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
a
x
y
z
y
x
x
y
;
2
3
1
2
2
2
3
1
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
3
3
2
c
a
a
V
a
a
a c
π
π
π
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 Spring '08
 WORMER
 Crystallography, Work, Reciprocal lattice, Brillouin zone, reciprocal lattice vectors

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