The fossil bone has a `.^(14)C` : `.^(12)C` ratio, which is `[(1)/(16)]` 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 :

Options

A. 11460 years
B. 17190 years
C. 45840 years
D. 22921 years

Answer

D

Explanation

(d) After `n` half-lives (`Le`., at `t = nT`) the number of nuclides left undecayed,
`N = N_0 ((1)/(2))^n`
Given, `(N)/(N_0) = (1)/(16)`
`:. (1)/(16) = ((1)/(2))^n`
or `((1)/(2))^4 = ((1)/(2))^n`
Equating the powers, we obtain
`n= 4`
i.e., `(t)/(T) = 4`
or `t = 4 T`
or `t = 4 xx 5730 = 22920 years`
`(because T = 5730 years)`.