Presentation on theme: "IAEA International Atomic Energy Agency Radioactivity -2 Decay Chains and Equilibrium Day 1 – Lecture 5."— Presentation transcript:
IAEA International Atomic Energy Agency Radioactivity -2 Decay Chains and Equilibrium Day 1 – Lecture 5
IAEA Objective To discuss radioactive decay chains (parent and single decay product) and equilibrium situations 2
IAEA Content Secular equilibrium Transient equilibrium Case of no equilibrium Radioactive decay series Ingrowth of decay product from a parent radionuclide 3
IAEA Types of Radioactive Equilibrium SecularHalf-life of parent much greater (> 100 times) than that of decay product 4
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
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
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.
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
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)
IAEA Buildup of a Decay Product under Secular Equilibrium Conditions Secular Equilibrium A Y (t) = (1 - e - t ) Y A Sr 10
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
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
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
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
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 = * 7 = 4.85 Rn t = (0.693/T Rn ) x 7 T Rn = * 7 = 4.85 e = A Rn (7 half-lives) = 100 µCi * ( )
IAEA Solution to Sample Problem 100 µCi 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:
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.
IAEA Transient Equilibrium A D = D - P A P D Expressing it in terms of activities of parent and decay product.
IAEA Time for Decay Product to Reach Maximum Activity Transient Equilibrium t mD = D - P ln D P
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 hr half life I 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 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
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
IAEA 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)
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
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) ( ) = 67.6 mCi of 99m Tc A Tc (t) = (e - t - e - t ) Mo Tc Tc - Mo Tc A Mo (see slide 10)
IAEA Solution to Sample Problem The maximum activity of 99m Tc is achieved at 21.9 hours which is nearly 1 day.
IAEA Types of Radioactive Equilibrium No EquilibriumHalf-life of parent less than that of decay product 27
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
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
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