A base-centered unit cell is shown in Fig. 8.4. Each base-centered atom contributes half to the unit cell. Therefore, two base-centered atoms will contribute
(2 × 1/2 = 1) one atom while other atom is contributed from 8 corners. This makes
that there are two atoms in the unit cell. The fractional coordinates of atom 1 is (0,
0, 0) and atom 2 is (1/2, 1/2, 0), respectively. Substituting these values in Eq. 8.1,
we obtain
F hkl
ð Þ ¼ f: exp2pi h:0 þ k:0 þ l:0
ð
Þ þ exp2pi h:
1
2
þ k:
1
2
þ l:0
!
¼ f:½1 þ exppi ðh þ kފ
So that, for a base-centered structure
F hkl
ð Þ ¼ f 1 þ exppi h þ k
ð
Þ
½
Š
It can be seen that if h and k both are odd or both are even integers (i.e., when h
and k are unmixed indices), then the sum (h + k) is an even integer. In this case,
F hkl
ð Þ ¼ 2f and I / F(hkl)
j
j
2 ¼ 4f
2
In case of mixed indices (say one is odd and another is even), then the sum
(h + k) will be an odd integer. Therefore,
F hkl
ð Þ = f 1À1
½
Š = 0, and I / jF(hkl)j
2 ¼ 0
Example 7 Determine the general form of structure factor and intensity corresponding to a face-centered cubic unit cell. Calculate the structure factor for Copper
(Cu) whose lattice parameter a = 3.615 Å and Z = 29.
Solution Given: Face-centered cubic unit cell, F hkl
ð Þ ¼ ?; I ¼ ?
Fig. 8.4 Base-centered
monoclinic structure
8.2 Determination of Phase Angle, Amplitude …
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