A radioactive nuclide A of half-life 3 h decays to a radioactive nuclide B of half-life 6 h, which decays to a stable nuclide C. At t=0 a sample…

The number of $B$ nuclei is $N_B(t)=\dfrac{\lambda_AN_0}{\lambda_B-\lambda_A}\left(e^{-\lambda_At}-e^{-\lambda_Bt}\right)$ with $\lambda_A=2\lambda_B$ since the half-life of $A$ is half that of $B$. Hence $N_B=2N_0\left(e^{-\lambda_Bt}-e^{-2\lambda_Bt}\right)$. It is maximum when $dN_B/dt=0$, i.e. when $\lambda_AN_A=\lambda_BN_B$, equivalently when $e^{-\lambda_Bt}=1/2$, which occurs at $t=6\ \mathrm{h}$ (one half-life of $B$). Putting $x=e^{-\lambda_Bt}=\tfrac12$ gives $N_B=2N_0\left(\tfrac12-\tfrac14\right)=N_0/2$. As a check, $N_A=N_0e^{-\lambda_At}=N_0/4$ and $N_B/N_A=\lambda_A/\lambda_B=2$, consistent with the transient-equilibrium condition. At that instant the activities of A and B are equal, so the activity plot of B crosses that of A at its peak. $N_0/4$ is the number of $A$ nuclei at that instant, a common slip.