For X-ray diffraction we would like to have monochromatic x-rays otherwise we will have multiple peaks for each plane. To attain this goal requires [1] a pure element target and a 'K beta' filter. Metal foils have absorption edges for x-rays which can be used to filter out most of the unwanted white or continuous radiation as well as the 'K beta' peak. In particular, Zr foil has an absorption edge that lies between the 'K beta' and 'K alpha' peaks for Mo. Ni foil works for Cu.
Because crystals are ordered 3-D periodic structures, the interaction of an x-ray beam with this array of planes of atoms will lead to scattering and can lead to cooperative scattering which we know as diffraction. Read over pages 278-80 in your text for the development of the Laue equations, the simplification made by Bragg allowing us to visualize diffraction using the ideas of reflection, and the origin of the Bragg equation. Remember that the treatment of diffraction as reflection is a simplification - diffraction is really a complex scattering problem. However the reflection approach works and we will use it.
Bragg recognized that the diffracted x-rays act as though they were "reflected" from planes of atoms in the structure. Unlike the continuous reflection of light from a mirror however, the x-ray "reflection" took place only at certain angles that were controlled by the spacings between atomic planes and the wavelength of the radiation. He showed that this "reflection" took place only when the equation:
n(lambda) = 2d sin (theta)
was satisfied where n is an integer, lambda is the wavelength of the radiation, d is the interplanar spacing, and theta is the incident angle relative to the plane of atoms. For "reflection" to occur from a set of parallel planes of atoms in a structure, these "reflections" must be in phase so that they constructively reinforce each other and generate a measurable signal. The geometry required for constructive interference can be seen in the following figure:
If the path difference (ABC) for the diffracted rays 1 and 2 is exactly one wavelength then they will reinforce each other by constructive interference. Under these same conditions the path difference A'B'C') for rays 1 and 3 will be two wavelengths and they also will constructively interfere.
The powder x-ray diffractometer uses Bragg's Law and the fact that we know the wavelength (lambda) and by varying the detector location (theta) can determine the interplanar spacings 'd'. This allows us to identify the mineral from information about its structure.
The electron microprobe uses Bragg's Law in a different way. In this instrument, electrons are accelerated and impact upon an unknown mineral causing x-rays to be emitted by the various elements constituting the unknown, These x-rays may be collected and analyzed using a crystal with a known d-spacing and a detector for counting the x-rays. Effectively we set up a geometry such that we know 'd' and theta and thus can solve for the wavelength lambda. The measured characteristic x-rays identify the elements present - the use of standards of known composition allows the amounts of the elements to be determined as well. Thus we can identify an unknown based on its chemistry.
These two techniques are complementary to one another and both rely on Bragg's Law and the diffraction of x-rays.
X-ray fluorescence analysis is similar to the electron microprobe in that both instruments collect and count x-rays emitted from a sample using Bragg's Law and a crystal of known d-spacing to separate out specific wavelengths of radiation. The difference between the techniques is that in the electron probe individual grains of a largely intact sample are excited using an accelerated electron beam while in x-ray fluorescence, a ground up sample is excited using the same kind of x-rays used in powder diffraction studies. The electron probe is used for individual mineral analyses, x-ray fluorescence is used for whole rock analyses.
See page 146 in your text for the example of NaCl.
What do we need to know?
What the atoms are and where they are = the 'solution' of the crystal structure.
What is the composition? (Link
to Periodic Table)
Density of halite is 2.165 g/cm3
What is the structure?
Return to NaCl
Consider the view below of a face-centered cubic array of atoms and the interaction of x-rays with the planes of atoms in the structure. At some angle theta, constructive interference will occur from the 'reflections' from the top and the bottom (100) planes of atoms related by d(100), the unit cell translation. At this angle, AB + BC = lambda. However there is another equally dense plane of atoms half way between each pair of (100) planes and for these (200) planes, A'B + BC' = lambda/2. Thus the (200) reflection will be 180 degrees out of phase with the (100) reflection and cancel it out - we say that the (100) reflection is extinct. This is a general truth for all face-centered lattices irrespective of the crystal system and it can be generalized by saying that all observed (non-extinct) reflections have (hkl) values where all the digits are either odd or even. (The (200) reflection will be present at the appropriate angle.) The proof of this is beyond this course but rules like this prove the power of understanding crystallography. There are other rules governing the extinctions in c-centered lattices; no systematic extinctions occur in primitive lattices. Think about the indices of the lines you observe in halite in the lab exercise.
Orienting a single crystal of a mineral such that even a single plane satisfies Bragg's Law is tedious work and thus 2 possibilities arise:
See your book for a detailed description of the powder camera and the strips of film it produces. In lab you will see an automated powder diffractometer which does away with the film and makes the determination of intensity and location of the various diffracted beams much easier. The intensity of a particular diffracted ray is proportional to the density of atoms defining the plane in the structure. This means that in general the intense reflections are caused by planes with simple indices and relatively large d-spacing. Each of the construction lines in the figure below is the same length - count the number of atoms per line and note the relationship to d-spacing.
