# Effusion Question about Uranium enrichening

1. Feb 11, 2013

### CAF123

1. The problem statement, all variables and given/known data
A gas with equal number densities of uranium hexafluoride (UF6) molecules containing two isotopes of uranium, atomic masses 235amu and 238amu, is passed through a porous membrane containing very small holes. Use Graham's Law of Effusion to calculate:

How many successive passages through similar membranes would be needed to produce one isotope contaminated by less than 1% of the other isotope?

2. Relevant equations
Grahams' Law: $$\frac{v_x}{v_y} = \sqrt{\frac{m_y}{m_x}}$$

3. The attempt at a solution
So we want $v_x > 99$%$v_y$. I calculated that per passage, $v_x$ is greater than $v_y$ by about 1.006. Then I divided this by 100 to get the percentage increase and divided this by 99. This gives about 9840. I think the correct answer is about 1070.

Many thanks.

2. Feb 11, 2013

### Staff: Mentor

Two passes mean 1.0062.

3. Feb 11, 2013

### Redbelly98

Staff Emeritus
For starters, have you considered the fluorine atoms' mass in your calculation? (It appears that you haven't)

Also, the 1.006 figure (or whatever you actually get after you account for the fluorine) is a ratio. Simply dividing it by 100 is not a way to get a percent. Instead, let me ask you this: if 1.006 is the isotope ratio after passing through one membrane, how would you calculate the ratio after passing through two membranes?

(Hint: see Borek's post )

4. Feb 11, 2013

### CAF123

In that case, per passage, the ratio would be $$\sqrt{\frac{238 + 19(6)}{235 + 19(6)}}$$

(1.004)2, where 1.004 is the new ratio I calculated.

We want the ratio of isotopes to be such that there is (in fact more than) 99 times more than the other. So we find $n$ such that $(1.004)^n > 99 \Rightarrow n$ is roughly 1151. Is this correct?