1. Apr 29, 2014

### kelvin490

For diffraction to occur in crystal, Bragg equation must be satisfied and h,k,l must be of certain combination. I would like to ask

1. How to distinguish a second order diffraction of (100) and first order diffraction of (200)? (they occurs at the same angle.)

2. How can a (200) plane diffraction of simple cubic possible? It seems that there is no atom from this plane because atom only at the four corners of a unit cell.

2. Apr 30, 2014

### kelvin490

Some may say that the (200) diffraction is actually the second order (100), but it seems that the (200) plane simply doesn't exist. FCC and BCC also have this problem.

3. May 1, 2014

### UltrafastPED

The Bragg reflections for first order: λ=2 d sin(Θ);
for higher orders (=n) we have: nλ=2 d sin(Θ).

The diffraction angle increases with the increase in n; thus the (222) reflection appears further out than the (111) reflection - however, in general it will be much weaker.

When you look at the families of possible crystal planes the simple cubic (100) includes every plane parallel to this surface; the (200) includes every other such plane; the (600) includes every 6th plane - these higher orders cover the cases where the intervening planes are "missed" by the incoming radiation; this must be possible for the Bragg equation to work: the model is based on "partially reflecting" surfaces, which are the planes.

In reciprocal space these planes are closer and closer together by the reciprocal ratio: the (200) family is twice as close together as the (100) family.

Kittel's "Introduction to Solid State Physics" goes over most everything you would need to know in an easy to understand fashion.

4. May 1, 2014

### kelvin490

Thanks. I understand that every higher order (n) diffraction of (hkl) can be represented by (nh nk nl) diffraction. The existence of (nh nk nl) diffraction can also be proved even the limitation of h,k,l combination are considered (e.g. h+k+l must be even for BCC).

What I still wondering is that how can we tell whether a plane exist? Does it necessarily have a set of atoms whose centers lying on that plane (like (100) in simple cubic) or if diffraction can occur then the plane exist even there is no atoms lying on that plane (like (200) in simple cubic)?

5. May 1, 2014

### UltrafastPED

Do you mean the selection rules, or the physical presence of the plane?

For the (100) plane you can consider just the unit cell; for the (200) plane you simply stack two unit cells, etc.

You won't run out of planes until you run out of atoms in the crystal ... there are usually more than enough, even in tiny crystallites. However, most of the beam energy is diffracted away by the lower order planes, so the higher orders become fainter and fainter - it often requires a time exposure to make them visible.

6. May 1, 2014

### kelvin490

If we stack two unit cells the interplanar distance is still that of (100) for simple cubic. The (200) plane has half the interplanar distance of (100), but there is no atom lying on (200) because the atoms only locate on the eight corner of unit cell. So I wonder how the (200) plane can cause diffraction.

7. May 1, 2014

### UltrafastPED

The distance between (200) planes is 1/2 that of the (100) planes only in the reciprocal space; in the configuration space of the crystal they are twice as far apart as the (100) planes.

And they are easy to find because they only require two layers of the crystal.

You would do well to read a text book like Kittel in order to improve your visualization; it does take a while to get used to crystallography and Miller indices and Bragg diffraction.

Perhaps this: http://www.slideshare.net/joserabelo/x-ray-diffraction-basics