A rock is `1.5 xx 10^(9)` years old. The rock contains `.^(238)U` which disintegretes to form `.^(236)U`. Assume that there was no `.^(206)Pb` in the rock initially and it is the only stable product fromed by the decay. Calculate the ratio of number of nuclei of `.^(238)U` to that of `.^(206)Pb` in the rock. Half-life of `.^(238)U` is `4.5 xx 10^(9). years. `(2^(1/3) = 1.259)` .

Answer

`3.861`

Explanation

Let `N_(0)` be the initial number of nuclei of `.^(238)U`.
After time t, `N_(U)=N_(0) (1/2)^(n)`
Here n= number of half-lives
`=t/t_(1//2)=(1.5xx10^(9))/(4.5xx10^(9))=1/3 implies N_(U)=N_(0)(1/2)^(1/3)` and
`N_(Pb)=N_(0)-N_(U)=N_(0) [1-(1/2)^(1//3)]`
`:. N_(U)/N_(Pb)=((1/2)^(1//3))/(1-(1/2)^(3))=3.861`