# How many atoms does one need to get reliable crystal information with X ray or electron diffraction?

I know that, in theoretical derivations of X ray diffraction (XRD) or electron diffraction (ED), we always assume an infinite lattice, and, as you make the crystal smaller, edge effects start to matter more.

As you make it even smaller, you start to lose information about actual orientation of the crystal and such (obviously, the diffraction from two atoms really tells you literally nothing).

I know it's a continuum, so "reliable" is a little subjective, but at what point can you not get any useful information from XRD or ED? I've seen ED images inset in TEM pictures of ~10nm structures, which comes out to about 100 atoms.

• Are you sure the insets were X-ray diffraction and not electron diffraction? Since TEMs can routinely do diffraction measurements it seems more like they were electron diffraction. – John Rennie Jul 17 '14 at 15:37
• Sorry, you're absolutely correct. Let me edit my post. Stupid of me because I've done it myself, SAED. – YungHummmma Jul 17 '14 at 16:00
• The wiki page on SAED actually answers my question a little, but not entirely: en.wikipedia.org/wiki/Selected_area_diffraction It says "It is similar to X-ray diffraction, but unique in that areas as small as several hundred nanometers in size can be examined, whereas X-ray diffraction typically samples areas several centimeters in size", but I know I've seen both of those techniques applied to much smaller areas. – YungHummmma Jul 17 '14 at 16:03

You say:

obviously, the diffraction from two atoms really tells you literally nothing

but that isn't true. Two atoms will give you a Young's slits type diffraction pattern and you could measure the fringe spacing to determine how far apart the two atoms are. In practice you'd run into some severe experimental difficulties! Still, in principle you'd still get a measurable diffraction pattern.

Leaving aside instrumental limitations the pattern you get is the Fourier transform of the lattice convolved with the Fourier transform of the scattering scattering area. Suppose you have some lattice with a repeat of $d$, and you're illuminating an area of this lattice of size $\ell$. The diffraction spots will have a spacing of approximately:

$$\theta \approx \frac{\lambda}{d}$$

give or take some numerical constants. Similarly the angular width of the diffraction pattern of the scattering area will be something like:

$$\theta' \approx \frac{\lambda}{\ell}$$

The end result is that you get spots separated by $\lambda/d$ and with a width of $\lambda/\ell$. So the ratio of the spot spacing to the spot width is of order $\ell/d$.

You don't say what your 10nm particles are made of, but let's say the lattice spacing is 0.5nm then $\ell/d \approx 20$ so the ratio of spot spacing to spot width would be 20:1. You'd still get pretty sharp spots. Well, sharp by electron diffraction standards - I suspect the powder X-ray chaps would sneer as such low resolution.

In practice I suspect the limitation would be that the lattice is deformed at the surface, because the atoms at the surface are in an asymmetrical environment, so the interatomic spacings will change at the surface. As you make your particles smaller the surface area to volume ratio increases, and your diffraction pattern broadens because your crystal is effectively getting more disordered. In principle you could go down to crystal sizes of the order of the lattice spacing.

• Thank you for the response. I say that two atoms tell you almost nothing because they can't really have a crystal orientation, or even crystal structure (amorphous would have two adjacent atoms as well). – YungHummmma Jul 17 '14 at 20:33
• @YungHummmma: yes, quite true, but you would still (in principle) measure a diffraction pattern from your two atoms. In fact for amorphous systems you do get a diffraction peak corresponding to the nearest neighbour distance i.e. basically diffraction from pairs of atoms. – John Rennie Jul 18 '14 at 5:17

in theoretical derivations of X ray diffraction (XRD) or electron diffraction (ED) we always assume an infinite lattice

Not always. Crystalline structure and infinite lattice are assumed for example in Bragg's Law (which gives positions of reflections: 2d sinθ = nλ). If the grain size is small (in nanometers range) applicability of the Bragg's Law is doubtful. But there are other ways of analysing diffraction patterns. For example, Debye's scattering formula tells you how intensities depend on interatomic distances without any assumptions about the structure.

For a long time Debye's formula has been used only (mostly) for non-crystalline systems, but in recent years or decades it has become popular also for nano-sized crystalline grains.

Going back to the Bragg's Law, when the grain size decreases, the peaks are getting not only wider and overlapping, but also they are not centered at the positions predicted by the law. It can be shown by calculating diffraction pattern from atomistic model and then checking if the analysis of such pattern gives results consistent with the model. (As a side note, I worked with people who promoted term apparent lattice parameters for "lattice parameters" calculated from peak positions in nanomaterials).

When we decrease the size of scattering crystal, the diffractogram peaks broaden. The broadening $\beta$ is related with the crystal size $\tau$ by the Scherrer equation: $$\tau = {{K\lambda}\over{\beta \cos \theta}}, \beta = {{K\lambda}\over{\tau \cos \theta}}$$ ($K$ depends on the form of the crystal, ~0.9 for spheres). So, the first note would be: depending on the initial (large crystal) pattern peaks will be unresolvable at different sizes and with decreasing of the crystal size peaks with larger $\theta$ become unresolved earlier.

If you want a numerical answer, you should take some diffractogram you have and estimate $\tau$ for which the peaks will get unresolved. For instance, if you have a peak at $2\theta = 12°$ and the next one at $2\theta = 14°$, then it is enough to have $\beta = 2°$ which means $$\tau = {{K\lambda}\over{\beta \cos \theta}} \sim {{\lambda}\over{\beta \cos \theta}} \sim 29 \lambda$$

Second note is scattering on two atoms holds large portion of information (including distance between them and their $\vec r_1-\vec r_2$ vector) but this is not called diffraction and facilities built to detect diffraction usually can not extract this information. There are general formulae for this case (looks like (64) here is sufficient).