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The half life of a radioactive substance is 20minutes. The approximate time interval $(t_{2} - t_{1})$ between the time $t_{2}$ when $\frac{2}{3}$ of it has decayed and time $t_{1}$ and $\frac{1}{3}$ of it has decayed is

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20min

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7min

answer is B.

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Detailed Solution

At time
$t_{2} = N_{0} - \frac{2}{3} N_{0} = \frac{N_{0}}{3} = N_{0} e^{- λ t_{2}} \to (2)$

At time

$\begin{aligned} t_{1} = N_{0} - \frac{N_{0}}{3} = \frac{2 N_{0}}{3} = N_{0} e^{- λ t_{1}} \to (2) \\ \frac{(1)}{(2)} \Rightarrow 2 = e^{λ (t_{2} - t_{1})} \\ \Rightarrow t_{2} - t_{1} = \frac{\ln_{2}}{λ} = T_{1 / 2} = 20 \min \end{aligned}$

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