The half life for radioactive decay of `.^(14)C` is 5730 years. An archaeological artifact containing wood had only `80%` of the `.^(14)C` found in a living tree. Estimat the age of the sample.

Explanation

Decay constant (k) `= (0.693)/(t_(1//2)) = 0.693/5730 yr^(-1)`
Radioactive decays follow first order kinetics
`t= 2.303/k log [A_(0)]/[A], k= (0.693)/(5730) yr^(-1) [A_(0)]=100%, [A]=80%`
`t = (2.303 xx i5730)/(0.693) xx log(100/80)`
`(2.303 xx 5730 xx 0.0969)/(0.693)` yr = 1845 years.