Uranium has two naturally occurring isotopes. $$238U^{238}$$\text{U} has a natural abundance of 99.3%99.3\% and $$235U^{235}$$\text{U} has an abundance of 0.7%0.7\%. It is the rarer $$235U^{235}$$\text{U} that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, UF6\text{UF}_6, which is a gas, then allowing it to diffuse through a series of porous membranes. $$235UF6^{235}UF_6$ has a slightly larger rms speed than 238UF6$$^{238}$$UF_6$$ and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of $$235UF6^{235}UF_6$ to that of 238UF6$$^{238}$$UF_6$$?

Ch 20: The Micro/Macro Connection

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 20: The Micro/Macro Connection

Problem 53

Chapter 20, Problem 53

Uranium has two naturally occurring isotopes. $to the 238th power cap U$ has a natural abundance of $99.3 percent sign$ and $to the 235th power cap U$ has an abundance of $0.7 percent sign$ . It is the rarer $to the 235th power cap U$ that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, $UF sub 6$ , which is a gas, then allowing it to diffuse through a series of porous membranes. $to the 235th power U F sub 6$ has a slightly larger rms speed than $to the 238th power U F sub 6$ and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of $to the 235th power U F sub 6$ to that of $to the 238th power U F sub 6$ ?

Verified step by step guidance

Step 1: Recall the formula for the root mean square (rms) speed of a gas molecule, which is given by $ v_{rms} = \sqrt{\frac{3RT}{M}} $, where $ R $ is the gas constant, $ T $ is the temperature, and $ M $ is the molar mass of the gas.

Step 2: Note that the temperature $ T $ and the gas constant $ R $ are the same for both isotopes, so the ratio of their rms speeds depends only on the molar masses of $ ^{235}UF_6 $ and $ ^{238}UF_6 $.

Step 3: Calculate the molar mass of $ ^{235}UF_6 $ and $ ^{238}UF_6 $. The molar mass of $ UF_6 $ is the sum of the molar mass of uranium and six fluorine atoms. For $ ^{235}UF_6 $, the molar mass is $ 235 + 6 \times 19 $, and for $ ^{238}UF_6 $, the molar mass is $ 238 + 6 \times 19 $.

Step 4: Write the ratio of the rms speeds as $ \frac{v_{rms, ^{235}UF_6}}{v_{rms, ^{238}UF_6}} = \sqrt{\frac{M_{^{238}UF_6}}{M_{^{235}UF_6}}} $, where $ M_{^{235}UF_6} $ and $ M_{^{238}UF_6} $ are the molar masses calculated in Step 3.

Step 5: Substitute the molar masses into the ratio formula and simplify. This will give the numerical ratio of the rms speeds of $ ^{235}UF_6 $ to $ ^{238}UF_6 $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Root Mean Square Speed

The root mean square (rms) speed is a measure of the average speed of particles in a gas, calculated as the square root of the average of the squares of the speeds of the particles. It is directly related to the temperature and molar mass of the gas, providing insight into the kinetic energy of the particles. For two different gases, the rms speed can be compared using the formula: v_rms = sqrt(3RT/M), where R is the gas constant, T is the temperature, and M is the molar mass.

Diffusion

Diffusion is the process by which particles spread from areas of high concentration to areas of low concentration, driven by the random motion of particles. In the context of gases, lighter molecules diffuse faster than heavier ones due to their higher average speeds. This principle is crucial in the separation of isotopes, as the different masses of uranium hexafluoride isotopes lead to different diffusion rates through porous membranes.

Isotope Separation

Isotope separation is the process of concentrating specific isotopes of an element, which can have different physical properties, such as mass. In the case of uranium, the separation of ²³⁵U from ²³⁸U is essential for nuclear applications. Techniques like gas diffusion exploit the slight differences in rms speeds of the isotopes to achieve separation, allowing for the enrichment of the desired isotope for use in nuclear reactors.

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