Question

${}_{92}{\text{U}}^{238}$ by successive radioactive decays changes to ${}_{82}{\text{Pb}}^{206}$. A sample of uranium ore was analyzed and found to contain 1.0 g of $U^{238}$ and 0.1 g of $P{b}^{206}$. Assuming that all the $P{b}^{206}$ had accumulated due to decay of $U^{238}$, find out the age of the ore. (Half life of $U^{238}=4.5\times {10}^{9}years$)

• A. $t=3.5\times {10}^{8}years$
• B. $t=7.1\times {10}^{8}years$
• C. $t=14.2\times {10}^{8}years$
• D. $t=7.1\times {10}^{7}years$

Answer 1

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

Total number of moles of $U^{238}$ present at time $t=\frac{1}{238}=0.0042$ moles.

Total number of moles of $P{b}^{206}$ present at time $t=\frac{0.1}{206}=0.000485$

$a=0.0042+0.000485=0.004687$

$a-x=0.0042$

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times {10}^9}=1.54\times {10}^{-10}$

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

$t=\frac{2.303}{1.54\times {10}^{-10}}log\frac{0.004687}{0.0042}=7.1\times {10}^{8}years$