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(D) Let $N_0$ be the initial number of nuclei for both materials $M_1$ and $M_2$.The number of nuclei remaining at time $t$ is given by the radioactiv

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$\frac{1}{8 \lambda}$

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(D) Let $N_0$ be the initial number of nuclei for both materials $M_1$ and $M_2$.The number of nuclei remaining at time $t$ is given by the radioactive decay law: $N(t) = N_0 e^{-\lambda t}$.For $M_1$,the number of nuclei at time $t$ is $N_1(t) = N_0 e^{-(9 \lambda) t}$.For $M_2$,the number of nuclei at time $t$ is $N_2(t) = N_0 e^{-\lambda t}$.The ratio of the number of nuclei of $M_1$ to that of $M_2$ is given as $\frac{N_1(t)}{N_2(t)} = \frac{1}{e}$.Substituting the expressions: $\frac{N_0 e^{-9 \lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e}$.This simplifies to $e^{-9 \lambda t + \lambda t} = e^{-1}$.$e^{-8 \lambda t} = e^{-1}$.Equating the exponents: $-8 \lambda t = -1$.Therefore,$t = \frac{1}{8 \lambda}$.

Two radioactive materials $M_1$ and $M_2$ have decay constants $9 \lambda$ and $\lambda$ respectively. Initially,they have the same number of nuclei. The ratio of the number of nuclei of $M_1$ to that of $M_2$ will be $\frac{1}{e}$ after a time interval of:

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