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 rock. (For \({ }^{238} \mathrm{U}\), \(t_{1 / 2}=4.5 \times 10^{9}\) years. \()\)

Short answer

The age of the rock is approximately 3.64 billion years.

Step-by-step solution

Write down the half-life formula

The half-life formula relates the amount of the original substance (N₀) remaining after some time (t) to the half-life time (\(t_{1/2}\)) and the decay constant (λ) for the substance. The formula is: \[ N_t = N_0 \cdot e^{-\lambda t} \] Here, \(N_t\) is the amount of the substance remaining after time \(t\). Note that the decay constant, λ, can be calculated from the half-life time using the formula: \[ \lambda = \frac{ln(2)}{t_{1/2}} \]

Set up an equation involving the given data

We are given that the ratio of \({ }^{206} \mathrm{~Pb}\) to \({ }^{238} \mathrm{U}\) in the rock is \(0.688:1\). We also know that \({ }^{238} \mathrm{~U}\) eventually decays into \({ }^{206} \mathrm{~Pb}\), and the half-life of \({ }^{238} \mathrm{U}\) is \(4.5 \times 10^9\) years. Given that there was no lead initially present in the rock, let's denote the initial amount of \({ }^{238} \mathrm{~U}\) as \(N_0\). Then, \(N_t = N_0 - 0.688N_0\), as the \(0.688N_0\) amount of \({ }^{238} \mathrm{~U}\) has decayed into \({ }^{206} \mathrm{~Pb}\). Now we can write an equation involving λ, \(N_0\), \(N_t\), and \(t\). \[ N_t = N_0 \cdot e^{-\lambda t} \] \[ N_0 - 0.688N_0 = N_0 \cdot e^{-\lambda t} \]

Find the decay constant λ

Using the half-life formula for decay constant, we can calculate λ: \[ \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 \times 10^9} \]

Solve the equation for t

Now we can substitute the λ value into the equation and solve for time, \(t\): \[ N_0 - 0.688N_0 = N_0 \cdot e^{-\frac{ln(2)}{4.5 \times 10^9} t} \] \[ 1 - 0.688 = e^{-\frac{ln(2)}{4.5 \times 10^9} t} \] \[ ln \left( \frac{312}{1000} \right) = -\frac{ln(2)}{4.5 \times 10^9} t \] Now, solve for \(t\): \[ t = \frac{ln \left( \frac{312}{1000} \right)}{-\frac{ln(2)}{4.5 \times 10^9}} \]

Calculate the age of the rock

Finally, perform the calculation to determine the age of the rock: \[ t \approx \frac{ln \left( \frac{312}{1000} \right)}{-\frac{ln(2)}{4.5 \times 10^9}} \] \[ t \approx 3.64 \times 10^9 \mathrm{~years} \] The age of the rock is approximately 3.64 billion years.

Key concepts

Half-life Formula

The half-life of an isotope is the time it takes for half of the substance to decay. It is a crucial concept in understanding radiometric dating, which helps us determine the age of materials such as rocks or fossils.

When examining the age of a rock sample, we utilize the half-life formula, which is expressed as:
\[ N_t = N_0 \cdot e^{-\lambda t} \]
In this formula, \(N_t\) is the amount of the original isotope remaining after time \(t\), \(N_0\) is the initial amount of the isotope, and \(\lambda\) is the decay constant, related to the half-life. The formula shows how the amount of a radioisotope decreases over time due to radioactive decay. A key aspect that students often struggle with is the exponential nature of decay. The substance doesn't decrease by the same amount over each time period. Instead, it decreases by half over each half-life. By solving the half-life formula for a particular isotope in a rock sample, we can estimate the sample's age.

Decay Constant

The decay constant, symbolized by \(\lambda\), is central to the calculations that underpin radiometric dating. It reflects the probability of a single atom decaying per unit time, which allows us to calculate how rapidly a substance decays.

The relationship between the decay constant and the half-life of an isotope is given by:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Here, \(\ln(2)\) is the natural logarithm of 2 (approximately 0.693), and \(t_{1/2}\) is the half-life of the substance. A higher decay constant indicates a faster rate of decay. Understanding this concept is essential when determining the age of a rock. It's a common mistake to forget that the decay constant is not just a different expression of the half-life, it is a representation of the physics underlying the decay process. For students, visualizing decay as a constant percentage of elements disintegrating over time can help in grasping how the decay constant operates.

Uranium-Lead Dating

Uranium-Lead (U-Pb) dating is one of the oldest and most refined methods of radiometric dating. It makes use of the fact that uranium isotopes radioactively decay to form lead isotopes. Specifically, uranium-238 decays to lead-206, a transition that can be tracked to reveal the timing of geological events.

When performing U-Pb dating, we make a critical assumption that no lead was originally present in the sample, which means all detected lead (\(^{206}Pb\)) must have arisen from radioactive decay. Finding out the proportion of uranium to lead gives us a tool to calculate the time elapsed since the formation of the rock.

In the textbook exercise, the calculation involves the currently measured ratio of lead to uranium in a rock to ascertain its age. This precision of U-Pb dating comes from the long half-lives involved, and it's especially useful for putting dates on rocks that are billions of years old. However, accounting for the lead that might have been present from the start or subsequent contamination is crucial, as such factors can introduce errors that complicate age determinations.

Isotopes in Geochronology

Isotopes play a vital role in geochronology, which is the science of determining the ages of rocks, fossils, and sediments. Different isotopes have different half-lives, which lends them to dating different types of geological materials and time spans.

Geochronologists select an isotope to study based on the age and the kind of material involved. For ancient rocks, isotopes with longer half-lives, such as uranium-238, are used. For more recent materials, isotopes like carbon-14 might be appropriate because it has a relatively short half-life.

In the exercise example, uranium-238 and lead-206 are used because their half-life span suits the age estimation of old geological formations. An understanding of isotopes and their decay products informs which radiometric dating method to use. However, not all isotopes are useful for all dating scenarios due to their varying half-lives, and some might not be present in significant amounts in certain samples or could have been affected by environmental factors. Hence, deciding the appropriate isotopes for geochronology and understanding their properties are essential steps in the accurate dating of geological samples.