Question:
The nuclear activity of a radioactive element becomes $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial value in 30 years. The half-life of radioactive element is __________ years.
Solution:
$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
$\frac{\mathrm{A}_{0}}{8}=\mathrm{A}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \Rightarrow \lambda \mathrm{t}=\ln 8$
$\lambda t=3 \ln 2$
$\frac{\ln 2}{\lambda}=\frac{\mathrm{t}}{3}=\frac{30}{3}=10$ years
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