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(D) According to the law of radioactive decay,the activity $R$ at time $t$ is given by $R = R_0 e^{-\lambda t}$.Given that at $t = 0$,$R_0 = N_0$ and

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$5 \log_e 2$

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(D) According to the law of radioactive decay,the activity $R$ at time $t$ is given by $R = R_0 e^{-\lambda t}$.Given that at $t = 0$,$R_0 = N_0$ and at $t = 5$ minutes,$R = N_0/e$.Substituting these values into the decay equation:$N_0/e = N_0 e^{-5\lambda}$$e^{-1} = e^{-5\lambda}$Comparing the exponents,we get $5\lambda = 1$,so $\lambda = 1/5$ per minute.The half-life $T_{1/2}$ is the time when the activity reduces to half its initial value,i.e.,$R = R_0/2$.Using $R = R_0 e^{-\lambda t}$,we have $R_0/2 = R_0 e^{-\lambda T_{1/2}}$.$1/2 = e^{-\lambda T_{1/2}}$$2 = e^{\lambda T_{1/2}}$Taking the natural logarithm on both sides:$\ln(2) = \lambda T_{1/2}$$T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{\log_e 2}{1/5} = 5 \log_e 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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