Question

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

Answer 1

Radioactive decay follows first order kinetics,

Decay constant (k) = $\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}y{r}^{-1}$

According to question, living tree has 80% ${}^{14}C$,

$k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}$ $k=\frac{2.303}{k}log\frac{[A{]}_0}{\left[A\right]}$

$t=\frac{2.303\times 5730}{0.693}log\frac{100}{80}$

$t=\frac{2.303\times 5730\times 0.0969}{0.693}$

$t=1845$ years

Hence the age of sample is 1845 years.