DETERMINATION OF THE ZONE AXIS (BEAM DIRECTION)
•
•
In the diffraction pattern considered as an
example for indexing the spots, it is noticed that
the l index for all diffraction spots = 0
beam
h 3k 3l 3
This means that the planes (which diffract the
beam) do not intersect the z-axis (or all the
planes are parallel to the z-axis)
•
The centre spot is formed by those electrons
which are not scattered and pass straight
through the crystal.
crystal.
•
Meaning that the diffraction pattern is oriented
with the electron beam parallel to the z-axis
axis..
•
The orientation of the specimen is defined by
stating that the zone axis (ZA) of the diffraction
pattern is [001
001]]
h 1k 1l 1
h 2k 2l 2
ZA
Zone of reflecting planes
ZA: is a zone axis
DETERMINATION OF THE ZONE AXIS (BEAM
DIRECTION)
•
The zone axis [UVW] is determined by using the zone axis equation:
U = k1l2 - k2l1
V = l1h2 - l2h1
W = h1k2 - h2k1
•
Where h1k1l1 and h2k2l2 are the Miller indices of any two spots in the DP.
•
In a simple way we can use:
h1 k1
l1
x
h2
k2
h1
x
l2
k1
l1
k2
l2
x
h2
1
INDEXING A PATTERN WITH MANY DIFFRACTION SPOTS
FORBIDDEN ELECTRON PATTERN SPOTS
•
Forbidden spots are the expected reflections which may not occur in the
diffraction patterns from certain crystals
crystals..
•
In a simple cubic crystal
crystal,, all reflections occur and there is no forbidden spots
spots..
•
Consider, for example, the diffraction pattern formed from Ni (FCC crystal) as
shown in Figure 4.8. (the beam direction is [001
001])
])..
•
The lattice parameter of Ni is aNi = 0.35 nm, λ = 0.004 nm and L = 800 mm.
mm. After
measuring the R values, we have
have::
2
a
d h1 k 1 l1 = 0 . 176 =
(h12 + k12 + l12 )
d h 2 k 2 l 2 = 0 . 124 =
(
a
h 22 + k 22 + l 22
)
Which gives:
h 12 + k 12 + l 12 = 4
h 22 + k 22 + l 22 = 8
•
So the only possible Miller indices are:
•
h 1 + k 1 + l1
of {200}
•
h2 + k 2 + l2
of {220}
•
The diffraction pattern is indexed as shown in Figure 4.9
•
Note the absence of alternate diffraction spots
3
To explain why some expected spots are being absent (or forbidden),
let us consider the FCC lattice structure
In this lattice, it is possible to draw an extra plane PQRS, containing
the face
face--centering atoms, in which the number and arrangement of
the atoms is exactly the same as in the original (100)
100) planes OAGF
and CBHE.
CBHE.
This plane (PQRS) is indexed (200)
200).
The electron beam cannot distinguish between planes made up of
the corner atoms of the unit cell and those made up of the face
face-centering atoms.
atoms.
4
•
There are twice the number of planes indexed (200)
200) at a/
a/2
2 lattice
spacing and (100)
100) planes do not exist any more.
more.
•
Therefore, diffraction spots from these planes (100)
100) will not appear in
the diffraction pattern
•
The selection rules for the cubic system allow reflections to appear in
the diffraction pattern only when:
when:
FCC crystal
h, k, l are all even
BCC crystal
h + k + l = even
RECIPROCAL LATTICE
•
The reciprocal lattice is directly related to the real lattice and is
commonly used in diffraction.
diffraction.
•
To construct the reciprocal lattice, a normal to each set of planes in the
real crystal lattice is drawn.
drawn.
•
Then, make off the points along these normals at distances 1/d from the
origin..
origin
•
For example, the (200
200)) planes of the FCC lattice give rise to a point
•200
200,, in the reciprocal lattice at a distance 1/d200, from the origin • 000
in a direction normal to the (200)
200) planes as shown in Figure 4.15.
15.
•
In a similar way, the (020
020),
), (002)
002) and (111)
111) planes as shown
shown..
5
When these points are combined they give the basic array of the FCC
reciprocal lattice, which can be extended by the addition of points for all the
other planes.
planes.
Further points are also added to represent the negative side of the planes as
shown in Figure
Figure..
The full reciprocal lattice for FCC is as shown in Figure 4.16
You must note that the FCC reciprocal lattice is actually the BCC lattice.
lattice.
The BCC reciprocal lattice is the FCC lattice.
lattice.
6
THE EWALD SPHERE
The relationship between the
diffraction pattern and reciprocal
lattice can be demonstrated by
the Ewald sphere construction
The Ewald sphere passes
through a reciprocal lattice point,
which is a distance 1/d from the
origin..
origin
incident
beam
θ
Ewald
circle
A
2θ
1/
1/λ
λ
diffracted
beam
1/d
B
0
7
From geometry, we find that:
1
sin θ =
1
d =
λ
λ
2d
or
λ = 2 d sin θ
That is: Bragg’s Law is satisfied.
INDEXING A PPATERN FROM AN UNKNOWN SPECIMEN
Indexing a diffraction pattern from an unknown specimen is not an
easy task.
task.
A trial and error approach is used, together with some intuition and
knowledge about the specimen.
specimen.
The correct formula of the d-spacings and interplanar angles must be
found..
found
The symmetry of the diffraction pattern can also help in determining
the type of crystal from which the diffraction pattern was obtained.
obtained.
8
h2+k2+l2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Primitive cubic
100
110
111
200
210
211
220
221/300
310
311
222
320
321
400
Face-centred cubic
111
200
220
311
222
400
Body-centred cubic
110
200
211
220
310
222
321
400
9