Question

In the early days of nuclear power 235U and 238U were separated by repeatedly
passing uranium hexafluoride gas through a filter. Calculate the ratio between the
effusion times for 235UF6 and 238UF6. Explain why it is necessary to undergo
several thousand passes through the filter to attain a high concentration of 235U

Answer 1

$\frac{rate \ of \ effusion \ of \ A}{rate \ of \ effusion \ of \ B} = (\frac{Molar \ mass \ of \ B}{Molar \ mass \ of \ A})^{1/2}$

Molar mass of 238UF6 = 238 + 19 * 6 = 238 + 114 = 352 g/mol

Molar mass of 235UF6 = 235 + 19 * 6 = 235 + 114 = 349 g/mol

$\frac{rate \ of \ effusion \ of \ 235UF_6}{rate \ of \ effusion \ of \ 238UF_6} = (\frac{Molar \ mass \ of \ 238UF_6}{Molar \ mass \ of \ 235UF_6})^{1/2} = (352/349)^{1/2} = 1.004$

Since the ratio between effusion time is pretty close to 1, hence it is difficult to separate the two compounds, hence we need to several thousand passes through the filter to attain high concentration of 235 U

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