Chapter 1: 1.2. (page 13)

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

Expert verified

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,

$m_{U F_{6}} = m_{U} + 6 m F G i v e n t h a t f o r U^{238}, m_{U} = 238 a m u a n d m_{F} = 19 a m u T h u s m_{U F_{6}} = 238 + 6 (19) = 352 a m u S i m i l a r l y f o r U^{235}, m_{U} = 235 a m u a n d m_{F} = 19 a m u m_{U F_{6}} = 235 + 6 (19) = 349 a m u$

Getting the mass in kg for each isotope

The mass of each UF₆ atom is calculated as,

$m = \frac{m_{U F_{6}}}{N_{A}} w h e r e N_{A} = 6.023 \times 10^{23} / m o l e T h u s f o r U^{238}, m = \frac{352 \times 10^{- 3} k g / m o l}{6.023 \times 10^{23}} m = 5.844 \times 10^{- 25} k g A n d f o r U^{235}, m^{1} = \frac{349 \times 10^{- 3} k g / m o l}{6.023 \times 10^{23}} m^{1} = 5.794 \times 10^{- 25} k g$

Analysis of faster isotope

The rms speed of a molecule is given by,

v_rms= $\sqrt{\frac{3 K T}{m}}$

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

Now for $U^{238}, v_{r m s} = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times \times 300}{5.844 \times 10^{- 25}}} v_{r m s} = 145.78 m / s A n d f o r U^{235}, {v^{1}}_{r m s} = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 300}{5.794 \times 10^{- 25}}} {v^{1}}_{r m s} = 146.4 m / s$

Thus UF₆of U²³⁵ isotope is faster than the UF₆of U²³⁸isotope.

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Short Answer

Step by step solution

Calculation of atomic mass of each isotope .

Getting the mass in kg for each isotope

Analysis of faster isotope

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