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(D) The number of radioactive nuclei $N$ at any time $t$ is given by the law of radioactive decay: $N(t) = N_0 e^{-\lambda t}$.Here,$N_0$ is the initi

Vedclass · Accepted Answer

$\frac{1}{7\lambda}$

Vedclass · Answer

(D) The number of radioactive nuclei $N$ at any time $t$ is given by the law of radioactive decay: $N(t) = N_0 e^{-\lambda t}$.Here,$N_0$ is the initial number of nuclei at $t=0$ and $\lambda$ is the decay constant.Given for material $A$: $\lambda_A = 8\lambda$ and $N_{0A} = N_0$.Given for material $B$: $\lambda_B = \lambda$ and $N_{0B} = N_0$.The number of nuclei remaining at time $t$ are:$N_A(t) = N_0 e^{-8\lambda t}$$N_B(t) = N_0 e^{-\lambda t}$The ratio of the number of nuclei of $B$ to $A$ is given by:$\frac{N_B(t)}{N_A(t)} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-8\lambda t}} = e^{8\lambda t - \lambda t} = e^{7\lambda t}$.We are given that this ratio is $\frac{1}{e} = e^{-1}$.So,$e^{7\lambda t} = e^{-1}$.Comparing the exponents: $7\lambda t = -1$.Since time $t$ must be positive,let's re-evaluate the ratio requirement. If the ratio $\frac{N_B}{N_A} = \frac{1}{e}$,then $e^{7\lambda t} = e^{-1}$ implies $t = -\frac{1}{7\lambda}$. However,if the question implies the ratio of $A$ to $B$ is $\frac{1}{e}$,then $e^{-7\lambda t} = e^{-1}$,which gives $t = \frac{1}{7\lambda}$. Given the options,the intended answer is $\frac{1}{7\lambda}$.

Radioactive material $A$ has decay constant $8 \lambda$ and material $B$ has decay constant $\lambda$. Initially,they have the same number of nuclei. After what time will the ratio of the number of nuclei of material $B$ to that of $A$ be $\frac{1}{e}$?

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