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A quick question about U238 vs U235

  • Thread starter Kites
  • Start date
  • #1
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Homework Statement



Uranium has two naturally occurring isotopes. U_238 has a natural abundance of 99.3% and U_235has an abundance of 0.7%. It is the rarer U_235 that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride UF_6, which is a gas, then allowing it to diffuse through a series of porous membranes. 235UF_6 has a slightly larger rms speed than 238UF_6 and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of 238UF_6? to that of 238UF_6 ?



Homework Equations



The relevant equation is only this.

V_rms = ([3*k_b*T]/m)^(1/2)



The Attempt at a Solution



My attempt thus far is

V_rms_1/V_rms_2 = solution ; Simple right?

Let's go further...

1u= 1.6691729*10^-27
([3*k_b*T]/(238*1u)^(1/2)/([3*k_b*T]/(235*1u)^(1/2)

This has come out wrong several times. I believe my error lies in converting the masses somehow since it's the only possible variable. Sadly I don't have a strong background in chem or working with these. Help would be very appreciated.
 

Answers and Replies

  • #2
147
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Since 3*k_b*T will be the same for both of the uranium isotopes you could ignore it and leave it as a variable 'A' or just completely remove it.

Then you could get an easier formula V_rms = (A/m)^1/2

When you set the ratio of V_rms_1 / V_rms_2 just go ahead and cancel the A values out and the formula you get will look better. (And again, to form a ratio you will not need to convert amu units to kg, a ratio is unitless and as long as you have the same mass units for both isotopes the equation will yield the same results)
 

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