KmvmkTmkTvpp,==FHGIKJ=1212222which is twice as large as that found in part (a). Thus, at T= 1.5 ×107K we have Kv,p= 1.3 keV, which is indicated in Fig. 43-10 by a single vertical line. 54. (a) Rather than use P(v) as it is written in Eq. 19-27, we use the more convenient nKexpression given in Problem 43-32. The n(K) expression can be derived from Eq. 19-27, but we do not show that derivation here. To find the most probable energy, we take the derivative of n(K) and set the result equal to zero: dn KdKnkTKKkTeKKKkTpp().,////=−==−FHGIKJ=1131203212which gives KkTp=12. Specifically, for T= 1.5 ×107K we find
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.