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

`0.259`

Explanation

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