^ 238 U decays to ^ 206 Pb with a half-life of 4.47 × 10^ 9 y...

Let $N$ be the present number of ^{238}\text{U}$atoms and$N_{Pb}$be the present number of ^{206}\text{Pb}$ atoms.

The number of moles of ^{238}\text{U}$is$n_U = \dfrac{2.00 \times 10^{-3}}{238}$<br><br>The number of moles of ^{206}\text{Pb}$ is $n_{Pb} = \dfrac{0.600 \times 10^{-3}}{206}$

Since all the lead originated from the uranium, the initial number of moles of ^{238}\text{U}$was:<br><br>$n_0 = n_U + n_{Pb}$<br><br>The ratio of the initial amount of uranium to the present amount is:<br><br>$\dfrac{n_0}{n_U} = 1 + \dfrac{n_{Pb}}{n_U} = 1 + \dfrac{\dfrac{0.600}{206}}{\dfrac{2.00}{238}}$<br><br>$\dfrac{n_0}{n_U} = 1 + \dfrac{0.600 \times 238}{2.00 \times 206} = 1 + \dfrac{142.8}{412} = 1 + 0.3466 = 1.3466$<br><br>According to the radioactive decay law,$n_U = n_0 e^{-\lambda t}$, which gives:<br><br>$t = \dfrac{1}{\lambda} \ln \left(\dfrac{n_0}{n_U}\right) = \dfrac{T_{1/2}}{\ln 2} \ln\left(\dfrac{n_0}{n_U}\right)

Substituting the given values:

$t =$\dfrac{4.47 \times 10^{9}}{0.693} \ln($1.3466)$

$t =$\dfrac{4.47 \times 10^{9}}{0.693} \times$0.2976 \approx 1.919 \times 10^{9} \text{ y}$

Rounding to three significant figures, the life of the rock is $1.92 \times 10^{9} \text{ y}$.