Electron Diffraction Patterns

At the end of this lecture you should be able to
index SAED patterns if the cell parameters are known
determine the possible space groups from SAED
patterns
determine possible point groups from CBED patterns
Combine (3) and (4) to find thé space group.

Reflections: what do they represent? What
is their origin? What information can they
give us?  cf. lectures Mark De Graef

Constructive vs. destructive interference
reflection – no reflection
position  d-values of the planes
intensity  occupation in the planes
both symmetry of the structure

Single
phase?
Crystalline or
amorphous?
Orientation
crystals/film
?
Targeted
phase?
Domain
formation?
Crystal
parameters?

Selected Area Electron
Diffraction
SAED

One atom type A
a
b
b>a
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(100)
100
000
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(100)
100
000
1/(3Å)
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(100)
100
a*
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(010)
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(010)
010
a=3 Å
b=5 Å

One atom type A
a
b
b>a
(010)
010
a=3 Å
b=5 Å
1/(5Å)

One atom type A
a
b
b>a
(010)
010b*
a=3 Å
b=5 Å

One atom type A
010
100
a
b
[001]
b>a a*>b*
a*
b*
a=3 Å
b=5 Å

One atom type A
010
100
a
b
[001]
b>a a*>b*
a*
b*
Central reflection
is always 000.
000
a=3 Å
b=5 Å

One atom type A
010
100
a
b
[001]
a=3 Å
b=5 Å

One atom type A
010
100
a
b
[001]
1/(3Å)
1/(5Å)
a=3 Å
b=5 Å

010
100
[001]
Indices all other reflections:
vector addition
110
020
200 210
120

1/(3Å)
1/(5Å)
Experimentally: the other way around:
010
100
[001]
*

*How?
You know the structure, thus all cell parameters.
Make a list of all reflections with hkl and their d-values.
• use excell to make the list yourself (e.g. cubic: )
• use free software like Powdercell
• ...
d hkl
8.00000 001
5.00000 010
4.23999 011, 01-1
4.00000 002
3.12348 012,01-2
3.00000 100
... ...
𝑑 =
𝑎
ℎ2 + 𝑘2 + 𝑙2

3Å
5Å
Experimentally: the other way around:
010
100
[001]
*
Let's agree to label the reflections
with the d-value instead of 1/(xÅ):
d hkl
8.00000 001
5.00000 010
4.23999 011, 01-1
4.00000 002
3.12348 012,01-2
3.00000 100
... ...

010
100
Zone-index: [001]
obtained by vector multiplication.
Circle ACW around 000.
1 0 0 1 0 0
0 1 0 0 1 0
0.0-0.10.0-1.0 1.1-0.0
[ 0 0 1 ]

(010)
(100)
(110) (𝟏 𝟏𝟎)
[001]
010
100 110
(𝟏 𝟏𝟎)
[001]

Exercise:
index the given patterns taken
from a CaF2 mineral
Solution slides (plus all other temporarily deleted slides)
are available after the lecture on
slideshare/johader

h k l d I F
1 1 1 3.15349 83.73 61.89
2 0 0 2.731 0.11 3.07
2 2 0 1.93111 100 96.55
3 1 1 1.64685 31.44 46.49
2 2 2 1.57674 0.2 6.81
4 0 0 1.3655 12.69 74.25
3 3 1 1.25307 11.35 38.65
4 2 0 1.22134 0.54 8.67
4 2 2 1.11493 23.75 61.87
5 1 1 1.05116 6.88 34.2
3 3 3 1.05116 2.29 34.2
You need this table made for CaF2
𝑑 =
𝑎
ℎ2 + 𝑘2 + 𝑙2
with a=5.4620 Å
SG. Fm3m

We are going to index these patterns:
Start with easiest:
highest symmetry or smallest interreflection distances
= usually lower zone indices (“main zones”)
(Online version:
workpage can be
found at the end.)
Tilt
series

First pattern:
Apparent symmetry: 4-fold

a
b
c
Along which direction does the 4-old axis lie in a cubic
system? <001>
<011>
<111>
b
a
c
CaF2

probably this is <001>
(Cubic: [100], [010], [001] equivalent = <001>)

