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1.2.

Expert-verifiedFound in: Page 13

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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.

Speed of the lighter isotope of UF_{6} is more than the speed of heavier isotope of UF_{6}

Atomic mass of UF_{6} is calculated as,

${m}_{U{F}_{6}}={m}_{U}+6mF\phantom{\rule{0ex}{0ex}}Giventhatfor{U}^{238},{m}_{U}=238amuand{m}_{F}=19amu\phantom{\rule{0ex}{0ex}}Thus{m}_{U{F}_{6}}=238+6\left(19\right)=352amu\phantom{\rule{0ex}{0ex}}Similarlyfor{U}^{235},{m}_{U}=235amuand{m}_{F}=19amu\phantom{\rule{0ex}{0ex}}{m}_{U{F}_{6}}=235+6\left(19\right)=349amu$

The mass of each UF_{6} atom is calculated as,

$m=\frac{{m}_{U{F}_{6}}}{{N}_{A}}where{N}_{A}=6.023\times {10}^{23}/mole\phantom{\rule{0ex}{0ex}}Thusfor{U}^{238},m=\frac{352\times {10}^{-3}kg/mol}{6.023\times {10}^{23}}\phantom{\rule{0ex}{0ex}}m=5.844\times {10}^{-25}kg\phantom{\rule{0ex}{0ex}}Andfor{U}^{235},{m}^{1}=\frac{349\times {10}^{-3}kg/mol}{6.023\times {10}^{23}}\phantom{\rule{0ex}{0ex}}{m}^{1}=5.794\times {10}^{-25}kg$

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 ${U}^{238},{v}_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times \times 300}{5.844\times {10}^{-25}}}\phantom{\rule{0ex}{0ex}}{v}_{rms}=145.78m/s\phantom{\rule{0ex}{0ex}}Andfor{U}^{235},{{v}^{1}}_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 300}{5.794\times {10}^{-25}}}\phantom{\rule{0ex}{0ex}}{{v}^{1}}_{rms}=146.4m/s\phantom{\rule{0ex}{0ex}}$

Thus UF_{6 }of U^{235} isotope is faster than the UF_{6} of U^{238} isotope.

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