An old piece of wood has 25.6% as much `C^(14)` as ordinary wood today has. Find the age of the wood. Half-life period of `C^(14)` is 5760 years?

Explanation

Suppose the amount of `C^(14)` present in the woodk originally (i.e., the which the wood today has) `= a`
Then the amount of `C^(14)` present now in the old wood
`(a -x) = (25.6)/(100) a = 0.256a`
The time `t` which `C^(14)` changed from `a` to `0.256a` will then be given by
`t = (2.303)/(K) log (a)/(0.256a)`
But ` K = (0.693)/(t_(1//2)) = (0.693)/(5760) = 1.203 xx 10^(-4) "year"^(-1)`
`:. t = (2.303)/(1.202 xx 10^(-4)) log (1)/(0.256)`
`= 11329` years