Nuclei - Result Question 69
70. 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
(a) $3 N_0$
(b) $\frac{9 N_0}{2}$
(c) $\frac{5 N_0}{2}$
(d) $2 N_0$
[2011M]
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Answer:
Correct Answer: 70. (b)
Solution:
- (b) Initially $P \to 4 N_0$
$Q \to N_0$
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
$\frac{4 N_0}{2^{t / 1}}=\frac{N_0}{2^{t / 2}}$
$\Rightarrow \frac{4 N}{2^{t / 1}}=\frac{1}{2^{t / 2}}$
$\Rightarrow 2^{t / 1}=4.2^{t / 2}$
$2^{2} \cdot 2^{t / 2}=2^{(2+t / 2)}$ $\Rightarrow \frac{t}{1}=2+\frac{t}{2} \Rightarrow \frac{t}{2}=2$
$\Rightarrow t=4 min$
$N_P=\frac{(4 N_0)}{2^{4 / 1}}=\frac{N_0}{4}$
at $t=4 min$
$N_0=\frac{N_0}{4}=\frac{N_0}{4}$
or population of $R$
$(4 N_0-\frac{N_0}{4})+(N_0-\frac{N_0}{4})$
$=\frac{9 N_0}{2}$
