The age of wine is determined by Tritium dating. HLT is 12.5 yrs. and radioactivity of wine sample is found to be 3%. The age of wine is approximately

$M=\frac{3}{100}X{M}_0;\frac{3}{100}{M}_0=M_0{e}^{-\lambda t}$

$e^{\lambda t}=\frac{{10}^2}{3}$

Take log on both sides

${\log }_e{e}^{\lambda t}={\log }_e\frac{{10}^2}{3}$

$\lambda t={\log }_e{10}^2-{\log }_{e}3$

$=\left(2{\log }_{10}10-{\log }_{10}3\right)\times 2.303$

$t=\frac{1}{\lambda }\left(2{\log }_{10}10-{\log }_{10}3\right)\times 2.303$

$=\frac{T}{0.693}(2-0.4771)2.303$

$=\frac{12.5}{0.693}\times 1.5229\times 2.303\mid ;\mathrm{t}=62.5\mathrm{yrs}$