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

Vedclass · Accepted Answer

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

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(A) Let the initial number of nuclei for both materials be $N_0$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $X_1$: $N_1(t) = N_0 e^{-(5\lambda)t}$.For material $X_2$: $N_2(t) = N_0 e^{-\lambda t}$.The ratio is given as $\frac{N_1(t)}{N_2(t)} = \frac{1}{e} = e^{-1}$.Substituting the expressions: $\frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.$e^{-5\lambda t + \lambda t} = e^{-1}$.$e^{-4\lambda t} = e^{-1}$.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. 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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