The structure factor for Cr is simplified by substituting N = 2, coordinates of two
atoms: x 1 ; y 1 ; z 1 ¼ 0; 0; 0 Á Á Á and x 2 ; y 2 ; z 2 ¼ 1=2; 1=2; 1=2; we obtain
F hkl
ð Þ ¼
X 2
i¼1
f Cr
1
2d hkl
e
2piH hkl
ð Þ:r j ¼ f Po
1
2d hkl
1 þ exppi h þ k þ l
ð
Þ
½
Š
Now, substituting different hkl values along with Cromer–Mann coefficients in
Eq. 8.5, the required structure factor can be calculated. The calculation for the
second value of hkl = 011, when h þ k þ l
ð
Þis an even integer, is
F 011
ð
Þ ¼ 2f 0:2452
ð
Þ¼2
X 4
i¼1
a i e
Àbi 0:2452
ð
Þ
2 þ c
¼ 2 a 1 e
Àb 1 0:2452
ð
Þ
2 þ a 2 e
Àb 2 0:2452
ð
Þ
2 þ a 3 e
Àb 3 0:2452
ð
Þ
2 þ a 4 e
Àb 4 0:2452
ð
Þ
2 þ c
¼ 2ð10:6406e
À6:1038 0:2452
ð
Þ
2 þ 7:3537e
À0:392 0:2452
ð
Þ
2 þ 3:324e
À20:2626 0:2452
ð
Þ
2
þ 1:4922e
À98:7399 0:2452
ð
Þ
2 þ 1:1832Þ
¼ 2 7:372065 þ 7:18241269 þ 0:9830648 þ :003941 þ 1:1932
ð
Þ
¼ 2 Â 16:73468 ¼ 33:46
Calculations for other reflections can be made similarly. They are shown below
in tabulated form for the initial 12 hkl values.
Structure factor calculation for Cr (Chromium)
hkl
h
2 þ k
2 þ l
2
ð
Þ
1=2
2a
F Cr hkl
ð Þ¼ 2f Cr ; ðelectronÞ
000
0
48
001
0.1734
0
011
0.2452
33.46
020
0.3467
27.19
021
0.3877
0
101
0.2452
33.45
111
0.3003
0
120
0.3877
0
121
0.4247
23.32
220
0.4904
20.70
221
0.5201
0
300
0.5201
0
301
0.5482
18/85
Example 6 Determine the general form of structure factor and intensity corresponding to a base-centered unit cell.
Solution Given: Base-centered unit cell, F hkl
ð Þ ¼ ?; I ¼ ?
308
8 Structure Factor Calculations