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.

Answer

A

Explanation

`(U^238)/(Pb^206)=3/1`
`N_0=3+1=4`
`N=3`
`N=N_0e^(-lambdat)` …(i)
`lambda=(1n2)/(t_(1//2)` …(ii) From Eqs. (i) and (ii), we get
`t=1.88xx10^9 yr`