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5. Diffraction of X-rays

We will examine in turn what happens when X-rays strike a single atom, when a plane wave front of X-rays strikes a line of atoms, when a beam of X-rays strikes a two-dimensional net of atoms and finally a 3-dimensional lattice of atoms.

 

Figure 12

When a wavefront of X-rays strikes an atom, the electrons in that atom interact with the X-rays and immediately re-emit the X-radiation, normally without change of wavelength, and the X-radiation that is emitted by the atom is emitted as a spherical wavefront (Fig. 12). This, of course, is an ideal situation which we cannot observe in practice. Consider a line of identical atoms, distance a apart. Let a beam of wavelength $\lambda$strike this line of atoms at an angle $\bar \mu$. Each of these atoms immediately begins to emit radiation in the form of spherical wavefronts (Fig. 13). If one observes the scattered radiation in the plane of the incident beam and scans all possible angles, the requirement for seeing a beam of enhanced intensity is that the path-length difference between the advancing incident wavefront and the advancing diffracted wavefront shall be a whole number of wavelengths.

 

Figure 13

 

Figure 14

Consider Fig. 14. The length PR-OQ should be equal to n$\times$wavelength $\lambda$. If that situation holds, then the observer looking along the lines of the arrows will see that the scattered radiation is very intense at that angle: there is constructive interference. If the observer moves away from that particular angle there will be no enhanced scattered radiation. It is important to notice that, while the sketch is two-dimensional, each atom is giving off a spherical wave of radiation. The directions of scattering thus constitute the surface of a series of cones (see Fig. 15). The largest cone angle corresponds to a difference of one wavelength between the incident and the diffracted radiation. The angle of deviation increases and the cone angle decreases as the integer becomes 1, 2, 3 and so on. It is fairly obvious that for cosine $\phi = 1, a = n \times \lambda$and it is this condition which tells you the maximum number of cones that will be observable from a line of atoms of spacing d for radiation of wavelength $\lambda$. Consider now a second line of atoms at some angle to the first line: a two-dimensional net. Consider the second line of atoms quite independently from the first line and it is fairly obvious that the second line of atoms when irradiated by the X-rays will generate a series of cones also obeying the same criteria. Thus the two-dimensional net will behave as if it were simply two lines of atoms and produce two families of intersecting cones. It should be fairly obvious that for two sets of cones with a common origin but with their axes non-collinear, the intersection of these cones will consist of a series of lines. Thus, for a two-dimensional net, very strong constructive interference will be seen only along certain well defined directions in space, and no longer anywhere along the surface of the cone as was observed for one line of atoms (see Fig. 16).

 

Figure 15

 

Figure 16

To extend this idea into three dimensions you need only add another line of atoms non-coplanar with the first net. Irradiate this line of atoms with the X-ray beam and, as before, a series of cones is formed. This family can only have a line of intersection common with those of the first two under the special condition that UV and W are coincident (see Fig. 17). The result is that, for a three dimensional lattice of atoms irradiated by X-radiation, strong constructive interference will only occur in specific directions and for specific conditions of incidence. In other words one will not observe constructive interference unless one stands exactly in the right place in space. This type of construction was done by Laue about 1912. Although it is mathematically very straightforward, it is very difficult to picture in three-dimensions what is going on. Fortunately for us, Bragg in 1913 saw that the conditions for constructive interference of X-rays were equivalent to that of a simple plane reflecting the X-ray radiation with the condition that the plane could be described by a triplet of Miller indices. We can take any plane (hkl) in our crystal and we will now consider not spacings between atoms or lattice points, but spacings between the planes.

 

Figure 17

 

Figure 18: The condition of `reflection' by a crystal lattice. Difference of path $= GY + YH = 2d \sin \theta$, which must $=n\lambda$.

