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Two radioactive nuclei $P$ and $Q,$ in a given sample decay into a stable nucleus $R.$ At time $t = 0,$ number of $P$ species are $4\,\, N_0$ and that of $Q$ are $N_0$. Half-life of $P$ (for conversion to $R$) is $1$ minute where as that of $Q$ is $2$ minutes. Initially there are no nuclei of $R$ present in the sample. When number of nuclei of $P$ and $Q$ are equal, the number of nuclei of $R$ present in the sample would be
$2N_0$
$3N_0$
$\frac{{3{N_0}}}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
$\;\frac{{9{N_0}}}{2}$
Solution
$P$ $Q$
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 $R$
${=\left(4 N_{0}-\frac{N_{0}}{4}\right)+\left(N_{0}-\frac{N_{0}}{4}\right)}$
${=\frac{15 N_{0}}{4}+\frac{3 N_{0}}{4}=\frac{9 N_{0}}{2}}$
