The half-life period of `C^(14)` is 5760 years. A piece of woods when buried in the earth had `1% C^(14)`. Now as charcoal it has only `0.25% C^(14)`. How long has the piece of wood been buried?

Explanation

`t_(1//2)` of `C^(14) = 5760`
`:. Lambda = (0.693)/(5760) "year"^(-1)`
`N_(0C)^(14) = 1%`
`N_(C^(14) = 0.25%`
`:.t = (2.303)/(lambda) "log" (N_(0))/(N)`
`= (2.303 xx 5760)/(0.693) "log" (1)/(2) = (2.303 xx 5760)/(0.693) "log" 2^(2)`
`t = 11520` year