Solution Given: Body-centered cubic unit cell, a = 2.884 Å and Z = 24
F hkl
ð Þ ¼ ?; I ¼ ?
A simple calculation will give us two atoms in a body-centered cubic unit cell
(Fig. 8.3). The fractional coordinates of atom 1 is (0, 0, 0) and atom 2 is (1/2, 1/2, 1/
2), respectively. Substituting these values in Eq. 8.1, we obtain
F hkl
ð Þ ¼ f: exp 2pi h:0 þ k:0 þ l:0
ð
Þ þexp 2pi h:
1
2
þ k:
1
2
þ l:
1
2
!
¼ f:½1 þ exp pi h + k + l
ð
Þ Š
But exppi ¼ cosp þ i sinp ¼ À1 þ 0 ¼ À1
However in general; exp pin
ð Þ ¼ À1
ð Þ
n ; where n is an integer
So that, for a bcc structure
F hkl
ð Þ ¼ f½1 þ exppi h þ k þ l
ð
Þ Š¼f½1 þ 1Š
¼ 2f; if h þ k þ l
ð
Þ is an even integer
and
I / F(hkl)
j
j
2 ¼ 4f
2
Similarly,
F hkl
ð Þ ¼ f 1 À 1
½
Š¼0; if h þ k þ l
ð
Þ is an odd integer:
and
I / FðhklÞ
j
j
2 ¼ 0
Thus for a body-centered lattice, the intensities of the diffracted beams coming
from the planes such as (100), (300), (111), (210), (221), etc. whose indices add up
to odd integers are zero, hence absent (this is known as extinction). On the other
hand, the intensities of the diffracted beams coming from the planes such as (110),
(200), (220), (222), etc. whose indices add up to even integers are proportional to
4f
2
; hence present.
Fig. 8.3 BCC structure of
chromium (Cr)
8.2 Determination of Phase Angle, Amplitude …
307