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(D) The number of nuclei remaining after time $t$ is given by $N = N_0 e^{-\lambda t}$.For substance $A$,$N_A = N_0 e^{-5 \lambda t} \dots (i)$.For su

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$1/(2 \lambda)$

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(D) The number of nuclei remaining after time $t$ is given by $N = N_0 e^{-\lambda t}$.For substance $A$,$N_A = N_0 e^{-5 \lambda t} \dots (i)$.For substance $B$,$N_B = N_0 e^{-\lambda t} \dots (ii)$.Dividing equation $(i)$ by equation $(ii)$,we get:$\frac{N_A}{N_B} = \frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-5 \lambda t + \lambda t} = e^{-4 \lambda t}$.Given that the ratio $\frac{N_A}{N_B} = (1/e)^2 = e^{-2}$.Equating the exponents: $-4 \lambda t = -2$.Therefore,$t = \frac{2}{4 \lambda} = \frac{1}{2 \lambda}$.

Two radioactive substances $A$ and $B$ have decay constants $5 \lambda$ and $\lambda$ respectively. At $t=0$,they have the same number of nuclei. The ratio of the number of nuclei of $A$ to those of $B$ will be $(1/e)^2$ after a time interval of:

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