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(C) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given $A(0) = N_0$ and $A(3) = \frac{N_0}{e}$.Substituting these

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$3 \ln 2$

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

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

Similar Questions

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