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Unformatted text preview: (see also Sample Problem 206). Since the mass m of the particle is related to M by the Avogadro constant, then v p = r 2 RT M = r 2 RT mN A = r 2 kT m using Eq. 207. With T = 1 . 5 10 7 K and m = 1 . 67 10 27 kg, this yields v p = 5 . 10 5 m/s. (c) The corresponding kinetic energy is K v,p = 1 2 mv 2 p = 1 2 m r 2 kT m ! 2 = kT which is twice as large as that found in part (a). Thus, at T = 1 . 5 10 7 K we have K v,p = 1 . 3 keV, which is indicated in ig. 4410 by a single vertical line....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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