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

The specific rate constant for the first-order radioactive decay of `.^(14)C` can be found from the half-life :
`k = (0.693)/(t_(1)//2)= (0.693)/(5730 yr)`
`= 1.20xx10^(-4) yr^(-1)`
Since k, `[R]_(0)` and `[R]_(t)` are know, the time elapse since the death of the tree can be found by the first order equation.
`t= (2/303)/(k) log(([R]_(0))/([R]_(t)))`
Since `[R]_(0) prop [W]_(0)` and `[R]_(t) prop [W_(t)]`, we have
`t= (2.303)/(k)log (([W]_(0))/([W]_(t)))`
Taking `[W_(0)]` as `100% [W]_(t)` becomes `80%. Thus
`T= (2.303)/(1.20xx10^(-4)yr(-1)) log (100)/(80)`
`= (2.303)/(1.20xx10^(-4)yr^(-1)) (0.0969)`
`= 1859` years .