceutical?
4. If the radioactivity of
111
In (t
=2.8days) is 200mCi (7.4GBq) on Monday,
1/2
noon, what is the activity (a) at 10:00a.m., the Friday before and (b) at 1:00p.m.,
the Wednesday after?
5. How long will it take for a 10-mCi (370-MBq) sample of
a 50-mCi (1.85-MBq) sample of
6. What is the time interval during which
99m
Tc (t
=6h) to possess the same activity?
1/2
111
In (t
=2.8days) decays to 37% of
1/2
123
I (t
1/2
=13.2h) and
the original activity?
7. For antibody labeling, 5mCi of
111
InCl3 is required. What amount of
111
InCl3
should be shipped if transportation takes one day?
8. What are the specic conditions of transient equilibrium and secular
equilibrium?
9. How long will it take for the decay of 7/8 of a 18F (t
=110min) sample?
1/2
10. What fraction of the original activity of a radionuclide has decayed in a period
equal to the mean life of the radionuclide?
11. A radioactive sample initially gives 8564cpm and 2h later gives 3000cpm.
Calculate the half-life of the radionuclide.
12. If N atoms of a sample decay in one half-life, how many atoms would decay in
the next half-life?
13. The 99Mo (t
=66h) and
1/2
99m
Tc (t
=6h) are in transient equilibrium in a Moly
1/2
generator. If 600mCi (22.2GBq) of 99Mo is present in the generator, what
would be the activity of
99m
Tc?
to
99m
Tc 132h later, assuming that 87% of 99Mo decays
14. A radionuclide decays with a half-life of 10days to a radionuclide whose halflife is 1.5h. Approximately how long will it take for the daughter activity to
reach a maximum?
Suggested Readings
Cherry SR, Sorensen JA, Phelps ME. Physics in Nuclear Medicine. 3rd ed. Philadelphia: W.B
Saunders; 2003.
Friedlander G, Kennedy JW, Miller JM. Nuclear Chemistry and Radiochemistry. 3rd ed. NewYork:
Wiley; 1981.
Saha GB. Fundamentals of Nuclear Pharmacy. 7th ed. NewYork: Springer; 2018.
Statistics ofRadiation Counting
As mentioned in previous chapters, radioactive decay is a random process, and
therefore one can expect uctuations in the measurement of radioactivity. The
detailed discussion of the statistical treatment of radioactive measurements is
beyond the scope of this book. Only the salient points of statistics related to radiation counting are given here.
4.1 Error, Accuracy, andPrecision
In the measurement of any quantity, an error in or deviation from the true value is
likely to occur. Errors can be two types: systematic and random. Systematic errors
appear as constant deviations and arise from malfunctioning instruments and inappropriate experimental conditions. These errors can be eliminated by rectifying the
incorrect situations. Random errors are variable deviations and arise from the uctuations in experimental conditions such as high-voltage uctuations or statistical
uctuations in a quantity such as radioactive decay.
The accuracy of a measurement of a quantity indicates how closely it agrees with
the “true” value. The precision of a series of measurements describes the reproduc-
ibility of the measurement, although the measurements may differ from the “average” or “mean” value. The closer the measurement is to the average value, the
higher the precision, whereas the closer the measurement is to the true value, the
more accurate the measurement. Remember that a series of measurements may be
quite precise, but their average value may be far from the true value (i.e., less accurate). Precision can be improved by eliminating the random errors, whereas better
accuracy is obtained by removing both the random and systematic errors.
4
© The Author(s), under exclusive license to Springer Science+Business Media, LLC,
part of Springer Nature 2025
G. B. Saha, Physics and Radiobiology of Nuclear Medicine,
https://doi.org/10.1007/978-1-0716-4816-2_4
35
36
n nN
i
/
n
t
n ct=
t
t
4 Statistics ofRadiation Counting
4.2 Mean andStandard Deviation
When a series of measurements is made on a radioactive sample, the most likely
value of these measurements is the average, or mean value, which is calculated by
adding the values of all measurements divided by the number of measurements.
Thus, the mean or average value n of a number of measurements of an item, in our
case radioactive decay rates, is given by
(4.1)
where ni is the value of individual measurements and N is the total number of measurements. Since radioactive decay rates are random, some measurements will be
lower than n and some will be higher than n. This deviation from the mean value is
called the standard deviation, denoted by σ. The standard deviation σ is given by
2
nn
i
i
N
1
(4.2)
We can express the mean value of a series of measurements as
4.3 Standard Deviation ofCount Rate
The standard deviation of a count rate is given by
c
where σ is the standard deviation of the total count n obtained in time t. Since the
total count n is the product of count rate and counting time,
Then,
cttc
n
c
(4.3)
A
xy
22
xy
22
M
xy
Division
xy
C
8640
12
c
//
4.4 Propagation ofErrors
Problem 4.1
A radioactive sample is counted for 12min and gives 8640 counts. Calculate
the count rate and its standard deviation for the sample.
Answer
37
ount rate countsper minutecpmc
720
Standard deviation,
ct
720 12 8
Therefore, the count rate is 720±8 cpm.
4.4 Propagation ofErrors
There are situations when two quantities, say x and y, having the standard deviations, σx and σy, are added, subtracted, multiplied, and divided. The standard deviations of the products of these arithmetic operations are given by the following
expressions:
ddition
ubtraction
ultiplicationxy
xy
xy
xy
xy
xy
/
/
(4.4)
(4.5)
2
x
2
y
2
x
2
y
(4.6)
(4.7)
Problem 4.2
A radioactive sample and the background were counted each for 5min and
found to give 8000 counts and 3000 counts, respectively. Calculate the net
count rate, its standard deviation, and percent standard deviation.