To do: measure the distances, compare to list d-hkl, index
consistently.
Step 1: Use the scalebar for the conversion
factor to 1/d-values.
Scalebar = R (in mm)
equal to 1/0.08 nm
R.d=L
L = R . 0.8Å = ?
Write down and
use this in the rest
of the exercise
Measure the scalebar

Step 2: measure the distance of two reflections, not on
the same line, calculate the corresponding d-value
Point 1
d
5.46 Å
3.15 Å
2.73 Å
Point 2
d
5.46 Å
3.15 Å
2.73 Å
1
2
To do: measure the distances, compare to list d-hkl, index
consistently.

To do: measure the distances, compare to list d-hkl, index.
Step 3: look up in the table to which
reflection this corresponds
100
110
200
Point 1
d
Point 2
d
Point 1
hkl
Point 2
hkl
1
2
5.46 Å
3.15 Å
2.73 Å
5.46 Å
3.15 Å
2.73 Å
100
110
200

Keep in mind: d-values for all equivalent {hkl}!
100
010
1
2
If point 1 is 200 then point 2 is 020 or 002.
Choose and stick to your choice.
Righthanded system.
Step 4: make the indexation consistent

Step 5: calculate the zone-index
[100]
[010]
[001]
010
200020

Next zone
Which one would be easiest next?
2
3
4

Next zone: with reflections closest to the central beam.
Because reflections far from the central beam:
lower d-values
larger amount of possible matches of hkl to this d
difficult to conclude which one is correct index!
1
3
5

Measure the distance of two reflections, not on the
same line, calculate the corresponding d-value
Point 1
d
2.57 Å
2.75 Å
3.15 Å
Point 2
d
1 2
2.57 Å
2.73 Å
3.15 Å

Look up in the table to which reflection
this corresponds
110
200
111
110
200
111
Point 1
d = 3.15 Å
Point 2
d = 2.73 Å
hkl hkl
1 2

Make the indexation in a consistent manner.
1 2
Point 2 should be indexed as
200
020
200
all are correct
-

Consistency:
This is a tilt series...
...so the common row needs to have the
same indices in all patterns
200 200
200 200

Consistency:
1
3
5
2001
Point 1 should be indexed as
111
111
111
all of the above are allowed because
-
--
𝑑 =
𝑎
ℎ2 + 𝑘2 + 𝑙2

Consistency:
200111
111
-
1
200
3
1 and 3 have the same d-value
+
relation between 1 and 3 = vector 200
you need two indices such that
h1 k1 l1 = h3+2 k3 l3
(also possible 111 and 111, make a choice and stick to it for the following patterns)
- - - - -

Consistency:
200111
111
-
1
200
3
If you would have taken 111 for point 1
-
311
-
Point 3 is
311
111

Consistency:
200111
111
-
1
200
3
point 3 = 311-
d311 ≠ dpoint3-
but
If you would have taken 111 for point 1
-

Calculate the zone-index
The zone-index is: [011]
[011]
200
111
111
-
022
-

...this helps to index the two remaining patterns!!!
Consistency:
This is a tilt series...

a
b
c
The crystallite is threedimensional.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0] So, the reciprocal lattice is threedimensional.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
ED patterns are sections of reciprocal space.
[001]
This section is the [001] zone.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[011]
-
This section is the [011] zone:
-

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[001]
[011]
-
We tilt from [001] to [011]:
-

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[001]
[011]
-
So the tilt series gives pattern of consecutive sections
between these two end zones
Closest is 020.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[001]
[015]
-
[011]
-
x
051
Closest is 151.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[011]
[001]
[013]
-x
031
Closest is 131.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[011]
[001]
x
021
[012]
-
042
Closest is 042.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[011]
[001]
[035]
-
x
053
Closest is 153.

