Question

Question: Consider again the decay chain in Problem 2.39 in which the nuclide A is produced at the constant rate of R atoms/sec. Derive an expression for the activity of B as a function of time.

Here is #2.39 for reference:

Domy ICICIKG 39. Consider the chain decay ABC, with no atoms of B present at t = 0. 1. Show that the activity of Brises to a maximum value at the time to given by at which time the activities of A and B are equal. 2. Show that, fort<tm, the activity of B is less than that of A, whereas the reverse is the case fort > tm.

Answer 1

consider a radioactive nuclide. A decay into anomer radioactive nuclide B , the latter again decays into a c o Ad B 22 Let ,,d be the decay constants of nuclides A and B respectively and Mi sN ₂ M3 be the number of atoms of the three kinds at any instant 6 stol to dustinas wizin Now the and nuclide would be formed at the rate an, by the delay of parent alom and disappear by its own decay at the rate of d2 M2. so we have decay equations dN2 = dini-42 42 orange von A2 = din, 22M2 - 2 d13 2 22N2 - © solving equation 6 we get in (N) 2 - Mio e it where Mio is the number of not aloms of nuclide e, at tao. so from out" @ we have dN2 - d. No e nit – 22 N ote -dit sidor or, dne 2 M = a, no e n tours or multiplying both sides by e Art, the integrating factor we get

etzt ( don + A 2 M2) - A Mine (AA) or, d (M2 e Azt) = de Moe A2 Integrating we get tante Më e azt a 21 No (27) + Ano 22-20 A r is the constant of integration. I to Now at t=0, (N2) = N20 = 0 since only first nuclide is present 03070 z cz. So, Az - A No - No da-di bob was on .:(N2) 2 2 9 No rendit e Azt) nba (22-21) Now from equation & and ③ we have -Azt MIA to e - t2 Moetit AB darditoupe evivlel At f20, (M3) + = 130 20 oil (m. so we get a mio anche a to at what a (M2), = No (1+ weer dit) 10 so equation 6 ③ and 6 constitute the Solution of the problem.

equation gives how Ne varies with t. If shows that No = 0 at 6=0 It increases with increasing t and attains a maximum at t=.tm. mis tm can be determined by the condition dN2 - at t=bm. at at = 0 health À , Mo (-a, e tage det) . IN 2 = 0 ) de destma, e nism a, e Corde) time or (21-22) 6m = ln a 12 or tm = ca en 11 nis m corresponds to masimum N.