How Is Uranium Isotope Enrichment Achieved Through Effusion?

[Pouyan]

1. In a gas UF6 (uranium hexafluoride) are uranium atoms of both the fissionable uranium isotope 235U , and 238U. To enrich the fissile isotope can let a gas UF6 with natural isotopic composition (0.7% 235U, 99.3% 238U) undergo effusion process. The process is then repeated in many steps in the effusion process, so that 235U progressively enriched, from the initial concentration of 0.7%. After enrichment in the first step of the process, the concentration of 235U increased from 0.7% to k * 0.7%. After the enrichment step n is the concentration (k^n) * 0.7%. What is the value of n when the concentration increased to 0.8% ?! we assume that the system is built so that molecules can not pass back to the original volume of gas, once again gone through the small holes (channels) that give rise to effusion. Relative atomic mass of fluorine is 19

Homework Equations

I know the formula :

v*= p/(sqrt(2pi*m*k*T))

p= pressrue
k= Boltzman constant
m= mass
T = temperature

The Attempt at a Solution

I know k is the number of collision for (235)UF6 and (238)UF6. In this case :

v* for (235)UF6 / v* (238)UF6
which will be
k= sqrt(m for (238)UF6 / m for (235)UF6) = sqrt((238 + 6*19)/235+6*19) = 1.004

and (1.004) ^n = 0.8/0.7

n= 33.5

This is my attempt and the original answer is a number near to it. My question is :

Should I think that pressure and temperature are constants in this case and the only variable is m ?!

 

[Pouyan]

Can somebody help me ?!