In an ore containing uranium, the ratio of `.^238U` to `.^206Pb` nuclei is `3`. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of `.^238U`. Take the half-life of `^238U` to be `4.5xx10^9` years.

Explanation

Given, `(M_(U))/(M_(Pb)) =3`
Let the initial mass of uranium be `M_(0)` .
Final mass of uranium after time t,`M =(3)/(4) M_(0)`
According to law of disitegration,
`(M)/(M_(0)) =((1)/(2))^(t//T) rArr (M_(0))/(M) =(2)^(t//T)`
`because n log_(10) (M_(0))/(M) =(t)/(T) log_(10) 2`
`t=T(log_(10).(M_(0))/(M))/(log_(10)2)`
`=4.5 xx10^(9) (0.1249/0.3010)`
`=1.867 xx 10^(9) years`.