There are 0.618 `mu` g of `""^(206)Pb` and 0.238 `mu` g of `""^(238)U` in a rock. If `T_(50)` of `""^(238)` U is `1.5 xx 10^(9)` yr. Age of the rock is

Options

A. `1.5 xx 10^(9)` yr
B. `3.0 xx 10^(9)` yr
C. `4.5 xx 10^(9)` yr
D. `0.75 xx 10^(9)` yr

Answer

B

Explanation

Basd on rock-dating ` (1 + ([Pb])/([U]) = (2)^(n)), (1 + ((0.628xx10^(-6))/(206))/((0.238xx10^(-6))/(238))) = 2n^(2)`
`(2)^(2) = (2)^(n) " "therefore n = 2, " " 2 = (t)/(T_(50))" " t = 2 xx T_(50) = 3 xx 10^(9)` yr
OR
`t = (2.303)/(0.693) xx 1.5 xx 10^(9) log ""((0.952)/(0.238))`