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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5545_Библиотеки_им_академика_М_И_Перельмана.pdf
22
A N
N Ne
t
t
0
A Ae
t
t
0
Ao
Time (hours)
Activity
3 Kinetics ofRadioactive Decay
Now from the preceding discussion, the following equation can be written:
(3.2)
From a knowledge of the decay constant and radioactivity of a radionuclide, one
can calculate the total number of atoms or mass of the radionuclides present (using
Avogadro’s number 1g·atom=6.02×1023 atoms). Because Eq. (3.1) is a rstorder differential equation, the solution of this equation by integration leads to
(3.3)
where N0 and Nt are the number of radioactive atoms at t=0 and time t, respectively.
Equation (3.3) is an exponential equation indicating that the radioactivity decays
exponentially. By multiplying both sides of Eq. (3.3) by λ, one obtains
The factor e
−λt
is called the decay factor. The decay factor becomes e
+λt
if the
(3.4)
activity at time t before t=0 is to be determined. The plot of activity versus time on
a linear graph gives an exponential curve, as shown in Fig.3.1. However, if the
activity is plotted against time on semilogarithmic paper, a straight line results, as
shown in Fig.3.2.
Fig. 3.1 Plot of
radioactivity versus time
on a linear graph indicating
an exponential curve
Ao
2
Ao
4
100
Time (half-lives)
6
Activity
t
0 693/.
AA
2
2
12
3.1 Radioactive Decay Equation
23
Fig. 3.2 Plot of
radioactivity against time
on a semilogarithmic graph
50
indicating a straight line.
The half-life of the
radionuclide can be
determined from the slope
of the line, which is given
as the decay constant λ.
Alternatively, an activity
20
10
5
and half its value and their
corresponding times are
read from the plot. The
2
difference in the two time
readings gives the half-life
1
23
45
3.1.2 Half-Life
Every radionuclide is characterized by a half-life, which is dened as the time
required to reduce its initial activity to one half. It is usually denoted by t
unique for a radionuclide. It is related to the decay constant λ of a radionuclide by
and is
1/2
12
(3.5)
From the denition of half-life, it is understood that A0 is reduced to A0/2in one
half-life; to A0/4, that is, to A0/22 in two half-lives; to A0/8, that is, to A0/23 in three
half-lives; and so forth. In n half-lives of decay, it is reduced to A0/2n. Thus, the
radioactivity At at time t can be calculated from the initial radioactivity A0 by
/
tt
00
A
t
n
where t is the time of decay. Here, t/t
t and t
. For example, a radioactive sample with t
1/2
A
/
tt
can be an integer or a fraction depending on
1/2
0
12
/
/
05
.
=3.2 days decaying at a rate of
1/2
(3.6)
10,000 disintegrations per minute would give, after seven days of decay,
10,000/2
(7/3.2)
= 10,000/2
2.2
= 10,000/4.59 = 2178 disintegrations per minute. It
should be noted that ten half-lives of decay reduce the radioactivity by a factor of
about 1000(210=1024), or to 0.1% of the initial activity.
The half-life of a radionuclide is determined by measuring the radioactivity at
different time intervals and plotting them on semilogarithmic paper, as shown in
Fig.3.2. An initial activity and half its value are read from the line, and the corresponding times are noted. The difference in time between the two readings gives the
half-life of the radionuclde. For a very long-lived radionuclide, the half-life is determined by Eq. (3.2) from a knowledge of its activity and the number of atoms
24
Time (hours)
Acvity (log scale)
70
50
N
W
A
23
1/
tt
12
//
Fig. 3.3 A composite
radioactive decay curve for
a sample containing two
radionuclides of different
half-lives. The long-lived
component (a) has a
half-life of 27h and the
short-lived component (b)
has a half-life of 5.8h
3 Kinetics ofRadioactive Decay
20
10
5
2
a + b
5.8 hr)=
Bt
2/1
a(t
= 27 hr)
2/1
2010 40
30
6050
present. The number of atoms N can be calculated from the weight W of the radionuclide with atomic weight A and Avogadro’s number 6.02×1023 atoms per g⋅atom
as follows:
60210
.
(3.7)
When two or more radionuclides are present in a sample, the measured count of
such a sample comprises counts of all individual radionuclides. A semilogarithmic
plot of the activity of a two-component sample versus time is shown in Fig.3.3. The
half-life of each of the two radionuclides can be determined by what is called the
peeling or stripping method. In this method, rst, the tail part (second component)
of the curve is extrapolated as a straight line up to the ordinate, and its half-life can
be determined as mentioned previously (e.g., 27h). Second, the activity values on
this line are subtracted from those on the composite line to obtain the activity values
for the rst component. A straight line is drawn through these points, and the halflife of the rst component is determined (e.g., 5.8h). The stripping method can be
applied to more than two components in the similar manner.
