Natural uranium is mostly nonfissionable \(^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \(^{235}\) U. For uranium to be useful as a nuclear fuel, the relative amount of \(^{235}\) U must be increased to about \(3 \% .\) This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \(^{238} \mathrm{UF}_{6}(g)\) and \(^{235} \mathrm{UF}_{6}(g) .\) The fluoride mixture is then enriched through a multistage diffusion process to produce a \(3 \%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter \(8,\) Section \(8-7\) ). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

Short Answer

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Graham's law of effusion can be used for uranium enrichment by taking advantage of the difference in effusion rates between the lighter isotope \(^{235}\mathrm{UF}_{6}(g)\) and the heavier isotope \(^{238}\mathrm{UF}_{6}(g)\). In the gaseous diffusion process, uranium hexafluoride (a mixture of both isotopes) is forced through a porous barrier, and the lighter \(^{235}\mathrm{UF}_{6}\) moves through the barrier slightly faster. This results in a higher concentration of \(^{235}\mathrm{UF}_{6}\) on the other side. The process is repeated through multiple stages until the desired enrichment level is achieved, thus allowing natural uranium to be enriched through gaseous diffusion using Graham's law of effusion.

Step by step solution

01

Recall Graham's law of effusion

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as: \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] where Rate₁ and Rate₂ are the effusion rates of two gases, and M₁ and M₂ are their respective molar masses.
02

Calculate the molar masses of 238UF6 and 235UF6

To be able to apply Graham's law, we need to know the molar masses of \(^{238}\mathrm{UF}_{6}\) and \(^{235}\mathrm{UF}_{6}\). We can calculate this by adding up the atomic masses of each constituent atom: \(^{238}\)U: 238.02891 amu \(^{235}\)U: 235.04393 amu F: 18.99840 amu \(^{238}\mathrm{UF}_{6}\): 238.02891 + 6*(18.99840) = 348.02891 amu \(^{235}\mathrm{UF}_{6}\): 235.04393 + 6*(18.99840) = 349.05001 amu
03

Apply Graham's law of effusion to the uranium isotopes

Using Graham's law, we can calculate the ratio of effusion rates for the two uranium isotopes in the form of uranium hexafluoride, \(^{238}\mathrm{UF}_{6}(g)\) and \(^{235}\mathrm{UF}_{6}(g)\): \[\frac{Rate_{^{235}\mathrm{UF}_{6}}}{Rate_{^{238}\mathrm{UF}_{6}}} = \sqrt{\frac{M_{^{238}\mathrm{UF}_{6}}}{M_{^{235}\mathrm{UF}_{6}}}}\] By plugging in the molar masses obtained in step 2, we get: \[\frac{Rate_{^{235}\mathrm{UF}_{6}}}{Rate_{^{238}\mathrm{UF}_{6}}} = \sqrt{\frac{348.02891}{349.05001}}\]
04

Explain how Graham's law of effusion allows for uranium enrichment

Graham's law of effusion can be used for uranium enrichment by taking advantage of the difference in effusion rates of the lighter isotope \(^{235}\mathrm{UF}_{6}(g)\) and the heavier isotope \(^{238}\mathrm{UF}_{6}(g)\). In the gaseous diffusion process, uranium hexafluoride (a mixture of both isotopes) is forced through a porous barrier. The lighter \(^{235}\mathrm{UF}_{6}(g)\) moves through the barrier slightly faster than the heavier \(^{238}\mathrm{UF}_{6}(g)\), due to its higher effusion rate. This results in a higher concentration of \(^{235}\mathrm{UF}_{6}\) on the other side of the barrier. The gaseous diffusion process is repeated through multiple stages until the desired enrichment level (in this case, 3% \(^{235}\)U) is achieved. Hence, Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process by exploiting the differences in effusion rates between the isotopes of uranium hexafluoride.

