Question

The ${\beta }^{-}-activity$ of a sample of $C{O}_2$ prepared from a contemporary wood gave a count rate of $25.5$ counts per minute (c.p.m.). The same mass of $C{O}_2$ from an ancient wooden statue gave a count rate of $20.5$ cpm, in the same counter condition. Calculate its age to the nearest $50year$ taking $t_{1/2}$ for ${}^{14}C$ as $5770$ year. What would be the expected count rate of an identical mass of $C{O}_2$ from a sample which is $4000$ year old?

Answer 1

$r=20.5cpm,r_0=25.5cpm$.

$\because r_0\propto N_0$ and $r\propto N$

$\therefore \frac{r_0}{r}=\frac{N_0}{N}$

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

$t=\frac{2.303\times 5770}{0.693}\log \frac{25.5}{20.5}=1817.5years$

Also, if $t=4000yr$ then

$4000=\frac{2.303\times 5770}{0.693}\log \frac{25.5}{r}$

$\therefore r=15.77cpm$.