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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 WignerSeitz 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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 Spring '08
 SOUZA
 Solid State Physics

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