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

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IAEA Objective To discuss radioactive decay chains (parent and single decay product) and equilibrium situations 2

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IAEA Content Secular equilibrium Transient equilibrium Case of no equilibrium Radioactive decay series Ingrowth of decay product from a parent radionuclide 3

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IAEA Types of Radioactive Equilibrium SecularHalf-life of parent much greater (> 100 times) than that of decay product 4

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IAEA Types of Radioactive Equilibrium Half-life of parent only (only 10 times greater) than that of decay product TransientHalf-life of parent only greater (only 10 times greater) than that of decay product 5

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IAEA 90 Sr 90 Y 90 Zr Sample Radioactive Series Decay where 90 Sr is the parent (half-life = 28 years) and 90 Y is the decay product (half-life = 64 hours) 6

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IAEA Differential Equation for Radioactive Series Decay = Sr N Sr - Y N Y dN Y 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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IAEA Parent and Single Decay Product Differential Equation for Radioactive Series Decay N Y (t) = (e - t - e - t ) Sr Y Sr N Sr Y - Sr o Recall that Sr N o Sr = A o Sr which equals the initial activity of 90 Sr at time t = 0

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IAEA General Equation for Radioactive Series Decay Y N Y (t) = (e - t - e - t ) Sr Y Y - Sr Y Sr N Sr o Activity of 90 Sr at time t = 0 Activity of 90 Y at time t or A Y (t)

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IAEA Buildup of a Decay Product under Secular Equilibrium Conditions Secular Equilibrium A Y (t) = (1 - e - t ) Y A Sr 10

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IAEA Secular Equilibrium Sr N Sr = Y N Y A Sr = A Y At secular equilibrium the activities of the parent and decay product are equal and constant with time 11

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IAEA Decay of 226 Ra to 222 Rn Secular Equilibrium A Rn (t) = A o (1 - e - t ) Rn 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 12

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IAEA 226 Ra (half-life 1600 years) decays to 222 Rn (half- life 3.8 days). If initially there is 100 µCi of 226 Ra in a sample and no 222 Rn, calculate how much 222 Rn is produced: a.after 7 half-lives of 222 Rn b.at equilibrium Sample Problem 1

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IAEA The number of atoms of 222 Rn at time t is given by: Solution to Sample Problem = Ra N Ra - Rn N Rn dN Rn dt Solving: N Rn (t) = (1 - e - t ) Rn Ra N Ra Rn

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IAEA Multiplying both sides of the equation by Rn : A Rn (t) = A Ra (1 - e - t ) Rn Solution to Sample Problem = 100 * (0.992) = 99.2 µCi of 222 Rn Let t = 7 T Rn Rn t = (0.693/T Rn ) x 7 T Rn = 0.693 * 7 = 4.85 Rn t = (0.693/T Rn ) x 7 T Rn = 0.693 * 7 = 4.85 e -4.85 = 0.00784 A Rn (7 half-lives) = 100 µCi * (1 - 0.00784 )

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IAEA Solution to Sample Problem 100 µCi + 100 µCi = 200 µCi Rn N Rn = Ra N Ra or A Rn = A Ra = 100 µCi Rn N Rn = Ra N Ra or A Rn = A Ra = 100 µCi Note that the total activity in this sample is: Rn N Rn + Ra N Ra or A Rn + A Ra = Rn N Rn + Ra N Ra or A Rn + A Ra = Now, at secular equilibrium:

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IAEA Transient Equilibrium D N D = D - P D P N P For the case of transient equilibrium, the general equation for radioactive series decay reduces to the above equation.

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IAEA Transient Equilibrium A D = D - P A P D Expressing it in terms of activities of parent and decay product.

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IAEA Time for Decay Product to Reach Maximum Activity Transient Equilibrium t mD = D - P ln D P

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IAEA Example of Transient Equilibrium 132 Te Decays to 132 I 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 life I 132 - 2.2 hr half life 20

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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 99 Mo (half-life 66 hours) and no 99m Tc (half-life 6 hours) calculate the: a. time required for 99m Tc to reach its maximum activity b. activity of 99 Mo at this time, and c. activity of 99m Tc at this time Sample Problem

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

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IAEA Tc = 0.693/(6 hr) = 0.12 hr -1 Tc = 0.693/(6 hr) = 0.12 hr -1 Mo = 0.693/(66 hr) = 0.011 hr -1 Mo = 0.693/(66 hr) = 0.011 hr -1 Solution to Sample Problem t mTc = Tc - Mo ln Tc Mo t mTc = 0.12 – 0.011 ln 0.12 0.011 = 21.9 hrs a)

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IAEA (b) The activity of 99 Mo is given by A(t) = A o e - t = 100 mCi e (-0.011/hr * 21.9 hr) = 100 * (0.79) = 79 mCi Solution to Sample Problem

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IAEA c) The activity of 99m Tc at t = 21.9 hrs is given by: Solution to Sample Problem A Tc (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 99m Tc A Tc (t) = (e - t - e - t ) Mo Tc Tc - Mo Tc A Mo (see slide 10)

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IAEA Solution to Sample Problem The maximum activity of 99m Tc is achieved at 21.9 hours which is nearly 1 day.

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IAEA Types of Radioactive Equilibrium No EquilibriumHalf-life of parent less than that of decay product 27

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IAEA 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. 28

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IAEA 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 29

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IAEA 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) 30

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