3.1.3 Mean Life
Another relevant quantity of a radionuclide is its mean life, which is the average
lifetime of a group of radionuclides. It is denoted by τ and is related to the decay
constant λ and half-life t
as follows:
1/2
(3.8)
In one mean life, the activity of a radionuclide is reduced to 37% of its initial value.
0 693 144
/. .
12
(3.9)
epb
111
TT
epb
T
TT
TT
pb
pb
10
22210
7
millicurie mCi dps
dpm
.
.
22210
4
microcurieCi dps
dpm
.
.
3.2 Units of Radioactivity
25
3.1.4 Effective Half-Life
As already mentioned, a radionuclide decays exponentially with a denite half-life,
which is called the physical half-life, denoted by Tp (or t
a radionuclide is independent of its physicochemical conditions. Analogous to
physical decay, radiopharmaceuticals administered to humans disappear exponentially from the biological system through fecal excretion, urinary excretion, perspiration, or other routes. Thus, after invivo administration, every radiopharmaceutical
has a biological half-life (Tb), which is dened as the time needed for half of the
radiopharmaceutical to disappear from the biologic system. It is related to decay
constant λb by λb=0.693/Tb.
Obviously, in any biologic system, the loss of a radiopharmaceutical is due to
both the physical decay of the radionuclide and the biologic elimination of the
radiopharmaceutical. The net or effective rate (λe) of loss of radioactivity is then
related to λp and λb by
Because λ=0.693/t
, it follows that
1/2
). The physical half-life of
1/2
(3.10)
T
(3.11)
or,
e
(3.12)
The effective half-life, Te, is always less than the shorter of Tp or Tb. For a very
long Tp and a short Tb, Te is almost equal to Tb. Similarly, for a very long Tb and short
Tp, Te is almost equal to Tp.
3.2 Units ofRadioactivity
The unit of radioactivity is a curie. It is dened as
curieCidisintegrationspersecond dps.
22210
710
12
.ddisintegrations per minute dpm
9
6
26
11
becquerelBq dps Ci.
38
kilobecquerelkBq dps Ci.
65
megabecquerelMBq dps Ci.
92
gigabecquerelGBq dps Ci.
12
terabecquerelTBq dps Ci
10
Ci Bq GBq.
7
mCiBqMBq.
4
Ci Bq kBq.
3 Kinetics ofRadioactive Decay
The System Internationale (SI) unit for radioactivity is the becquerel (Bq), which
is dened as 1 dps. Thus,
Similarly,
3.3 Specific Activity
The presence of “cold,” or nonradioactive, atoms in a radioactive sample always
induces competition between them in their chemical reactions or localization in a
body organ, thereby compromising the concentration of the radioactive atoms in the
organs. Thus, each radionuclide or radioactive sample is characterized by specic
activity, which is dened as the radioactivity per unit mass of a radionuclide or a
radioactive sample. For example, suppose that a 200-mg
antibody sample contains 350-mCi (12.95-GBq)
123
I radioactivity, its specic activity would be 350/200=1.75mCi/mg or 64.75MBq/mg. Sometimes, it is confused
with concentration, which is dened as the radioactivity per unit volume of a sample. If a 10-ml radioactive sample contains 50mCi (1.85GBq), it will have a concentration of 50/10=5mCi/ml or 185MBq/ml.
Specic activity is at times expressed as radioactivity per mole of a labeled com-
pound, for example, mCi/mole (MBq/mole) or mCi/μmole (MBq/μmole) for
14
C‐, and 35S-labeled compounds.
The specic activity of a carrier-free (see Chap. 5) radionuclide sample is related
to its half-life and mass number A: the shorter the half-life and the smaller the A, the
higher the specic activity. The specic activity of a carrier-free radionuclide with
mass number A and half-life t
in hours can be calculated as follows:
1/2
123
I-labeled monoclonal
3
H‐,
N
110
60210
3
20
AA
.
disintegration rate DN
12
/
.
1 1589 10
17
12
/
3.4 Calculation
Suppose 1mg of a carrier-free radionuclide is present in the sample.
27
umber of atoms in the sample
60210
0 693
Decay constant
t
12
/
.
60 60
23
.
1
s
Thus,
60 60
17
10
dps
20
0 693 60210
..
tA
12
/
1 1589
.
At
Thus,
mCimg/
At
At
9
00
.
3131
where A is the mass number of the radionuclide, and t
/
12
is the half-life of the radio-
1/2
.
37 10
7
(3.13)
nuclide in hours.
From Eq. (3.13), specic activities of carrier-free
99m
Tc and
131
I can be calculated
as 5.27×106 mCi/mg (1.95×105 GBq/mg) and 1.25×105 mCi/mg (4.6×103 GBq/
mg), respectively.
3.4 Calculation
Some examples related to the calculation of radioactivity and its decay follow:
28
61
.s
N
8
s
201
14
0
0 693
iG
.
?
A
3 Kinetics ofRadioactive Decay
Problem 3.1
Calculate the total number of atoms and total mass of
(370MBq) of
201
Tl (t
=3.04d).
1/2
Answer.
