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3/2003 Rev 1 I.2.8 – slide 1 of 31 Session I.2.8 Part I Review of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session.

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Presentation on theme: "3/2003 Rev 1 I.2.8 – slide 1 of 31 Session I.2.8 Part I Review of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session."— Presentation transcript:

1 3/2003 Rev 1 I.2.8 – slide 1 of 31 Session I.2.8 Part I Review of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 8Decay Chains and Equilibrium IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

2 3/2003 Rev 1 I.2.8 – slide 2 of 31 Introduction  Radioactive serial decay and equilibrium will be discussed  Students will:  learn the differences between secular and transient equilibrium  identify when no equilibrium is possible  understand how series decay works  calculate ingrowth of a decay product from a radioactive parent

3 3/2003 Rev 1 I.2.8 – slide 3 of 31 Content  Secular equilibrium  Transient equilibrium  Case of no equilibrium  Radioactive decay series  Ingrowth of decay product from a parent radionuclide

4 3/2003 Rev 1 I.2.8 – slide 4 of 31 Overview Radioactive decay chains (parent and single decay product) and equilibrium situations will be discussed

5 3/2003 Rev 1 I.2.8 – slide 5 of 31 Types of Radioactive Equilibrium SecularHalf-life of parent much greater (> 100 times) than that of decay product

6 3/2003 Rev 1 I.2.8 – slide 6 of 31 Types of Radioactive Equilibrium TransientHalf-life of parent only a little greater than that of decay product

7 3/2003 Rev 1 I.2.8 – slide 7 of 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)

8 3/2003 Rev 1 I.2.8 – slide 8 of 31 Differential Equation for Radioactive Series Decay = Sr N Sr - Y N Y dN Y dt Parent and Single Decay Product

9 3/2003 Rev 1 I.2.8 – slide 9 of 31 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

10 3/2003 Rev 1 I.2.8 – slide 10 of 31 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)

11 3/2003 Rev 1 I.2.8 – slide 11 of 31 Buildup of a Decay Product under Secular Equilibrium Conditions Secular Equilibrium A Y (t) = (1 - e - t ) Y A Sr

12 3/2003 Rev 1 I.2.8 – slide 12 of 31 Secular Equilibrium Sr N Sr = Y N Y A Sr = A Y

13 3/2003 Rev 1 I.2.8 – slide 13 of 31 Decay of 226 Ra to 222 Rn Secular Equilibrium A Rn (t) = A o (1 - e - t ) Rn Ra

14 3/2003 Rev 1 I.2.8 – slide 14 of Ra (half-life 1600 years) decays to 222 Rn (half-life 3.8 days). If initially there is 4000 kBq 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

15 3/2003 Rev 1 I.2.8 – slide 15 of 31 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

16 3/2003 Rev 1 I.2.8 – slide 16 of 31 Multiplying both sides of the equation by Rn : A Rn (t) = A Ra (1 - e - t ) Rn Solution to Sample Problem = 4000 x (0.992) = 3968 kBq of 222 Rn Let t = 7 T Rn Rn t = (0.693/T Rn ) x 7 T Rn = x 7 = 4.85 Rn t = (0.693/T Rn ) x 7 T Rn = x 7 = 4.85 e = A Rn (7 half-lives) = 4000 kBq x ( )

17 3/2003 Rev 1 I.2.8 – slide 17 of 31 Solution to Sample Problem 4000 kBq kBq = 8000 kBq Rn N Rn = Ra N Ra or A Rn = A Ra = 4000 kBq Rn N Rn = Ra N Ra or A Rn = A Ra = 4000 kBq 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:

18 3/2003 Rev 1 I.2.8 – slide 18 of 31 Transient Equilibrium D N D = D - P D P N P

19 3/2003 Rev 1 I.2.8 – slide 19 of 31 Transient Equilibrium A D = D - P A P D

20 3/2003 Rev 1 I.2.8 – slide 20 of 31 Time for Decay Product to Reach Maximum Activity Transient Equilibrium t mD = D - P ln D P

21 3/2003 Rev 1 I.2.8 – slide 21 of 31 Example of Transient Equilibrium 132 Te Decays to 132 I Transient Equilibrium

22 3/2003 Rev 1 I.2.8 – slide 22 of 31 The principle of transient equilibrium is illustrated by the Molybdenum-Technetium radioisotope generator used in nuclear medicine applications. Given that the generator initially contains 4000 MBq 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

23 3/2003 Rev 1 I.2.8 – slide 23 of 31 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

24 3/2003 Rev 1 I.2.8 – slide 24 of 31 Tc = 0.693/(6 hr) = 0.12 hr -1 Tc = 0.693/(6 hr) = 0.12 hr -1 Mo = 0.693/(66 hr) = hr -1 Mo = 0.693/(66 hr) = hr -1 Solution to Sample Problem t mTc = Tc - Mo ln Tc Mo t mTc = 0.12 – ln = 21.9 hrs a)

25 3/2003 Rev 1 I.2.8 – slide 25 of 31 (b) The activity of 99 Mo is given by A(t) = A o e - t = 4000 x e (-0.011/hr x 21.9 hr) = 4000 x (0.79) = 3160 MBq Solution to Sample Problem

26 3/2003 Rev 1 I.2.8 – slide 26 of 31 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)(4000 MBq)(0.86) = (3787) ( ) = 2704 MBq of 99m Tc A Tc (t) = (e - t - e - t ) Mo Tc Tc - Mo Tc A Mo (see slide 10)

27 3/2003 Rev 1 I.2.8 – slide 27 of 31 Solution to Sample Problem The maximum activity of 99m Tc is achieved at 21.9 hours which is nearly 1 day.

28 3/2003 Rev 1 I.2.8 – slide 28 of 31 Types of Radioactive Equilibrium No EquilibriumHalf-life of parent less than that of decay product

29 3/2003 Rev 1 I.2.8 – slide 29 of 31 No Equilibrium

30 3/2003 Rev 1 I.2.8 – slide 30 of 31 Summary  Secular equilibrium was defined  Transient equilibrium was defined  Case of no equilibrium was defined  Series decay equations were developed  Decay examples were discussed  Problems in secular and transient equilibrium were solved

31 3/2003 Rev 1 I.2.8 – slide 31 of 31 Where to Get More Information  Cember, H., Johnson, T. E., Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2008)  Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012)  Jelley, N. A., Fundamentals of Nuclear Physics, Cambridge University Press, Cambridge (1990)  Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999)


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