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Unformatted text preview: Solutions to problem set 8 (141A Sp07) 1 1. (ISSP 7.1). (a) At the center of a face, ~ k = ( / a , 0) and so E face = ~ 2 k 2 2 m = ~ 2 2 2 ma 2 At the corner, ~ k = ( / a ,/ a ) and so k = 2( / a ), giving E corner = ~ 2 k 2 2 m = 2 ~ 2 2 2 ma 2 Therefore the energy is twice as high at the corner than it is at the face. (b) At the corner of the 3D lattice, ~ k = ( / a ,/ a ,/ a ) and so k = 3( / a ), giving E corner = ~ 2 k 2 2 m = 3 ~ 2 2 2 ma 2 (c) Recall that there are 2 N states in every band. Therefore you might expect that the electrons in the divalent element would fill up the lowest band and it would be a band insulator. However, if the gap at ~ k = ( / a , , 0) to the next highest band is smaller than 3 ~ 2 2 2 ma 2 ~ 2 2 2 ma 2 = 2 ~ 2 2 2 ma 2 , then this band will be partially occupied, the lowest band will be partially empty and the element will be a metal....
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This note was uploaded on 08/01/2008 for the course PHYSICS 141A taught by Professor Souza during the Spring '08 term at University of California, Berkeley.
 Spring '08
 SOUZA
 Energy, Solid State Physics

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