Similarly, for the second set of fractional coordinates, we have
L 2 ¼
0:410 À 0:210
ð
Þ
2 Â3
2
þ 0:050 þ 0:150
ð
Þ
2 Â4
2
þ
2 0:410 À 0:210
ð
Þ0:050 þ 0:150
ð
Þ Â 3 Â 4cos115
"
# 1=2
¼ 0:2
ð Þ
2 Â9 þ 0:2
ð Þ
2 Â16 þ 2 0:2
ð Þ À0:1
ð
ÞÂ12 Â À0:423
ð
Þ
h
i 1=2
¼ 0:04 Â 9 þ 0:04 Â 16 þ 24 Â 0:04
ð
ÞÂ À0:423
ð
Þ
½
Š
1=2
¼ 0:36 þ 0:64 À 0:406
½
Š
1=2
¼ 0:594
½
Š
1=2 ¼ 0:770 ˚
A
Example 2 A unit cell has the following dimensions: a = 5 Å, b = 8 Å and
c = 50°. Calculate the length of the vectors with the help of following components:
(i) [0.50, 1.0] and (ii) [0.75, −0.25].
Solution: Given: a = 5 Å, b = 8 Å and c = 50°. Vector components:
(i) x 2 À x 1
ð
Þ¼0:50; y 2 À y 1
ð
Þ¼1:00: (ii) x 2 À x 1
ð
Þ¼0:75; y 2 À y 1
ð
Þ¼À0:25:
Making use of Eq. 3.3 in slightly modified form, we can obtain the length of the
vectors for the given components as:
U ¼ u 1 a
ð Þ
2 þ u 2 b
ð Þ
2 þ 2u 1 u 2 ab cos50
h
i 1=2
¼ 5 Â 0:5
ð
Þ
2 þ 8 Â 1
ð
Þ
2 þ 2 Â 0:5 Â 1 Â 5 Â 8 Â 0:643
h
i 1=2
¼ 6:25 þ 64 þ 25:72
½
Š
1=2 ¼ 95:97
½
Š
1=2 ¼ 9:8 ˚
A
Similarly,
V ¼ v 1 a
ð
Þ
2 þ v 2 b
ð Þ
2 þ 2v 1 v 2 ab cos50
h
i 1=2
¼ 0:75 Â 5
ð
Þ
2 þ À0:25 Â 8
ð
Þ
2 þ 2 Â 0:75 Â À0:25
ð
ÞÂ5 Â 8 Â 0:643
h
i 1=2
¼ 14:06 þ 4 À 9:645
½
Š
1=2 ¼ 8:42
½
Š
1=2 ¼ 2:9 ˚
A
Example 3 A unit cell has the following dimensions: a = 2 Å, b = 3 Å and
c = 90°. Calculate the distance between the points with fractional coordinates:
(i) 0.100, 0.250 and 0.200, 0.050, (ii) 0.100, −0.250 and 0.210. 0.050.
Solution: Given: a = 2 Å, b = 3 Å and c = 90°. Fractional coordinates: (i) 0.100,
0.250 and 0.200, 0.050, (ii) 0.100, −0.250 and 0.210. 0.050.
106
3 Unit Cell Calculations