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Radioactivity -2
Decay Chains and Equilibrium
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International Atomic Energy Agency
Day 1 – Lecture 5
Objective
To discuss radioactive decay chains
(parent and single decay product)
and equilibrium situations
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Content

Secular equilibrium

Transient equilibrium

Case of no equilibrium

Radioactive decay series

Ingrowth of decay product from a parent
radionuclide
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Types of Radioactive Equilibrium
Secular
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Half-life of parent much greater
(> 100 times) than that of decay
product
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Types of Radioactive Equilibrium
Transient
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Half-life of parent only greater
(only 10 times greater) than that
of decay product
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Sample Radioactive Series Decay
90Sr

90Y

90Zr
where 90Sr is the parent (half-life = 28 years)
and 90Y is the decay product (half-life = 64 hours)
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Differential Equation for
Radioactive Series Decay
Parent and Single Decay Product
dNY
= Sr NSr - Y NY
dt
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The instantaneous rate of change of Y-90 is made up
of two terms: the production rate, which is equal to
the Sr-90 decay rate; and the rate of loss, which is
the decay rate of Y-90.
Differential Equation for
Radioactive Series Decay
Parent and Single Decay Product
SrNSr

t

t
Sr
(e
- e Y)
NY(t) =
Y - Sr
o
Recall that Sr NoSr = AoSr which equals the
initial activity of 90Sr at time t = 0
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General Equation for
Radioactive Series Decay
Activity of 90Sr at time t = 0
Y SrNSr - Srt - Yt
(e
-e )
YNY(t) =
Y - Sr
o
Activity of 90Y at time t or AY(t)
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Secular Equilibrium
Buildup of a Decay Product under
Secular Equilibrium Conditions
AY(t) = ASr (1 - e-Yt)
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Secular Equilibrium
SrNSr = YNY
ASr = AY
At secular equilibrium the activities of the parent and decay
product are equal and constant with time
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Secular Equilibrium
Decay of
226Ra to 222Rn
ARn (t) = Ao (1 - e-Rn t )
Ra
Beginning with zero activity, the activity of
the decay product becomes equal to the
activity of the parent within 7 or so half-lives
of the decay product
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Sample Problem 1
226Ra
(half-life 1600 years) decays to 222Rn (halflife 3.8 days). If initially there is 100 µCi of 226Ra
in a sample and no 222Rn, calculate how much
222Rn is produced:
a.
b.
after 7 half-lives of 222Rn
at equilibrium
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Solution to Sample Problem
The number of atoms of 222Rn at time t is given by:
dNRn
dt
= Ra NRa - Rn NRn
Solving:
NRn(t) =
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RaNRa
Rn
(1 - e-Rnt)
Solution to Sample Problem
Multiplying both sides of the equation by Rn:
ARn(t) = ARa (1 - e-Rn t)
Let t = 7 TRn
Rnt = (0.693/TRn) x 7 TRn = 0.693 * 7 = 4.85
e-4.85 = 0.00784
ARn (7 half-lives) = 100 µCi * (1 - 0.00784 )
= 100 * (0.992) = 99.2 µCi of 222Rn
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Solution to Sample Problem
Now, at secular equilibrium:
RnNRn = RaNRa
or ARn = ARa = 100 µCi
Note that the total activity in this sample is:
RnNRn + RaNRa
or ARn + ARa =
100 µCi + 100 µCi = 200 µCi
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Transient Equilibrium
 D  P NP
 D ND =
D - P
For the case of transient equilibrium, the general
equation for radioactive series decay reduces to
the above equation.
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Transient Equilibrium
AD =
AP  D
D - P
Expressing it in terms of
activities of parent and
decay product.
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Transient Equilibrium
Time for Decay Product
to Reach Maximum Activity
tmD
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D
ln
P
=
D - P
Transient Equilibrium
Example of
Transient Equilibrium
132Te Decays to 132I
Te-132 - 78.2 hr half life
I 132 - 2.2 hr half life
Note that:
I-132 reaches a maximum activity,
after which it appears to decay with
the half-life of the parent Te-132.
the activity of the decay product
can never be higher than the initial
activity of its parent
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Sample Problem
The principle of transient equilibrium is illustrated by the
Molybdenum-Technetium radioisotope generator used in
nuclear medicine applications.
Given initially that the generator contains 100 mCi of 99Mo
(half-life 66 hours) and no 99mTc (half-life 6 hours) calculate
the:
a. time required for 99mTc to reach its maximum activity
b. activity of 99Mo at this time, and
c. activity of 99mTc at this time
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Sample Problem
Note that only
86% of the 99Mo
transformations
produce 99mTc.
The remaining
14% bypass the
isomeric state and
directly produce
99Tc
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Solution to Sample Problem
ln
tmTc =
a)
Tc
Mo
Tc - Mo
Tc = 0.693/(6 hr) = 0.12 hr-1
Mo = 0.693/(66 hr) = 0.011 hr-1
ln
tmTc =
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0.12
0.011
0.12 – 0.011
= 21.9 hrs
Solution to Sample Problem
(b) The activity of 99Mo is given by
A(t) = Ao e-t = 100 mCi e(-0.011/hr * 21.9 hr)
= 100 * (0.79) = 79 mCi
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Solution to Sample Problem
c) The activity of 99mTc at t = 21.9 hrs is given by:
TcAMo
t)
ATc(t) =
(e- Mot - e- Tc
Tc - Mo
ATc(t) =
(0.12)(100 mCi)(0.86)
(0.12 – 0.011)
(see slide 10)
(e-(0.011)(21.9) - e-(0.12)(21.9))
= (94.7) (0.785 - 0.071) = 67.6 mCi of 99mTc
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Solution to Sample Problem
The maximum
activity of 99mTc is
achieved at 21.9
hours which is
nearly 1 day.
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Types of Radioactive Equilibrium
No Equilibrium
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Half-life of parent less than
that of decay product
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No Equilibrium
In this case, the half-life of the
parent is less than that of the decay
product and no equilibrium can be
established.
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Summary
 Activity defined and units discussed
 Decay constant defined
 Half-life defined - relationship to decay
constant
 Radioactive decay equation derived
 Mean life derived - relationship to half-life
 Secular equilibrium was defined
 Transient equilibrium was defined
 Case of no equilibrium was defined
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Where to Get More Information
 Cember, H., Johnson, T. E, Introduction to
Health Physics, 4th Edition, McGraw-Hill, New
York (2009)
 International Atomic Energy Agency,
Postgraduate Educational Course in Radiation
Protection and the Safety of Radiation Sources
(PGEC), Training Course Series 18, IAEA,
Vienna (2002)
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