Answer to Question #118488 in Atomic and Nuclear Physics for Yussif Bodombie

Question

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.

Answer 1

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

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

The solution is:

$$N_A = N_{A0}e^{-(B+C)t} = N_{A0}e^{-At}$$

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):

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

The solution is:

$$N_C(t) = [N_{A0}\dfrac{A}{A-C}\cdot (e^{-At} - e^{-Ct}) ] \times \dfrac{C}{A}+ N_{C0}e^{-Ct},$$

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

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

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