In a Uranium ore the ratio of $U^{238}$ nuclei to $P b^{206}$ nuclei is $η = 2.8$ . Evaluate the age of the ores, assuming all the lead $P b^{206}$ to be a final decay product of the uranium series. The half life of $U^{238}$ nuclei is $4.5 \times 1 0^{9}$ years.

Question

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Answer 1

What this implies is that in the time since the ore was formed, $\frac{η}{1 + η} U^{238}$ nuclei have remained undecayed. Thus

$\frac{η}{1 + η} = e^{- t \times \frac{l n 2}{T _{1 / 2}}}$

Taking logarithm both side , we get

or $t = T_{\frac{1}{2}} \frac{l n \frac{1 + η}{η}}{l n 2}$

Substituting $T_{\frac{1}{2}} = 4.5 \times 1 0^{9}$ years , $η = 2.8$

we get $t = 1.98 \times 1 0^{9}$ years

Answer 2

What this implies is that in the time since the ore was formed,

\frac{η}{1 + η} U^{238}

nuclei have remained undecayed. Thus

\frac{η}{1 + η} = e^{- t \times \frac{l n 2}{T _{1 / 2}}}

Taking logarithm both side , we get

or

t = T_{\frac{1}{2}} \frac{l n \frac{1 + η}{η}}{l n 2}

Substituting

T_{\frac{1}{2}} = 4.5 \times 1 0^{9}

years ,

η = 2.8

we get

t = 1.98 \times 1 0^{9}

years