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The half life of a radioactive substance is $20$ minutes. The approximate time interval $(t_2 - t_1)$ between the time $t_2$ when $\frac{2}{3}$ of it had decayed and time $t_1$ when $\frac{1}{3}$ of it had decayed is ..........$min$
$14$
$20$
$28 $
$7 $
Solution
Number of undecayed atom after time $t_{2};$
$\frac{N_{0}}{3}=N_{0} e^{-\lambda t_{2}}$ …. $(i)$
Number of undecayed atom after time $t_{1};$
$\frac{2 N_{0}}{3}=N_{0} e^{-\lambda t_{1}}$ …. $(ii)$
From $(i)$, $e^{-\lambda t_{2}}=\frac{1}{3}$
$\Rightarrow \quad-\lambda t_{2}=\log _{e}\left(\frac{1}{3}\right)$ …. $(iii)$
From $(ii)$ $-e^{-\lambda t_{2}}=\frac{2}{3}$
$\Rightarrow \quad-\lambda t_{1}=\log _{\mathrm{e}}\left(\frac{2}{3}\right)$ …. $(iv)$
Solving $(iii)$ and $(iv)$, we get $t_{2}-t_{1}=20 \mathrm{\,min}$
