# A note on Bragg’s law for X-ray scattering

In 1912, at the age of only 22, William L. Bragg wrote a paper which essentially began the field now known as X-ray crystallography. Earlier that year, Max von Laue and others had passed X-rays through crystals of copper sulphate and zinc sulphide, obtaining patterns of dots on a photographic film. These patterns were the first evidence that X-rays could be diffracted by the regularly arranged atoms in crystals. The X-ray diffraction pattern produced by zinc sulphide was particularly sharp and clear, and seemed to show a four-fold symmetry.

Laue attempted to explain the patterns of dots mathematically but was not completely successful. He had tried to set up the maths based on overly complicated and partially incorrect models of what was going on at the atomic level. At this time the young William L. Bragg was just finishing a mathematics degree at Cambridge University. He heard about Laue’s work from his father, William H. Bragg, who was a mathematician and physicist himself and had been working on X-rays at the University of Leeds. When thinking about the problem, the young William Bragg envisaged a much simpler model than Laue’s in which the planes in the crystal lattice acted like mirrors. The incoming X-ray beams, interpreted as electromagnetic waves with a particular wavelength $\lambda$, would be reflected by the planes in the crystal lattice. The outgoing beams would then interfere with one another to give either constructive or destructive interference, depending on whether the crests of the waves were in phase or out of phase. The diagram below shows the standard set-up for deriving Bragg’s law that appears in every textbook and article on this topic.

The incident beams are at an angle $\theta$ to the planes in the lattice and are reflected at the same angle. The extra distance travelled by the lower beam is the sum of the lengths of the two black arrows in the diagram. Elementary trigonometry shows that the length of each black arrow is $\text{d} \sin \theta$, so the extra distance travelled by the lower beam is $2 \text{d} \sin \theta$. A black dot is produced on the photographic film only if the extra distance travelled by the lower beam is a whole number of wavelengths, since in this case there will be constructive interference of the outgoing waves and the reinforced beam will register strongly. If the extra distance travelled by the lower beam is not exactly equal to a whole number of wavelengths, there will be some X-ray beam nearby that will interfere destructively with it and they will cancel each other out. For example, the diagram below shows how the beams cancel each other out when the extra distance travelled by the lower beam is an odd multiple of half a wavelength. In this case, the scattering of the X-ray beams will not be detectable and no black spot will appear in the corresponding position on the photographic film.

It follows that it is possible to calculate the positions of the black spots on the photographic film using the equation

$2 \text{d} \sin \theta = \text{n} \lambda$

where n is an integer. This is Bragg’s law. A black spot appears on the photographic film only at those positions where the three quantities d, $\theta$ and $\lambda$ satisfy the equation of Bragg’s law simultaneously; otherwise no diffraction spot will appear on the film. This is the law that the young William Bragg published in his 1912 paper. Together, the two William Braggs continued to investigate crystal structures using X-ray diffraction, and father and son were both jointly awarded the Nobel Prize in physics for this work in 1915. The young William Bragg was only twenty-five and remains the youngest ever Nobel winner for science.

This is a lovely story and the standard textbook set-up above for deriving Bragg’s law seems straightforward enough. However, a bright A-level student of mine expressed dissatisfaction with it and asked if there is some alternative approach to deriving Bragg’s law that involves an explicit subtraction operation, i.e., an approach showing that a specific distance travelled by the lower X-ray beam minus a specific distance travelled by the upper X-ray beam equals $2 \text{d} \sin \theta$. I felt this was an interesting question but when we tried to find an alternative approach in the literature we noticed that every textbook and every source online that we looked at had exactly this same set-up, or something very similar to it that the student also did not like! We eventually managed to come up with a calculation of the kind the student wanted to see by varying the standard textbook set-up slightly, and this is what I want to record in the present note.

We altered the standard set-up by moving the two short perpendiculars to the X-ray beams symmetrically away from the point of incidence of the upper beam, as indicated in the following diagram.

A convenient place to put the blue perpendiculars is at the points of intersection of the lower X-ray beam with the upper crystal plane, which is what I have done in the diagram above. This then allows the type of explicit subtraction operation to be carried out that the student wanted to see. All we needed to do was to demonstrate that the distance shown in red in the diagram below, minus the distance shown in green, exactly equals $2 \text{d} \sin \theta$.

From the diagram we see immediately that

$\text{h} = \frac{\text{d}}{\sin \theta}$

$\frac{\text{j}}{\text{k}} = \cos \theta$

$\frac{\text{d}}{\text{k}} = \tan \theta$

Therefore

$\text{j} = \text{k} \cos \theta = \frac{\text{d}}{\tan \theta} \cos \theta$

and the desired subtraction operation is then

$2 \text{h} - 2 \text{j} = \frac{2 \text{d}}{\sin \theta} - \frac{2 \text{d}}{\tan \theta} \cos \theta = 2 \text{d} \bigg( \frac{1}{\sin \theta} - \frac{\cos^2 \theta}{\sin \theta} \bigg) = 2 \text{d} \sin \theta$

as required. One can now imagine the blue perpendiculars sliding back together towards the centre of the diagram. The difference between the red and the green distances will always remain equal to $2 \text{d} \sin \theta$ as the blue perpendiculars slide inwards but the green distance will eventually shrink to zero. We will then have arrived back at the standard textbook set-up and the extra distance travelled by the lower X-ray beam will (of course) remain $2 \text{d} \sin \theta$.