Question

Two radioactive nuclei A and B decay into C with decay constant $\lambda$ and $2\lambda$ respectively. Initially, both A and B have the same number of nuclei $N_0$. The rate of production of C after time $\frac{\tau }{\lambda }$ is $\alpha \lambda N_0$, where $\tau$ is the half life of A. Find $\alpha$.

Answer 1

We have the decay process happening for $A\to C$ and $B\to C$

By law of radioactivity,
$N_A=N_0{e}^{-\lambda t}$
$N_B=N_0{e}^{-2\lambda t}$

Rate of production of $C$,
$\frac{d{N}_C}{dt}=-\frac{d{N}_A}{dt}-\frac{d{N}_B}{dt}$
$=\lambda \times N_0{e}^{-\lambda t}+2\lambda \times N_0{e}^{-2\lambda t}$
$=\lambda N_A\left(t\right)+2\lambda N_B\left(t\right)$

At $t=\frac{\tau }{\lambda }$; where $\tau =\frac{\ln 2}{\lambda }$
$N_A=N_0{e}^{-\ln 2}=\frac{N_0}{2}$;
$N_B=N_0{e}^{-\ln 4}=\frac{N_0}{4}$

$\Rightarrow \frac{d{N}_C}{dt}=\lambda N_A\left(t\right)+2\lambda N_B\left(t\right)$
$=\lambda \frac{N_0}{2}+2\lambda \frac{N_0}{4}$
$=\lambda N_0$
Hence, $\alpha =1$