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# 12hw - Solutions to problem set 12(141A Sp07 1 1(ISSP 9.3(a...

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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 ~τ = (2 / 3) ~ a 1 + ~ a 2 / 3 + ~ a 3 / 2. Then, since U ~ G ∼ 1 + e- i ~ G · τ (see problem 7.4 from problem set 9), we have U ~ G c ∼ 1 + e- i (2 π/ c )( c / 2) = 1 + e- i π = 0. (b) U ~ 2 G c ∼ 1 + e- i (4 π/ c )( c / 2) = 1 + e- 2 i π = 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 π k 2 F (2 π/ L ) 2 ⇒ k 2 F 2 π = n = 2 a 2 ⇒ k F...
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12hw - Solutions to problem set 12(141A Sp07 1 1(ISSP 9.3(a...

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