# A rock contains 0.688 mg of "^206Pb of every 1000 mg of "^238U. How do you calculate the age of the rock?

## Assume that no lead was originally present, that all the "^206Pb formed over the years has remained in the rock, and that the number of nuclides in the intermediate stages of decay between "^206Pb and "^238U is negligible. The exact mass of "^206Pb is 205.974449 g/mol and the exact mass of "^238U is 238.050784 g/mol.

Jan 11, 2017

You can do it like this:

#### Explanation:

The ratio of the amounts of U 238 and Pb 206 in a rock sample enables the age of the rock to be estimated using the technique of radiometric dating.

U 238 forms a decay chain in which it undergoes a sequence of 8 $\alpha$ and 6 $\beta$ decays: It moves back in the periodic table until the isotope falls in the band of stability at Pb 206.

Each step has its own individual half - life but the first decay to Th 234 is about 20,000 times slower than the other decay steps.

Those of you who are familiar with chemical kinetics will know that it is the slowest step in a mechanism which determines the overall rate of reaction, the so - called "rate determining step".

This is the case here in the conversion of U 238 to Pb 206.

As the U 238 decays exponentially, the amount of Pb 206 grows correspondingly: The half - life of U 238 is about 4.5 billion years. As time passes, the ratio of Pb 206 to U 238 will increase and it is this which enables the age of the rock to be estimated.

We need to do some "math" to show this:

Radioactive decay is a first order process:

$\textsf{{U}_{t} = {U}_{0} {e}^{- \lambda \text{t}}}$

$\textsf{{U}_{0}}$ is the number of undecayed U 238 atoms initially.

$\textsf{{U}_{t}}$ is the number of undecayed U 238 after time $\textsf{t}$.

$\textsf{\lambda}$ is the decay constant.

Since the decay of 1 U 238 atom will result in the formation of 1 atom of Pb 206 we can say that:

$\textsf{{U}_{0} = {U}_{t} + P {b}_{t}}$

Where $\textsf{P {b}_{t}}$ is the number of Pb 206 atoms formed after time $\textsf{t}$.

The decay equation can therefore be written:

$\textsf{{U}_{t} = \left({U}_{t} + P {b}_{t}\right) {e}^{- \lambda \text{t}}}$

$\therefore$$\textsf{\frac{{U}_{t}}{\left({U}_{t} + P {b}_{t}\right)} = {e}^{- \lambda \text{t}}}$

$\therefore$sf((U_(t)+Pb_(t))/(U_(t))=e^(lambdat)

$\therefore$$\textsf{1 + \frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}}}$

$\therefore$$\textsf{\frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}} - 1}$

The half - life, $\textsf{{t}_{\frac{1}{2}}}$, of U 238 is $\textsf{4.468 \times {10}^{9}}$ years.

We can get the value of the decay constant from the expression:

sf(lambda=0.693/(t_(1/2))

$\textsf{\lambda = \frac{0.693}{4.468 \times {10}^{9}} = 0.1551 \times {10}^{- 9} {\text{ " "yr}}^{- 1}}$

We can get the number of moles of $\textsf{n}$ each isotope by dividing the mass by the mass of one mole, which is the $\textsf{{A}_{r}}$ in grams:

$\textsf{{n}_{P {b}_{t}} = \frac{0.688 \times {10}^{- 3}}{205.974449} = 3.33402 \times {10}^{- 6}}$

$\textsf{{n}_{{U}_{t}} = \frac{1000 \times {10}^{- 3}}{238.050784} = 4.20078 \times {10}^{- 3}}$

There is no need to convert these into a number atoms by multiplying by the Avogadro Constant, since we are interested in the Pb 206 / U 238 ratio so this would cancel anyway.

$\therefore$$\textsf{\frac{3.33402 \times {10}^{- 6}}{4.20078 \times {10}^{- 3}} = {e}^{\lambda \text{t}} - 1}$

$\therefore$$\textsf{0.000793666 = {e}^{\lambda \text{t}} - 1}$

$\therefore$$\textsf{{e}^{\lambda \text{t}} = 1.000793666}$

Taking natural logs of both sides:

$\textsf{\lambda \text{t} = \ln \left(1.000793666\right) = 7.935512 \times {10}^{- 4}}$

$\therefore$$\textsf{t = \frac{7.935512 \times {10}^{- 4}}{0.1551 \times {10}^{- 9}} \text{ " "yr}}$

$\textsf{t = 5.11 \times {10}^{6} \text{ ""yr}}$

or 5.11 million years.