Ratio of rms speeds of two isotopes

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Uranium has two isotopes, 238U and 235U, with natural abundances of 99.3% and 0.7%, respectively, with 235U being essential for nuclear reactors. The isotopes are separated by forming uranium hexafluoride gas, which diffuses through porous membranes, with 235UF6 having a slightly larger rms speed than 238UF6. The ratio of the rms speeds can be calculated using the formula v_rms = sqrt[3k_B*T/(m)], where the constants can be ignored for a ratio. An initial calculation of the ratio of v_rms for 235U to 238U yielded an incorrect result of 1.0064, but the user later corrected their approach. The discussion emphasizes the importance of accurate calculations in isotope separation processes.
Linus Pauling
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1. Uranium has two naturally occurring isotopes. 238U has a natural abundance of 99.3% and 235U has an abundance of 0.7%. It is the rarer 235U that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, which is a gas, then allowing it to diffuse through a series of porous membranes. has a slightly larger rms speed than and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of 235UF6 to that of 238UF6?

Express answer to five significant figures.




2. v_rms = sqrt[3k_B*T/(m)]



3. Since I need a ratio, I simply ignored the 2*k_B*T as well as the hexfluoride, and computed the ratio of v_rms 235U/235U, obtaining an asnwer of 1.0064, which is incorrect.
 
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Nevermind I got it.
 
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