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, $\frac{N}{N_0} = \frac{1}{16}$
$∴ \frac{1}{16}= (\frac{1}{2})^n$
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 $t = 4 × 5730= 22920\,$ years
$(\because T = 5730\,{\text{ years}})$