This is a tutorial on indexing diffraction patterns, deriving reflection conditions from SAED, derving point groups from CBED and combining both to find the space group. The slides contain exercises, the page to work on is at the end of the presentation and should be printed first to be able to measure on that page.

1. Electron Diffraction

2. At the end of this lecture you should be able to
(1) index SAED patterns in case the cell parameters
are already known
(2) determine the possible space groups from SAED
patterns
(3) determine possible point groups from CBED
patterns
Combine (2) and (3) to find the space group.

3. Reflections: what do they represent?
What is their origin? What information
can they give us?

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

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

6. Selected Area Electron
Diffraction
SAED

7. One atom type A
a
b
b>a
a=3 Å
b=5 Å

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

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

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

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

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

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

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

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

16. 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

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

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

19. 010
100
[001]
Indices all other reflections:
vector addition
110
020
200 210
120

20. 1/3Å
1/5Å
Experimentally: the other way around:
010
100
[001]
*

21. *How?
Make a list of all reflections with hkl and their d-values.
• use Excel to make the list yourself
• use free software like Powdercell
• ...
d hkl
7.68 001
5.64 010
5.46 100
4.55 011
4.45 101
3.92 110
... ...

22. 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*1 0*0-1*0 1*1-0*0
[ 0 0 1 ]
0 0
1 0
0 1
0 0
1 0
0 1

23. (010)
(100)
(110)
[001]
010
100 110
[001]
110
-
(110)
-

24. Exercise:
index the given patterns taken
from a CaF2 mineral

25. a
b
c
b
a
c
CaF2
cubic
a=5.46 Å
Fm3m-

26. 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

27. We are going to index these patterns.
They are obtained by tilting around the diagonal row.
(Online version:
working page can
be found at the
end.)

28. We are going to index these patterns.
They are obtained by tilting around the diagonal row.
(Online version:
workpage can be
found at the end.)

29. Start with easiest:
highest symmetry or smallest interreflection distances
= usually lower zone indices (“main zones”)
(Online version:
workpage can be
found at the end.)

30. 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
Why go for smaller interreflection distances?
=higher d = less choices

31. First pattern:
Apparent symmetry: 4-fold
Go for highest symmetry

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

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

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

35. To do: measure the distances, compare to list d-hkl, index
consistently.
Scalebar = R (in mm)
Step 1: Use the scalebar for the conversion
factor to 1/d-values.
equal to 1/0.08 nm
R.d=L
L
42
43
44
34.4 mmÅ
53.8 mmÅ
0.02 mmÅ

36. To do: measure the distances, compare to list d-hkl, index
consistently.
Scalebar = R (in mm)
Step 1: Use the scalebar for the conversion
factor to 1/d-values.
equal to 1/0.08 nm
R.d=L
L
42
43
44
34.4 mmÅ
53.8 mmÅ
0.02 mmÅ

37. To do: measure the distances, compare to list d-hkl, index
consistently.
Scalebar = R (in mm)
Step 1: Use the scalebar for the conversion
factor to 1/d-values.
equal to 1/0.08 nm
R.d=L
L
42
43
44
34.4 mmÅ
53.8 mmÅ
0.02 mmÅ

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

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

45. 100
010
200020

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

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

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

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

50. Next zone: with reflections closest to the central beam.
Reflections closer to the central beam:
higher d-values
smaller amount of possible matches of hkl to this d
1
3
5

51. 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

52. 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

53. 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

54. 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

55. 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

56. 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

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

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

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

60. Consistency:
1
3
5
2001
Point 1 should be indexed as
111
111
111
all of the above are allowed
-
--

61. Consistency:
1
3
5
2001
Point 1 should be indexed as
111
111
111
all of the above are allowed
-
--

62. Consistency:
200111
111
-
If 111 for point 1-
311
-
Point 3 is
311
111

63. Consistency:
If 111 for point 1-
311
-
Point 3 is
311
111
2001

64. Consistency:
200111
111
-
1
200
3
If 111 for point 1-
point 3 = 311-

65. Consistency:
200111
111
-
1
200
3
If 111 for point 1-
point 3 = 311-
d311 ≠ dpoint3-
but

66. Consistency:
200111
111
-
1
200
3
If 111 for point 1-
point 3 = 311-
d311 ≠ dpoint3-
but

67. 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
h3+2 = h1
k3+0 = k1
l3+0 = l1
(also possible 111 and 111, make a choice and stick to it for the following patterns)
- - -
= h1 k1 l1

68. Consistency:
200
111
111
-
Sum = 022.
Indeed consistent: if 𝑔 perpendicular to 𝑔200
then type of reflection needs to be 0kl.
022 200
022

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

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

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

72. a
b
c
The crystallite is threedimensional.

73. 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.

74. 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.

75. 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:
-

76. 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]:
-

77. 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.

78. 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.

79. 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.

80. 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.

81. 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.

82. 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.

83. 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

84. 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
We can also see this in projection

85. 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]
We can also see this in projection

86. 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]
We can also see this in projection
which is easier to draw manually...

87. 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]

88. 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]
-

89. 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]
-

90. 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]
-

91. 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]
-

92. 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]
-

93. Right upper zone:
Point 2
d
1.22 Å
1.11 Å
1.05 Å
Measure the distance of two reflections, not on the
same line, calculate the corresponding d-value
We already know the
first point: 200.
200
2

94. Right upper zone:
Point 2
d
1.22 Å
1.11 Å
1.05 Å
Measure the distance of two reflections, not on the
same line, calculate the corresponding d-value
We already know the
first point: 200.
200
2

95. 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

96. 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

97. If this were not a tilt series...
Point 2 could have been at first sight
both 115 and 333...
In this case:
Can compare the experimental angles between reflections
to the theoretical angles
-either formulas from any standard crystallography work
-or simply simulate the different zones calculated for the
different options (JEMS, CrystalKit, Carine,...) to check this
Or in this particular case of 333: you would need to see 111 and 222 at 1/3 and 2/3 of the distance.

