`U^(238)` is found to be in secular equilibrium with `Ra^(226)` on its ore. If chemical analysis shows 1 nuclei of `Ra^(226)` per `3.6xx10^(-6)` nuclei of `U^(238)`, find the half-life of `U^(238)`. Given the half-life is 1500 years.

Explanation

According to secular equilibrium.
`lamda_(U)N_(U)=lamda_(Ra)N_(Ra)`
where, `lamda_(U)=` decay constant of uranium
`lamda_(Ra)=` decay constant of Ra
`N_(U)`= number of nuclei of U
`N_(Ra)` = number of nuclei of Ra
We know,
`Decay constant (lamda)=(In2)/("Half time"(t_(1//2)))`
`rArr" "(In2)/((t_(1//2))_(U))N_(U)=(In2)/((t_(1//2))Ra)N_(Ra)`
`rArr" "(3.6xx10^(6))/((t_(1//2))_(U))=(1)/(1500)`
`rArr" "(t_(1//2))_(U)=3.6xx10^(6)xx1500`
`" "=5400xx10^(6)=5.4xx10^(9)yr`