p44_038 - (see also Sample Problem 20-6). Since the mass m...

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38. (a) Rather than use P ( v ) as it is written in Eq. 20-27, we use the more convenient nK expression given in problem 37 of this chapter [44]. The n ( K ) expression can be derived from Eq. 20-27, but we do not show that derivation here. To Fnd the most probable energy, we take the derivative of n ( K ) and set the result equal to zero: dn ( K ) dK ¯ ¯ ¯ ¯ K = K p = 1 . 13 n ( kT ) 3 / 2 µ 1 2 K 1 / 2 K 3 / 2 kT e K/kT ¯ ¯ ¯ ¯ K = K p =0 , which gives K p = 1 2 kT . SpeciFcally, for T =1 . 5 × 10 7 K we Fnd K p = 1 2 kT = 1 2 (8 . 62 × 10 5 eV / K)(1 . 5 × 10 7 K) = 6 . 5 × 10 2 eV or 0 . 65 keV, in good agreement with ±ig. 44-10. (b) Eq. 20-35 gives the most probable speed in terms of the molar mass M , and indicates its derivation
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Unformatted text preview: (see also Sample Problem 20-6). 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. 20-7. 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. 44-10 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.

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