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

Answer 1

Answer: (C)

After $n$ half-lives (i.e., at $t=nT$) the number of nuclides left undecayed,
$N=N_0(\frac{1}{2}{)}^n$
Given, $\frac{N}{N_0}=\frac{1}{16}$
$\therefore \frac{1}{16}=(\frac{1}{2}{)}^n$
or $(\frac{1}{2}{)}^4=(\frac{1}{2}{)}^n$
Equating the powers, we obtain
$n=4$
i.e., $\frac{t}{T}=4$
or $\frac{t}{T}=4$
or $t=4\times 5730=22920$ years
$(\because T=5730\text{ years})$