IAEA International Atomic Energy Agency Radioactivity -2 Decay Chains and Equilibrium Day 1 – Lecture 5.

IAEAInternational Atomic Energy Agency

Radioactivity -2

Decay Chains and Equilibrium

Day 1 – Lecture 5

IAEA 2

Objective

To discuss radioactive decay chains (parent and single decay product) and equilibrium situations

IAEA 3

Content

Secular equilibrium

Transient equilibrium

Case of no equilibrium

Radioactive decay series

Ingrowth of decay product from a parent radionuclide

IAEA 4

Types of Radioactive Equilibrium

Secular Half-life of parent much greater (> 100 times) than that of decay product

IAEA 5

Types of Radioactive Equilibrium

Transient Half-life of parent only greater (only 10 times greater) than that of decay product

IAEA 6

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)

IAEA

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.

IAEA

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

IAEA

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)

IAEA 10

Buildup of a Decay Product underSecular Equilibrium Conditions

Secular Equilibrium

AY(t) = (1 - e- t)YASr

IAEA 11

Secular Equilibrium

SrNSr = YNY

ASr = AY

At secular equilibrium the activities of the parent and decay product are equal and constant with time

IAEA 12

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

IAEA

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

IAEA

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

IAEA

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 )

IAEA

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:

IAEA

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.

IAEA

Transient Equilibrium

AD = D - P

AP D

Expressing it in terms of activities of parent and decay product.

IAEA

Time for Decay Productto Reach Maximum Activity

Transient Equilibrium

tmD = D - P

lnD

P

IAEA

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

IAEA

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

IAEA

Note that only 86% of the 99Mo transformations produce 99mTc. The remaining 14% bypass the isomeric state and directly produce 99Tc

Sample Problem

IAEA

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)

IAEA

(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

IAEA

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)

IAEA

Solution to Sample Problem

The maximum activity of 99mTc is achieved at 21.9 hours which is nearly 1 day.

IAEA 27

Types of Radioactive Equilibrium

No Equilibrium Half-life of parent less than that of decay product

IAEA 28

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.

IAEA 29

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

IAEA 30

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)