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# 06hw - Solutions to problem set 6(141A Sp07 1 1(ISSP 6.2(a...

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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 0 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 . 0 × 10 57 Therefore there will be 10 57 electrons in the star. The electron density is then n = 10 57 4 3 π (2 × 10 7 ) 3 = 3 32 π 2 × 10 36 = 1 . 0 × 10 34 m - 3 Then k F == (3 π 2 n ) 1 / 3 (30 π 2 10 33 ) 1 / 3 = 6 . 7 × 10 11 . Then v F = ( / m ) k F 7 × 10 7 m / s. We’re on the border of the relativistic region: v / c . 2. Therefore we can use the non-relativistic limit but only trust our answer up to ( v / c ) 2 . 01.

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06hw - Solutions to problem set 6(141A Sp07 1 1(ISSP 6.2(a...

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