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Unformatted text preview: Solutions to problem set 12 (141A Sp07) 1 1. (ISSP 9.3). (a) Referring to page 16 of ISSP, we see that the hcp structure has two atoms per unit cell, one at say ~ 0, the other at ~ = (2 / 3) ~ a 1 + ~ a 2 / 3 + ~ a 3 / 2. Then, since U ~ G 1 + e i ~ G (see problem 7.4 from problem set 9), we have U ~ G c 1 + e i (2 / c )( c / 2) = 1 + e i = 0. (b) U ~ 2 G c 1 + e i (4 / c )( c / 2) = 1 + e 2 i = 2, so it is not zero. (c) The electrons from the divalent metal can fill up the lowest band as far as the Brillouin zone boundary. For the simple hexagonal lattice without a basis we expect U ~ G , 0, so the gap can make the material an insulator. (d) For the hcp structure we know that U ~ G c = 0. Therefore there will be no gap at this part of the brillouin zone boundary and the material will be a metal. 2. (ISSP 9.4). Let a be the lattice spacing. Then the radius of the fermi circle is defined by N = 2 k 2 F (2 / L ) 2 k 2 F 2 = n = 2 a 2 k F...
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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
 Solid State Physics

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