When a monochromatic beam of x-rays is shone upon a regular crystalline material then the beam will be scattered from the material at definite angles. This is produced by an interference effect called diffraction between the X-rays from different layers within the crystal.

Crystals consist of regular arrangements of atoms that can be considered to be in layers with specific separations. These layers are separated by distances in the order of Angstroms (0.1nm) apart. This is comparable to the wavelength of x-rays. If x-rays of a known frequency are emitted at a crystalline solid then under particular circumstances of layer separation and angle of incidence to layer, the scatterings from the different layers will constructively interfere (add up), at most angles they will destructively interfere (cancel out).

This effect is characterised by the Bragg Equation, which is the basis of X-ray Crystallography. It was developed by W.H. and W.L. Bragg (Father and Son) who received the 1915 Nobel Prize for Physics for this work.

A method by which one can determine the structure, atomic arrangement, purity, and identity of unknown crystaline substances.

An X-ray Diffractometer consists of the following components:

    1. x-ray tube: consists of a metal casing, a tungsten filament, copper plate, cooling devices, and up to 4 focal points for the x-ray beam.

    2. target: stand upon which your unknown (or known) substance is placed

    3. detector: like a camera for x rays. It records the patterns of the x-ray "diffraction" or reflections off the substance's surface.

    4. Lead casing: umm, keeps out radiation, duh!

The tungsten fillament is supplied with a current. When the filament is excited it gives off heat. This heat excites the copper target into giving off electrons (x rays). A cooling device is present so that the copper target does not melt. These x-rays are in turn projected through the focus point of the x-ray tube towards the studied substance. The atoms absorb x-rays and emit them in all directions. Interference of these waves cancel each other out, except at the diffraction angle. The waves at those angles are read by the detector. The x-ray tube is then rotated around the subject so that a reading may be taken from multiple angles (denoted as 2θ). The readings from the reciever are then (finally) transposed into a readable graph by PC software (ie JADE).
X-ray diffraction can also be used to determine the degree of crystallinity of a solid. To do this you need a graph on which the angle of diffraction is on the horizontal axis and the intensity on the vertical axis.

When a diffraction graph is made of an amorphous solid, there will be no peaks in the graph (since there is no lattice to produce interference). However, the material will still scatter the radiation and the result will be a smooth halo of scattered radiation in the graph.
A material that is partly amorphous and partly crystalline will produce a graph that consists of the amorphous halo with peaks superimposed.

To determine the fractions of both phases, you can determine the relative intensity of the halo and the peaks by calculating the surface area under the peaks and under the halo, and then calculating the ratio.

For quantitative determinations you'll also need samples of which you already know the degree of crystallinity (preferably a completely amorphous and a completely crystalline sample), for calibration.

X-ray diffraction is a phenomenon that can be used to help understand the structure of crystals. There are great writeups already on this topic. This writeup discusses the physical basis and gives the mathematical condition for x-ray diffraction. See also x-ray crystallography.

Suppose there is a system of two identical isolated atoms or molecules separated by a vector d. Now suppose that a plane wave (for our purposes an x-ray but this could be a neutron or electron too) is incident on the system, as shown below. The two atoms or molecules respond to the incident light by reradiating light in all directions. We will make the reasonable assumption that photon energy, and thus wavelength and wavevector magnitude, is conserved in the scattering process (i.e. the scattering is elastic). The question is, from what directions will we see waves scattered from the crystal?


                 
                
              _O----------->
              /^
           d / |            Reradiated wave
            /  | d.a        with wavevector k' = ka'
           /   | 
          O----|----------->
          ^d.a'|  
          |    |  
          |    |   
          |    |   
          |    |   
          |    |
          |    |
          |    |
       Incident plane wave with
       wavevector k = ka

Note: In the diagram, the 90-degree difference between incident and reradiated waves was chosen only to simplify the ASCII art. The vectors a and a' are arbitrary unit vectors. The points O correspond to atoms or molecules. The incident and reradiated waves have the same wavevector magnitude k. Also note how the tendency for us to want to draw from left to right gets in the way of logic.

For there to be constructive interference of the reradiated x-rays, the difference in the lengths of the paths travelled by the two incident waves must be an integer multiple of the wavelength.

The difference in the lengths of the paths is given by

d.a - d.a' = d.(a-a').

Setting the path difference equal to an integer n times the wavelength λ and using the fact that k = 2π/λ, the condition for constructive interference is

d.(k-k') = 2πn.

In a crystal there are identical scattering basis atoms at every point in the Bravais lattice. For there to be constructive interference from the radiation of all of the Bravais lattice sites, the condition for constructive interference must hold true when d is any Bravais lattice vector R. The constructive interference equation is equivalent to

ei(k-k').R = 1 for all R.

As mentioned in the writeup reciprocal lattice, the above holds true whenever (k-k') is some reciprocal lattice vector K. Noting that k and k' have the same magnitude and that if K is a reciprocal lattice vector then so is -K, we can write

k = |k - K|,

and after squaring both sides we get

k.K = K/2,

where K is the magnitude of a reciprocal lattice vector. The equation above defines Bragg planes! The geometrical method for constructing these planes is discussed in the writeup Brillouin zone.

The approach I followed above is called the Von Laue formulation. It turns out to be identical to the formulation done by Bragg, who showed that the condition for constructive interference can also be written as

nλ = 2dsinθ,

where d is the distance between two parallel planes of atoms in a crystal's real-space lattice and θ is the angle of incidence between the incident wave and a plane (not the normal to the plane).

Reference: Ashcroft/Mermin Solid State Physics

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