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(D) The number of nuclei remaining at time $t$ is given by the radioactive decay law: $N(t) = N_0 e^{-\lambda t}$.For material $X_1$ with decay consta

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$1 / (9 \lambda)$

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(D) The number of nuclei remaining at time $t$ is given by the radioactive decay law: $N(t) = N_0 e^{-\lambda t}$.For material $X_1$ with decay constant $10 \lambda$: $N_1 = N_0 e^{-10 \lambda t}$.For material $X_2$ with decay constant $\lambda$: $N_2 = N_0 e^{-\lambda t}$.Given that the ratio $N_1 / N_2 = 1 / e$ at time $t$:$\frac{N_1}{N_2} = \frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = e^{-10 \lambda t + \lambda t} = e^{-9 \lambda t}$.We are given $\frac{N_1}{N_2} = e^{-1}$.Equating the exponents: $-9 \lambda t = -1$.Therefore, $t = \frac{1}{9 \lambda}$.

Two radioactive materials $X_1$ and $X_2$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1 / e$ after a time:

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