Question

The $\beta ^{-}$ activity of a sample of $CO_{2}$ prepared from a contemporary wood gave a count rate of $\text{25.5}$ counts per minute $\left(\right.cpm\left.\right)$ . The same mass of $CO_{2}$ 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 $\left(log\right)_{10}\left(\frac{255}{205}\right)\approx0.095$ ]

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

Answer 1

Answer: (A)

The activity of $CO_{2}$ prepared from contemporary wood is
$r_{0}=25.5 \, cpm$ .
This can be considered as the initial activity of the $CO_{2}$ 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 $t=\frac{2 .303}{\lambda }log \left(\frac{N_{0}}{N}\right)=\frac{2 .303}{\lambda }log⁡\left(\frac{r_{0}}{r}\right)$
$t=\frac{2 .303 \times 5770}{0 .693}\text{log} \left(\frac{25 .5}{20 .5}\right)=1822years$