Question

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{Pb}\) for every \(1.000 \mathrm{mg}\) \(^{238} \mathrm{U}\) present Assuming that no lead was originally present, that all the \(^{206} \mathrm{Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206} \mathrm{Pb}\) is negligible, calculate the age of the \(\text { rock. For }^{238} \mathbf{U}, t_{1 / 2}=4.5 \times 10^{9} \text { years. }\)

Short answer

The age of the rock is approximately \(2.96 \times 10^9\) years.

Step-by-step solution

Write down the given information

We are given the following information: $$\frac{^{206}\text{Pb}}{^{238}\text{U}} = 0.688$$ Half-life of \(^{238}\text{U}\): \(t_{1/2}=4.5 \times 10^9\) years

Use half-life formula to find the ratio between remaining isotopes

The formula relating the amount of remaining isotopes to elapsed time and half-life is: $$N_t = N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}$$ Here, \(N_t\) is the amount of remaining isotope at time \(t\), \(N_0\) is the initial amount, and \(t_{1/2}\) is the half-life. We want to find the ratio between the remaining amount of \(^{238}\text{U}\) and the decayed \(^{206}\text{Pb}\), so we can write: $$ \frac{N_\text{Pb}(t)}{N_\text{U}(t)} = \frac{N_0 - N_\text{U}(t)}{N_\text{U}(t)}$$

Solve for elapsed time

Now we can plug in the given information into the equation: $$ \frac{N_\text{Pb}(t)}{N_\text{U}(t)} = 0.688 = \frac{N_0 - N_\text{U}(t)}{N_\text{U}(t)}$$ Using the half-life formula, we can rewrite the equation in terms of remaining \(^{238}\text{U}\): $$ 0.688 = \frac{N_0-N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}}{N_0\left(\frac{1}{2}\right)^{t/t_{1/2}}}$$ At this point, the initial amount of isotope \(N_0\) cancels out: $$ 0.688 = \frac{1-\left(\frac{1}{2}\right)^{t/t_{1/2}}}{\left(\frac{1}{2}\right)^{t/t_{1/2}}}$$ Now we can solve for \(t\), the age of the rock: $$ 0.688\left(\frac{1}{2}\right)^{t/t_{1/2}} + \left(\frac{1}{2}\right)^{t/t_{1/2}} = 1$$ $$\left(\frac{1}{2}\right)^{t/t_{1/2}} = \frac{1}{1.688}$$ Taking the logarithm on both sides: $$\frac{t}{t_{1/2}}\log\left(\frac{1}{2}\right) = \log\left(\frac{1}{1.688}\right)$$ And finally, solving for \(t\): $$t = t_{1/2}\frac{\log\left(\frac{1}{1.688}\right)}{\log\left(\frac{1}{2}\right)}$$

Calculate the age of the rock

Now we can plug in the half-life of \(^{238}\text{U}\) and find the age of the rock: $$t = 4.5 \times 10^9 \text{ years }\frac{\log\left(\frac{1}{1.688}\right)}{\log\left(\frac{1}{2}\right)}$$ We find that: $$t \approx 2.96 \times 10^9 \text{ years}$$ So, the age of the rock is approximately \(2.96 \times 10^9\) years.

Key concepts

Uranium-Lead Dating

Uranium-Lead dating is a classic method of geochronology, which is the study of the age of geological formations. This technique involves comparing the ratio of uranium isotopes to the quantity of their lead decay products in a sample. Uranium-lead dating is reliable because it is based on the predictable decay rates of radioactive isotopes of uranium. Uranium-238 decays to lead-206, and as this decay occurs at a known rate, it provides a ‘clock’ that can be used to measure the age of rocks.

When using this method, assumptions are made such as: all the lead present has come from the decay of uranium, no lead was present when the rock was formed, and the system has remained closed to both uranium and lead since formation. These assumptions are critical to the precision of the dating method. One way the age of a rock is calculated is by measuring the ratio of lead-206 to uranium-238 and using the known half-life of uranium-238.

Half-Life Calculation

The concept of a half-life is essential in the field of radiometric dating, which includes uranium-lead dating. The half-life of a radioactive isotope is the time required for half of the isotope in a sample to decay. For uranium-238, this half-life is quite long, about 4.5 billion years, which makes it suitable for dating extremely old rocks.

The calculation of a rock’s age using half-life involves complex mathematics, including logarithms, as shown in the solution for the exercise. Understanding half-life calculations helps us obtain a specific measure of elapsed time since the rock's formation by indicating how many half-lives have passed and, therefore, how much original uranium remains now compared to its decay products.

Geochronology

Geochronology is the science of determining the age of rocks, fossils, and sediments using signatures inherent in the rocks themselves. The primary tools used in this field are methods of radiometric dating which measure the concentrations of radioactive isotopes and their decay products. Through these methods, geochronologists can construct a timeline of Earth’s history and provide a window into past climates, biological evolution, and the formation and alteration of rock units.

Uranium-lead dating plays a significant role in this field by dating zircon crystals found in sedimentary strata. These tiny zircon crystals resist weathering and can be dated to provide an accurate age of the strata in which they are found, thus giving clues about Earth's geological history.

Isotope Ratio

The isotope ratio is an expression of the relative abundance of isotopes in a sample. This ratio can provide a wealth of information about the origin and age of rocks and minerals. In radiometric dating, scientists measure the ratio of different isotopes, using devices like mass spectrometers, to infer ages of various specimens.

For instance, the exercise illustrates the use of the uranium-lead isotope ratio to determine the age of a rock. By observing the current ratios and knowing the rates at which uranium isotopes decay into lead isotopes, researchers can backtrack to find out how long the decay has been occurring. This method, when applied correctly, allows us to peak back into geological time and determine the absolute ages of rocks with considerable accuracy.