The half-life for decay of `""^(14)C` by `beta`-emission is 5730 years. The fraction of `""^(14)C` decays, in a sample that is 22,920 years old, would be

Options

A. `1//8`
B. `1//16`
C. `7//8`
D. `15//16`

Answer

D

Explanation

`N/(N_0) = (1/2)^(n)`
Where, `N_0` = Initial amount, N = Amount left
`n = t/(t_(1//2))`
Given , `t_(1//2) = (22,920)/(5730) = 4 , N/(N_0) = (1/2)^(4) = 1/16 implies N = (N_0)/(16)`
The fraction of `""^(14)C` that decayed = `N_0 - N`
`= N_0 - (N_0)/(16) = 15/16 N_0`