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

`{:(N_(0),0,t=0),(U^(238)rarr,Pb^(206),),(N,N_(0)-N,t=t):}`
At `t = t, (N)/(N_(0)-N) = 3 rArr N = 3N_(0) - 3N`
`N = (3N_(0))/(4)`
`N_(0)e^(-lambdat) = (3N_(0))/(4) rArr e^(-lambdat) = (3)/(4)`
`e^(lambdat) =(3)/(4) rArr lambdat = ln((4)/(3))`
`((ln2))/(T_(1//2))t = ln((4)/(3))`
`t = (ln(4//3))/(ln2)T_(1//2) = (0.288xx4.5xx10^(9))/(0.693)`
`= 1.87 xx 10^(9)` year