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Unformatted text preview: Physics 240B, Spring 2008 Homework 1 Solutions Jesse Noffsinger February 14, 2008 Solutions are in general not the original work of the author. Problem 1) Chapter 6 of Kittel is very helpful for this problem. Portions not worked out explicitly here appear in the recommended texts. a) The total number of electrons up to wavevector k is: N e = 2 k 3 (2 /L ) 3 = V 3 2 k 3 (1) The density of states, D ( k ) is: D ( k ) = dN dk = V 2 k 2 (2) while: = h 2 k 2 2 m (3) Yielding D ( ) = dN d = V 2 2 parenleftbigg 2 m h 2 parenrightbigg 3 / 2 1 / 2 (4) b) The total number of electrons is: N = V 3 2 parenleftbigg 2 m F h 2 parenrightbigg 3 / 2 1 / 2 F = 3 N F bracketleftBigg V 2 parenleftbigg 2 m h 2 parenrightbigg 3 / 2 bracketrightBigg 1 (5) Entering this expression for 1 / 2 F into Eqn (4) gives: D ( F ) = 3 N 2 F (6) 1 c) C v = U T (7) The total energy of the sytem being U = integraldisplay f ( ,T ) D ( ) d (8) which depends on temperature only through the Fermi-Dirac distribution. The Sommerfield expansion [Eqn 2.70 in Ashcroft and Mermin] can be applied, or a direct following of Kittels derivation gives: C v = 2 3 k 2 b TD ( F ) (9) d) If we associate an energy H with spin up/down electrons, we can find the paramagnetic susceptibility of the free electron gas, defined as = M H (10) The total magnetic moment per volume is M = ( n + n ), arising from the difference in the number of spin up and spin down electrons. M = 1 2 integraldisplay [ f ( H ) f ( + H )] D ( ) d (11) The 1/2 accounts for the spin in the standard density of states formula. Taking H to be small with respect to the electronic energies [Ziman pg.332], and the energy derivative of the Fermi-Dirac distribution to be a delta function peaked at F : M = 2 HD ( F ) (12) or = 2 D (...
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This note was uploaded on 08/01/2008 for the course PHYSICS 240B taught by Professor Cohen during the Spring '08 term at University of California, Berkeley.
- Spring '08