where f Cs and f Cl are the scattering factors of Cs
þ and Cl
À ions, respectively.
Simplifying the above equation, we obtain
F hkl
ð Þ ¼ f Cs þ f Cl :exppi h þ k þ l
ð
Þ
¼ f Cs þ f Cl ; if h þ k þ l
ð
Þ is an even integer;
¼ f Cs À f Cl ; if h þ k þ l
ð
Þ is an odd integer
Similarly,
F hkl
ð Þ
j
j
2 ¼ f Cs þ f Cl :cosp h þ k þ l
ð
Þ
f
g
2 þ f Cl :sinp h þ k þ l
ð
Þ
f
g
2
¼ f Cs þ f Cl
ð
Þ
2 ; if h þ k þ l
ð
Þ is an even integer;
¼ f Cs À f Cl
ð
Þ
2 ; if h þ k þ l
ð
Þ is an odd integer
Although, this structure resembles with bcc, but the results suggest that unlike
bcc the reflections in CsCl structure are present (with a diminished intensity) even if
(h + k+l) is an odd integer.
Example 12 Determine the general form of structure factor and intensity corresponding to the unit cell of NaCl structure.
Solution Given: Unit cell of NaCl structure, F hkl
ð Þ ¼ ?; I ¼ ?
NaCl structure consists of equal number of sodium and chlorine ions placed at
alternate points of a simple cubic lattice as shown in Fig. 8.10. There are four NaCl
molecules per unit cell. The fractional coordinates of sodium and chlorine ions are:
Na : 0; 0; 0
ð
Þ; 1=2; 1=2; 0
ð
Þ ; 0; 1=2; 1=2
ð
Þ ; 1=2; 0; 1=2
ð
Þ
Cl : 1=2; 1=2; 1=2
ð
Þ ; 0; 0; 1=2
ð
Þ ; 0; 1=2; 0
ð
Þ ; 1=2; 0; 0
ð
Þ
Now, substituting these values in Eq. 8.1 with the assumption that f Na and f Cl
are the scattering factors of Na
þ and Cl
À ions, respectively, we obtain
Fig. 8.10 NaCl crystal
structure
318
8 Structure Factor Calculations