Solved Examples
Example 1 In a rotation photograph, seven layer lines are observed, three above
and three below the zero layer line. A millimeter scale placed next to the developed
film provides the following readings:
Layer line
Height (mm)
Height w.r.t zero layer
3
05.40
34.00
2
22.44
16.96
1
31.84
07.56
0
39.40
–
-1
46.96
07.56
-2
56.35
16.95
-3
73.20
33.80
Determine the value of the cell height, if the diameter of the camera is 57.3 mm
and the wavelength of the X-ray used is 1.542 Å.
Solution: Given: mm scale reading of three upper and three lower layer lines,
D = 57.3 mm = 5.73 cm, so that R = 2.865 cm, k = 1.542 Å = 1.542 Â 10
–8 cm.
From the given data, we obtain the height of the layer lines: S 1 ¼ 7:56 mm ¼
0:756 cm, S 2 ¼ 16:96 mm ¼ 1:696 cm, S 3 ¼ 34 mm ¼ 3:4 cm. Now, to determine the cell parameter along a particular axis, make use of the equation (say)
a ¼
nk
S n
R
2
þ S
2
n
À
Á 1=2
For n = 1,
a 1 ¼
1 Â 1:542 Â 10
À8
0:756
2:865
ð
Þ
2 þ 0:756
ð
Þ
2
h
i 1=2
¼ 6:044 Â 10
À8 cm ¼ 6:044 ˚
A
Similar calculations will give us
a 2 ¼ 6:054 ˚
A
Therefore, the average value of a = 6.05 Å.
Example 2 In a rotation photograph, first and second layer lines are observed (w.r.
t zero layer line) at 18 mm and 40 mm, respectively. If the wavelength of the
MoKa radiation is 0.71 Å and the film diameter 57.3 mm, determine the lattice
parameter of the sample.
Solution: Given: S 1 = 18 mm and S 2 = 40 mm, D = 57.3 mm, so that
R = 28.65 mm, k = 0.71 Å, a = ?
Case I: We can determine the value of “a” corresponding to the first layer line by
using the formula
362
9 Determination of Crystal Structure Parameters
Example 1 In a rotation photograph, seven layer lines are observed, three above
and three below the zero layer line. A millimeter scale placed next to the developed
film provides the following readings:
Layer line
Height (mm)
Height w.r.t zero layer
3
05.40
34.00
2
22.44
16.96
1
31.84
07.56
0
39.40
–
-1
46.96
07.56
-2
56.35
16.95
-3
73.20
33.80
Determine the value of the cell height, if the diameter of the camera is 57.3 mm
and the wavelength of the X-ray used is 1.542 Å.
Solution: Given: mm scale reading of three upper and three lower layer lines,
D = 57.3 mm = 5.73 cm, so that R = 2.865 cm, k = 1.542 Å = 1.542 Â 10
–8 cm.
From the given data, we obtain the height of the layer lines: S 1 ¼ 7:56 mm ¼
0:756 cm, S 2 ¼ 16:96 mm ¼ 1:696 cm, S 3 ¼ 34 mm ¼ 3:4 cm. Now, to determine the cell parameter along a particular axis, make use of the equation (say)
a ¼
nk
S n
R
2
þ S
2
n
À
Á 1=2
For n = 1,
a 1 ¼
1 Â 1:542 Â 10
À8
0:756
2:865
ð
Þ
2 þ 0:756
ð
Þ
2
h
i 1=2
¼ 6:044 Â 10
À8 cm ¼ 6:044 ˚
A
Similar calculations will give us
a 2 ¼ 6:054 ˚
A
Therefore, the average value of a = 6.05 Å.
Example 2 In a rotation photograph, first and second layer lines are observed (w.r.
t zero layer line) at 18 mm and 40 mm, respectively. If the wavelength of the
MoKa radiation is 0.71 Å and the film diameter 57.3 mm, determine the lattice
parameter of the sample.
Solution: Given: S 1 = 18 mm and S 2 = 40 mm, D = 57.3 mm, so that
R = 28.65 mm, k = 0.71 Å, a = ?
Case I: We can determine the value of “a” corresponding to the first layer line by
using the formula
362
9 Determination of Crystal Structure Parameters
