Solution: Given: Sides of primitive rectangle, a = 2Å, b = 3Å, coordinates of new
cell edges: 2, 0 and 0, 3. (a) Area of the primitive rectangle cell = a  b = 2
3 = 6 Å
2 .
(a) Lengths of the edges from the origin:
(i) For coordinates 0, 0 and 2, 0
a
0
¼ ð0 À 2Þ
2 Â 2
2
þ 0
h
i 1=2 ¼ ½16Š
1=2 ¼ 4 ˚
A
(i) For coordinates 0, 0 and 0, 3
b
0
¼ 0 þ 0 À 3
ð
Þ
2
 3
2
Â
à 1=2 ¼ 81
½ Š
1=2
¼ 9 ˚
A
Angle between the lines with the end coordinates 2, 0 and 0, 3
cos c ¼
0
ffiffiffiffi ffi
2
2
p
ffiffiffiffi ffi
3
2
p ¼ 0 ) c ¼ 90
(b) Area of the new unit cell a
0
 b
0
¼ 4 Â 9 ¼ 36 ˚
A
2
(c) Ratio of the two areas =
36
6 ¼ 6
(d) The new unit cell is non-primitive and the number of lattice points in it is 6.
Example 6 The side of a primitive square unit cell is, a = 2Å. A new unit cell is
chosen with the edges defined by the vectors from the origin to the points with
coordinates 1, 0 and 0, 3. Determine (a) area of the original unit cell (b) lengths of
the two edges and angle between them (c) area of the new unit cell (d) the number
of the lattice points in the new unit cell.
Solution: Given: Sides of primitive square unit cell, a = 2Å, coordinates of new
cell edges: 1, 0 and 0, 3. We know that for a square lattice, c = 90°. Therefore,
(a) Area of the primitive square unit cell = a
2 = 4 Å
2
(b) Lengths of the edges from the origin:
(i) For coordinates 0, 0 and 1, 0
a
0
¼ ð0 À 1Þ
2 Â 2
2
þ 0
h
i 1=2 ¼ ½4Š
1=2 ¼ 2 ˚
A
(ii) For coordinates 0, 0 and 0, 3
b
0
¼ 0 þ ð0 À 3Þ
2 Â 2
2
h
i 1=2 ¼ ½36Š
1=2 ¼ 6 ˚
A
Angle between the lines with the end coordinates 1, 0 and 0, 3
1.2 Choice of Axes and Unit Cells
13