For a substance, fraction of its initial quantify $\left(\mathrm{N}_0\right)$ which will disintegrate in its…
For a substance, fraction of its initial quantify $\left(\mathrm{N}_0\right)$ which will disintegrate in its average life time is about $(\mathrm{e}=2.71)$
$\left(\frac{1}{3}\right) \mathrm{N}_0$
$\left(\frac{1}{2}\right) \mathrm{N}_0$
$\left(\frac{2}{3}\right) \mathrm{N}_0$
$(0.9) \mathrm{N}_0$
Solution
For a radioactive substance, $\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda t}$
Given the mean life, $\mathrm{t}=\frac{1}{\lambda}$
$\therefore \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_0}{\mathrm{e}}=\frac{\mathrm{N}_0}{2.71}=0.37 \mathrm{~N}_0$
$0.37 \mathrm{~N}_0$ is the amount of radioactive substance left after time $\tau$.
$\therefore$ The amount disintegrated in time $\tau$
$=\mathrm{N}_0-0.37 \mathrm{~N}_0=0.63 \mathrm{~N}_0=\frac{2}{3} \mathrm{~N}_0$