In a Uranium ore the ratio of $$U^{238}$$ nuclei to $$Pb^{206}$$ nuclei is $$ \eta = 2.8$$. Evaluate the age of the ores, assuming all the lead $$Pb^{206} $$ to be a final decay product of the uranium series. The half life of $$U^{238} $$ nuclei is $$4.5 \times 10^{9} $$ years.
What this implies is that in the time since the ore was formed, $$ \dfrac{\eta}{1 + \eta} U^{238} $$ nuclei have remained undecayed. Thus
$$ \dfrac{\eta}{1 + \eta} = e^{-t \times \dfrac{ln\ 2}{T_{1/2}}} $$ Taking logarithm both side , we get
or $$ t = T_{\dfrac{1}{2}} \dfrac{ln \dfrac{1 + \eta}{\eta}}{ln\ 2} $$
Substituting $$ T_{\dfrac{1}{2}} = 4.5 \times 10^{9} $$ years , $$ \eta = 2.8 $$
we get $$ t = 1.98 \times 10^{9} $$ years