Example 13 Determine the structure factor and intensity corresponding to the unit
cell of ZnS having zinc blende structure.
Solution Given: Unit cell of ZnS structure, F hkl
ð Þ ¼ ?; I ¼ ?
When two interpenetrating lattices (as considered for the formation of diamond
structure) are of two different elements such as Zn and S, they still produce a
tetrahedral arrangement (like carbon atoms in diamond) as shown in Fig. 8.11a.
A ZnS (zinc blende) structure has equal number of zinc and sulfur ions distributed
on a diamond lattice so that each ion has four opposite ions as the nearest neighbors. The fractional coordinates of Zn and S ions are:
Zn : 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
S : 1=4; 1=4; 1=4
ð
Þ ; 3=4; 3=4; 1=4
ð
Þ ; 1=4; 3=4; 3=4
ð
Þ ; 3=4; 1=4; 3=4
ð
Þ
Substituting these values in Eq. 8.1, we obtain structure factor expression similar
to that of diamond, that is,
F hkl
ð Þ ¼ f Zn : 1 þ exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
½
þ f S :exp
pi h þ k þ l
ð
Þ
2
: 1 þ exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
½
¼ f Zn þ f S :exp
pi h þ k þ l
ð
Þ
2
: 1 þ exppi h þ k
ð
Þþ exp pi k þ l
ð
Þþexppi l þ h
ð
Þ
½
The terms within the bracket [ ] are identical as in fcc and reduces to zero for
mixed hkl indices and 4 for unmixed indices. Now considering the complete
equation for unmixed hkl indices, we can write
Fig. 8.11 ZnS a Zinc blende b Wurtzite structure
320
8 Structure Factor Calculations
cell of ZnS having zinc blende structure.
Solution Given: Unit cell of ZnS structure, F hkl
ð Þ ¼ ?; I ¼ ?
When two interpenetrating lattices (as considered for the formation of diamond
structure) are of two different elements such as Zn and S, they still produce a
tetrahedral arrangement (like carbon atoms in diamond) as shown in Fig. 8.11a.
A ZnS (zinc blende) structure has equal number of zinc and sulfur ions distributed
on a diamond lattice so that each ion has four opposite ions as the nearest neighbors. The fractional coordinates of Zn and S ions are:
Zn : 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
S : 1=4; 1=4; 1=4
ð
Þ ; 3=4; 3=4; 1=4
ð
Þ ; 1=4; 3=4; 3=4
ð
Þ ; 3=4; 1=4; 3=4
ð
Þ
Substituting these values in Eq. 8.1, we obtain structure factor expression similar
to that of diamond, that is,
F hkl
ð Þ ¼ f Zn : 1 þ exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
½
þ f S :exp
pi h þ k þ l
ð
Þ
2
: 1 þ exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
½
¼ f Zn þ f S :exp
pi h þ k þ l
ð
Þ
2
: 1 þ exppi h þ k
ð
Þþ exp pi k þ l
ð
Þþexppi l þ h
ð
Þ
½
The terms within the bracket [ ] are identical as in fcc and reduces to zero for
mixed hkl indices and 4 for unmixed indices. Now considering the complete
equation for unmixed hkl indices, we can write
Fig. 8.11 ZnS a Zinc blende b Wurtzite structure
320
8 Structure Factor Calculations