Question
In a mean life of a radioactive sample
✓
Answer
(b) By using $N = {N_0}{e^{ - \lambda t}}$ and $t = \tau = \frac{1}{\lambda }$
Substance remains $ = N = \frac{{{N_0}}}{e} = 0.37{N_0}\simeq \,\frac{{{N_0}}}{3}$
$\therefore$ Substance disintegrated $ = {N_0} - \frac{{{N_0}}}{3} = \frac{{2{N_0}}}{3}$
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