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(D) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.Given that at $t = 5 \, minutes$,the activity is $N = N_0/e$.Subs

Vedclass · Accepted Answer

$5 \log_e 2$

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(D) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.Given that at $t = 5 \, minutes$,the activity is $N = N_0/e$.Substituting these values into the equation: $N_0/e = N_0 e^{-\lambda(5)}$.This simplifies to $e^{-1} = e^{-5\lambda}$,which implies $5\lambda = 1$,so $\lambda = 1/5 \, min^{-1}$.The half-life $T_{1/2}$ is defined as the time required for the activity to reduce to half its initial value,given by $T_{1/2} = \frac{\ln 2}{\lambda}$.Substituting $\lambda = 1/5$,we get $T_{1/2} = \frac{\ln 2}{1/5} = 5 \ln 2 \, minutes$.Therefore,the correct option is $D$.

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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