Problem 61 - Fundamentals of Physics

Question

The isotope \({ }^{238} \mathrm{U}\) decays to \({ }^{206} \mathrm{~Pb}\) with a half-life of \(4.47 \times 10^{9} \mathrm{y}\). Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, $${ }^{238} \mathrm{U} \rightarrow{ }^{206} \mathrm{~Pb}+\text { various decay products. }$$ A rock is found to contain \(4.20 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(2.135 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) \({ }^{238} \mathrm{U}\) and (b) \({ }^{206} \mathrm{~Pb}\) does the rock now contain? (c) How many atoms of \({ }^{238} \mathrm{U}\) did the rock contain at formation? (d) What is the age of the rock?

Step-by-Step Solution

Answer

The rock is approximately 3.15 billion years old.

Step 1: Calculate the number of atoms in 238U and 206Pb currently in the rock

To find the number of atoms, we use the formula:\[\text{Number of atoms} = \frac{\text{mass} \times N_A}{\text{molar mass}}\]where \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\) is Avogadro's number.**For \(^{238}\text{U}\):**- Molar mass of \(^{238}\text{U} = 238 \text{ g/mol}\)- Mass of \(^{238}\text{U} = 4.20 \text{ mg} = 4.20 \times 10^{-3} \text{ g}\)\[\text{Number of atoms of } ^{238}\text{U} = \frac{4.20 \times 10^{-3} \times 6.022 \times 10^{23}}{238} \approx 1.06 \times 10^{19}\]**For \(^{206}\text{Pb}\):**- Molar mass of \(^{206}\text{Pb} = 206 \text{ g/mol}\)- Mass of \(^{206}\text{Pb} = 2.135 \text{ mg} = 2.135 \times 10^{-3} \text{ g}\)\[\text{Number of atoms of } ^{206}\text{Pb} = \frac{2.135 \times 10^{-3} \times 6.022 \times 10^{23}}{206} \approx 6.24 \times 10^{18}\]

Step 2: Calculate the initial number of 238U atoms in the rock

Since all \(^{206}\text{Pb}\) was formed from \(^{238}\text{U}\) decay, the initial number of \(^{238}\text{U}\) atoms should include current \(^{238}\text{U}\) and \(^{206}\text{Pb}\):\[\text{Initial number of } ^{238}\text{U} = 1.06 \times 10^{19} + 6.24 \times 10^{18} \approx 1.684 \times 10^{19}\]

Step 3: Calculate the age of the rock using the decay formula

The decay of \(^{238}\text{U}\) is described by:\[N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}\]where:- \(N\) is the number of \(^{238}\text{U}\) atoms now,- \(N_0\) is the initial number of \(^{238}\text{U}\) atoms,- \(T_{1/2}\) is the half-life of \(^{238}\text{U} = 4.47 \times 10^9 \text{ years}\).Thus,\[\left( \frac{1}{2} \right)^{t/T_{1/2}} = \frac{N}{N_0} = \frac{1.06 \times 10^{19}}{1.684 \times 10^{19}} \approx 0.63\]To solve for \(t\), take the logarithm on both sides:\[\log_{10} \left( \frac{1}{2} \right) \cdot \frac{t}{T_{1/2}} = \log_{10}(0.63)\]\[t = \frac{\log_{10}(0.63)}{\log_{10}(0.5)} \times T_{1/2} \approx \frac{-0.201}{-0.301} \times 4.47 \times 10^9 \approx 3.15 \times 10^9 \text{ years}\]

Key Concepts

Nuclear Decay

Nuclear decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. In the context of uranium-lead dating, it involves the transformation of uranium-238 (\(^{238}\text{U}\)) to lead-206 (\(^{206}\text{Pb}\)). This transformation occurs through a series of steps, but the long half-life of \(^{238}\text{U}\) allows us to simplify this process in calculations.- **Decay Process**: The nucleus of \(^{238}\text{U}\), over time, breaks down through radioactive decay, releasing alpha particles (helium nuclei), and ultimately forming \(^{206}\text{Pb}\).- **Chain of Decay**: Despite there being multiple decay processes for \(^{238}\text{U}\) to reach \(^{206}\text{Pb}\), the model can often consider it as a single-step transition due to the much longer half-life of the first decay series in comparison to subsequent steps.Through nuclear decay, we can estimate the age of rocks in which these radioactive isotopes are found, by measuring the proportions of the remaining radioactive isotope and the stable decay product. This method provides critical insights into geological and archaeological dating.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry and physics, which provides the ratio of entities per mole. The value is approximately \(6.022 \times 10^{23} \) entities per mole. It's crucial for converting between atomic scale quantities and macroscopic amounts of a substance. - **Definition**: Avogadro's number is the number of atoms, molecules, or particles in one mole of a substance.- **Applications**: In uranium-lead dating, Avogadro's number allows us to calculate the number of uranium-238 or lead-206 atoms present in a material sample from its mass.For instance, if a rock contains 4.20 mg of \(^{238}\text{U}\), knowing its molar mass (238 g/mol) and using Avogadro's number enables the determination of the actual number of atoms: \[\text{Number of atoms of } ^{238}\text{U} = \frac{4.20 \times 10^{-3} \times 6.022 \times 10^{23}}{238} \approx 1.06 \times 10^{19}\] This conversion helps in understanding the amount of uranium initially present or remaining in a sample, making it possible to gauge the decay progress and, consequently, the age of the rock.

Half-Life Calculation

The half-life of a radioactive isotope is the time required for half of the radioactive atoms in a sample to decay. This concept is central to nuclear dating techniques, such as uranium-lead dating, as it directly influences age determination.- **Half-Life of \(^{238}\text{U}\)**: The half-life of uranium-238 is approximately \(4.47 \times 10^{9}\) years. This long half-life makes \(^{238}\text{U}\) ideal for dating geological samples over a wide range of ages.- **Decay Formula**: The decay of uranium-238 is modeled using the formula: \[N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}\]where: - \(N\) is the number of uranium-238 atoms currently present. - \(N_0\) is the initial number of uranium-238 atoms. - \(t\) is the time that has passed. - \(T_{1/2}\) is the half-life.By rearranging and solving this equation, one can estimate the time elapsed since the rock formation. This calculation hinges on the precise measurements of current \(^{238}\text{U}\) and \(^{206}\text{Pb}\) quantities:\[\left( \frac{1}{2} \right)^{t/T_{1/2}} = \frac{1.06 \times 10^{19}}{1.684 \times 10^{19}} \approx 0.63\]Using the logarithmic form of the equation, one finds the age of the rock as approximately \(3.15 \times 10^{9}\) years. Understanding half-life calculations allows scientists to trace back the history of geological formations accurately.