Consider Fig. 18. The incident X-radiation strikes the planes (hkl) at an angle $\theta$. The spacing between these planes is d. We will assume that the X-rays are reflected in exactly the same way as light will be reflected from a mirror so that the reflected beam leaves the plane at angle $\theta$. The requirement for constructive interference is very simple. Once again it is that the path length difference between the incoming and the outgoing beams should be a whole number of wavelengths. It is very easy to show that the path difference is equal to $2 \times d \times \sin \theta$and the final result is $n\lambda = 2d \sin \theta$. This is called Bragg's law, and it is of considerable importance in X-ray crystallography. The maximum value that $\sin \theta$ will ever have is 1. For $\sin \theta = 1, \theta = 90^\circ$ and the X-ray strikes perpendicular to the face of the crystal and is reflected back along the incident path. For this case, $2d = n\lambda$; the minimum d-spacing that we can ever observe with X-rays in any crystal will be equal to one half of the incident wavelength of the X-rays. What is interesting here, is that Bragg included the numerical value n (the integer) because he chose planes whose Miller indices (hkl) were prime numbers. In other words, Bragg's planes would be called (111) but never (222) or (333). This followed the normal conventions used in geology. He would talk about the plane (111) and then he would consider the n in the equation to be the 1st, 2nd, 3rd, 4th and so on order of reflection from that plane. For the single crystal X-ray crystallographer the numerical value of n is normally said to be equal to 1 and then we would simply use the equation $2d \sin \theta = \lambda$ but the d in the equation would be for a plane whose Miller indices can be either prime numbers or non-prime. Notice something else about the Bragg equation. Theta is a variable, in other words it is an angle which you can choose simply by rotating the crystal relative to the X-ray beam. The wavelength has a fixed value and d is obviously a fixed value determined by the size of the unit cell and the Miller indices. If we write down the equation as $2\sin \theta = \lambda/d$ and fix $\lambda$ in our experiment, we see that the experimentally observed value of $\sin \theta$ is a direct measure not of the d spacing but of the reciprocal of the d spacing of the planes. Notice also that somehow or other we are looking at perpendiculars to these planes because this d spacing is the perpendicular distance between planes of the crystal. For this reason, Ewald (1912) constructed what he called a reciprocal lattice. A reciprocal lattice is merely a construction that consists of normals drawn to all possible lattice planes whose indices are (hkl). These normals radiate from a common origin (the point 0, 0, 0 in the unit cell) and each normal terminates at a distance from the origin, proportional to the reciprocal of the d-spacing of the plane (hkl). Consider Fig. 19. What is interesting about it, is that each family of planes whose Miller indices are (hkl) is described by only one point in the reciprocal lattice: you may have an infinite number of planes of indices (hkl) but they correspond to one point (hkl) in the reciprocal lattice. The geometrical relationship between real lattices and unit cells and these reciprocal lattices and reciprocal cells are fairly complex. They will not help us directly with the applications that we want to consider in this article, therefore it is best that we stop the discussion of reciprocal lattices at this point and proceed to some other aspects of X-rays and interactions of X-radiation with crystals.

 

Figure 19: The relation between the direct lattice (heavy lines, solid circles) and the reciprocal lattice.

We accept at this stage that any set of planes in a crystal will cause a reflection of the X-ray beam if the set of planes is set at the right angle to the incident X-ray beam. The question now is how strongly will this set of planes reflect the X-rays? Will they necessarily reflect the incoming beam strongly or will they reflect it fairly weakly? The intensity of the reflected beam will be proportional to the product of the intensity of the incident beam and the concentration or density of electrons in the plane that is reflecting the beam. Note: It is the concentration of electrons , not of atoms, because it is the electrons surrounding the atoms that cause the scattering of the X-rays. It should be obvious that, if we know the size of the unit cell and if we know exactly where all the atoms are in that unit cell and if we know the atomic number of each of these atoms (in other words if we know how many electrons are associated with each atom in that cell), we should be able to calculate for any chosen plane with Miller indices (hkl) exactly what the concentration of electrons in that plane will be. In other words, if we know the structure of the unit cell, we should be able to calculate the intensity with which any chosen plane in that cell will scatter X-rays. In fact, this is very easy to do and the name given to such a calculated value is the Structure Factor.

Consider the reverse situation. Imagine that we know the size of the unit cell and that we can measure the intensities of reflection from all possible planes. It appears that from this information we should be able to calculate the positions of the atoms in the cell and not only that, but also the relative number of electrons per atom. The problem of working out where the atoms are in the unit cell from the observed d-spacings and intensities of reflection is called `solving a crystal structure'. This is one thing we will not attempt to do in this course, because what may sound very easy is, in fact, extremely complicated under some circumstances. All that is important at this stage, is to bear in mind that the intensity with which any family of planes can reflect an X-ray beam is directly proportional to the concentration of electrons in those planes. It is obvious that all compounds whose formulae are different, or whose unit cells are different, must have a different collection of possible d-spacings and of different intensities of reflection. The combination of different d-spacings and different intensities is characteristic of any crystalline material and we can use the observed pattern of spacings and intensities of the reflections as a way of identifying an unknown compound in a specific crystalline phase. It will be as characteristic of a particular crystal structure as a fingerprint is of a specific person.

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