# X-ray and neutron powder diffraction: peak broadening due to crystal size

When analysing powder diffraction patterns, the broadening of peaks can be used to estimate crystal sizes. Smaller crystal size gives larger broadening according to the Scherrer equation:

$$\beta = {{K\cdot\lambda}\over{D\cdot\cos \theta}}.$$

What is the physical origin of this effect?

Edit: Added homework tag as this is related to my thesis on soil studies using diffraction methods.

The X-ray diffraction pattern is the Fourier transform of whatever is doing the diffracting. If you had an infinite plane of atoms then the spots (rings in a powder pattern) would be infinitely sharp because the Fourier transform of an infinite wave is a delta function. However a real crystal is the product of an infinite plane with an envelope function, where the envelope is the size of the crystal, so the spot is the convolution of a delta function with the Fourier transform of the envelope function. In a powder pattern we have many crystals of differing sizes, so the average Fourier transform of the crystal size ends up looking something like a gaussian and the spots have a roughly gaussian profile.

Re your formula, suppose the average crystal ends up looking like a sphere, i.e. a disk in profile, then it's Fourier transform is going to be an Airy disk (but blurred out by the variation in particle size). The half angle subtended by the Airy disk, $\beta$, is (for small angles):

$$\beta \propto \frac{\lambda}{d}$$

which is the Scherrer equation for small $\theta$ (I'd have to go away and look at the derivation of the Scherrer equation to remember why there's a factor of $\cos\theta$, but for small $\theta$ this factor is approximately $1$ anyway).

Response to comment:

The incoming X-rays are scattered by the atoms in the crystal, so each atom acts as an X-ray source. We get the diffraction pattern by summing up the X-rays emitted (i.e. scattered) by all the atoms in the crystal. If you start with one atom then the scattered X-rays will be just be a spherical wave. Add a second atom and now the pattern will be the same as the Young's slits experiment. As you add more and more atoms the pattern will tend towards the pattern we expect from a large crystal, however to get an infinitely sharp spot would require an infinite number of atoms. When we have a finite number of atoms the spot will have a finite width.

It's a bit like a Fourier synthesis (which is where we came in). Each atom adds a term to the Fourier sum, but to get a perfect transform of the lattice requires contributions from an infinite number of atoms. With a finite number of scatterers the diffraction pattern will only be an approximation to the FT of the lattice.

• Thank you for the reply. However my question was more about the physics, i.e. why does smaller crystals give broader peaks? Your answer is explaining it in mathematical terms. I was more looking for an explanation of what happened to the interaction between wave and crystal when crystals are small. – SRJ Aug 14 '13 at 21:18
• @SRJ: I've extended my answer to (I hope) make it clearer what the physical origin is. – John Rennie Aug 15 '13 at 6:20
• I like your thought experiment. If we consider the case with just one atom and place a detector at some distance (large compared to the atom), what patttern will we then detect? Since X-rays have an wavelength of 1.5-2 Å and the electron cloud is of the same order, I would expect that it would be similar to single slit diffraction. With the diffraction minima at $$\sin \theta = \lambda / a$$ where a is the size of the electron cloud?? – SRJ Aug 28 '13 at 10:12