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(A) The number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.Given initial number of nuclei $N_0$ is the same for both material

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

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(A) The number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.Given initial number of nuclei $N_0$ is the same for both materials.Let $N_1$ and $N_2$ be the number of nuclei of $X_1$ and $X_2$ at time $t$ respectively.$N_1 = N_0 e^{-5\lambda t}$ and $N_2 = N_0 e^{-\lambda t}$.The ratio is $\frac{N_1}{N_2} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t}$.We are 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. 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 $\frac{1}{e}$ after a time:

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