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Topic: 2 step decay HW help  (Read 3063 times)

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Offline asdfomg12

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2 step decay HW help
« on: May 02, 2012, 08:59:40 PM »
My attempted work will be in italics

If  234Pa is the end product of the decay sequence of 238U    (actually it’s  206Pb), that is:

238:rarrow: 234Th  :rarrow: 234Pa
where the half-life of 238U to 234Th is 4.5x109yr and 234Th to 234Pa is 24.1day
 
Therefore: k1 = (ln2)/4.5x109yr = 1.54x10-10yr-1 & k2 = (ln2)/24.1day = 2.876x10-2day-1

   (a)   If we start with 1 mg of  238U , how much 238U  will be present after 60 days? 

[238U] = [238U]0e-k1t = .99999999997468mg

   (b)   How much 234Th will be present after 60 days?   
[234Th] = [238U](k1/(k2 - k1))(e-k1t - e-k2t) = 4.40118x10-9mg


   (c)   How much 234Pa will be formed after 60 days?   
This is the step i don't know how to derive the proper equation for calculation

Offline Enthalpy

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Re: 2 step decay HW help
« Reply #1 on: May 03, 2012, 08:44:28 AM »
(a) Agree

(b) Impossible. 2.5*10-11 mg less 238U can't create 4.4*10-9 mg 234Th.
60 days nearly reach the equilibrium, where the amount of 234Th would be 24.1 days divided by 4.5 billion years - corrected by 234/238 - equating 1.4*10-11 mg. After only 60 days, you may get 4/5 of this asymptotic value approximately.

(c) If you know how much 238U is gone and how much 234Th is left, 234Pa is the difference. At least for the formal computation, because numbers like 1-10-11 need careful handling.

(d) For cases more complicated, with more nuclides or a wide initial mix, electrical engineering would have all the necessary math tools, without any limit. Some redundent methods: state variables, matrix exponentials, Laplace transformation, fluence graphs... Probably a too heavy investment for your needs, I guess some software does this for you.

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