These slides follow up on the slides with a tutorial on how to index ED patterns, 5 10a. From the result achieved in those slides, we now derive the reflection conditions and corresponding possible space groups.

1. Exercise:
Determine the diffraction symbol from
the patterns of CaF2.

2. You will need this table (from IT volume A)
Figure out reflection
conditions for these
sets.
For shortening the exercise, let's suppose we already know that it is cubic (the fact that a=b=c is not enough to declare
that it is cubic, it needs the right symmetry elements! If we do not know this, the difference will be that we will also take
the tabes for orthorhombic, etc and will just end up with more possibilities to check further.

3. 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]
-
We indexed these patterns in an earlier exercise in the lecture "How to index SAED patterns". Which general reflection
condition can you derive from these patterns?

4. hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
Step 1: determine the reflection conditions from the patterns.
-
For these patterns
both would be
good....!?
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
Both the second and third answer are in agreement with these patterns! This means you cannot determine the general
condition from these patterns. In real life you would need to go back to the microscope and get extra patterns from other
zones.

5. 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
[001]
[015]
-
[013]
-
[012]
-
[035]
-
[011]
-
We already had the patterns that are circled in red. We could add to these for example [0-12] by a new experiment or if you
are in control, simply make sure to record all zones you encounter right in the first session.

6. This means we do not have sufficient information!
we miss [012], which will make the difference.
-
By coincidence
200
042
For these patterns
both would be
good....!?
hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
An extra pattern that can bring the solution here is for example the [01-2] shown in this slide. Determine which of the two
conditions still agrees with this pattern.

7. 200
042 hkl:
h+k+l=2n
h+k, k+l, h+l=2n
h+k=2n
200
042
This means we do not have sufficient information!
we miss [012], which will make the difference.
-
By coincidence
For these patterns
both would be
good....!?
Now only hkl: h+k=2n, k+l=2n, h+l=2n fits. Which means that this is the correct reflection condition. The correct reflection
conditions should fit all zones of the same material!

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

9. 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
After the general condition, determine the special conditions.

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

11. 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

12. 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

13. 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
And afterwards, the serial conditions. The International table asks for the condition 00l, and we cannot see these reflections.
But remember, in cubic there is equivalence: h00, 0k0 and 00l are interchangeable.
Again, if we did not know it was cubic, we would need to also check the tables of the other crystal classes, check more
combinations and need for that more zones. This however all goes in the same way, so to learn the concepts now, assuming
that cubic is known is perfectly ok.

14. 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
Possible! It
might be double
diffraction!
Do not forget about the possibility of double diffraction!

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

16. ?
?

17. 200 and 020 could be due to double diffraction...
200
131
200
151
200111
200020
[001] [015]
-
[013]
-
[011]
-
 Tilt around 200 until all other reflections gone
except h00 axis:

18. 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

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

20. From the reflection conditions you get
the diffraction 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