Question

The fossil bone has a ${}^{14}C:{}^{12}C$ ratio, which is $\left(\frac{1}{16}\right)$ of that in a living animal bone. If the half-left time of ${}^{14}C$ is $5730$ years, then the age of the fossil bone is:

• A. $11460$ years
• B. $17190$ years
• C. $22920$ years
• D. $45840$ years

Answer 1

The correct option is C $22920$ years

The fossil bone has a ${}^{14}C{:}^{12}C$ ratio, which is $\left(\frac{1}{16}\right)$ of that in a living animal bone.
After one half life period the ratio will be $\left(\frac{1}{2}\right)$
After $n$ half life periods, the ratio will be $\left(\frac{1}{2^n}\right)$
Hence,
$\left(\frac{1}{16}\right)=\left(\frac{1}{2^n}\right)$
But $\left(\frac{1}{16}\right)=\left(\frac{1}{2^4}\right)$
Hence, $\left(\frac{1}{2^4}\right)=\left(\frac{1}{2^n}\right)$
or $n=4$
One half life period corresponds to 5730 years.
4 half life periods corresponds to $4\times 5730=22920$ years.
The age of the fossil bone is $22920$ years.