The final product of `U^(238)` is `Pb^(206)`. A sample of pitchblende contains `0.0453 g` of `Pb^(206)` for every gram of `U^(238)` present in it. Supposing that the mineral pitchblende formed at the time of formation of the earth did not contain any `Pb^(206)`, calculate the age of the earth (half-life period of `U^(238) = 4.5 xx 10^(9)` years).

Explanation

Amount of `U^(238)` now present, i.e., `a - x = 1 g`
Amount of `U^(238)` originally present
= Amount of `U^(238)` now present + Amount of `U^(238)` changed into `0.0453 g` of `Pb^(206)`
`1 g + (238)/(206) xx 0.05234` of `U^(238)` changed into `1 g` of `U^(238)`, then
`t = (2.303)/(K) "log" (a)/(a - x)`
But `K = (0.693)/(t_(1//2)) = (0.693)/(4.5 xx 10^(9)) = 1.54 xx 10^(10) "year"^(-1)`
`:. t = (2.303)/(1.54 xx 10^(-10)) "log" (1.05234)/(1) = 3.29 xx 10^(8)` years