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(D) Let $N_0$ be the initial number of nuclei for both materials.After time $t$,the number of nuclei remaining for $X_1$ is $N_1 = N_0 e^{-5 \lambda t

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$\frac{1}{4 \lambda}$

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(D) Let $N_0$ be the initial number of nuclei for both materials.After time $t$,the number of nuclei remaining for $X_1$ is $N_1 = N_0 e^{-5 \lambda t}$.After time $t$,the number of nuclei remaining for $X_2$ is $N_2 = N_0 e^{-\lambda t}$.The ratio of the number of nuclei is given by $\frac{N_1}{N_2} = \frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-5 \lambda t + \lambda t} = e^{-4 \lambda t}$.Given that $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.Equating the exponents: $-4 \lambda t = -1$.Therefore,$t = \frac{1}{4 \lambda}$.

Two radioactive materials $X_1$ and $X_2$ have decay constants $5 \lambda$ and $\lambda$ respectively. Initially,they have the same number of nuclei. After time $t$,the ratio of the number of nuclei of $X_1$ to that of $X_2$ is $\frac{1}{e}$. Then $t$ is equal to:

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