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Centrifuging may be used for the separation of isotopes. To do this a mixture of two gases is placed inside a cylindrical vessel rotating at a high speed. Because of the action of centrifugal forces, the isotope concentration near the cylinder wall will be different from that in the contre. Compare the concentrations of the light and the heavy uranium isotopes near the centrifuge walls, if the diameter of the cylinder is 10 cm, the rotation speed is 2.0 xx 10^3 r. p.s., the temperature of uranium hexafluoride is 27^@C. For concentrations in normal conditions see xi 25.6. Find the enrichment factor in the mixture of the heavy isotope near the walls of the vessel. The term enrichment factor applies to the quotient obtained by dividing the concentration ratio during rotation by the initial concentration ratio: x = ((n_2)/(n_1)) ((n_(02))/(n_(01))) |
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Answer» `n_(02) n_(01) = 99:28 : 0.715 = 139:1`. During the rotation of the centrifuge, the isotope of greater mass DUE to the action of centrifugal forces CONCENTRATES near the walls of the vessel. Using the result obtained in the previous problem, we have `n_1 = n_(01) exp ((m_1 omega^2 r^2)/(2kT)) = n_(01) exp ((M_1 omega^2 r^2)/(2RT))` `n_2 = n_(02) exp ((M_2 omega^2 r^2)/(2RT))` The concentration ratio close to the vessel.s walls is `(n_2)/(n_1) = (n_(02))/(n_(01)) exp [((M_2 - M_1)omega^2 r^2)/(2RT)]` Taking logarithms , we obtain `"log" ((n_2)/(n_1)) = "log" ((n_(02))/(n_(01))) +(0.434 (M_2 - M_1)omega^2 r^2)/(2RT)` `= "log " 139+ (0.434 xx 3 xx 4 pi^2 xx 4 xx 10^(6) xx 25 xx 10^(-4))/(2 xx 8.3 xx 10^3 xx 3 xx 10^2)` `2.143 + 0.103 = 2.246` Hence `n_2 n_1 = 176` 1. The enrichment factor is `x = ((n_2)/(n_1)) ((n_(02))/(n_(01))) = exp [((M_2 - M_1) omega^2 r^2)/(2RT)]` The obvious result is `log x = 0.103 `, given x = 1.27. |
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