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

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

100. [001]
[015]-
[013]
-
[012]-
[035]-
[011]-
010
031 051
053
What if you didn’t know the material?
You would just need to check
more possibilities:
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

101. Pattern bottom left:
Point 2
d
Measure the distance of two reflections, not on the
same line, calculate the corresponding d-value
200
2
1.65 Å
1.58 Å
1.37 Å

102. Pattern bottom left:
Point 2
d
Measure the distance of two reflections, not on the
same line, calculate the corresponding d-value
200
2
1.65 Å
1.58 Å
1.37 Å

103. 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 042
131
153
Point 2
d
Point 2
hkl
200
2

104. 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
Point 2
d
Point 2
hkl
200
2
042
131
153

105. The indexation is indeed consistent.
200
131062
131
-

106. 200
131062
131
-
[013]-

107. 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”

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

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

110. What if you do not have any prior
knowledge when you have to
index?
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.

111. If we do not know the space group, the next step would
be to determine it!
(maybe you started from 0 or you had only cell parameters from XRD or...)
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

112. Reflection conditions
(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
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
+

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

114. Be careful: forbidden reflections can occur because of
dynamical diffraction
Incident electron wave

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

116. 020
100
F(100)≠0
F(110) ≠0
F(010)=0
Need to tilt to remove these paths...

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

118. You will need this table (from IT volume A)
Figure out reflection
conditions for these
sets.

119. hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

120. hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
For these patterns
both would be
good....!?

121. This means we do not have sufficient information.
we missed [012], which will make the difference.
-
By coincidence
200
042

122. 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]
-

123. This means we do not have sufficient information.
we missed [012], which will make the difference.
-
By coincidence
200
042 hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n

124. This means we do not have sufficient information.
we miss [012], which will make the difference.
-
By coincidence
200
042 hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
For only h+k=2n there is no reason why
021 would be absent.

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

126. 0kl:
k=2n
k,l=2n
k+l=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

127. 0kl:
k=2n
k,l=2n
k+l=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

128. hhl:
h=2n
h,l=2n
h+l=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

129. hhl:
h=2n
h,l=2n
h+l=2n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

130. 00l:
no condition
l=2n
l=4n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

131. 00l:
no condition
l=2n
l=4n
Step 1: determine the reflection conditions from the patterns.
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
200
042

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

133. 200 and 020 could be due to double
diffraction...
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

134. 200 and 020 could be due to double
diffraction...
 Tilt around 200 until all other
reflections gone except h00 axis:
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-

135. 200 and 020 could be due to double
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

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

137. From the reflection conditions you get
the extinction symbol:
F - - -

138. From the reflection conditions you get
the extinction symbol:
F - - -
This still leaves 5 possible space groups
F23 Fm3 F432 F43m Fm3m
-- -

139. From the reflection conditions you get
the extinction symbol:
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 CBED
- -

140. Convergent Beam Electron Diffraction
CBED

141. 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

142. 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

143. For CBED you need sufficiently thick crystals:
10 nm 20 nm 30 nm
40 nm 50 nm

144. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

145. Symmetry is 4mm.
Whole pattern projection symmetry

146. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

147. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

148. 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

149. [111]

150. 6mm
[111]

151. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

152. 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

153. 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

154. 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

157. CBED
SAED

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

159. Try it yourself on example SnO2

160. SAED CBED
Example: rutile-type SnO2

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

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

163. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992
Projection WP: 4mm
Projection diffraction group:
Table Eades
4
4RmmR
4mm1R

164. Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992
Projection WP: 4mm
Projection diffraction group:
Table Eades
4
4RmmR
4mm1R

165. Projection diffraction group:
4mm1R
Possible diffraction groups:
4mRmR
4mm
4RmmR
4mm1R
Table from: J.A. Eades,
Convergent beam diffraction, in:
Electron Diffraction Techniques,
volume 1, ed. J. Cowley, Oxford
University Press, 1992

166. 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

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

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

169. Whole
Pattern
projection
symmetry
Whole
pattern
symmetry

170. WP symmetry
4
4mm
2mm

171. WP symmetry
4
4mm
2mm

172. [...] [...]

173. 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

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

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

176. Possible projection
diffraction group:
21R
m1R
2mm1R
Projection
whole pattern:
2mm

177. Possible projection
diffraction group:
21R
m1R
2mm1R
Projection
whole pattern:
2mm

178. Possible projection
diffraction group:
21R
m1R
2mm1R
Projection
whole pattern:
2mm

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

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

181. Diffraction group:
2mm
2RmmR
2mm1R
Whole pattern
m

182. Diffraction group:
2mm
2RmmR
2mm1R
Whole pattern
m

183. 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

184. Possible point groups
4/mmm
m3m

185. What would make a difference further?
For example:
-cell parameters
-look for a third zone etc.
-SAED for reflection conditions

186. For example, if you need
cell parameters a=b= 4.72 Å, c=3.16 Å
to be able to index all patterns,
the point group is
4/mmm
m3m

187. For example, if you need
cell parameters a=b= 4.72 Å, c=3.16 Å
to be able to index all patterns,
the point group is
4/mmm
m3m

188. CBED
SAED
Space Group P42/mnm
Then you would combine this again with reflection
conditions (not derived in this exercise) to get the
space group.

189. At the end of this lecture you should be able to
(1) index SAED patterns in case the cell parameters
are already known
(2) determine the possible space groups from SAED
patterns
(3) determine possible point groups from CBED
patterns
Combine (2) and (3) to find the space group.

190. Working page for indexing