Question

Uranium has two common isotopes, with atomic masses of 238 and 235. one way to separate these isotopes is to combine the uranium with fluorine to make uranium hexafluoride gas, UF_6, then exploit the difference in the average thermal speeds of molecules containing the different isotopes. Calculate the rms speed of each molecule at room temperature, and compare them.

Short answer

Speed of the lighter isotope of UF₆is more than the speed of heavier isotope of UF₆

Step-by-step solution

Calculation of atomic mass of each isotope .

Atomic mass of UF₆ is calculated as,

$$\begin{gathered} m_{U{F}_6}=m_U+6mF \\ Giventhatfor{U}^{238},m_U=238amuand{m}_F=19amu \\ Thus{m}_{U{F}_6}=238+6\left(19\right)=352amu \\ Similarlyfor{U}^{235},m_U=235amuand{m}_F=19amu \\ {m}_{U{F}_6}=235+6\left(19\right)=349amu \end{gathered}$$

Getting the mass in kg for each isotope 

The mass of each UF₆ atom is calculated as,

$$\begin{gathered} m=\frac{m_{U{F}_6}}{N_A}where{N}_A=6.023\times {10}^{23}/mole \\ Thusfor{U}^{238},m=\frac{352\times {10}^{-3}kg/mol}{6.023\times {10}^{23}} \\ m=5.844\times {10}^{-25}kg \\ Andfor{U}^{235},m^1=\frac{349\times {10}^{-3}kg/mol}{6.023\times {10}^{23}} \\ {m}^1=5.794\times {10}^{-25}kg \end{gathered}$$

Analysis of faster isotope

The rms speed of a molecule is given by,

v_rms=$\sqrt{\frac{3KT}{m}}$

Where K = Boltzman constant & T is absolute temperature = 300k

Now for $$\begin{gathered} U^{238},v_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times \times 300}{5.844\times {10}^{-25}}} \\ {v}_{rms}=145.78m/s \\ Andfor{U}^{235},{v^1}_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 300}{5.794\times {10}^{-25}}} \\ {v^1}_{rms}=146.4m/s \end{gathered}$$

Thus UF₆of U²³⁵ isotope is faster than the UF₆of U²³⁸isotope.