$P$                    $Q$

No. of nuclei, at $t=0$                    $4N_0$            $N_0$

Half – life                                        $1\,min$           $2\,min$

No. of nuclei after time $t$            $N_P$                  $N_Q$

Let after $t$ min the number of nuclei of $P$ and $Q$ are equal. $\therefore \quad N_{P}=4 N_{0}\left(\frac{1}{2}\right)^{t / 1}$ and $N_{Q}=N_{0}\left(\frac{1}{2}\right)^{t / 2}$

As $N_{p}=N_{0}$

$\therefore \quad 4 N_{0}\left(\frac{1}{2}\right)^{t / 1}=N_{0}\left(\frac{1}{2}\right)^{t / 2}$

$\frac{4}{2^{t / 1}}=\frac{1}{2^{t / 2}} \text { or } 4=\frac{2^{t}}{2^{t / 2}}$

or $4=2^{t/2}$ or $2^{2}=2^{t/2}$

or $\frac{t}{2}=2 \quad$ or $t=4 \mathrm{min}$

After $4$ minutes, both $P$ and $Q$ have equal number of nuclei.

$\therefore$ Number of nuclei of $R$

${=\left(4 N_{0}-\frac{N_{0}}{4}\right)+\left(N_{0}-\frac{N_{0}}{4}\right)}$

${=\frac{15 N_{0}}{4}+\frac{3 N_{0}}{4}=\frac{9 N_{0}}{2}}$