# I Why XRD peaks are so sharp?

#### Livio Arshavin Leiva

I was reviewing an old topic for me that's x-ray diffraction, and one doubt I always had in my mind arised again.

When introducing the Bragg's law, the typical explanation is that the x-ray waves reflecting in two adjacent planes interfere with each other, leading to a fully constructive interference when the difference between the lengths of each wave's paths is a multiple of the wavelength, as it's shown in the image below from Wikipedia.

But in the case of this model being correct, one would had a maximum when the waves are coherent and zero intensity when the phase is 180°, but in between if one calculates the integral of the square of the resulting wave after interference it leads to a cosine law with the phase angle... so between two peaks of the same family (for example 100 and 200 Miller's indexes, that means that 1 and 2 wavelengths are exactly contained in the difference of path length, respectively) one should see a cosine dependence in the pattern (more or less, I know that the 2theta angle is not lineal with the phase between the waves (there's a sine relation)) but clearly the peaks of a real pattern are way sharper.

In fact, it's always said that according to theory, they should be deltas. But that's not Bragg's theory I guess, right?

I know that this explanation is simplified and a more accurate (or more real) model is the von Laue's approach that considers spherical waves re-radiated by all atoms right after receiving the x-ray excitation wave. If one considers this, are the peaks sharper? I always heard that the two formulations were equivalent, but maybe they were equivalent only when looking at the constructive interference condition.

So my question is, to sum up, how can you explain that the peaks are so sharp?
Thank you in advance, and sorry if this is a noob question...

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#### Lord Jestocost

Gold Member
2018 Award
In order to understand how destructive interference takes place, you have to consider not only two scattering planes, but all the contributions from planes deeper in the crystal. The "sharpness" of the diffraction peaks depens on the thickness of the crystal measured in a direction perpendicular to a given set of reflecting planes.

If the path difference between rays scattered by the first two planes differs only slightly from an integral number of wavelengths, then the plane scattering a ray exactly out of phase with the ray from the first plane will lie deep within the crystal. If the crystal is so small that this plane does not exist, then complete cancellation of all the scattered rays will not result. It follows that there is a connection between the amount of "out-of-phaseness" that can be tolerated and the size of the crystal. We will find that very small crystals cause broadening (a small angular divergence) of the diffracted beam, i.e., diffraction (scattering) at angles near to, but not equal to, the exact Bragg angle. We must therefore consider the scattering of rays incident on the crystal planes at angles deviating slightly from the exact Bragg angle.
(B. D. Cullity in “Elements of X-ray Diffraction, Second Edition”)

#### ZapperZ

Staff Emeritus
2018 Award
I was reviewing an old topic for me that's x-ray diffraction, and one doubt I always had in my mind arised again. When introducing the Bragg's law, the typical explanation is that the x-ray waves reflecting in two adjacent planes interfere with each other, leading to a fully constructive interference when the difference between the lengths of each wave's paths is a multiple of the wavelength, as it's shown in the image below from Wikipedia. But in the case of this model being correct, one would had a maximum when the waves are coherent and zero intensity when the phase is 180°, but in between if one calculates the integral of the square of the resulting wave after interference it leads to a cosine law with the phase angle... so between two peaks of the same family (for example 100 and 200 Miller's indexes, that means that 1 and 2 wavelengths are exactly contained in the difference of path length, respectively) one should see a cosine dependence in the pattern (more or less, I know that the 2theta angle is not lineal with the phase between the waves (there's a sine relation)) but clearly the peaks of a real pattern are way sharper. In fact, it's always said that according to theory, they should be deltas. But that's not Bragg's theory I guess, right?
View attachment 243153
I know that this explanation is simplified and a more accurate (or more real) model is the von Laue's approach that considers spherical waves re-radiated by all atoms right after receiving the x-ray excitation wave. If one considers this, are the peaks sharper? I always heard that the two formulations were equivalent, but maybe they were equivalent only when looking at the constructive interference condition.
So my question is, to sum up, how can you explain that the peaks are so sharp?
Thank you in advance, and sorry if this is a noob question...
It's the same reason why, keeping the slit width and slit separation lengths the same, interference pattern from a diffraction grating is so much sharper than interference pattern from 2 slits. There are more "sources" that need to be satisfied for when and where the constructive interference can occur.

