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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 $t=5$ min. The time (in minute) at which the activity reduces to half its value is:

A

\log_e 2 / 5

B

5 \log_e 2

C

5 \log_{10} 2

D

5 \log_e 2

Step-by-Step Solution

Decay Law: The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$ , where $\lambda$ is the decay constant .
Determine Decay Constant ( $\lambda$ ):

Given: At $t = 5$ min, $A = N_0/e$ (where $A_0 = N_0$ ).
Substitute into the equation: $\frac{N_0}{e} = N_0 e^{-\lambda(5)}$ .
$e^{-1} = e^{-5\lambda}$ .
Equating exponents: $-1 = -5\lambda \implies \lambda = \frac{1}{5} \text{ min}^{-1}$ .

Calculate Half-Life ( $T_{1/2}$ ):

The half-life is the time required for the activity to reduce to half its initial value.
Formula: $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\log_e 2}{\lambda}$ .
Substitute $\lambda = 1/5$ : $T_{1/2} = \frac{\log_e 2}{1/5} = 5 \log_e 2$ min.

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