Question

${}_{92}^{238}U$ by successive radioactive decays changes to ${}_{82}^{260}Pb$. A sample of uranium ore was analysed and found to contain 1.0g ${}^{238}U$ and 0.1 g ${}^{206}Pb$. Assuming that $6^{260}Pb$ has accumulated due to decay of ${}^{238}U$, find out the age of the ore? ($T_{50}$ of ${}^{238}U$ is $4.5\times {10}^9$ years).

• A. $2\times {10}^9$ years
• B. $7.1\times {10}^8$ years
• C. $4\times {10}^9$ years
• D. $6.2\times {10}^9$ years

Answer 1

The correct option is B $7.1\times {10}^8$ years

The half life period is $4.5\times {10}^9$ years. The decay constant is $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times {10}^9}=1.54\times {10}^{-10}/yr$

1.0g 238U corresponds to $\frac{1.0}{238}=0.0042$ moles.

0.1 g 206Pb corresponds to $\frac{0.1}{206}=0.000485$ moles.

$\frac{a}{a-x}=\frac{0.0042+0.000485}{0.0042}=1.1155$

The age of the ore $t=\frac{2.303}{\lambda }log\frac{a}{a-x}$

$t=\frac{2.303}{1.54\times {10}^{-10}}log1.1155=7.1\times {10}^{8}year$.