Two radioactive nuclei $P$ and $Q$, in a given sample decay into a stable nucleous $R$. At time $t=0$,…
Two radioactive nuclei $P$ and $Q$, in a given sample decay into a stable nucleous $R$. 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 min where as that of $Q$ is $2 \mathrm{~min}$. 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
$3 N_0$
$\frac{9 N_0}{2}$
$\frac{5 N_0}{2}$
$2 N_0$
Solution
Initially $P \rightarrow 4 N_0$
$Q \rightarrow N_0$
Half life $T_p \rightarrow 1 \mathrm{~min}$
$T_Q \rightarrow 2 \min$
Let after time $t$ number of nuclei of $P$ and $Q$ are equalie, $\quad \frac{4 N_0}{2^{t / 1}}=\frac{N_0}{2^{t / 2}}$
$\begin{aligned}
4 & =2^{t / 2} \\
2^2 & =2^{t / 2} \\
\frac{t}{2} & =2 \\
t & =4 \mathrm{~min}
\end{aligned}$
Disactive nucleus or Nuclei of $R$
$\begin{aligned}
& =\left(4 N_0-\frac{4 N_0}{2^4}\right)+\left(N_0-\frac{N_0}{2^2}\right) \\
& =4 N_0-\frac{N_0}{4}+N_0-\frac{N_0}{4} \\
& =5 N_0-\frac{N_0}{2} \\
& =\frac{9}{2} N_0
\end{aligned}$