Question

Estimate the age of a rock which is found to contain $5\times {10}^{-5}$ mL of helium at STP and $3\times {10}^{-7}$ g of uranium. ($T_{50}$ of ${}_{92}^{238}U$ is $4.51\times {10}^9$ yr and each ${}_{92}^{238}U$ is converted into ${}_{82}^{206}Pb$
Plan ${}_{92}^{238}U{⟶}_{82}^{206}Pb+6\beta +8\alpha \left(helium\right)$
We take the equation $\left[\frac{\left[Pb\right]}{\left[U\right]}+1\right]=(2{)}^y$
where $\frac{\left[Pb\right]}{\left[U\right]}$ is the molar ratio, $y=\frac{t}{T_{50}}$

• A. $1.30\times {10}^9$ yr
• B. $7.20\times {10}^9$ yr
• C. $5.30\times {10}^9$ yr
• D. None of these

Answer 1

The correct option is A $1.30\times {10}^9$ yr

Number of moles of He at STP
$=\frac{5\times {10}^{-5}}{22400}$ mol
$\left[Pb\right]=\frac{1}{8}\left[He\right]=\frac{5\times {10}^{-5}}{8\times 22400}$ mol
$=2.79\times {10}^{-10}$ mol
$\left[U\right]=\frac{3\times {10}^{-7}}{238}=12.6\times {10}^{-10}$ mol
$[\frac{2.79\times {10}^{-10}}{12.6\times {10}^{-10}}+1]=(2{)}^y$
$1.2214=(2{)}^y$
taking log, we have $y=0.2886$
but $y=totalnumberofhalf-life=\frac{t}{T_{50}}$
$t=y\times T_{50}$
$=0.2886\times 4.5\times {10}^9$
$=1.30\times {10}^9$ yr