p44_041 - 41(a From H = 0.35 = np mp we get the proton...

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41. (a) From ρ H =0 . 35 ρ = n p m p ,wegettheprotonnumberdens ity n p : n p = 0 . 35 ρ m p = (0 . 35)(1 . 5 × 10 5 kg / m 3 ) 1 . 67 × 10 27 kg =3 . 14 × 10 31 m 3 . (b) From Chapter 20 (see Eq. 20-9), we have N V = p kT = 1 . 01 × 10 5 Pa (1 . 38 × 10 23 J / K)(273 K) =2 . 68 × 10 25 m
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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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