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

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$\frac{1}{6 \lambda}$

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(B) Let $N_0$ be the initial number of nuclei for both materials $A$ and $B$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $A$: $N_A(t) = N_0 e^{-(7 \lambda) t}$.For material $B$: $N_B(t) = N_0 e^{-\lambda t}$.We are given that the ratio $\frac{N_B(t)}{N_A(t)} = e$.Substituting the expressions: $\frac{N_0 e^{-\lambda t}}{N_0 e^{-7 \lambda t}} = e$.Simplifying the ratio: $e^{-\lambda t + 7 \lambda t} = e^1$.$e^{6 \lambda t} = e^1$.Equating the exponents: $6 \lambda t = 1$.Therefore,$t = \frac{1}{6 \lambda}$.

Two radioactive materials $A$ and $B$ having decay constants $7 \lambda$ and $\lambda$ respectively,initially have the same number of nuclei. The time taken for the ratio of the number of nuclei of material $B$ to that of $A$ to be $e$ is:

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