Question

In an ore containing uranium, the ratio of $U^{238}$ to $P{b}^{206}$ is $3$. Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of $U^{238}$. Take the half-life of $U^{238}$ to be $4.5\times {10}^9$ yr.

• A. $1.6\times {19}^{3}yr$
• B. $1.5\times {10}^{4}yr$
• C. $1.867\times {10}^{9}yr$
• D. $2\times {10}^{5}yr$

Answer 1

Answer: (C)

$1.867\times {10}^{9}yr$

Let the initial mass of uranium be $M_0$
Final mass of uranium after time $t,M=\frac{3}{4}{M}_0$
According to the law of radioactive disintegration.
$\frac{M}{M_0}={\left(\frac{1}{2}\right)}^{t/T}\Rightarrow \frac{M_0}{M}=(2{)}^{t/T}$
$\therefore {\log }_{10}\left(\frac{M_0}{M}\right)=\frac{t}{T}{\log }_{10}\left(2\right)$
$t=T\frac{{\log }_{10}\left(\frac{M_0}{M}\right)}{{\log }_{10}\left(2\right)}=\frac{T{\log }_{10}\left(\frac{4}{3}\right)}{{\log }_{10}\left(2\right)}$
$=\frac{T{\log }_{10}\left(1.333\right)}{{\log }_{10}\left(2\right)}=4.5\times {10}^9\left(\frac{0.1249}{0.3010}\right)$
$\Rightarrow t=1.867\times {10}^{9}yr$.