0,4,0
0,4,2
1,5,1
0,4,4
1,5,3
0,2,0
0,2,2
1,3,1
0,2,4
2,4,0
1,3,3
2,4,2
3,5,1
0,0,2
2,4,4
1,1,1
3,5,3
0,0,4
2,2,0
1,1,3
2,2,2
3,3,1
2,2,4
4,4,0
3,3,3
2,0,0
4,4,2
5,5,1
2,0,2
4,4,4
3,1,1
5,5,3
2,0,4
4,2,0
3,1,3
4,2,2
5,3,1
4,2,4
5,3,3
4,0,0
4,0,2
5,1,1
4,0,4
5,1,3
Zone axis : [0,0,0]
[001]
[011]
-
Closest is 111. Perpendicular is 022.
200
111
111
-
022

0,4,0
0,4,2
0,4,4
0,2,0
1,5,1
0,2,2
1,5,3
0,2,4
1,3,1
0,0,2
1,3,3
0,0,4
2,4,0
1,1,1
2,4,2
1,1,3
2,4,4
2,2,0
3,5,1
2,2,2
3,5,3
2,2,4
2,0,0
3,3,1
2,0,2
3,3,3
2,0,4
4,4,0
3,1,1
4,4,2
3,1,3
4,4,4
4,2,0
5,5,1
4,2,2
5,5,3
4,2,4
4,0,0
5,3,1
4,0,2
5,3,3
4,0,4
5,1,1
5,1,3
We can also see this in projection

0,4,4
0,4,2
0,4,0
1,5,3
1,5,1
0,2,4
0,2,2
0,2,0
1,3,3
1,3,1
0,0,4
0,0,2
2,4,4
2,4,2
2,4,0
1,1,3
1,1,1
3,5,3
3,5,1
2,2,4
2,2,2
2,2,0
3,3,3
3,3,1
2,0,4
2,0,2
2,0,0
4,4,4
4,4,2
4,4,0
3,1,3
3,1,1
5,5,3
5,5,1
4,2,4
4,2,2
4,2,0
5,3,3
5,3,1
4,0,4
4,0,2
4,0,0
5,1,3
5,1,1

0,4,4
0,4,2
0,4,0
0,2,4
0,2,2
0,2,0
0,0,4
0,0,2
1,5,3
1,5,1
1,3,3
1,3,1
1,1,3
1,1,1
2,4,4
2,4,2
2,4,0
2,2,4
2,2,2
2,2,0
2,0,4
2,0,2
2,0,0
3,5,3
3,5,1
3,3,3
3,3,1
3,1,3
3,1,1
4,4,4
4,4,2
4,4,0
4,2,4
4,2,2
4,2,0
4,0,4
4,0,2
4,0,0
5,5,3
5,5,1
5,3,3
5,3,1
5,1,3
5,1,1
Zone axis : [0,0,0]

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
which is easier to draw manually...

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
Let's look only at the indices of 2 layers
020
040
111
002 022 042
131 151
113
004 024 044
133 153
at h=1
at h=0
at h=1
at h=0
at h=0

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
020
040
111
002 022 042
131 151
113
004 024 044
133 153

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
[013]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
[013]
-
[012]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
[013]
-
[012]
-
[035]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
[013]
-
[012]
-
[035]
-
[011]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153

Right upper zone:
Point 2
d
1.22 Å
1.11 Å
1.05 Å
We already know the
first point: 200.
200
2

Look up in the table to which reflection this corresponds:
We know already it is either 151 or 131 or 042 or 153
1.05 Å 151
131
042
Point 2
d
Point 2
hkl
200
2

If this were not a tilt series...
could have
been at
first sight
both 151
and 333...
Or in this particular case of 333: you would need to see 111 and 222 at 1/3 and 2/3 of the distance.
1.05 Å
Point 2
d
h k l d I F
1 1 1 3.15349 83.73 61.89
2 0 0 2.731 0.11 3.07
2 2 0 1.93111 100 96.55
3 1 1 1.64685 31.44 46.49
2 2 2 1.57674 0.2 6.81
4 0 0 1.3655 12.69 74.25
3 3 1 1.25307 11.35 38.65
4 2 0 1.22134 0.54 8.67
4 2 2 1.11493 23.75 61.87
5 1 1 1.05116 6.88 34.2
3 3 3 1.05116 2.29 34.2

