Question

${}_{92}^{238}U$ is radioactive and it emits $\alpha$ and $\beta$-particles to form ${}_{82}^{206}Pb$. An ore of ${}_{92}^{238}U$ is found to contain ${}_{92}^{238}U$ and ${}_{82}^{206}Pb$ in the weight ratio of 1 : 0.1. The half-life period of ${}_{92}^{238}U$ is $4.5\times {10}^9$ years. Calculate the age of the ore?

• A. $=7.1\times {10}^8$ years
• B. $=7.7\times {10}^8$ years
• C. $=9.1\times {10}^8$ years
• D. None of these

Answer 1

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

$[1+\frac{\left[pb\right]}{\left[U\right]}]=1+\left(\frac{\frac{0.1}{206}}{\frac{1}{238}}\right)=1.1155=(2{)}^y$
$y=\frac{t}{t_{50}}$
y log 2 = log 1.1155$y=0.1578$
$\therefore t=4.5\times {10}^9\times 0.1578$
$=7.1\times {10}^8$ years