Question

Suppose you have a sample of with 100 mCi of 82Sr. When will the activity of 82Rb reach over 99% of the activity of 82Sr? min You currently have 7 submissions for this question. Only 10 submission are allowed. You can make 3 more submissions for this question. Help: Since the daugther's half-life is much smaller than the parent's half-life, the daugther reaches the parent's activity only asymptotically over several half-lives. Activity of daugther after n half-lives = (1/2)n. You need to find the value of n such that ARb/ASr>=0.99. Note: You do not have to find the exact time when ARb/ASr=0.99. Instead find the smallest value of the integer n such that, after n half-lives ARb/ASr>=0.99. 3) How much daughter activity will there be after 20 minutes? mCi You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 4) How much daughter activity (Question 1) will there be after 20 days? mCi

Answer 1

Let us assume we have initially only N_o nuclides of ⁸²Sr . ⁸²Srundergoes decay with half life 25 days giving ⁸²Rb and ⁸²Rb decays with half life 75 s.

Under equilibrium, if N₂ is number of ⁸²Rb nuclides after time t , then we have

..........................(1)

where is decay constant of ⁸²Sr and is decay constant of ⁸²Rb

Since activity is proportional to number of nuclides present ,

we have        or   .........................(2)

Number of ⁸²Sr nuclides present at time t ,

Activity of ⁸²Sr at time t is given by ,

Hence we get .......................(3)

By substituting N₂ from eqn.(2) and N_o from eqn.(3) , we rewrite eqn.(1) as

....................(4)

Hence

We rewrite eqn.(2) as

.........................(5)

if activity of ⁸²Rb equals 99% of activity of ⁸²Sr , then we have

   or or t = 4.605   or t = 4.605 / (9.242 10^-3 )

Hence time t = 498 s = 8.3 min

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Since ⁸²Sr has long half life , we assume after 20 minutes its activity is still 100 mCi.

Then from eqn.(5) , we get activity of ⁸²Rb at 20 minutes as

This is because if we substitiute t = 20 minutes = 1200 s in

we get

Hence at t= 20 minutes , activity of ⁸²Rb will be 100 mCi