201
For
Tl,
0 693
.
304246060
.
10 37 10 37 10
78
..A dps
2 638 10
Using Eq. (3.2),
Ax
37 10
2 638 10
.
Because 1 g · atom
201
Tl = 201 g
(Avogadro’s number), mass of
Therefore, 10mCi of
201
Tl contains 1.4×1014 atoms and 46.7ng.
.
x
201
Tl in 10mCi (370MBq).
14010
.
.
46 710
.gng
46 7
6
.
6 0210
14010
021
23
9
14
.x atom
Tl = 6.02 × 1023 atoms of
201
Tl present in 10mCi
201
Tl
Problem 3.2
At 10:00a.m., Wednesday, the
(5.55GBq). What was the activity at 6a.m. and 3p.m. on the same day (t
99m
Tc=6 h)?
99m
Tc radioactivity was measured as 150mCi
of
1/2
Answer.
Time from 6a.m. to 10a.m. is 4h:
99 1
m
Tc h
6
A
t
mC
150 555
.
0 1155
.
Bq
Using Eq. (3.4).
(continued)
0.3
.%
A
t
150=
mCi
3.4 Calculation
Problem 3.2 (continued)
150
Ae
0
Time from 10a.m. to 3p.m. is 5h:
Using Eq. (3.4).
.
0 1155 4
Ae
0
.
0 462
150
150 1 5872
.
238 1881
..mCiGBq attam6..
0
A
=
?
29
Ae
150
t
150
150 0 5613
84 231
..mCiGBg attpm3..
0 1155 5
0 5775..
e
.
Problem 3.3
If a radionuclide decays at a rate of 30% per h, what is its half-life?
Answer.
-1
0.3
h
0.693
t
1/2
0.693 0.693
t
1/2
= 2.31 h
Problem 3.4
If 11% of
99m
Tc-labeled diisopropyliminodiacetic acid (DISIDA) is eliminated via renal excretion, 35% by fecal excretion, and 3.5% by perspiration in
5h from the human body, what is the effective half-life of the radiopharmaceutical (Tp=6 h for
99m
Tc)?
Answer.
Total biological elimination
.%
in h
49 55
11 35 35
%%
(continued)
30
dN
dt
NN
dd
ee
dp
tA
tt
pt
//
Problem 3.4 (continued)
Therefore, Tb≈5h.
Tp=6h
3 Kinetics ofRadioactive Decay
T
e
TT
bp
TT
bp
56
563011
27. h
3.5 Successive Decay Equations
3.5.1 General Equation
In the preceding section, we derived equations for the activity of any radionuclide
that is decaying. Here, we shall derive equations for the activity of a radionuclide
that is growing from another radionuclide and at the same time is itself decaying.
If a parent radionuclide p decays to a daughter radionuclide d, which in turn
decays to another radionuclide (i.e., p→d→), then the rate of growth of d becomes
d
pp
By integration, Eq. (3.14) becomes
A
AN
d
dd
t
dp
t
0
t
p
d
Equation (3.15) gives the activity of the daughter nuclide d at time t as a result of
growth from the parent nuclide p and also due to the decay of the daughter itself.
(3.14)
(3.15)
3.5.2 Transient Equilibrium
If λd>λp, that is, (t
1/2)d
<(t
1/2)p
when t is sufciently long. Then Eq. (3.15) becomes
, then e
A
d
−λdt
in Eq. (3.15) is negligible compared to e
A
dp
0
t
dp
dp
12
/
12 12
e
dp
A
t
p
p
t
−λpt
t
p
(3.16)
(3.17)
Time (hours)
RADIO ACTIVITY
3.5 Successive Decay Equations
Fig. 3.4 Plot of activity
versus time on a
semilogarithmic graph
illustrating the transient
equilibrium. Note that the
daughter activity reaches a
maximum, then transient
equilibrium, and follows
an apparent half-life of the
parent. The daughter
activity is higher than the
parent activity at
equilibrium
100
10
31
Parent
Daughter
1
40 8 12
16
This relationship is called the transient equilibrium. This equilibrium holds good
when (t
this equilibrium equation is shown in Fig.3.4. The daughter nuclide initially builds
up as a result of the decay of the parent nuclide, reaches a maximum, and then
achieves the transient equilibrium decaying with an apparent half-life of the parent
nuclide. In equilibrium, the ratio of the daughter to parent activity is constant. It can
be seen from Eq. (3.17) that the daughter activity is always greater than the parent
activity, because (t
Problem 3.5
A radionuclide A with a half-life of 30-h decays to a radionuclide B with a
half-life of 2h. What would be the activity of B, 10h. later, from a sample A
whose initial activity is 60mCi?
(t
(t
1/2)p
and (t
differ by a factor of about 10–50. The semilogarithmic plot of
1/2)d
/((t
1/2)p
−(t
1/2)p
Answer.
=30h Activity =60mCi.
1/2)A
=2h
1/2)B
) is always greater than 1.
1/2)d
(continued)