A radioactive nucleus X decays into a nucleus Y which itself is radioactive. Y decays into a stable nucleus Z.

$X\underset{=\frac{3}{2}hour}{\overset{HalfLife}{\to }}Y\underset{=\frac{3}{8}hour}{\overset{HalfLife}{\to }}Z$

Initially the sample has only nuclei X & number count of Y is zero. Find time (in minutes) when number of count for Y is maximum.

$$\begin{gathered} X\stackrel{{\lambda }_1}{\to }Y\stackrel{{\lambda }_2}{\to }Z \\ {N}_X=N_0{e}^{-{\lambda }_{1}t} \\ \frac{d{N}_y}{dt}={\lambda }_1{N}_x-{\lambda }_2{N}_Y \\ \frac{d{N}_y}{dt}+{\lambda }_2{N}_y=N_0{\lambda }_1{e}^{-{\lambda }_{1}t} \\ {N}_y{e}^{{\lambda }_{2}t}=N_0{\lambda }_1\int e^{-{\lambda }_{1}t}{e}^{{\lambda }_{2}t}dt+C \\ {N}_y{e}^{{\lambda }_{2}t}=\frac{N_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}{e}^{({\lambda }_2-{\lambda }_1)t}+C \\ at t = 0, N_Y=0 \\ C=-\frac{N_0{\lambda }_1}{{\lambda }_2-{\lambda }_1} \\ {N}_Y{e}^{{\lambda }_{2}t}=\frac{N_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}{e}^{({\lambda }_2-{\lambda }_1)t}-\frac{N_0{\lambda }_1}{{\lambda }_2-{\lambda }_1} \\ {N}_y=\frac{N_0{\lambda }_1}{{\lambda }_2-{\lambda }_1}[e^{-{\lambda }_{1}t}-e^{-{\lambda }_{2}t}] \\ \frac{d{N}_Y}{dt}=0 \\ -{\lambda }_1{e}^{-{\lambda }_{1}t}+{\lambda }_2{e}^{-{\lambda }_{2}t}=0 \\ {e}^{({\lambda }_2-{\lambda }_1)t}=\frac{h_2}{h_1}\Rightarrow t=\frac{\mathcal{l}n\frac{{\lambda }_2}{{\lambda }_1}}{{\lambda }_2-{\lambda }_1}=1 hour \end{gathered}$$