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(A) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given at $t = 0$,$A = N_0$,so $A_0 = N_0$.At $t = 5$ minutes,$A =

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$5 \ln 2$

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(A) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given at $t = 0$,$A = N_0$,so $A_0 = N_0$.At $t = 5$ minutes,$A = N_0/e$.Substituting these values into the equation: $\frac{N_0}{e} = N_0 e^{-\lambda (5)}$.This simplifies to $e^{-1} = e^{-5\lambda}$,which implies $5\lambda = 1$,or $\lambda = \frac{1}{5} 	ext{ min}^{-1}$.The half-life $T_{1/2}$ is given by the formula $T_{1/2} = \frac{\ln 2}{\lambda}$.Substituting $\lambda = \frac{1}{5}$,we get $T_{1/2} = \frac{\ln 2}{1/5} = 5 \ln 2$ 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 its value is

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