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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 $4N_0$ and that of Q are $N_0$. Half-life of P (for conversion to R) is 1 minute whereas 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
$\frac{9N_0}{2}$
Detailed Solution
After time t (in minutes): $N_P = 4N_0\left(\frac{1}{2}\right)^{t/1}$ and $N_Q = N_0\left(\frac{1}{2}\right)^{t/2}$
Setting $N_P = N_Q$: $4\times2^{-t} = 2^{-t/2}$
$2^{2 - t} = 2^{-t/2}$, so $2 - t = -\frac{t}{2}$
$t = 4$ minutes
At t = 4 min: $N_P = \frac{4N_0}{2^4} = \frac{N_0}{4}$ and $N_Q = \frac{N_0}{2^2} = \frac{N_0}{4}$
Nuclei of P decayed = $4N_0 - \frac{N_0}{4} = \frac{15N_0}{4}$; nuclei of Q decayed = $N_0 - \frac{N_0}{4} = \frac{3N_0}{4}$
Every decayed nucleus becomes R: $N_R = \frac{15N_0}{4} + \frac{3N_0}{4} = \frac{18N_0}{4}$
$N_R = \frac{9N_0}{2}$
Setting $N_P = N_Q$: $4\times2^{-t} = 2^{-t/2}$
$2^{2 - t} = 2^{-t/2}$, so $2 - t = -\frac{t}{2}$
$t = 4$ minutes
At t = 4 min: $N_P = \frac{4N_0}{2^4} = \frac{N_0}{4}$ and $N_Q = \frac{N_0}{2^2} = \frac{N_0}{4}$
Nuclei of P decayed = $4N_0 - \frac{N_0}{4} = \frac{15N_0}{4}$; nuclei of Q decayed = $N_0 - \frac{N_0}{4} = \frac{3N_0}{4}$
Every decayed nucleus becomes R: $N_R = \frac{15N_0}{4} + \frac{3N_0}{4} = \frac{18N_0}{4}$
$N_R = \frac{9N_0}{2}$