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

Vedclass · Accepted Answer

$\frac{1}{9 \lambda}$

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(D) Let $N_0$ be the initial number of nuclei for both materials.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $x_1$,$N_1(t) = N_0 e^{-(10\lambda)t}$.For material $x_2$,$N_2(t) = N_0 e^{-\lambda t}$.Given the ratio $\frac{N_1}{N_2} = \frac{1}{e}$,we have:$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$$e^{-10\lambda t + \lambda t} = e^{-1}$$e^{-9\lambda t} = e^{-1}$Comparing the exponents: $-9\lambda t = -1$Therefore,$t = \frac{1}{9\lambda}$.

The 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 $t$ equal to:

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