If this were not a tilt series...
Point 2 could have been at first sight
both 115 and 333...
Again, consistency check:
Or in this particular case of 333: you would need to see 111 and 222 at 1/3 and 2/3 of the distance.
200
151
200
333
151
115 has same d-
value as 115
-
-
133 does not have
same d-value as 333
133

Calculate the zone-index
The zone-index is:
[0 2 10]
[0 1 5]
[0 1 5]
200
151
-

[001]
[015]-
[013]
-
[012]-
[035]-
[011]-
010
031 051
053
What if you don’t know the material and thus don't know
the reflection conditions?
You might have different reflection conditions.
You might have more possibilities for in-between
zones:
043
032
041021
[025]-
052 [014]-
[023]-
[034]-
When indexed correctly, the patterns in between
have to give you one of these as zone-index.
011

Last pattern, pattern bottom left:
Point 2
d
200
2
1.65 Å
1.58 Å
1.37 Å

Look up in the table to which reflection this corresponds.
We know already it is either: 151 or 131 or 042 or 153
1.65 Å 113
131
311
Point 2
d
Point 2
hkl
200
2
Make sure the vector addition is consistent.

Look up in the table to which reflection this corresponds.
We know already it is either: 151 or 131 or 042 or 153
1.65 Å 113
131
311
Point 2
d
Point 2
hkl
200
2
Make sure the vector addition is consistent.
At this point (table+vector
addition) also 113 is still
possible. However, when
considering the other zones
( next slides), you will see
that only 131 remains.

Calculate the zone-index = [013]
= in agreement with sections scheme (pdf p. 62-82)
-
200
131
062 200
131062
131
-
The indexation is indeed consistent
concerning vector addition for both.
Calculate the zone-index = [031]
= not in agreement with sections scheme
200
131
062 200
113026
113
-
-
CORRECT
WRONG

Make your analysis easier by not taking
ED patterns from separate crystals, but
taking different ED patterns from the
same crystallite, if possible.
=“Tilt series”

So now you have indexed these four patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

...indexed patterns give you info on phase,
orientation, cell parameters,...
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

What if you do not know the cell
parameters beforehand?
Analyse the patterns  try to propose basis vectors
(For example reflections closest to the central beam)
Same system as previous slides:
can you index all reflections?
If not, adapt your choice of
basis vectors and try again.

What if you need to know if
anything changed in the
symmetry?
Need to determine the space group
Determine the
extinction symbol
from the spot patterns
Determine the point
group from the CBED
patterns

Let's say we need to know if the CaF2 in the example
is still the Fm3m type CaF2.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
-
Part 1. You index the patterns using just the cell parameters.
Part 2. You determine the space group.

1. Extinction symbol(SAED)
Space group?
P no reflection conditions
F h+k=2n, k+l=2n, h+l=2n
I h+k+l=2n
A/B/C k+l=2n/h+k=2n/h+k=2n
2. Point Group (CBED)
glide planes conditions on hk0/h0l/0kl
screw axes conditions on h00/0k0/00l
mirror planes, inversion
centre, rotation axes
no extra conditions
CBED
SAED

Reflection conditions can be looked up in tables in
International Tables for Crystallography Vol. A
Or using freeware such as Space Group Explorer

Be careful: forbidden reflections can be present
because of multiple diffraction
Incident electron wave

Can see this:
When reflection
conditions say this:
For example possible
010
100
020
100

020
100
F(100)≠0
F(110) ≠0
F(010)=0
need to tilt.

Destroy double diffraction paths by tilting.
If becomes
If stays
then extinct, was due to DD
then not extinct.

Determine the general reflection condition from the patterns.
hkl:Centering Reflection condition
P no conditions for hkl
I hkl: h+k+l=2n
F hkl: h+k=2n, k+l=2n,
h+l=2n
A hkl: k+l=2n
B hkl: h+l=2n
C hkl: h+k=2n

hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
Determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
(remember it is cubic
so h, k and l are
interchangeable)

hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
For these patterns
both would be
good....!?
Only disagreement
with hkl:h+k=2n is
in [001] where also
110 is absent.
However, this
could be due to a
special reflection
condition on hk0.

