Question #118488
A radioactive material A ( decay constant A) decays into a material B( decay constant B) and into material C ( decay constant C) which is also radioactive. Determine the amount of C material remaining after a time t.
1
Expert's answer
2020-05-27T10:23:45-0400

The equation for th e radioactive material A decay (trough two decay modes):


dNAdt=BNACNA\dfrac{dN_A}{dt} = -B\cdot N_A - C\cdot N_A

The solution is:


NA=NA0e(B+C)t=NA0eAtN_A = N_{A0}e^{-(B+C)t} = N_{A0}e^{-At}

The equation for the radioactive material C decay (as a daughter of the parent A) is given by the Bateman equation (see https://en.wikipedia.org/wiki/Bateman_equation):


dNCdt=CNCCANA=CNCCANA0eAt\dfrac{dN_C}{dt} = -C\cdot N_C -\dfrac{C}{A}\cdot N_A = -C\cdot N_C - \dfrac{C}{A}\cdot N_{A0}e^{-At}

The solution is:


NC(t)=[NA0AAC(eAteCt)]×CA+NC0eCt,N_C(t) = [N_{A0}\dfrac{A}{A-C}\cdot (e^{-At} - e^{-Ct}) ] \times \dfrac{C}{A}+ N_{C0}e^{-Ct},

where NA0N_{A0} and NC0N_{C0} are amount of the materials A and C at the initial moment of time and the multiplier CA\dfrac{C}{A} represents branching ratio for decay to the material C.

As far as A=B+CA = B+C, the final solution will be:


NC(t)=NA0CB(eAteCt)+NC0eCt.N_C(t) = N_{A0}\cdot \dfrac{C}{B}\cdot (e^{-At} - e^{-Ct}) + N_{C0}e^{-Ct}.

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Comments

Yussif Bodombie
27.05.20, 11:42

Thank you for making assignments very easy...,

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