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Key Concepts

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

Understanding Graham's Law of Effusion
Graham's Law of Effusion is a principle in chemistry that describes the relationship between the effusion rates of gases and their molar masses. Specifically, it states that the rate at which a gas escapes through a small hole into a vacuum is inversely proportional to the square root of the molar mass of its particles. This fundamental law can be represented mathematically as \[\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\] where \(Rate_1\) and \(Rate_2\) are the effusion rates of two different gases, and \(M_1\) and \(M_2\) are their respective molar masses.
When applying this law to the process of uranium enrichment, it becomes clear why different isotopes of uranium hexafluoride effuse at different rates. The lighter \(^{235}\mathrm{UF}_6\) will effuse more rapidly than the heavier \(^{238}\mathrm{UF}_6\), allowing for a separation based on mass and ultimately leading to the enrichment of the fissionable \(^{235}\mathrm{U}\).
Uranium Hexafluoride in Nuclear Fuel Preparation
Uranium Hexafluoride (\(\mathrm{UF}_6\)) is a key compound used in the nuclear fuel preparation process. This chemical is the form in which uranium is typically processed for enrichment, due to its unique properties. It is a volatile substance, which allows it to be in a gaseous state at relatively low temperatures; this is crucial for the enrichment process.
Uranium in its natural state is composed mainly of the isotope \(^{238}\mathrm{U}\), with only about 0.7% of the more reactive \(^{235}\mathrm{U}\) isotope, which is capable of sustaining a nuclear chain reaction. To become viable as nuclear fuel, the proportion of \(^{235}\mathrm{U}\) must be increased. The first step in this process is combining natural uranium with fluorine to create \(\mathrm{UF}_6\), consisting of both \(^{238}\mathrm{UF}_6\) and \(^{235}\mathrm{UF}_6\). The uranium hexafluoride is then subjected to gaseous diffusion to separate these isotopes, leveraging the principles outlined in Graham's law of effusion.
Isotopic Separation for Nuclear Fuel
Isotopic separation is the cornerstone of creating usable nuclear fuel. It involves increasing the percentage of the fissionable \(^{235}\mathrm{U}\) isotope within uranium to make it suitable for use in nuclear reactors. This process hinges on the slight differences in physical properties between isotopes, most notably their mass.
The gaseous diffusion method is one example of isotopic separation. It takes advantage of the different effusion rates of \(^{238}\mathrm{UF}_6\) and \(^{235}\mathrm{UF}_6\) as predicted by Graham's Law. Another method is centrifugation, where isotopes are separated due to their different masses under the force of a rapidly spinning rotor. These techniques aim to increase the proportion of \(^{235}\mathrm{U}\) from 0.7% to approximately 3-5% for reactor-grade nuclear fuel, an essential step for powering nuclear energy facilities.
The Role of Nuclear Fuel in Energy Production
Nuclear fuel is a material that can be 'burned' by nuclear fission to derive energy. It primarily consists of the isotope \(^{235}\mathrm{U}\), which is capable of sustaining a controlled chain reaction. To harness this energy, nuclear fuel must first be processed through enrichment to increase the concentration of \(^{235}\mathrm{U}\).
Once enriched to the proper levels, the nuclear fuel can be fabricated into fuel rods or assemblies and placed into a nuclear reactor. Inside the reactor, atoms of \(^{235}\mathrm{U}\) are bombarded with neutrons, causing them to split and release a tremendous amount of energy in the form of heat. This heat can then be used to produce steam, which in turn drives turbines to generate electricity. The careful and detailed process of creating this fuel, starting from mining uranium ore to the isotopic separation and enrichment, illustrates the intricate work behind the power that many take for granted when they flip on a light switch.

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Most popular questions from this chapter

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?

A proposed system for storing nuclear wastes involves storing the radioactive material in caves or deep mine shafts. One of the most toxic nuclides that must be disposed of is plutonium- 239 which is produced in breeder reactors and has a half-life of 24,100 years. A suitable storage place must be geologically stable long enough for the activity of plutonium-239 to decrease to \(0.1 \%\) of its original value. How long is this for plutonium-239?

The bromine- 82 nucleus has a half-life of \(1.0 \times 10^{3}\) min. If you wanted \(1.0 \mathrm{g}^{82} \mathrm{Br}\) and the delivery time was 3.0 days, what mass of NaBr should you order (assuming all of the Br in the NaBr was \(\left.^{82} B r\right) ?\)

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

Strontium-90 and radon-222 both pose serious health risks. \(^{90}\) \(\mathrm{Sr}\) decays by \(\beta\) -particle production and has a relatively long half-life (28.9 years). Radon-222 decays by \(\alpha\) -particle production and has a relatively short half-life (3.82 days). Explain why each decay process poses health risks.

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