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)` .

Explanation

`.^(238)U` atoms decay to form `.^(206)Pb` atoms. Let `N_(0)` be initial number of `U `atoms. After time `t`, let `N_(U)` be the number of (U) atoms left. Then
`(N_(U))/(N_(0))=((1)/(2))^(n)`, where `n` is number of half-lives
`n=(t)/(T) =(1.5 xx 10^(9) years)/(4.5 xx 10^(9)years) - (1)/(3)`
`because N_(U) =((1)/(2))^(1//3) N_(0)`
Numbers of Pb atoms, `N_(Pb) =N_(0)-N_(U)`
Hence, required ratio is
`R=(N_U /N_Pb) = (1/(2^(1//3)) -1)=(1/1.259 -1)=(1/0.259)=3.86`.