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-life of ${}^{14}C$ is $5730$ years, then the age of the fossil bone is

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

Answer 1

Answer: (C)

$22920years$

$\frac{{}^{14}C}{{}^{12}C}=\frac{1}{16}=\frac{N}{N_0}$

$\because$ $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow \frac{1}{16}={\left(\frac{1}{2}\right)}^n$ $\Rightarrow {\left(\frac{1}{2}\right)}^4={\left(\frac{1}{2}\right)}^n$

$\Rightarrow n=4$
$\Rightarrow \left(\frac{1}{T}\right)=4$
$t=4\times T=4\times 5730=22920years$