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(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.At $t = 0$, activity is $N_0$.At $t = 5\, 	ext{minutes}$, activi

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$5\, \log_e 2$

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(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.At $t = 0$, activity is $N_0$.At $t = 5\, 	ext{minutes}$, activity is $N(5) = N_0/e$.Substituting these values: $N_0/e = N_0 e^{-\lambda(5)}$.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 activity becomes $N_0/2$.$N_0/2 = N_0 e^{-\lambda T_{1/2}} \Rightarrow 1/2 = e^{-\lambda T_{1/2}}$.Taking natural logarithm on both sides: $\ln(1/2) = -\lambda T_{1/2} \Rightarrow -\ln 2 = -\lambda T_{1/2}$.$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\ln 2}{1/5} = 5 \ln 2\, 	ext{minutes}$.

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 $t = 5\, \text{minutes}$. The time (in minutes) at which the activity reduces to half of its initial value is

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