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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
A
$5\log_e2$
B
$\log_e\frac{2}{5}$
C
$\frac{5}{\log_e2}$
D
$5\log_{10}5$
Detailed Solution
Activity decays as $A = A_0e^{-\lambda t}$
At t = 5 min: $\frac{N_0}{e} = N_0e^{-5\lambda}$, so $e^{-1} = e^{-5\lambda}$
$5\lambda = 1$, so $\lambda = \frac{1}{5}\ min^{-1}$ (the mean life is 5 minutes)
Half-life: $T_{1/2} = \frac{\log_e2}{\lambda}$
$T_{1/2} = 5\log_e2$ minutes
At t = 5 min: $\frac{N_0}{e} = N_0e^{-5\lambda}$, so $e^{-1} = e^{-5\lambda}$
$5\lambda = 1$, so $\lambda = \frac{1}{5}\ min^{-1}$ (the mean life is 5 minutes)
Half-life: $T_{1/2} = \frac{\log_e2}{\lambda}$
$T_{1/2} = 5\log_e2$ minutes