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(D) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $X_1$,$N_1 = N_0 e^{-10\lambda t}$.For material $X

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$1/9\lambda$

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(D) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $X_1$,$N_1 = N_0 e^{-10\lambda t}$.For material $X_2$,$N_2 = N_0 e^{-\lambda t}$.Given the ratio $N_1/N_2 = 1/e$ at time $t$,we have:$\frac{N_1}{N_2} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t}$.Setting this equal to $1/e = e^{-1}$,we get:$e^{-9\lambda t} = e^{-1}$.Equating the exponents:$-9\lambda t = -1$.Therefore,$t = 1/(9\lambda)$.

Two radioactive materials $X_1$ and $X_2$ have decay constants $10\lambda$ 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 $1/e$ after a time:

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