Half life of a radio active substance A is two times the half life of another radio active substance B. Initially number of active nuclei of A and B are $N_{1} and N_{2}$ respectively. After three half lives of A, number of nuclei of both are equal. Then the ratio $\frac{N_{1}}{N_{2}}$ is

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answer is B.

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Detailed Solution

$T_{1} = {Half life of A, T}_{2} = Half life of B$

Now, $N_{1}^{1} = \frac{N_{1}}{{(2)}^{\frac{t}{T_{1}}}} = \frac{N_{1}}{{(2)}^{\frac{3 T_{1}}{T_{1}}}} = \frac{N_{1}}{8}$

and $N_{2}^{1} = \frac{N_{2}}{{(2)}^{\frac{t}{T_{2}}}} = \frac{N_{2}}{{(2)}^{\frac{3 T_{1}}{T_{2}}}} = \frac{N_{2}}{{(2)}^{3 \times 2}} = \frac{N_{2}}{64}$

Since $N_{1}^{1} = N_{2}^{1} \Rightarrow \frac{N_{1}}{8} = \frac{N_{2}}{64} \Rightarrow \frac{N_{1}}{N_{2}} = \frac{1}{8}$

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