0,0,4
0,0,2
0,2,4
0,2,2
0,4,4
0,2,0
0,4,2
0,4,0
1,1,3
1,1,1
1,3,3
1,3,1
1,5,3
1,5,1
2,0,4
2,0,2
2,2,4
2,0,0
2,2,2
2,4,4
2,2,0
2,4,2
2,4,0
3,1,3
3,1,1
3,3,3
3,3,1
3,5,3
3,5,1
4,0,4
4,0,2
4,2,4
4,0,0
4,2,2
4,4,4
4,2,0
4,4,2
4,4,0
5,1,3
5,1,1
5,3,3
5,3,1
5,5,3
5,5,1
xis : [0,0,0]
[001]
[015]
-
[013]
-
[012]
-
[035]
-
[011]
-
020
040
111
002 022 042
131 151
113
004 024 044
133 153
This means we do not have sufficient
information  Obtain extra zones, e.g. [012]
-

200
131
200
151
200111
200020
[001] [015]
[013]
-
[011]
-
200
042
hkl: h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n

200
042
hkl: h+k, k+l, h+l=2n
h+k=2n
242
x
x
x x
x
021 121
100
141
221
"hkl: h+k=2n"
should look like this:

It is possible to draw the wrong
conclusions if you do not have
enough zones!

Look up the matching extinction symbol
in the International Tables of
Crystallography.
?

200
131
200
151
200111
200020
[001] [015]
[013]
-
[011]
-
200
042
0kl: k=2n, l=2n
k+l=4n, k=2n, l=2n
hkl:h+k, k+l, h+l=2n

200
020
[001]
0kl: k=2n, l=2nhkl:h+k, k+l, h+l=2n
example is cubic,
so 0kl = hk0 = h0l
If it were 0kl:k+l=4n, k=2n, l=2n
it should look like:
200
020
[001]
040
400
220
240
440
420

200
131
200
151
200111
200020
[001] [015]
[013]
-
[011]
-
200
042
hhl: h+l=2n
h=2n,l=2n
0kl: k+l=2n

200111
[011]
-
0kl: k+l=2n
hhl: h+l=2n
311
511
222
422
111
-
h=2n,l=2n

Step 2: look up the matching extinction
symbol in the International Tables of
Crystallography.
?
?

200 and 020 could be due to double
diffraction...

diffraction...
 Tilt around 200 until all other
reflections gone except h00 axis:

diffraction...
Tilt around 200 until all other reflections
gone except h00 axis:
200 does not disappear
 It is not double diffraction
 00l: l=2n
not 00l: l=4n

Step 2: look up the matching extinction symbol in
the International Tables of Crystallography.

From the reflection conditions you get
the extinction symbol:
F - - -
Extinction symbol

F - - -
This still leaves 5 possible space groups
F23 Fm3 F432 F43m Fm3m
-
Extinction symbol

Point Group
F - - -
This still leaves 5 possible space groups
F23 Fm3 F432 F43m Fm3m
Only difference: rotation axes and mirror planes
cannot be derived from reflection conditions
 need
-
Extinction symbol

Symmetry-relation in real space
𝑟′ = 𝑅. 𝑟 + 𝑡
R = rotation matrix
t=translation vector
Symmetry-relation amplitudes in reciprocal space
ℎ′ 𝑘′ 𝑙′ = ℎ𝑘𝑙 . 𝑅

Within one point group, there are the same
theoretical equivalences between the reflections
For example have the same equivalent reflections:
𝑃4 𝑃41 𝑃42 𝑃43 𝐼4 𝐼41
But not
𝑃4 𝐼4
However! Friedel's Law adds an inversion symmetry
On an experimental ED pattern all of the above have
the same equivalences between reflections!
Can discern only Laue symmetries.

