Question

The ${\beta }^{-}$ activity of a sample of $C{O}_2$ prepared from a contemporary wood gave a count rate of $\text{25.5}$ counts per minute $(cpm)$ . The same mass of $C{O}_2$ from an ancient wooden statue gave a count rate of $\text{20.5}$ $cpm$ under the same conditions. If the half life of ${}^{14}C$ is $5770years$ , then the age of the statue is close to [Take ${\left(log\right)}_{10}\left(\frac{255}{205}\right)\approx 0.095$ ]

• A. $1822years$
• B. $182years$
• C. $822years$
• D. $18220years$

Answer 1

Answer: (A)

The activity of $C{O}_2$ prepared from contemporary wood is
$r_0=25.5cpm$ .
This can be considered as the initial activity of the $C{O}_2$ from the ancient wooden statue.
The present activity from the wooden statue is
$r=20.5cpm$ .
From the law of radioactive decay, we know that the activity of a sample is directly proportional to the number of active nuclei
$r\propto N\Rightarrow \frac{r_0}{r}=\frac{N_0}{N}$
The time taken by the activity of a sample to change from $r_0\text{to }r$ is $t=\frac{2.303}{\lambda }log\left(\frac{N_0}{N}\right)=\frac{2.303}{\lambda }log\left(\frac{r_0}{r}\right)$
$t=\frac{2.303\times 5770}{0.693}\text{log}\left(\frac{25.5}{20.5}\right)=1822years$