Question

A sample of $U^{238}$ ( half life $4.5\times {10}^9$ yr) ore is found to contain 23.8 g of $U^{238}$ and 20.6 g of $P{b}^{206}.$ Calculate the age of the ore?

• A. $4.5\times {10}^{9}years$
• B. $6.13\times {10}^{6}years$
• C. $61.3\times {10}^{6}years$
• D. None f these

Answer 1

The correct option is A $4.5\times {10}^{9}years$

3.8 g of $U-238$ corresponds to 0.1 mole.
20.6 g of $Pb-206$ corresponds to 0.1 mole
Thus at time $t$, 0.1 moles of U-238 and 0.1 moles of Pb-206 are present.
Initially 0.2 moles of U-238 and 0 moles of Pb-206 will be present. Hence in time t, one half of U-238 decays. Hence, the age of the ore is equal to the half life period. It is $4.5\times {10}^9$ years.