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. `22920` years
D. `45840` years

Answer

c

Explanation

After `n` half-lives (i.e., at `t=nT`) the number of nuclides left undecayed
`N=N_(0)((1)/(2))^(n)`
Given, `(N)/(N_(0))=(1)/(16)implies(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= 4T`
or `t=4xx5730= 22920 years`
`( because T= 5730 years)`