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

`(""^(14) C)/(""^(12) C) = (1)/(16) = (N)/(N_(0)) because (N)/(N_(0)) = ((1)/(2))^(n)`
`implies (1)/(16) = ((1)/(2))^(n) implies ((1)/(2))^(4) = ((1)/(2))^(n)`
or n = 4 or `(t)/(T) = 4`
or `t = 4 xx T = 4 xx 5730 = 22920` years