# Class 06_BB ```September 28, 2010
Class 6
Ewald construction
The Bragg or Laue condition is a rather restrictive one. A way to see the level of the coincidence
that has to happen for diffraction to occur, notice that since k  G  v1b1  v 2 b2  v3 b3 , the dot
product with our real space unit translation vectors gives,
a1.k=2v1
a2. k=2v2
a3. k=2v3
These equations tell us that k lies on a cone
around a1, but also on a cone around a2, and
a3. So k must be along the single line where
the three cones intersect. This is a very
rigorous restriction and it is difficult to find
out what that line is. So Ewald came up with
a construction that allow to determine those
wavevectors that will produce diffraction
A k vector is drawn from any origin but
ending in one of the reciprocal points. Then a
sphere is drawn centered on the origin of the k
vector. Diffraction will occur only if another
point in the reciprocal lattice lies on the
surface of the sphere then the reciprocal
vector G that connects these two points
determines k’=k+G
In fact, any two points which lay on the surface of a sphere with radius k, will represent a set of
planes which meet the diffraction condition. The figure above illustrates the severity of this
condition in two dimensions.
It is clear that if the k is fixed with respect to the reciprocal lattice, most of the time no
diffraction will arise. This is equivalent to an experiment where a crystal is positioned on a
holder fixed in place and monochromic x-rays are incident at a fixed angle. In order to measure
something several techniques exist.
1) The Laue Method: The crystal is fixed in place, the incoming radiation has a fix angle,
but the beam is not monochromatic. This is equivalent to draw a continuous of spheres
of increasing radius all coincident in the tip of k. any reciprocal point in the area between
the sphere with the minimum radius and the one with the maximum radius will produce
diffraction
2) The Rotating Crystal Method: in this case monochromatic radiation is used but the
crystal is placed on a rotating holder. As the crystal rotates both the direct and the
reciprocal lattice rotate about the same axis. As the reciprocal points rotate the intersect
the Ewald sphere
1
3) The Powder or Debye-Scherrer methods: A polycrystalline sample is used, thus small
crystals in the sample are oriented in all ld sphere producing diffraction directions, this is
equivalent to the rotating crystal method but the sample need not to be rotated.
Brillouin Zones
A geometrical interpretation of the Laue condition can also be obtained by dividing the Laue
equation by two, then
2 k.G=G2 =&gt; k. (&frac12;G)=(&frac12;G)2
So all the waves with wavevectors k whose projection over G is equal to &frac12; of the length of G
will produce diffraction.
Consider all the G vectors from a lattice points and draw the set of planes that are
perpendicular bisectors to these vectors, the intersection of all these planes will enclose portions
of the space. The one with the smallest volume is known as the first Brillouin zone.
Notice that this is the recipe to build the Weigner-Seitz primitive cell
Any k that connects the center point and the surface of this volume is a wavevector that will
produce diffraction. So giving a relationship between  and .
The Weigner-Seitz primitive cell of the reciprocal lattice is known as the first Brillouin Zone and
has very particular implication in the crystal property. We will find the name Brillouin Zone
applied to a number of properties in solid state, although we will not always be talking about the
same thing, in general the Brillouin zones refer to the first, second, third, etc period on a
periodical property. In this particular case, the first Brillouin zone contain ALL the wave vectors
that will produce first order diffraction (corresponding to n=1 in the Bragg condition) the second
zone will contain the wave vectors that satisfy Bragg for n=2, etc.
Fourier analysis of the Basis


 
Remember that F   dV n( r ) exp  ik  r (scattering amplitude)
When the diffraction condition is satisfied for a particular lattice vector G, then k=G and the
scattering amplitude for a crystal of N cells is


 
FG  N  dVcell n(r ) exp  iG  r  NSG
Where SG is called the structure factor.
It is convenient to express the crystal electron density in term of atomic electron density nj as
follows:
s

 
nr    n j r  rj  Where s is the number of atoms in the atomic basis in the cell and rj is the
j 1
vector position of each of the atoms with respect to origin in the cell.
2
With this substitution

s
s
s

 iG j
 
  iGrrj rj  s iGrj
iGrj


SG    dV n j (r  rj ) e iGr    dV n j (r  rj ) e
 e
dV
n

e

f
e
 j
 j j
j 1
j 1
j 1
j 1
  iG  j
f j   dV n j  j e
is the atomic factor. Notice that the j=r-rj (or r=+rj) is the coordinate
with origin at the center of the atom and sampling outwards towards the electron cloud of each
atom. SG is then the atomic factor for the basis
There is no unique way to partition the atomic electrons to assign them to particular atoms, thus
nj have no unique form. However the physics of the crystal, total atomic density, diffraction
spectrum, etc, is not altered by the choice of nj as long as it is consistent.
By writing rj, the position of each atom of the basis with respect to the origin in the cell in terms
of the direct lattice vectors
G.rj=(v1b1+v2b2+v3b3). (xja1+yja2+zja3)=2(xjv1+yjv2+zjv3)
Where now xj, yj, and zj are not necessarily integer numbers and are relative to the lattice
constants, then
s
SG  1 , 2 , 3 ,   f j e

i 2 x j v1  y j v2  z j v3

j 1
The structure factor does not need to be real since the scatter intensity, which is what is actually
measured, involves S2
Structure factor for the bcc lattice
Consider the conventional lattice for a bcc lattice. The basis consists of two identical atoms, 1 at
000 and a second one at &frac12;&frac12;&frac12;
So the structure factor for the bcc lattice is

S G  1 , 2 , 3 ,  f 1  e  i v1 v2 v3 
What means that

S=0 when v1+ v2+ v3=odd
S=2f when v1+ v2+ v3=even
This means that diffraction of planes such that (100), (300), (111), etc will not be present in the
diffraction spectrum (where the index refer to a cubic lattice). For instance for the (100) line, the
phase difference must be 2 (one full wave length) for a wave bouncing off planes at a distance a
from each other. For the bcc lattice, there exists an intermediate layer between the two that exist
on a cubic lattice. Waves bouncing off that layer are  out of phase and cancel the other waves.
For this to happen, the atoms must be of the same nature, the (100) does show in the CsCl
3
spectrum since Cs and Cl have different atomic factors but again, CsCl is NOT a bcc but a sc
lattice
Structure factor for the fcc lattice
Consider the conventional lattice for a fcc lattice. The basis consist of four identical atom, 1 at
000; 0&frac12;&frac12;; &frac12;0&frac12;; &frac12;&frac12;0
So the structure factor for the bcc lattice is

SG  1 , 2 , 3 ,  f 1  e  i v2 v3   e  i v1 v3   e i v1 v2 

What means that if all the indexes are even integers or odd integers, then SG=4f. However if one
has a different parity, SG=0
S=4f when v1, v2, v3 have the same parity
S=0 when v1, v2, v3 do have the same parity
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