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(B) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given at $t = 0$, $A(0) = N_0$.At $t = 3 	ext{ min}$, $A(3) = N_

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$3 \log_e 2$

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(B) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given at $t = 0$, $A(0) = N_0$.At $t = 3 	ext{ min}$, $A(3) = N_0/e$.Substituting these values into the equation: $N_0/e = N_0 e^{-\lambda(3)}$.This simplifies to $e^{-1} = e^{-3\lambda}$, which gives $3\lambda = 1$, so $\lambda = 1/3 	ext{ min}^{-1}$.The half-life $T_{1/2}$ is given by the formula $T_{1/2} = \frac{\ln 2}{\lambda}$.Substituting $\lambda = 1/3$, we get $T_{1/2} = \frac{\ln 2}{1/3} = 3 \ln 2 = 3 \log_e 2 	ext{ minutes}$.

The activity of a radioactive sample is measured as $N_0$ counts per minute at time $t = 0$ and $N_0/e$ counts per minute at time $t = 3 \text{ minute}$. The time (in minute) in which the activity reduces to half the value, is

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