Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
$^{38}S \xrightarrow{2.48 \ h} ^{38}Cl \xrightarrow{0.62 \ h} ^{38}Ar$
Assume that we start with $1000$ $^{38}S$ nuclei at time $t = 0$. The number of $^{38}Cl$ nuclei is zero at $t = 0$ and will again be zero at $t = \infty$. At what value of $t$ would the number of $^{38}Cl$ nuclei be a maximum?

(C) Let $N_1$ be the number of $^{38}S$ nuclei and $N_2$ be the number of $^{38}Cl$ nuclei.
The decay constants are $\lambda_1 = \frac{\ln 2}{2.48} \approx 0.2794 \ h^{-1}$ and $\lambda_2 = \frac{\ln 2}{0.62} \approx 1.118 \ h^{-1}$.
The rate of change of $N_2$ is given by $\frac{dN_2}{dt} = \lambda_1 N_1 - \lambda_2 N_2$.
For $N_2$ to be maximum,$\frac{dN_2}{dt} = 0$,which implies $\lambda_1 N_1 = \lambda_2 N_2$.
The number of nuclei at time $t$ is $N_1(t) = N_0 e^{-\lambda_1 t}$.
The solution for $N_2(t)$ is $N_2(t) = N_0 \frac{\lambda_1}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$.
Setting $\frac{dN_2}{dt} = 0$ leads to $t_{max} = \frac{\ln(\lambda_2 / \lambda_1)}{\lambda_2 - \lambda_1}$.
Substituting the values: $t_{max} = \frac{\ln(1.118 / 0.2794)}{1.118 - 0.2794} = \frac{\ln(4)}{0.8386} \approx \frac{1.386}{0.8386} \approx 1.65 \ h$.