Step 2: look up the matching extinction symbol in
the International Tables of Crystallography.
Need a different technique: CBED

P4/mmm P4
.mm clearly absent!
Sometimes the amplitudes àre useful...

For CBED you need sufficiently thick crystals:
t=10 nm t=20 nm t=30 nm
t=40 nm t=50 nm
CBED patterns
calculated using JEMS

Easiest* : "Eades method"
Obtain CBED patterns of high symmetry zones,
preferably showing HOLZ
Derive "diffraction group" of each zone
Look for the point group that can have all the
diffraction groups you found for the different zones
* purely personal opinion

You need two tables:
relation between symmetries in CBED patterns
and the diffraction groups
which point groups contain which diffraction groups
(both tables are included in these slides)
1
2

Table from: J.A. Eades,
Convergent beam
diffraction, in: Electron
Diffraction Techniques,
volume 1, ed. J. Cowley,
Oxford University Press,
1992
1
Each zone axis
CBED pattern
has its own
diffraction
group.

From: Atlas of Electron Diffraction, Jean-Paul Morniroli, freely available at
http://www.electron-diffraction.fr/Files/220_atlas_version_september_2015.pdf
Whole pattern, 3D symmetry
aka "whole pattern symmetry"
3D WP
Whole pattern, 2D symmetry
aka "whole pattern projection symmetry"
2D WPBright field, 3D symmetry
aka "bright field symmetry"
3D BF
no HOLZ lines? 2D BF = BF projection symm

Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992
.

6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
-1
1
2
m
2/m
222
mm2
mmm
4
-4
4/m
422
4mm
-42m
4/mmm
3
-3
32
3m
-3m
6
-6
6/m
622
6mm
-6m2
6/mmm
23
m3
432
-43m
m3m
Table redrawn
from B.F. Buxton,
J.A. Eades, J.W.
Steeds, G.M.
Rackham: Phil.
Trans. R. Soc.
London, 281
(1976) 171
Diffraction groups vs. Point groups
2
Leftover point groups
to decide between

Symmetry is 4mm.
Whole pattern projection symmetry

6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
-1
1
2
m
2/m
222
mm2
mmm
4
-4
4/m
422
4mm
-42m
4/mmm
3
-3
32
3m
-3m
6
-6
6/m
622
6mm
-6m2
6/mmm
23
m3
432
-43m
m3m
Table redrawn
from B.F. Buxton,
J.A. Eades, J.W.
Steeds, G.M.
Rackham: Phil.
Trans. R. Soc.
London, 281
(1976) 171

6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
-1
1
2
m
2/m
222
mm2
mmm
4
-4
4/m
422
4mm
-42m
4/mmm
3
-3
32
3m
-3m
6
-6
6/m
622
6mm
-6m2
6/mmm
23
m3
432
-43m
m3m
Table redrawn
from B.F. Buxton,
J.A. Eades, J.W.
Steeds, G.M.
Rackham: Phil.
Trans. R. Soc.
London, 281
(1976) 171
m3m
-

So, sometimes just whole pattern projection
symmetry is enough if you combine it with the
reflection conditions from SAED.

Exercise:
determine the point
group of SnO2

SAED CBED
Example: rutile-type SnO2

Projection whole pattern symmetry [001]
4
4mm
2mm

Projection WP: 4mm
Projection diffraction group:
Table Eades
4
4RmmR
4mm1R

Projection diffraction group:
4mm1R
Possible diffraction groups:
4mRmR
4mm
4RmmR
4mm1R

6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
-1
1
2
m
2/m
222
mm2
mmm
4
-4
4/m
422
4mm
-42m
4/mmm
3
-3
32
3m
-3m
6
-6
6/m
622
6mm
-6m2
6/mmm
23
m3
432
-43m
m3m
Table redrawn
from B.F. Buxton,
J.A. Eades, J.W.
Steeds, G.M.
Rackham: Phil.
Trans. R. Soc.
London, 281
(1976) 171

What will be useful to narrow it down further?
look at the bright field symmetry
look at the whole pattern symmetry
[...] [...]