(continued)
38
1000
Gross sample count ratec
8000
ec
22
21
1000
Problem 4.2 (continued)
Answer
Using Eqs. (4.3) and (4.5).
Background count rate
Netcount rat
1600 600 1000
4 Statistics ofRadiation Counting
pm
1600
5
3000
600
5
cpm
pm
c
5
1600
600
5
16005600
5
1
1600 600
5
1
2200
44
.
22
.
46 9
.
224
cpm
21
Thus, the count rate of the sample is 1000±21 cpm.
standard deviation of the count rate
Problem 4.3
A thyroid patient is given a
131
I−NaI capsule to measure the 24-h thyroid
100 21
uptake. The 2-min counts are as follows: standard, 90,000; room background,
1000; thyroid, 40,000; and thigh, 2000. Calculate the thyroid uptake and its
percent standard deviation.
Answer
Net standard count=90,000–1000=89,000.
90 000
s
,
302
(continued)
Px
xx!
38 000
89 000
,
,
.%
22
.
0 0027
100
4.5 Poisson Distribution
Problem 4.3 (continued)
Net thyroid count=40,000–2000=38,000.
40 000 2000
t
Percent thyroid uptake
Standard deviation in uptake (using Eq. (4.7)).
39
,
205
100 42 7
38 000
,
89 000
,, ,
0 427 000
..
0 427 0 006373
..
0 0027
.
standard deviation in uptake
302
89 000
00011514 0 000029103
063
.%
205
38 000
.
.
0 427
Thus, the thyroid uptake =42.7±0.63%.
It should be noted that although all counts were taken for 2min, count rates
(cpm) were not used in the calculations. One can do similar calculations using
count rates and obtain the same results.
4.5 Poisson Distribution
The Poisson distribution is a probability distribution that predicts how many events
will occur in a given period of time, provided the events occur independently at a
constant rate. It is given by
e
(4.8)
where P(x) is the probability that λ events will be detected in a time interval and x is
the mean number of events. Both the mean and variance of the Poisson distribution
are equal to λ. Here λ is equal to P x x. The graphic representation of Poisson distribution is not symmetrical and skewed in shape. If the mean is large, then the Poisson
distribution is approximately normal.
40
P3
2
4 Statistics ofRadiation Counting
An important application of Poisson distribution is to predict the probability of
an event to occur in the future based on current knowledge of occurences of the
same events. For example, Poisson distribution can predict the number of future
visitors in a museum based on current attendances for a given period. In nuclear
medicne, many instances such as radioactive decay, production of scintillation
pulse, etc., are dened by Poisson distribution.
Problem 4.4
The average number of hurricanes in a Florida city is 2 per year. What is the
probability that exactly 3 hurricanes will hit the city next year?
Answer
λ=2
x=3
e=2.718
So,
3
!
018
3
1 0827
.
6
.
2 1718 2
.
Therefore, the probability of having three hurricanes next year is 0.18, that
is, 18%.
4.6 Gaussian Distribution
If a series of measurements are made repeatedly on a radioactive sample giving a
mean count n, then the distribution of counts would normally follow a Poisson distribution. If the number of measurements is large, the distribution can be approximated by a Gaussian distribution, illustrated in Fig.4.1. It can be seen 68% of all
measurements fall within one standard deviation on either side of the mean, that is,
within the range n±σ; 95% of all measurements fall within the range n±2σ; and
99% fall within the range n±3σ. Also, the Gaussian curve shows that half of the
measurements are below the mean value, and the other half are above it.
The standard deviations in radioactive measurements indicate the statistical uc-
tuations of radioactive decay. For practical reasons, only single counts are obtained
on radioactive samples instead of multiple repeat counts to determine the mean
value. In this situation, if a single count n of a radioactive sample is large, then n can
be estimated as close to n; that is, n=n and σ=√n. It can then be said that there
is a 68% chance that the true value of the count falls within n±σ or that the count n
falls within one standard deviation of the true value (Fig.4.1). This is called the
68% condence level. That is, one is 68% condent that the count n is within one
Counts (n)
Number of Measurements
2
i
XX
X
4.7 Chi-Square Test
Fig. 4.1 A Gaussian
distribution of radioactive
measurements. Note the
68% condence level at
±1σ, 95% condence level
at ±2σ, and 99%
condence level at ±3σ
%68
95
%
%
99
σ
σ
3
-
2
-
σ
1
-
σ
2
1
n
41
σ
σ
3
standard deviation of the true value. Similarly, 95% and 99% condence levels can
be set at two standard deviations (2σ) and three standard deviations (3σ), respectively, of any single radioactive count.
4.7 Chi-Square Test
The chi-square (χ
surements are due to statistical randomness of the data or due to variations in entities, such as equipment, patients, and the like, used in the measurements. The latter
variations may be systematic, such as a xed voltage drop throughout the measurement or random, such as uctuations in voltage supply to the equipment. If there are
N measurements made of a parameter, then for a Gaussian distribution of the data,
which is true in radioactive measurement, the χ
where Xi is the value of the ith measurement and X is the average of N measurements. The value of N should be 10 or larger. The χ2 values are given in Table4.1 for
various probability (p) values as a function of degree of freedom, which is equal to
the number of measurements minus one, N–1.
In Table4.1, for a series of eight measurements, (degrees of freedom 7), there is
a 50% chance of the value of chi-square being greater than 6.35 and equal
2
) test is a useful test for verifying if the variations in a set of mea-
2
is given by
N
2
i
(4.9)