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IAEAInternational Atomic Energy Agency
Radioactivity -2
Decay Chains and Equilibrium
Day 1 – Lecture 5
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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 Half-life of parent much greater (> 100 times) than that of decay product
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Types of Radioactive Equilibrium
Transient Half-life of parent only greater (only 10 times greater) than that of decay product
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90Sr 90Y 90Zr
Sample Radioactive Series Decay
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 forRadioactive Series Decay
= Sr NSr - Y NY dNY
dt
Parent and Single Decay Product
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.
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Parent and Single Decay Product
Differential Equation forRadioactive Series Decay
NY(t) = (e- t - e- t)Sr YSrNSr
Y - Sr
o
Recall that Sr NoSr = Ao
Sr which equals the
initial activity of 90Sr at time t = 0
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General Equation forRadioactive Series Decay
YNY(t) = (e- t - e- t)Sr Y
Y - Sr
Y SrNSro
Activity of 90Sr at time t = 0
Activity of 90Y at time t or AY(t)
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Buildup of a Decay Product underSecular Equilibrium Conditions
Secular Equilibrium
AY(t) = (1 - e- t)YASr
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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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Decay of226Ra to 222Rn
Secular Equilibrium
ARn (t) = Ao (1 - e- t ) Rn
RaBeginning 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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226Ra (half-life 1600 years) decays to 222Rn (half-life 3.8 days). If initially there is 100 µCi of 226Ra in a sample and no 222Rn, calculate how much 222Rn is produced:
a. after 7 half-lives of 222Rnb. at equilibrium
Sample Problem 1
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The number of atoms of 222Rn at time t is given by:
Solution to Sample Problem
= Ra NRa - Rn NRn dNRn
dt
Solving:
NRn(t) = (1 - e- t)RnRaNRa
Rn
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Multiplying both sides of the equation by Rn:
ARn(t) = ARa (1 - e- t)Rn
Solution to Sample Problem
= 100 * (0.992) = 99.2 µCi of 222Rn
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 )
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Solution to Sample Problem
100 µCi + 100 µCi = 200 µCi
RnNRn = RaNRa or ARn = ARa = 100 µCi
Note that the total activity in this sample is:
RnNRn + RaNRa or ARn + ARa =
Now, at secular equilibrium:
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Transient Equilibrium
DND = D - P
D P NP
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 = D - P
AP D
Expressing it in terms of activities of parent and decay product.
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Example of Transient Equilibrium
132Te Decays to 132I
Transient Equilibrium
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.
Te-132 - 78.2 hr half lifeI 132 - 2.2 hr half life
20
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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 activityb. activity of 99Mo at this time, andc. activity of 99mTc at this time
Sample Problem
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Note that only 86% of the 99Mo transformations produce 99mTc. The remaining 14% bypass the isomeric state and directly produce 99Tc
Sample Problem
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Tc = 0.693/(6 hr) = 0.12 hr-1
Mo = 0.693/(66 hr) = 0.011 hr-1
Solution to Sample Problem
tmTc = Tc - Mo
lnTc
Mo
tmTc = 0.12 – 0.011
ln0.12
0.011= 21.9 hrs
a)
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(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
Solution to Sample Problem
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c) The activity of 99mTc at t = 21.9 hrs is given by:
Solution to Sample Problem
ATc(t) = (e-(0.011)(21.9) - e-(0.12)(21.9))(0.12 – 0.011)
(0.12)(100 mCi)(0.86)
= (94.7) (0.785 - 0.071) = 67.6 mCi of 99mTc
ATc(t) = (e- t - e- t )Mo TcTc - Mo
TcAMo(see slide 10)
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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 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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