Two radioactive materials X_1 and X_2 contain | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{2\lambda} \, s$

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(A) Let $N_0$ be the initial number of nuclei for both materials.The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For $X_1$,$N_1 = N_0 e^{-6\lambda t}$.For $X_2$,$N_2 = N_0 e^{-4\lambda t}$.The ratio is given as $\frac{N_1}{N_2} = \frac{1}{e}$.Substituting the expressions: $\frac{N_0 e^{-6\lambda t}}{N_0 e^{-4\lambda t}} = e^{-6\lambda t + 4\lambda t} = e^{-2\lambda t}$.Equating to $\frac{1}{e} = e^{-1}$,we get $e^{-2\lambda t} = e^{-1}$.Comparing the exponents: $-2\lambda t = -1$.Therefore,$t = \frac{1}{2\lambda} \, s$.

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