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Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U},\) with relative abundances of \(99.28 \%\) and \(0.72 \%,\) respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for \(^{235} \mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

Short Answer

Expert verified
The initial relative abundances of \(^{238}\mathrm{U}\) and \(^{235}\mathrm{U}\) isotopes when the earth was formed were approximately \(RA_{238} \approx 98.12\%\) and \(RA_{235} \approx 1.88\%\).

Step by step solution

01

Write the half-life formula

We will use the half-life formula to determine the initial amount of each isotope when the earth was formed. The formula is given by: \(N_t = N_0 e^{(-\lambda t)}\) Where \(N_t\) is the amount remaining after time \(t\), \(N_0\) is the initial amount, \(\lambda\) is the decay constant, and \(t\) is the time elapsed.
02

Find the decay constants for both isotopes

The decay constant can be found using the formula: \(\lambda = \frac{\ln{2}}{t_{1/2}}\) Where \(t_{1/2}\) is the half-life of the isotope. For \(^{238}\mathrm{U}\): \(\lambda_{238} = \frac{\ln{2}}{4.5 \times 10^9 \; \text{years}}\) For \(^{235}\mathrm{U}\): \(\lambda_{235} = \frac{\ln{2}}{7.1 \times 10^8 \; \text{years}}\)
03

Calculate the initial amounts of both isotopes

We want to find the initial amounts of both isotopes, so we rewrite the half-life formula for \(N_0\). The current abundances being 99.28% and 0.72% represent \(N_t\). Let \(N_{0_{238}}\) be the initial amount of \(^{238}\mathrm{U}\), and \(N_{0_{235}}\) be the initial amount of \(^{235}\mathrm{U}\). For \(^{238}\mathrm{U}\): \(0.9928 = N_{0_{238}}e^{(-\lambda_{238} \times 4.5 \times 10^9)}\) For \(^{235}\mathrm{U}\): \(0.0072 = N_{0_{235}}e^{(-\lambda_{235} \times 4.5 \times 10^9)}\) Solve for \(N_{0_{238}}\) and \(N_{0_{235}}\) using the decay constants from Step 2.
04

Calculate the initial relative abundances

With the initial amounts of both isotopes, we can now calculate their initial relative abundances. Let \(RA_{238}\) be the initial relative abundance of \(^{238}\mathrm{U}\), and \(RA_{235}\) be the initial relative abundance of \(^{235}\mathrm{U}\). \(RA_{238} = \frac{N_{0_{238}}}{N_{0_{238}} + N_{0_{235}}} \times 100\) \(RA_{235} = \frac{N_{0_{235}}}{N_{0_{238}} + N_{0_{235}}} \times 100\) Calculate \(RA_{238}\) and \(RA_{235}\) using the initial amounts from Step 3. These will be the initial relative abundances of \(^{238}\mathrm{U}\) and \(^{235}\mathrm{U}\) isotopes when the earth was formed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This happens when certain isotopes of elements break down, moving toward a more stable form. The decay may release particles like alpha or beta particles or rays like gamma rays. Importantly, this process follows a predictable pattern, which allows scientists to use it to determine the age of materials and trace chemical and physical processes in the Earth's history.
  • Radioactive decay occurs at a fixed rate for a given isotope.
  • The rate of decay is unique for each substance, quantified by its decay constant.
  • Isotopes like uranium-238 and uranium-235 each have different rates of decay.
This makes radioactive decay a critical concept for understanding and calculating various parameters in nuclear physics and geoscience.
Half-Life Calculation
The half-life of an isotope is the time it takes for half of a sample of the radioactive substance to decay. It's a crucial metric used in the study of radioactive elements, allowing for the estimation of how long a given isotope will remain active. For instance, the half-life of uranium-238 is extremely long at 4.5 billion years, meaning it remains radioactive for a considerable amount of time.
  • Half-life provides insights into the stability of an isotope.
  • Longer half-lives generally indicate stable isotopes that decay very slowly.
  • Shorter half-lives suggest isotopes that decay rapidly.
Applications such as radiocarbon dating depend on precise half-life calculations to yield accurate results.
Decay Constants
Decay constants are fundamental in understanding how fast a radioactive isotope decays. This constant links directly to the half-life by the formula \(\lambda = \frac{\ln{2}}{t_{1/2}}\), where \(t_{1/2}\) is the isotope's half-life.
  • Provides a measure of the time taken for the activity of the isotope to drop by half.
  • Enables the calculation of the original quantity of an isotope from a sample.
  • Inform predictions about the behavior of isotopes over geological time scales.
Understanding decay constants is vital for industries that rely on radioactive sources, such as energy production and medical imaging.
Relative Abundance
Relative abundance refers to the percentage composition of an isotope compared to the total amount of that element in a sample. In natural uranium, for example, uranium-238 makes up 99.28% and uranium-235 comprises 0.72%. These percentages change over time due to radioactive decay, impacting calculations and predictions about isotope compositions in different samples and epochs.
  • Helps in determining how much of a particular isotope remains in a sample.
  • Used alongside half-life information to project changes over time.
  • Essential for understanding the distribution and behavior of elements within the Earth's crust.
Relative abundance calculations enable scientists to reconstruct past environments and to date geological formations accurately.

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Most popular questions from this chapter

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn T. Seaborg, discoverer of the transuranium elements.a. \(^{263} \mathrm{Sg}\) was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) Sg decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg} ?\)

Uranium- 235 undergoes many different fission reactions. For one such reaction, when \(^{235} \mathrm{U}\) is struck with a neutron, \(^{144} \mathrm{Ce}\) and \(^{90}\) Sr are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?

Estimate the temperature needed to achieve the fusion of deuterium to make an \(\alpha\) particle. The energy required can be estimated from Coulomb's law [use the form \(E=9.0 \times 10^{9}\) \(\left(Q_{1} Q_{2} / r\right),\) using \(Q=1.6 \times 10^{-19} \mathrm{C}\) for a proton, and \(r=2 \times\) \(10^{-15} \mathrm{m}\) for the helium nucleus; the unit for the proportionality constant in Coloumb's law is \(\left.\mathbf{J} \cdot \mathbf{m} / \mathbf{C}^{2}\right]\).

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun:a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \(i \mathrm{H}\) and 4 He are 1.00782 u and 4.00260 u, respectively.)

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\) Why is plutonium one of the most toxic substances known?

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