# Diffraction question

1. Jun 16, 2015

### Coffee_

Take a cuboid unit cell structure where we put an atom on each of the vertices and a different atom in the center. We have a single crystal of that material.

Now take one of the cuboid axes to be the ''a'' basis vector. We send in a diffraction beam parallel to that direction. On the other side we will see a diffraction pattern that corresponds to the reciprocal lattice. If we identify each point of this lattice to a set of planes we will find that all these points share on thing. They have ''0'' on the first place. So things like (002).

This is according to one of the exercises I have made. However I have a conceptual question - a plane having a vector with a zero on the first place means that the normal vector of this plane always perpendicular to ''a'' or the direction of the incoming beam here. How can you have diffraction happening on planes that are parallel to the incoming beam? In Bragg law you need to always have an angle between the planes and the incoming beam. I'm clearly misunderstanding something big here so please some help.

2. Jun 17, 2015

### Simon Bridge

You are saying that the diffracting planes are those which are edge-on to the incoming beam?

3. Jun 17, 2015

### Coffee_

If by edge-on you mean that from the viewpoint of the beam all you see of the plane is a line, then yes. That's not the claim I'm trying to defend, I'm baffled by it myself but that is indeed how it seems to be from the exercises I've checked and the program CrystalDiffract if you kow it. So my confusion lies in the fact that I thought that to diffract and angle was supposed to exist between the beam and the planes. However if I send a beam in from the ''a'' direction in an orthorhombic crystal, and then index the diffraction pattern, the program I mentioned - or exercises in class seem to indicate that points like (0,k,l) are present in the pattern. As you see (0,k,l) planes lie edge-on to the incoming beam here so how can they diffract.

4. Jun 17, 2015

### DrDu

One guess: You are talking about a centered cubic lattice. The primitive cell which contains only one atom has its basis vectors not coinciding with the cubic vectors you are considering.

5. Jun 17, 2015

### Coffee_

By your answer I judge that what I'm describing is considered odd, right? That normally this shouldn't be the case?

6. Jun 17, 2015

### DrDu

7. Jun 17, 2015

### Coffee_

I can see why that is the case for monoclinic or triclinic structures but I was solely dealing with ortho/tetra/cubic crystals.

8. Jun 17, 2015

### DrDu

Just try to draw a primitive unit cell (i.e. one which contains only one atom). Is this cell rectangular?

9. Jun 17, 2015

### Coffee_

In general it obviously depends, but in my case yes, the cell would be rectangular and I would place one atom at each of the 8 vertices of the cell. This is equivalent to containing one atom since each one counts for 1/8 at the vertices I guess.

10. Jun 17, 2015

### DrDu

Ah, I just saw that you wrote "a different atom in the center". So you are not talking about a body centered lattice.

11. Jun 17, 2015

### Coffee_

Oh no that was not my intention. That was actually not even needed, for my understanding it would be enough to discuss a cubic primitive cell.

12. Jun 17, 2015

### DrDu

For a primitive cubic cell I don't understand it either!

13. Jun 17, 2015

### Coffee_

Example of what I mean:

Check slide 7 first picture,

The direction of the beam seems to be [001]. However there are diffraction dots present of planes that are parallel with the beam direction! Very strange. This is what I encountered in exercises and in the program (CrystalDiffract) for cubic,orthorhombic,tetagonal primitive stuff.

14. Jun 17, 2015

### DrDu

I would have taken [001] to mean that just to be the view along that direction on the reciprocal lattice. This does not mean that there is any ray incident along this direction.

15. Jun 17, 2015

### Coffee_

Now I'm quite confused, could you elaborate?

16. Jun 17, 2015

### DrDu

17. Jun 18, 2015

### DrDu

So this is peculiar to electron microscopy where samples are very thin. The finite size of the crystal leads to a broadening of the angular distribution of the scattered light so that even for incidence parallel to the planes, reflection can be observed.