Calculate the time interval between 33 % decay and 67 % decay if half-life of a substance is 20 minutes.

Solution:

(2)

$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$

$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{0}}=\mathrm{e}^{-\lambda t_{1}}$

$0.67=\mathrm{e}^{-\lambda t_{1}}$

$\ln (0.67)=-\lambda t_{1}$

$\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{2}}$

$\frac{\mathrm{N}_{2}}{\mathrm{~N}_{0}}=\mathrm{e}^{-\lambda t_{2}}$

$0.33=\mathrm{e}^{-\lambda t_{2}}$

$\ln (0.33)=-\lambda \mathrm{t}_{2}$

$\ln (0.67)-\ln (0.33)=\lambda \mathrm{t}_{1}-\lambda \mathrm{t}_{2}$

$\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\ln \left(\frac{0.67}{0.33}\right)$

$\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \cong \ln 2$

$\mathrm{t}_{1}-\mathrm{t}_{2} \simeq \frac{\ln 2}{\lambda}=\mathrm{t}_{1 / 2}$

Half life $=\mathrm{t}_{1 / 2}=20$ minutes.