Q. 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 $4N_{0}$ 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
A) $\frac{5N_{0}}{2}$
B) $2N_{0}$
C) $3N_{0}$
D) $\frac{9N_{0}}{2}$
Explanation
(d) Initially $P \rightarrow 4\ N_{0};Q \rightarrow N_{0}$
Half life $T_{p} = 1$ min; $T_{Q} = 2$ min
Let after time $t$ number of nuclei of $P$ and $Q$ are equal
That is $\frac{4N_{0}}{2^{t/1}} = \frac{N_{0}}{2^{t/2}}$
Or $\frac{4}{2^{t/2}} = 1$ or $t = 4$ min
So at $t = 4$ min
$N_{P} = \frac{(4N_{0})}{2^{4/1}} = \frac{N_{0}}{4}$
At $t = 4$ min. $N_{Q} = \frac{N_{0}}{2^{4/2}} = \frac{N_{0}}{4}$
Or no. of nuclei of R $= \left( 4N_{0} - \frac{N_{0}}{4} \right) + \left( N_{0} - \frac{N_{0}}{4} \right)$
$= \frac{9N_{0}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.