a 1 ¼
nk
sin tan À1 S n
R
h
i¼
1 Â 0:71
sin tan À1 18
28:65
À
Á
Â
Ã
¼
0:71
sin tan À1 0:628
ð
Þ
½
Š
¼
0:71
sin 32:14
¼
0:71
0:53
¼ 1:33 ˚
A
Case II: Similarly, we can determine the value of “a” corresponding to the second
layer line
a 2 ¼
nk
sin tan À1 S n
R
h
i¼
2 Â 0:71
sin tan À1 40
28:65
À
Á
Â
Ã
¼
1:42
sin tan À1 1:396
ð
Þ
½
Š
¼
1:42
sin 54:39
¼
1:42
0:813
¼ 1:74 ˚
A
Therefore, the average value of a = 1.54 Å.
Example 3 In a rotation photograph six layer lines are observed both above and
below the zero layer line. If the heights of these layer lines above (or below) the
zero layer are 0.29, 0.59, 0.91, 1.25, 1.65, and 2.2 cm, obtain the cell height of the
crystal along the axis of rotation. The radius of the camera is 3 cm and the
wavelength of the X-ray is 1.54 Å.
Solution: Given: S 1 ¼ 0:29 cm, S 2 ¼ 0:59 cm, S 3 ¼ 0:91 cm, S 4 ¼ 1:25 cm, S 5 ¼ 1:65 cm,
S 6 ¼ 2:2 cm, R = 3 cm, k = 1.54 Å = 1.54 Â 10
–8 cm, the unit cell height along
the axis of rotation = ?
Let us suppose that the given photograph is the a-axis photograph and then the
value of the lattice parameter “a” corresponding to the first layer line is
a 1 ¼
nk
sin tan À1 S n
R
h
i¼
1 Â 1:54 Â 10
À8
sin tan À1 0:29
3:0
À Á
Â
Ã
¼
1:54 Â 10
À8
sin tan À1 0:097
ð
Þ
½
Š
¼
1:54 Â 10
À8
sin 5:52
¼
1:54 Â 10
À8
0:096
¼ 16 Â 10
À8 cm ¼ 16 ˚
A
Making a similar calculation after substituting different values of n and S, we can
obtain the same value of the lattice parameter “a.” Therefore the average value of
the unit cell height of the crystal is 16 Å.
9.1 Steps in Crystal Structure Determinations
363