Question

Assuming a pre-thermonuclear test equilibrium value of 1.3 TU (tritium units) for tritium in a specific grape-growing region, calculate the age of a wine produced there that has a 3H/1H of 8.35 x10-20. 1 TU = 1 atom 3H/1 x 1018 atoms of 1H. The half-life of tritium is 12.33 years.

Answer 1

equilibrium value of is $X_0=1.3\,TU=\frac{1.3}{1\times10^{18}}$

value of in wine is = $8.35\times10^{-20}$

Half-life of Tritium $T_{1/2}=12.33\,years$

since $X=X_0e^{-\lambda{t}}=X_02^{-\frac{t}{T_{1/2}}}$

$8.35\times10^{-20}=\frac{1.3}{1\times10^{18}}2^{-\frac{t}{T_{1/2}}}$

$2^{-\frac{t}{T_{1/2}}}=\frac{8.35\times10^{-2}}{1.3}$

$2^{\frac{t}{T_{1/2}}}=\frac{1.3\times10^{2}}{8.35}=\frac{130}{8.5}$

$t=48.52\,years$