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Unformatted text preview: Solutions to problem set 6 (141A Sp07) 1 1. (ISSP 6.2) (a) p = U V ! S , N At T = 0 the entropy is a constant as V is varied. Therefore, since E F = CV 2 / 3 , p = 3 5 NC ( ) 2 3 V 5 / 3 = 3 5 NE F 2 3 V 1 = 2 3 U V (b) From part (a), p = C V 5 / 3 where C is a constant. Then B = V p V = 5 3 p (c) B = 10 9 U V = 2 3 nE F = 2 3 (1 . 4 10 28 )(2 . 12)(1 . 6 10 19 ) = 3 . 17 10 9 J / m 3 = 3 . 17 GPa Wikipedia gives the bulk modulus of potassium as 3.1 GPa (presumably at room tem perature). 2. (ISSP 6.4) (a) The sun is 70% hydrogen and 30% helium; therefore about 85% of its mass comes from protons, giving Number of electrons = Number of protons = . 85(2 10 30 ) 1 . 67 10 27 = 1 . 10 57 Therefore there will be 10 57 electrons in the star. The electron density is then n = 10 57 4 3 (2 10 7 ) 3 = 3 32 2 10 36 = 1 . 10 34 m 3 Then k F == (3 2 n ) 1 / 3 ' (30 2 10 33 ) 1 / 3 = 6 . 7 10 11 . Then v F = ( ~ / m ) k F ' 7 10...
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 Spring '08
 SOUZA
 Solid State Physics, Entropy

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