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(D) Let $N_0$ be the initial number of nuclei for both substances $A$ and $B$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\

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$1/2\lambda$

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(D) Let $N_0$ be the initial number of nuclei for both substances $A$ and $B$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For substance $A$: $N_A(t) = N_0 e^{-5\lambda t}$.For substance $B$: $N_B(t) = N_0 e^{-\lambda t}$.The ratio of the number of nuclei is $\frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-4\lambda t}$.We are given that this ratio is $(1/e)^2 = e^{-2}$.So,$e^{-4\lambda t} = e^{-2}$.Equating the exponents: $-4\lambda t = -2$.Solving for $t$: $t = \frac{-2}{-4\lambda} = \frac{1}{2\lambda}$.

Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$,a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(1/e)^2$ will be

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