A sample of `U^(238)` (half-line `= 4.5xx10^(9) yr`) ore is found to contain `23.8g` of `U^(238)` and `20.6g` of `Pb^(206)` Calculate the age of the ore.

Explanation

We know,
`._(92)U^(238) rarr ``._(82)Pb^(206) + 8_(2) He^(4) + 6 ``._(-1)e^(0)`
`Pb` present `= (20.6)/(206) = 0.1g` atom `= U` decayed
`U` present `= (23.8)/(238) = 0.1g` atom,
Thus, `N = 0.1g` atom
`N_(0) = U` present `+ U` decayed
`= 0.1+ 0.1 = 0.2g` atom
Now `t = (2.303)/(k) log_(10) (N_(0))/(N)`
`t = (2.303xx4.5xx10^(9))/(0.693) log_(10) (0.2)/(0.1)`
`t = 4.5xx10^(9)` year