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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 28 \(\mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

Short Answer

Expert verified
When the Earth was formed, the relative abundances of Uranium-238 and Uranium-235 isotopes were \(99.28\%\) and \(100\%\) respectively.

Step by step solution

01

Identify the decay formula

The decay formula is given by: \[ N_t = N_0 \times (1/2)^{t/T}\] Where \(N_t\) is the amount of the isotope at time \(t\), \(N_0\) is the initial amount of the isotope, \(t\) is the time elapsed, and \(T\) is the half-life of the isotope.
02

Set up the decay formulas for both isotopes

Let the initial relative abundance of Uranium-238 be \(X\%\) and that for Uranium-235 be \((100-X)\%\). Now, we can write the decay formulas for both isotopes as: \[N_t(^{238}U) = X \times (1/2)^{t/T_{238}}\] \[N_t(^{235}U) = (100-X) \times (1/2)^{t/T_{235}}\]
03

Substitute the known values

We know the present relative abundances and half-lives of both isotopes and the age of the Earth. We can substitute these values into the formulas we derived in Step 2: \[99.28 = X \times (1/2)^{4.5 \times 10^9 / 4.5 \times 10^9}\] \[0.72 = (100-X) \times (1/2)^{4.5 \times 10^9 / 7.1 \times 10^8}\]
04

Solve for the initial relative abundance of Uranium-238

Solving the first equation for \(X\), we find that the initial relative abundance of Uranium-238 is the same as its present relative abundance: \[X = 99.28\%\]
05

Solve for the initial relative abundance of Uranium-235

Now, we solve the second equation for the initial relative abundance of Uranium-235: \[0.72 = (100-99.28) \times (1/2)^{4.5 \times 10^9 / 7.1 \times 10^8}\] \[(1/2)^{4.5 \times 10^9 / 7.1 \times 10^8} = 0.0072 = \frac{(100-X)}{Y}\] \[Y = (100-X) / 0.0072\] \[Y \approx 100\] The initial relative abundance of Uranium-235 when the Earth was formed is approximately \(100\%\). So, when the Earth was formed, the relative abundances of Uranium-238 and Uranium-235 isotopes were \(99.28\%\) and \(100\%\) respectively.

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Key Concepts

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

Uranium Isotopes
Uranium isotopes, particularly Uranium-238 ( ^{238}U) and Uranium-235 ( ^{235}U), are fascinating components of nuclear chemistry. These isotopes differ in their atomic mass, which affects their physical properties and stability. ^{238}U and ^{235}U are naturally occurring isotopes, with ^{238}U being much more abundant at 99.28%, compared to ^{235}U at a mere 0.72%.
These isotopes are vital for different applications. ^{235}U is especially important in nuclear reactors and nuclear weapons due to its ability to sustain a nuclear chain reaction. This is because it can easily undergo fission when struck by a neutron. In contrast, ^{238}U is more stable and doesn't fission easily, serving more of a supportive role in these processes but is crucial for generating ^{239}Pu, another fissionable material.
Understanding the balance of these isotopes in the past, just after the Earth's formation, offers insights into geological processes and the Earth's history. This balance has shifted over billions of years due to radioactive decay, making isotopic analysis an essential tool in geology and archeology.
Half-Life
Half-life is a fundamental concept in understanding radioactive decay. It refers to the time required for half of the atoms in a radioactive material to decay. In nuclear chemistry, knowing the half-life of isotopes allows scientists to predict their behavior over time.
For ^{238}U, the half-life is extremely long, approximately 4.5 billion years. This means it decays very slowly, remaining prevalent in the Earth's crust to this day. In contrast, ^{235}U has a shorter half-life of about 710 million years. This faster decay rate explains its lower abundance as compared to ^{238}U.
Half-life calculations involve the decay formula: \[N_t = N_0 \times \left(\frac{1}{2}\right)^{t/T}\]where N_t is the remaining quantity of the isotope at time t, N_0 is the initial quantity, and T is the half-life. Understanding this formula helps calculate past abundances of isotopes by mapping the decay process backward, a useful tool in many scientific investigations.
Nuclear Chemistry
Nuclear chemistry focuses on reactions and processes at the atomic nucleus level. Radioactive decay, like the process involving ^{238}U and ^{235}U, is central to this field. It investigates changes in an atom's nucleus and how these changes affect the atom's properties.
This branch of chemistry explores phenomena such as nuclear reactions, fission, and fusion. In the context of uranium isotopes, nuclear chemistry helps explain how isotopes transform over time and contribute to larger processes, be it in nature or technology. ^{238}U's long half-life makes it significant for age dating geological samples, providing crucial data about Earth's history.
Knowledge in nuclear chemistry enables advancements in energy production, medical treatments, and understanding of cosmic events. The decay of uranium isotopes, for example, not only provides energy via nuclear power but also helps scientists determine the age of rocks and fossils, offering a timeline for Earth's history.

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Most popular questions from this chapter

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction $$ \mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad 8^{\circ}=-2.36 \mathrm{V} $$ a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{G} \circ, \Delta G^{\circ},\) and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred? If \(1.00 \times 10^{3} \mathrm{kg}\) Zreacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at 1.0 \(\mathrm{atm}\) and \(1000 .^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in \(1986,\) hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ It was not possible to prevent a chemical explosion at Chernobyl. In light of this, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

The radioactive isotope \(^{242} \mathrm{Cm}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \(^{242} \mathrm{Cm}\) through many transformations to end up as \(^{206} \mathrm{P} \mathrm{b}\) . In the complete decay series, how many \(\alpha\) and \(\beta\) particles are produced?

Rubidium- 87 decays by \(\beta\) -particle production to strontium- 87 with a half-life of \(4.7 \times 10^{10}\) years. What is the age of a rock sample that contains 109.7 \mug of \(^{87} \mathrm{Rb}\) and 3.1\(\mu \mathrm{g}\) of \(^{87} \mathrm{Sr} ?\) Assume that no \(^{87}\) Sr was present when the rock was formed. The atomic masses for \(^{87}\mathrm{Rb}\) and \(^{87} \mathrm{Sr}\) are 86.90919 \(\mathrm{u}\) and 86.90888 u, respectively.

Calculate the binding energy in J/nucleon for carbon-12 (atomic mass \(=12.0000\) u) and uranium-235 (atomic mass \(=\) 235.0439 u). The atomic mass of \(_{1}^{1} \mathrm{H}\) is 1.00782 \(\mathrm{u}\) and the mass of a neutron is 1.00866 u. The most stable nucleus known is \(^{56}\) Fe \((\text { see Exercise } 50)\) . Would the binding energy per nucleon for \(^{56} \mathrm{Fe}\) be larger or smaller than that of \(^{12} \mathrm{C}\) or \(^{235} \mathrm{U}\) ? Explain.

The earth receives \(1.8 \times 10^{14} \mathrm{kJ} / \mathrm{s}\) of solar energy. What mass of solar material is converted to energy over a \(24-\mathrm{h}\) period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? (Coal releases 32 \(\mathrm{kJ}\) of energy per gram when burned.)

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