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(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.Given at $t = 0$,$N = N_0$ and at $t = 5$ minutes,$N = N_0/e$.Sub

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$5 \ln 2$

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(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.Given at $t = 0$,$N = N_0$ and at $t = 5$ minutes,$N = N_0/e$.Substituting these values: $\frac{N_0}{e} = N_0 e^{-\lambda(5)}$.This simplifies to $e^{-1} = e^{-5\lambda}$,which gives $5\lambda = 1$,so $\lambda = 1/5 	ext{ min}^{-1}$.The half-life $T_{1/2}$ is the time when the activity reduces to half of its initial value,i.e.,$N = N_0/2$.Using $N = N_0 e^{-\lambda t}$,we have $\frac{N_0}{2} = N_0 e^{-\lambda t} \Rightarrow 2 = e^{\lambda t}$.Taking the natural logarithm on both sides: $\ln 2 = \lambda t$.Substituting $\lambda = 1/5$: $\ln 2 = \frac{t}{5} \Rightarrow t = 5 \ln 2$ minutes.

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half of its initial value is

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