The activity of radioactive sample is measured as N 0 counts per minute at time t = o , and e N 0 counts per minute at time t = 3 minute, The activity reduces to half its value in time (in minut

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The activity of radioactive sample is measured as N 0 ​ counts per minute at time t = o , and e N 0 ​ ​ counts per minute at time t = 3 minute, The activity reduces to half its value in time (in minute)

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The activity of a radioactive sample at time $t$ is given by $A = A_{0} e^{- λ t}$ .

Given that at $t = 0$ , $A_{0} = N_{0}$ .

At $t = 3$ minutes, $A = \frac{N _{0}}{e}$ .

Substituting these values into the activity equation:

$\frac{N _{0}}{e} = N_{0} e^{- 3 λ}$

$e^{- 1} = e^{- 3 λ}$

Equating the powers of $e$ , we get:

$3 λ = 1 \Rightarrow λ = \frac{1}{3} minute^{- 1}$

The time in which the activity reduces to half its value is the half-life $T_{1/2}$ , which is given by:

$T_{1/2} = \frac{lo g _{e} 2}{λ}$

Substituting the value of $λ$ :

$T_{1/2} = \frac{lo g _{e} 2}{\frac{1}{3}} = 3 lo g_{e} 2 minutes$

Answer: $3 lo g_{e} 2$

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