Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.
Calculate the time interval between $33 \%$ decay and $67 \%$ decay if half-life of a substance is 20 minutes.
Correct Option: 2,
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}_{1}}$
$0.67=\mathrm{e}^{-\lambda t_{1}}$
$\ln (0.67)=-\lambda t_{1}$
$\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$
$\frac{\mathrm{N}_{2}}{\mathrm{~N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}_{2}}$
$0.33=e^{-\lambda t_{2}}$
$\ln (0.33)=-\lambda t_{2}$
$\ln (0.67)-\ln (0.33)=\lambda t_{1}-\lambda t_{2}$
$\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)=\ln \left(\frac{0.67}{0.33}\right)$
$\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \cong \ln 2$
$\mathrm{t}_{1}-\mathrm{t}_{2} \simeq \frac{\ln 2}{\lambda}=\mathrm{t}_{1 / 2}$
Half life $=\mathrm{t}_{1 / 2}=20$ minutes.
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