Question

${}_{921}{U}^{238}$ by successive radioactivity decays changes to ${}_{82}P{b}^{206}$.A sample of uranium ore was analyzed and found to contain 1.0g 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 $U^{238}=4.5\ast {10}^{9}years).$

• A. $t=7.1\ast {10}^5$ years
• B. $t=7.1\ast {10}^8$ years
• C. $t=7.1\ast {10}^6$ years
• D. $t=7.1\ast {10}^3$ years

Answer 1

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

Number of moles of U-$238$ present $=\frac{1.0}{238}=0.0042$

Number of moles of Pb-$206$ present $=\frac{0.1}{206}=0.000485$

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

$t=\frac{2.303}{\lambda }log\frac{a}{a-x}$

$t=\frac{2.303}{0.693}\times 4.5\times {10}^9\times log\frac{0.0042+0.000485}{0.0042}$

$t=7.1\times {10}^8$ years