A sample of uranium mineral was found to contain `Pb^(208)` and `U^(238)` in the ratio of 0.008 : 1. Estimate the age of the mineral (half life of `U^(238)` is `4.51 xx 10^(9)` years).

Explanation

Given `t_(1//2)` of `U^(238) = 4.51 xx 10^(9)` years
Ratio by masss of `Pb^(206) : U^(238) = 0.008 : 1`
`:.` Ratio by moles of `Pb^(206) : U^(238) = (0.008)/(206) : (1)/(238) = 0.0092`
We know that
`K = (2.303)/(t) log (N_(0))/(N)`
`(0.693)/(t_(1//2)) = (2.303)/(t) log ((U^(238) + Pb^(206))/(U^(238)))` `[k = (0.693)/(t_(1//2))]`
`(0.693)/(t_(1//2)) = (2.303)/(t) log (1 + (Pb^(206))/(U^(238)))`
Substituting all values,
`(0.693)/(4.51 xx 10^(9)) = (2.303)/(t) log (1 + 0.0092)`
`:. t = (2.303 xx 4.51 xx 10^(9))/(0.692) log (1.0092)`
`= 0.05945 xx 10^(9)` years
`= 5.945 xx 10^(7)` years