1.2. - An Introduction to Thermal Physics

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.

Step-by-Step Solution

Answer

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

Step 1 : Calculation of atomic mass of each isotope .

Atomic mass of UF₆ is calculated as,

$$\begin{gathered} m_{U{F}_6}=m_U+6mF \\ Given that for U^{238}, m_U=238 amu and m_F=19amu \\ Thus m_{U{F}_6}=238+6\left(19\right)=352 amu \\ Similarly for U^{235}, m_U=235amu and m_F=19amu \\ {m}_{U{F}_6}=235+6\left(19\right)=349 amu \end{gathered}$$

Step 2 : Getting the mass in kg for each isotope

The mass of each UF₆ atom is calculated as,

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

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

Thus UF₆ of U²³⁵ isotope is faster than the UF₆ of U²³⁸ isotope.