Physics - Modern Physics Question with Solution | TestHub

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 $\tau$ $={\mathrm{N}}_0-0.37{\mathrm{N}}_0=0.63{\mathrm{N}}_0=\frac{2}{3}{\mathrm{N}}_0$