Question

There are $0.618\mu g$ of ${}^{206}Pb$ and $0.238\mu g$ of ${}^{238}U$ in a rock. If $T_{50}$ of ${}^{238}U$ is $1.5\times {10}^{9}yr$, age of the rock is:

• A. $1.5\times {10}^{9}yr$
• B. $7.5\times {10}^{9}yr$
• C. $15\times {10}^{9}yr$
• D. $3\times {10}^{9}yr$

Answer 1

The correct option is D $3\times {10}^{9}yr$

As given,

$\left[Pb\right]=\frac{0.618\times {10}^{-6}}{206}mol=3\times {106}^{-9}mol$

$\left[U\right]=\frac{0.238\times 106-6}{238}mol=1\times {10}^{-9}mol$

For rock-dating, $(1+\frac{\left[Pb\right]}{\left[U\right]})=(2{)}^y$

$\therefore y=2=\frac{t}{T_{50}}$

$\therefore t=3.0\times {10}^{9}years$