Now, join BD and CE and construct another unit cell. The required unit cell is
DBCE. Consider the triangle BCD, where the < BCD = 90°. Therefore,
BD ¼ 4
2
þ 3
2
À
Á 1=2 ¼ 25
ð Þ
1=2 ¼ 5 ˚
A
Further, the angle made by this side on BC is
c ¼ sin
À1 3
5
¼ 36.87
Example 2 A primitive cell of square lattice has a = 2 Å. A new unit cell is
chosen with edges defined by the vectors from the origin to the points with coordinates 1, 0 and 0, 3. Calculate: (i) area of the original unit cell, (ii) length of the
two edges and angle between them, (iii) area of the new unit cell, and (iv) number
of lattice points in the new unit cell.
Solution: Given: a = 2 Å, for a square lattice, c = 90°.
(i) Area of the original unit cell, a
2
¼ 4 ˚
A
2
(ii) Length of the edges from the origin:
(a) For coordinates 0, 0 and 1, 0
a ¼ 0 À 1
ð
Þ
2 Â2
2
þ 0
h
i 1=2 ¼ 4
ð Þ
1=2 ¼ 2 ˚
A
(b) For coordinates 0, 0 and 0, 3
b ¼ 0 þ 0 À 3
ð
Þ
2 Â2
2
h
i 1=2 ¼ 36
ð Þ
1=2 ¼ 6 ˚
A
Angle between the lines with end coordinates 1, 0 and 0, 3:
cos c ¼
0
ffiffi ffi
1
p ffiffi ffi
3
p ¼ 0; ) c ¼ 90
(iii) Area of the new unit cell, ab = 2 Â 6 = 12 ˚
A
2
(iv) Ratio of the two areas =
12
4 ¼ 3. Therefore, the number of lattice points in
the new unit cell is 3
Fig. 3.10 Rectangular unit
cells
112
3 Unit Cell Calculations