F hkl
ð Þ ¼ f Au :exp2pi h:0 þ k:0 þ l:0
ð
Þ
þ f Cu exp2pi h:
1
2
þ k:
1
2
þ l:0
þ exp2pi h:0 þ k:
1
2
þ l:
1
2
þ exp2pi h:
1
2
þ k:0 þ l:
1
2
!
¼ f Au þ f Cu ½exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
Exponential terms within the [ ] are identical with the fcc, hence similar criteria
will apply. Therefore, when h, k, l are all odd or all even, we obtain
F hkl
ð Þ ¼ ðf Au þ 3f Cu Þ and I / F(hkl)
j
j
2 ¼ ðf Au þ 3f Cu Þ
2
Similarly, if indices h, k, l are mixed, it can be seen that
F hkl
ð Þ ¼ ðf Au À f Cu Þ; and I / F(hkl)
j
j
2 ¼ ðf Au À f Cu Þ
2
In the disordered form, the atomic scattering factor for each atomic site is taken
as the weighted average of both the atoms, that is,
f Av ¼
1
4
ðf Au þ 3f Cu Þ
Hence, the structure factor for h, k, l all odd or all even is given by
F hkl
ð Þ ¼ 4:
1
4
f Au þ 3f Cu
ð
Þ¼ f Au þ 3f Cu
ð
Þ and I / F(hkl)
j
j
2 ¼ ðf Au þ f Cu Þ
2
and for mixed h; k; l
F hkl
ð Þ ¼ 0
Example 9 Determine the general form of structure factor and intensity corresponding to a diamond cubic unit cell.
Solution Given: Diamond cubic unit cell, F hkl
ð Þ ¼ ?; I ¼ ?
A diamond structure is supposed to be built up from two interpenetrating fcc
lattices, which are displaced with respect to one another along the body diagonal of
the cube by one-quarter of the length of the diagonal as shown in Fig. 8.7a.
Alternatively, it may be considered that each atom in the unit cell to be at the center
of a tetrahedron with its four nearest neighbors at the four corners of that tetrahedron as shown in Fig. 8.7b. Accordingly, there are 8 (carbon) atoms in the unit
cell. The fractional coordinates are: (0, 0, 0), (1/2, 1/2, 0), (0, 1/2, 1/2), (1/2, 0, 1/2),
(1/4, 1/4, 1/4), (3/4, 3/4, 1/4), (1/4, 3/4, 3/4), (3/4, 1/4, 3/4), respectively.
Substituting these values in Eq. 8.1, we obtain
8.2 Determination of Phase Angle, Amplitude …
313
ð Þ ¼ f Au :exp2pi h:0 þ k:0 þ l:0
ð
Þ
þ f Cu exp2pi h:
1
2
þ k:
1
2
þ l:0
þ exp2pi h:0 þ k:
1
2
þ l:
1
2
þ exp2pi h:
1
2
þ k:0 þ l:
1
2
!
¼ f Au þ f Cu ½exppi h þ k
ð
Þþexppi k þ l
ð
Þþexppi l þ h
ð
Þ
Exponential terms within the [ ] are identical with the fcc, hence similar criteria
will apply. Therefore, when h, k, l are all odd or all even, we obtain
F hkl
ð Þ ¼ ðf Au þ 3f Cu Þ and I / F(hkl)
j
j
2 ¼ ðf Au þ 3f Cu Þ
2
Similarly, if indices h, k, l are mixed, it can be seen that
F hkl
ð Þ ¼ ðf Au À f Cu Þ; and I / F(hkl)
j
j
2 ¼ ðf Au À f Cu Þ
2
In the disordered form, the atomic scattering factor for each atomic site is taken
as the weighted average of both the atoms, that is,
f Av ¼
1
4
ðf Au þ 3f Cu Þ
Hence, the structure factor for h, k, l all odd or all even is given by
F hkl
ð Þ ¼ 4:
1
4
f Au þ 3f Cu
ð
Þ¼ f Au þ 3f Cu
ð
Þ and I / F(hkl)
j
j
2 ¼ ðf Au þ f Cu Þ
2
and for mixed h; k; l
F hkl
ð Þ ¼ 0
Example 9 Determine the general form of structure factor and intensity corresponding to a diamond cubic unit cell.
Solution Given: Diamond cubic unit cell, F hkl
ð Þ ¼ ?; I ¼ ?
A diamond structure is supposed to be built up from two interpenetrating fcc
lattices, which are displaced with respect to one another along the body diagonal of
the cube by one-quarter of the length of the diagonal as shown in Fig. 8.7a.
Alternatively, it may be considered that each atom in the unit cell to be at the center
of a tetrahedron with its four nearest neighbors at the four corners of that tetrahedron as shown in Fig. 8.7b. Accordingly, there are 8 (carbon) atoms in the unit
cell. The fractional coordinates are: (0, 0, 0), (1/2, 1/2, 0), (0, 1/2, 1/2), (1/2, 0, 1/2),
(1/4, 1/4, 1/4), (3/4, 3/4, 1/4), (1/4, 3/4, 3/4), (3/4, 1/4, 3/4), respectively.
Substituting these values in Eq. 8.1, we obtain
8.2 Determination of Phase Angle, Amplitude …
313