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Unformatted text preview: 1 Homework # 2  Solution Problem #1 The primitive vectors in the simple cubic lattice are 1 2 3 ˆ ˆ ˆ ; ; a a a = = = a x a y a z . The reciprocal lattice vectors are 1 2 3 2 2 2 ˆ ˆ ˆ ; ; a a a π π π = = = b x b y b z . We can now build vectors G : 1 2 3 2 2 2 ˆ ˆ ˆ h k l h k l a a a π π π = + + = + + G b b b x y z . This gives an interplane distance: 2 2 2 2 a d h k l π = = + + G . Problem #2 The primitive vectors are 1 2 3 3 1 3 1 ˆ ˆ ˆ ˆ ˆ ; ; 2 2 2 2 a a a a c = + =  + = a x y a x y a z a) The volume of the primitive cell is given by ( 29 2 1 2 3 3 1 3 1 3 1 3 1 3 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 2 2 2 2 2 2 2 V a a a a c a a ac ac a c = × = + + × = + + = a a a x y x y z x y y x b) The reciprocal lattice vectors are defined as 1 2 3 2 π = × b a a V , 2 3 1 2 π = × b a a V , 3 1 2 2 π = × b a a V , where ( 29 1 2 3 = ⋅ × a a a V is the volume of the unit cell. Therefore, 1 2 3 2 2 2 3 1 4 3 1 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 2 2 3 3 3 2 a a c V a a a a c π π π π π = × = + × = + = + b a...
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This note was uploaded on 09/21/2011 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at Aarhus Universitet.
 Spring '08
 WORMER
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