Question

The half-life for radioactive decay of ${}^{14}C$ is 5730 yr. An archaeological artifact containing wood had only $80\%$ of the ${}^{14}C$ found in a living tree. Estimate the age of the sample.

Answer 1

Decay constant (k) $=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5730}y{r}^{-1}$
Radioactive decays follow first order kinetics
$t=\frac{2.303}{k}log\frac{[A{]}_0}{\left[A\right]}$
Given, $\left[A\right]=80\%$ It means that $[A{]}_0=100\%$
$t=\frac{2.303\times 5730}{0.693}yr\times log\frac{100}{80}$
$=\frac{2.303\times 5730\times 0.0969}{0.693}$
$=1845yr$
Estimated age = 1845 yr