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Chapter 19: Problem 43

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

Expert verified
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

01

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.
02

Write down the formula for root-mean-square speed

The formula for the root-mean-square speed, \(v_{rms}\), is given by: \(v_{rms}=\sqrt{\frac{3kT}{m}}\)
03

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: \(v_{rms,1}=\sqrt{\frac{3kT}{m_1}}\) \(v_{rms,2}=\sqrt{\frac{3kT}{m_2}}\)
04

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: \(\frac{v_{rms,1}}{v_{rms,2}}=\frac{\sqrt{\frac{3kT}{m_1}}}{\sqrt{\frac{3kT}{m_2}}}\)
05

Simplify the ratio

Now we can simplify the ratio, and both \(3kT\) terms will cancel: \(\frac{v_{rms,1}}{v_{rms,2}}=\sqrt{\frac{m_2}{m_1}}\)
06

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.

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