# 02hw - Solutions to problem set 2(141A Sp07 1 1(ISSP 2.2(a...

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Solutions to problem set 2 (141A Sp07) 1 1. (ISSP 2.2) (a) V c = ~ a 1 · ( ~ a 2 × ~ a 3 ) = a 2 c ( 3 2 ˆ x + 1 2 ˆ y ) · ( - 3 2 ( - ˆ y ) + 1 2 ˆ x ) = a 2 c 3 2 (b) We take the deﬁnitions of ~ b 1 , ~ b 2 and ~ b 3 from page 29 of ISSP. ~ b 1 = 2 π a 2 c 3 2 ± - ac 3 2 ( - ˆ y ) + ac 1 2 ˆ x ! = 2 π a ± 1 3 ˆ x + ˆ y ! ~ b 2 = 2 π a 2 c 3 2 ± ac 3 2 ( ˆ y ) + ac 1 2 ( - ˆ x ) ! = 2 π a ± - 1 3 ˆ x + ˆ y ! ~ b 3 = 2 π a 2 c 3 2 ± a 2 3 2 1 2 z ) + a 2 3 2 1 2 z ) ! = 2 π c ˆ z (c) The ﬁrst Brillouin zone is a hexagonal prism, ie., any horizontal slice (parallel to the xy plane) shows a hexagonal cross-section. 2. (ISSP 2.4) (a) Note that (1 - e - i φ )(1 - e i φ ) = 1 - e i φ - e - i φ + 1 = 2 - 2 cos φ = 4 sin 2 ( φ/ 2) Then | F | 2 = 4 sin 2 1 2 M ~ a · ~ Δ k 4 sin 2 1 2 ~ a · ~ Δ k = sin 2 1 2 M ~ a · ~ Δ k sin 2 1 2 ~ a · ~ Δ k (b) Note that when | F | 2 is a maximum, sin( M ~ a · ~ Δ k / 2) is equal to zero (the maximum comes from the fact that sin( ~ a · ~ Δ k / 2) is also zero at that point.) To get to the nearest zero of | F | 2 , we need to change the phase slightly to get to the next zero of sin( M ~ a · ~ Δ k / 2) (we should also check that sin( ~ a · ~ Δ k / 2) is not zero at that point). To move from one zero to the next of the sin function, we need to change the argument by π . This gives M 2 ± = π ± = 2 π M Also, sin( ~ a · ~ Δ k / 2 + π/ M ) , 0, and so we get a zero of | F | 2 for ± = 2 π/ M . 3. (ISSP 2.5)

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Solutions to problem set 2 (141A Sp07) 2 (a) Firstly, we have the four atoms associated with the fcc Bravais lattice: τ 1 = ~ 0, τ 2 = a 2 x + ˆ y ), τ 3 = a 2 ( ˆ y + ˆ z ), τ 4 = a 2 z + ˆ x ). To each of these we must add another atom at i + ~ Δ , where ~ Δ = a 4 x + ˆ y + ˆ z ). This gives a total of 8 terms in the structure factor.
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02hw - Solutions to problem set 2(141A Sp07 1 1(ISSP 2.2(a...

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