Question

Two isotopes of uranium, \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U},\) are separated by a gas diffusion process that involves combining them with flourine to make the compound \(\mathrm{UF}_{6} .\) Determine the ratio of the root-mean-square speeds of UF \(_{6}\) molecules for the two isotopes. The masses of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) are \(249 \mathrm{amu}\) and \(252 \mathrm{amu}\).

Short answer

The approximate ratio of the root-mean-square speeds of the UF$_{6}$ molecules for the two isotopes is 1.006.

Step-by-step solution

Identify the masses of the isotopes

We are given the masses of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\), denoted as \(m_1\) and \(m_2\), which are \(249 \mathrm{amu}\) and \(252 \mathrm{amu}\), respectively.

Write down the formula for root-mean-square speed

The formula for the root-mean-square speed, \(v_{rms}\), is given by: \(v_{rms}=\sqrt{\frac{3kT}{m}}\)

Calculate the root-mean-square speeds of the isotopes

Using the formula above, we can write the root-mean-square speeds of the two isotopes as follows: \(v_{rms,1}=\sqrt{\frac{3kT}{m_1}}\) \(v_{rms,2}=\sqrt{\frac{3kT}{m_2}}\)

Divide both root-mean-square speeds to find the ratio

To find the ratio, we can divide \(v_{rms,1}\) by \(v_{rms,2}\), which will give us: \(\frac{v_{rms,1}}{v_{rms,2}}=\frac{\sqrt{\frac{3kT}{m_1}}}{\sqrt{\frac{3kT}{m_2}}}\)

Simplify the ratio

Now we can simplify the ratio, and both \(3kT\) terms will cancel: \(\frac{v_{rms,1}}{v_{rms,2}}=\sqrt{\frac{m_2}{m_1}}\)

Substitute the mass values and calculate the ratio

We can now substitute the mass values of \(m_1\) and \(m_2\) into the formula: \(\frac{v_{rms,1}}{v_{rms,2}}=\sqrt{\frac{252 \mathrm{amu}}{249 \mathrm{amu}}}\) Finally, calculating the ratio: \(\frac{v_{rms,1}}{v_{rms,2}}\approx 1.006\) So the ratio of the root-mean-square speeds of UF\(_{6}\) molecules for the two isotopes is approximately 1.006.

Key concepts

Gas Diffusion Process

Gas diffusion is an interesting process where gases spread from an area of high concentration to an area of lower concentration. Picture it like the way the fragrance of a freshly baked cake spreads through a room.
In the context of isotope separation, like the uranium isotopes, the gas diffusion process is crucial because it takes advantage of the small differences in the mass of isotopes. Since lighter molecules move faster than heavier ones, this principle can help in separating isotopes.

Why it Matters - Used in industries to enrich uranium which is necessary for nuclear energy production. - Helps in research and medical fields where pure isotopes are needed.
By utilizing differences in root-mean-square speeds of gases, this process becomes efficient in separating different molecular species. It is a fundamental concept in chemistry and physics with wide applications.

Uranium Isotopes

Uranium is a heavy metal that has several isotopes, with different numbers of neutrons in the nucleus. The main isotopes you'll encounter are extsuperscript{235}U and extsuperscript{238}U. They are key players in nuclear chemistry due to their different respective properties.

Understanding Isotopes - ** extsuperscript{235}U**: This isotope is crucial for nuclear reactors because it can sustain a chain reaction. This means when it splits, it releases neutrons that can cause other nuclei to split too. - ** extsuperscript{238}U**: While not as reactive as extsuperscript{235}U, it is more abundant in nature. It can be converted into plutonium, another significant element in nuclear fuels.
The key to separating these isotopes lies in their slight mass difference. Using methods like the gas diffusion process efficiently isolates one from the other.

Molecular Mass

Molecular mass, often called molecular weight, is the sum of the masses of all the atoms in a given molecule. It is generally expressed in atomic mass units (amu).
Understanding molecular mass is essential for several reasons:

- It helps in calculating the root-mean-square speed of gas molecules.- Crucial in determining the behavior of molecules during chemical reactions and physical processes like diffusion.
When considering the separation of uranium isotopes combined with fluorine to form UF extsubscript{6}, molecular masses of extsuperscript{235}UF extsubscript{6} and extsuperscript{238}UF extsubscript{6} molecules (249 amu and 252 amu respectively) can be used to derive conclusions about their kinetic properties.
The root-mean-square speed formula (\(v_{rms}\)) illustrates the important relationship between the molecular mass and kinetic energy of molecules in a gas:\[v_{rms} = \sqrt{\frac{3kT}{m}}\\]In simpler terms, lighter molecules move faster, which is fundamental to processes like gas diffusion and the effective separation of isotopes.