The activity for a sample of radioactive element found in an experiment is $6400$ decays per minute. Repeating this experiment after $6$ days, activity was found to be $400$ decays per minute. Calculate half life of the given element.

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Solution

We know that
Activity $R = λ N$
where,
$N = N_{0} e^{- λ t}$
$R = λ N_{0} e^{- λ t}$
$R = R_{0} e^{- λ t}$
$\frac{R}{R_{0}} = e^{- λ t}$ .................... $(1)$
From Rutherford soddy law, $N = N_{0} e^{- λ t}$
$\frac{N}{N_{0}} = e^{- λ t}$ ...... $(2)$
From equation $(1)$ and $(2)$ ,
$\frac{R}{R_{0}} = \frac{N}{N_{0}}$
where, $N = \frac{N_{0}}{2^{n}}$ $\Rightarrow \frac{R}{R_{0}} = \frac{1}{2^{t / T_{1 / 2}}}$
$N = \frac{N_{0}}{2^{t / T_{1 / 2}}}$
$\Rightarrow \frac{1}{16} = \frac{1}{2^{6 / T_{1 / 2}}}$
$\frac{N}{N_{0}} = \frac{1}{2^{t / T_{1 / 2}}}$
$\Rightarrow \frac{1}{2^{4}} = \frac{1}{2^{6 / T_{1 / 2}}}$
Given,
$\frac{R}{R_{0}} = \frac{400}{6400}$ $\Rightarrow \frac{R}{R_{0}} = \frac{1}{16}$
$t = 6 d a y s$
Comparing powers
$\frac{6}{T_{1 / 2}} = 4$
$T_{1 / 2} = \frac{6}{4}$
$T_{1 / 2} = 1.5 d a y s$

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