h þ 2k
ð
Þ
3
þ
l
2
!
¼ n where n is an integer:
In such a situation,
cospn ¼ Æ1 and cos
2
pn ¼ 1
Therefore,
F(hkl)
j
j
2 ¼ 4f
2
:
The results obtained from considering all possible h, k and l values are summarized as follows:
h þ 2k
ð
Þ
l
FðhklÞ
j
j
2
3n
Odd
0
3n
Even
4f
2
3n Æ1
Odd
3f
2
3n Æ1
Even
f
2
Example 11 Determine the general form of structure factor and intensity corresponding to the unit cell of CsCl structure.
Solution Given: Unit cell of CsCl structure, F hkl
ð Þ ¼ ?; I ¼ ?
CsCl unit cell contains Cs
þ ions at all the 8 corners of the cube while one Cl
À
ion is situated at the body center of the cube (Fig. 8.9). Thus, the structure has one
Cs
þ ion and one Cl
À ion per unit cell. The fractional coordinates of Cs ion is (0, 0,
0) and that of the Cl ion is (1/2, 1/2, 1/2), respectively. Substituting these values in
Eq. 8.1, we obtain
F hkl
ð Þ ¼ f Cs :exp2pi h:0 þ k:0 þ l:0
ð
Þ þ f Cl :exp2pi h:
1
2
þ k:
1
2
þ l:
1
2
Fig. 8.9 CsCl crystal
structure
8.2 Determination of Phase Angle, Amplitude …
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