# 01hw - Solutions to problem set 1 (141A Sp07) 1 Figure 1:...

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Solutions to problem set 1 (141A Sp07) 1 Figure 1: Q1 (a). The Bravais lattice is triangular. Figure 2: Q1 (b). The primitive lattice vectors. This choice is not the only one possible. 1. (a) See ﬁgure 1 for the solution. You can see that the lattice is triangular. (b) See ﬁgure 2. I’ll assume that the distance between two black dots of the original lattice is a . Geometry gives that | ~ a 1 | and | ~ a 2 | equal a 3. ~ a 1 = a 3 1 2 ˆ x + 3 2 ˆ y ! ~ a 2 = a 3 1 2 ˆ x - 3 2 ˆ y ! (c) The Wigner-Seitz cell is shown in ﬁgure 3. The area of an equilateral triangle of side a is ( 3 / 4) a , and so the area of the hexagon is (3 3 / 2) a 2 . If you had used a as the distance between a red and black site (as I probably should have), then the area is 6 3 a 2 . (d) See ﬁgure 4. There are many choices possible. 2. (a) See ﬁgure 5. We are looking down on the hcp conﬁguration. On the bottom plane there is an atom at the vertices of the equilateral triangle. There is another atom in the center of triangle, raised a distance c / 2 relative to the centers of the three atoms below it. Imagine a line going from the center of one of the lower atoms to the center of the raised atom. Its length is a , and it forms the hypotenuse of a triangle with base x and height c / 2. From the geometry, cos 30

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## 01hw - Solutions to problem set 1 (141A Sp07) 1 Figure 1:...

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