In an ore containing Uranium, the ratio of `U^(238)` to `Pb^(206` nuceli is `3`. Calculate the age of the ore, assuming that alll the lead present in the ore is the final stable, product of `U^(238)`. Take the half-like of `U^(238)` to be `4.5 xx 10^(9)` years. In `(4//3) = 0.288`.

Explanation

Suppose x is the number of `Pb^(206)` nulei. The number of `U^(238)` nuclei will be 3x. Thus
`3x+x=N_(0)`
We know that `N=N_(0)e^(-lamdat)`
or `3x=4xe^(-lamdat)`
`:.e^(lamdat)=4/3`
or `t=(In 4//3)/(lamda)=(In 4//3)/((0.693//t_(1//2)))`
`=(In 4//3)/((0.693/4.5xx10^(9)))`
`=1.868xx10^(9)` year.