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

Note : The radio active decay follows first order kinetics. Here we take `N_(0) equiv C_(0) and N_(t) equiv C_(t) and lambda equiv k` The first order rate equation for radioactive decay is :
`lambda t = 2.303 log frac(N_(0))(N_(t))" Where "lambda= (0.693)/t_(1//2)`
`N_(0)` = initial nuclei (at t = 0)
`N_(i)` = final nuclei (at t)
`lambda` = decay constant or disintegration constant
`{:(,U^(238),rarr,Pb^(206)),(N_(0),x,,0),(N_(t),x-y,,y):}`
`rArr N_(0)/N_(t)=x/(x-y)`
Using `lambda t = 2.303 log frac(N_(0))(N_(t))`
Given : `(x-y)/y = 3 " " rArr x/(x-y) = 4/3`
`rArr t = (2.303)/(0.693)xx 4.5 xx 10^(9)log frac(4)(3)`
`rArr t = 1.85 xx 10^(9) ` years.