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A sample of radioactive material decays simultaneously by two processes $A$ and $B$ with half lives $\frac{1}{2}$ and $\frac{1}{4} \mathrm{hr}$ respectively. For first half $\mathrm{hr}$ it decays with the process $A$, next one hr with the process $B$ and for further half an hour with both $\mathrm{A}$ and $\mathrm{B}$. If originally there were $\mathrm{N}_{0}$ nuclei, find the number of nuclei after 2 hr of such decay.

(A) $\frac{N_{0}}{(2)^{8}}$

(B) $\frac{N_{0}}{(2)^{4}}$

(C) $\frac{N_{0}}{(2)^{6}}$

(D) $\frac{N_{0}}{(2)^{5}}$

After first half hrs $N=N_{0} \frac{1}{2}$

for $t=\frac{1}{2}$ to $t=1 \frac{1}{2}$

$N=\left(N_{0} \frac{1}{2}\right)\left[\frac{1}{2}\right]^{4}=N_{0}\left(\frac{1}{2}\right)^{5}$

For $t=1 \frac{1}{2}$ to $t=2$ hrs.

[for both $A$ and $B \frac{1}{t_{1 / 2}}=\frac{1}{1 / 2}+\frac{1}{1 / 4}$

$=2+4=6$

$t_{1 / 2}=1 / 6 \text { hrs.] }$

$N=\left[N_{0}\left(\frac{1}{2}\right)^{5}\right]\left(\frac{1}{2}\right)^{3}=N_{0}\left(\frac{1}{2}\right)^{8}$

So, The correct option of this question is (A).

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