cos c ¼
0
ffiffiffiffi ffi
1
2
p
ffiffiffiffi ffi
3
2
p ¼ 0 ) c ¼ 90
(c) Area of the new unit cell a′ x b′ = 2 Â 6 = 12 Å
2
(d) Ratio of the two areas =
12
4 ¼ 3
This implies that the new unit cell is non-primitive and the number of lattice
points in it is 3.
Example 7 A primitive orthorhombic unit cell has a = 5Å, b = 6Å and c = 7Å;
a = b = c = 90°. A new unit cell is chosen with the edges defined by the vectors
from the origin to the points with coordinates 3, 1, 0; 1, 2, 0 and 0, 0, 1. Determine
(a) volume of the original unit cell (b) lengths of the three edges and angles between
them (c) volume of the new unit cell (d) the number of the lattice points in the new
unit cell.
Solution: Given: Orthorhombic unit cell parameters, a = 5Å, b = 6Å and c = 7Å;
a = b = c = 90°, coordinates of the new cell edges: 3, 1, 0; 1, 2, 0 and 0, 0, 1.
(a) Volume of the primitive orthogonal unit cell,
V 0 ¼ abc
¼ 5 Â 6 Â 7 ¼ 210 ˚
A
3
(b) Length of the vector a′ between the coordinates: 0, 0, 0 and 3, 1, 0
a
0
¼ 0 À 3
ð
Þ
2 Â5
2
þ 0 À 1
ð
Þ
2 Â6
2
þ 0
h
i 1=2 ¼ 9 Â 25 þ 36
½
Š
1=2 ¼ 16.16 ˚
A
Similarly, the length of the vector b′ between the coordinates; 0, 0, 0 and 1, 2, 0
b
0
¼ 0 À 1
ð
Þ
2 Â5
2
þ 0 À 2
ð
Þ
2 Â6
2
þ 0
h
i 1=2 ¼ 25 þ 4 Â 36
½
Š
1=2 ¼ 13 ˚
A
and the length of the vector c
0 between the coordinates: 0, 0, 0 and 0, 0,1
c
0
¼ 0 þ 0 þ 0 À 1
ð
Þ
2 Â7
2
h
i 1=2 ¼ 7
½ Š
1=2 ¼ 7 ˚
A
Now the angle between the edges with the end coordinates 3, 1, 0 and 1, 2, 0 is
given by
cos c
0
¼
3 þ 2
9 þ 1 þ 0
ð
Þ 1=2 ð1 þ 4 þ 0Þ
1=2
¼
5
10
ð Þ
1=2 ð5Þ
1=2
¼
1
ffiffi ffi
2
p
) c
0
¼ 45
Similarly, the angle between the edges with the end coordinates 1, 2, 0 and 0, 0,
1 is given by
14
1 Unit Cell Composition