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, UF₆, then exploit the difference in the average thermal speeds of molecules containing the different isotopes. Calculate the rms speed of each type of molecule at room temperature, and compare them.

Short answer

The rms speed of uranium-235 and uranium-238 are

v₂₃₅=145.92 m/s

v₂₃₈=145.53 m/s

Step-by-step solution

Given information

Temp (room temp) T=25^oC = 298K
Mass of U₂₃₅= 235.04 u
Mass of U₂₃₈= 238.028 u
Mass of fluorine = 18.998 u

Explanation

Root mean square speed of molecules is calculated as

$v=\sqrt{\frac{3kT}{m}}..........................\left(1\right)$

Where,

m = mass

k = Boltzmann constant

T = temp

First find the mass of hexa- fluoride of both isotopes

$$\begin{gathered} m_{235}=m_{U_{235}}+6\times m_{F_{19}} \\ =(235.04+6\times 18.998)\mathrm{u} \\ =5.794\times {10}^{-25}\mathrm{kg} \\ \mathrm{similarly} \\ {\mathrm{m}}_{238}={\mathrm{m}}_{{\mathrm{U}}_{238}}+6\times {\mathrm{m}}_{{\mathrm{F}}_{19}} \\ =(238.02891+6\times 18.998)\mathrm{u} \\ =5.843\times {10}^{-25}\mathrm{kg} \end{gathered}$$

Now using equation (1) calculate rms speed for both isotopes.

$$\begin{gathered} v_{235}=\sqrt{\frac{3kT}{m_{235}}}=\sqrt{\frac{3\times (1.38\times {10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1})\times \left(298K\right)}{5.794\times {10}^{-25}kg}} \\ {v}_{235}=145.92{\mathrm{ms}}^{-1} \\ Similarly \\ {v}_{238}=\sqrt{\frac{3kT}{m_{238}}}=\sqrt{\frac{3\times (1.38\times {10}^{-23}{m}^{2}kg{s}^{-2}{K}^{-1})\times \left(298K\right)}{5.843\times {10}^{-25}kg}} \\ {v}_{238}=145.53{\mathrm{ms}}^{-1} \end{gathered}$$

The rms speed is V₂₃₅= 135.92 m/s and V₂₃₈ = 145.53 m/s