Zz.

#### SpinFlop

So my question is, to sum up, how can you explain that the peaks are so sharp?
Kittel "Intro to Solid State" has a simple exercise to walk you through demonstrating this yourself. It is problem 4 at the end of chapter 3. Assuming you have a copy, then you can work this out yourself. I will abbreviate this exercise even further to provide an immediate answer and a bit of insight.

In 1D, for scattering centers (atoms) a distance of $\mathbf{a}$ apart, the scattering amplitude is proportional to the sum of phase differences:

$$F = \sum_{n=1}^{M} e^{-i(n\mathbf{a})\cdot\mathbf{k}}$$

where $M$ is the total number of scattering centers. The exercise is to show that this can be written as

$$F = \frac{\sin(\frac{M}{2}\mathbf{a}\cdot\mathbf{k})}{\sin(\frac{1}{2}\mathbf{a}\cdot\mathbf{k})}$$

and from this it can then be shown that the width of the peak $|F|^2$ is proportional to $1/M$. For the macroscopic size crystals (which is what are almost always measured) $M$ is extremely large and so $|F|^2$ has vanishing linewidth, ie: it is a delta function. This directly confirms what Lord Jestocost wrote:

“... It follows that there is a connection between the amount of "out-of-phaseness" that can be tolerated and the size of the crystal...
(B. D. Cullity in “Elements of X-ray Diffraction, Second Edition”)
So in short, the sum of phase differences you had in mind was just two and in this case we are in full agreement that the linewidth will be very broad, but in reality there are lots and lots of planes and as you increase the sum you decrease the linewidth and this is why you get a delta function. And as ZapperZ pointed out this same relationship also holds for linewidth and slit number for a diffraction grating:

It's the same reason why, keeping the slit width and slit separation lengths the same, interference pattern from a diffraction grating is so much sharper than interference pattern from 2 slits. There are more "sources" that need to be satisfied for when and where the constructive interference can occur.
All together that should provide a solid explanation just thinking in terms of phase sums. However, there is a nice way to think about this in terms of the Fourier analysis of scattering. In general, scattering patterns are the Fourier map of a process in real space. For the x-ray scattering we are discussing now this process would be Bragg scattering from a crystal. The Fourier transform of something very long gives you something very small, and vice-verca. In the case of a long range ordered crystal, this is why you have a delta function. If you have some way to make your crystal short range ordered then the Fourier transform broadens. Above we did this by just considering really small crystals. Another way would be by introducing stacking faults in the crystal, for instance along the c-axis of some layered structure. If you mapped the peak out in reciprocal space $\mathbf{Q} = [H,K,L]$, then you would see that your Bragg spot smears out into a line along the $L$ direction, but remains sharp in $H$ and $K$. The length of this line would give information on the average distance between faults. The same sort of linewidth analysis can also be used for measuring short range order or glassiness of spins in a magnetically ordered state. So linewidths are often of importance. Indeed, in the full theory of scattering where you include both elastic and inelastic processes $(\mathbf{r},t) \rightarrow (\mathbf{Q},E)$ you find that a similar Fourier picture applies for the inelastic processes where now the linewidth encodes the lifetimes of the excitations produced in by the scattering. In fact, the reason linewidths are typically Lorentzian is because most orders/excitations decay exponentially in space/time and the Fouier transform of an exponential is a Lorentzian.

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#### Livio Arshavin Leiva

Everything is much more clear now. Thank you all for your answers, especially SpinFlop. I knew that the linewidth was a measure of the size of the crystal but I hadn't realize that there was such a simple, though elegant, explanation. Also your examples of application of Fourier analysis are bringing back to life a lot of concepts I had forgotten and everything is recovering sense now.

"Why XRD peaks are so sharp?"

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