6mm1R
3m1R
6mm
6mRmR
61R
31R
6
6RmmR
3m
3mR
6R
3
4mm1R
4RmmR
4mm
4mRmR
41R
4R
4
2mm1R
2RmmR
2mm
2mRmR
m1R
m
mR
21R
2R
2
1R
1
-1
1
2
m
2/m
222
mm2
mmm
4
-4
4/m
422
4mm
-42m
4/mmm
3
-3
32
3m
-3m
6
-6
6/m
622
6mm
-6m2
6/mmm
23
m3
432
-43m
m3m

Projection whole pattern [101]
2
m
2mm
(smaller cond.ap.)

Possible projection
diffraction group:
21R
m1R
2mm1R

Whole pattern [101]
2
m
2mm
(smaller cond.ap.)

Diffraction group:
2mm
2RmmR
2mm1R

Possible point groups
4/mmm
m3m

What would make a difference further?
For example:
-cell parameters
-look for a third zone etc.
(e.g. is there zone with 3 or 6 fold  m-3m)
-SAED for reflection conditions

If you need
cell parameters a=b= 4.72 Å, c=3.16 Å
to be able to index all SAED patterns,
the point group is
4/mmm
m3m

from: International Tables for Crystallography Volume A, ed. Theo Hahn, Kluwer Ac. Pub., 2002
CBED

Combine with information about
reflection conditions from SAED patterns
020
200
110
020
011
002
hkl:
no conditions
h+k+l=2n
002
110
-

Combine with information about
reflection conditions from SAED patterns
020
200
110
020
011
002
hk0:
no conditions
h+k=2n
0kl:
no conditions
k+l=2n
k=2n or l=2n
002
110
-
hhl:
no conditions
l=2n

International Tables
CBED
SAED
Space Group P42/mnm

Solving and refining a structure from
electron diffraction data
(very concisely)

Experimental data:
-electron diffraction
patterns
-targeted composition
A model
for the
structure
Trouble: dynamical diffraction
Direct methods
Simulated annealing
Charge flipping
...
Refined
coordinates
Rietveld refinement
...

Possible ways around dynamical diffraction:
• Precession electron diffraction (PED)
• Electron diffraction tomography (EDT)

Precession electron diffraction

/1, 00 kk
Ewald sphere
Diffraction pattern
PED technique:
R. Vincent, P. Midgley,
Ultramicroscopy,1994, 53, 271

Extract the intensities using
EDM (free)
EXTRAX (free)
PETS (free)
CRISP-EDT (not free)
...

• imperfect orientation
• not thin enough
• too small precession angle
• too large precession angle
• overlapping patterns (twins,
impurities, ...)
• …
Intensities will deviate from the correct values:
But good enough to solve structures.
Deviations have been proven to not destroy the possibility to solve structures. Even a rough division into weak - medium
- strong reflections is often good enough anyway [H. Klein&J.David, Acta Cryst. (2011). A67, 297–302]

Treat the data of the individual patterns:
• could apply Lorentz factor (no consensus whether
necessary)
𝐼′ = 𝐼. 𝑔 1 −
𝑔
2𝑅
2
(R=radius Laue circle)
• could symmetrize = replace reflections by average of
equivalent reflections (also gives Rsym)
• ...

Merge separate lists into one list
Scale factor based on common reflections
Possible pitfalls in merging:
• (wrong?) ratio of small amount of reflections is
imposed on all reflections
• merging is done in series, errors accumulate
• typical tilt series around main axis, but main axes
often stronger deviations by themselves
• should a weighing be used?
• ...

Electron diffraction tomography

Manually + treatment of
patterns with PETS (free)
Automated in ADT (not
free)
You obtain a list of hkl
and intensities
PETS: Palatinus et al. Acta Cryst. (2015). A71, 235-24

Direct methods, or optimisation methods made
for single crystal data
SIR2011 (free)
Fox (free)
Endeavour (not free)
Superflip (free)
...
Same concerning refinement e.g. Jana (free)
Right structure càn come out (GIGO).

Electron Diffraction Patterns

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