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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Time for Decay Productto Reach Maximum Activity

Transient Equilibrium

tmD = D - P

lnD

P

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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

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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)