Question

A piece of wood from an archeological source shows a ${}^{14}C$ activity which is $60\%$ of the activity found in fresh wood today. Calculate the age of the archeological sample. $(t_{1/2}\text{of}{}^{14}C=5770year)$.

Answer 1

We know for disintegration,

$t=\frac{2.303}{K}\log \frac{N_0}{N}$

Note: $r_0\propto N_0$ and $r\propto N$

$t=\frac{2.303\times 5770}{0.693}\log \frac{100}{60}[\because \frac{r_0}{r}=\frac{N_0}{N}=\frac{100}{60}]$

$t=4253.96yr$.