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Given an initial nucleus, A, which decays to B and then B to C. I want to calculate the time when the radioactive decay activity of A and B are equal to one another. I have tried to set the $\lambda_A * N_A = \lambda_B * N_B$. But by doing that the time variable cancels out on both side. So what is the correct approach for this problem?

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closed as off-topic by AccidentalFourierTransform, heather, Kyle Kanos, Jon Custer, rob Feb 7 '17 at 20:01

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  • $\begingroup$ You don't show any time variable. How can they cancel? $\endgroup$ – Bill N Feb 6 '17 at 22:00
  • $\begingroup$ Well, the activity is defined as λN = λN_0*exp(-λ*t) $\endgroup$ – Kane Billiot Feb 6 '17 at 22:08
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    $\begingroup$ Only for A, and the $\lambda$ terms could be different. The expression for activity of B is more complicated because it has the decay of A feeding into the population of B while B decays: $\dot{B}=-\lambda_B N_B(t) + \lambda_A N_A(t)$. $\endgroup$ – Bill N Feb 6 '17 at 22:16
  • $\begingroup$ If I set $B'$ = $A'$ then I would I end up with 0 = $−λ_BN_B(t)+2λ_AN_A(t)$. Move one of the product to the other side, then wouldn't t cancel out again? @BillN $\endgroup$ – Kane Billiot Feb 6 '17 at 22:28
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    $\begingroup$ No, don't do that. The activity of X is $\lambda_X X$, not $\dot{X}$. You have to solve the $\dot{B} differential equation, which is non-homogeneous, and use the boundary condition for the general solution. It's a non-trivial problem. $\endgroup$ – Bill N Feb 6 '17 at 22:33
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Once I applied all the hints from Bill N (from the comment section), I arrived at the solution of $t = [ln(λ_A/λ_B)]/(λ_A-λ_B)$.

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  • $\begingroup$ Glad you managed to add this before the question got closed... could you show the work (the solution is not as interesting as how you arrived at it). $\endgroup$ – Floris Feb 7 '17 at 20:04
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You are on the right track--you want to solve that equation, just remember that $N_A(t)$ and $N_B(t)$ are functions of time (with $N_B(0)=0$).

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