Question

On analysis a sample of uranium ore was found to contain 0.277 g of ${}_{82}P{b}^{206}$ and 1.667 g of ${}_{92}S{i}^{238}.$ The half-life period of $U^{238}is4.51\times {10}^9$ years. If all the lead were assumed to have come from the decay of ${}_{92}S{i}^{238},$ what is the age of earth?

• A.
  $5.20\times {10}^{3}years.$
• B.
  $1.14\times {10}^{9}years.$
• C. $1.75\times {10}^{6}years.$
• D. $1.09\times {10}^{10}years.$

Answer 1

The correct option is B

$1.14\times {10}^{9}years.$

Let time = t year

${}_{92}S{i}^{238}=1.667g=\frac{1.667}{238}mol$

${}_{82}P{b}^{206}=0.277g=\frac{0.277}{206}mol$

∵ All the lead have come from decay of U.

∴ Mole of Pb formed $=\frac{0.277}{206}$

Mole of U decayed $=\frac{0.277}{206}$

∴ Total mole of uranium before decay.

$N^0=\frac{1.667}{238}mol+\frac{0.277}{206}$

Also, N for $U^{238}=\frac{1.667}{238}$

For $U^{238},t=\frac{2.303}{\lambda }lo{g}_{10}\frac{N^0}{N}$

$=\frac{2.303\times 4.51\times {10}^9}{0.693}lo{g}_{10}\frac{\frac{1.667}{238}+\frac{0.277}{206}}{\frac{1.667}{238}}$

$=1.143\times {10}^{9}years.$