d
                                                        \(P\)                    \(Q\)

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 \(R\)

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

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