FAQ : X-Ray Diffraction [VIVA QUESTIONS]

The following are some of the frequently asked questions regarding X-ray diffraction(XRD) technique.

  1. What is the principle behind X-ray crystallography?
    A. The main principle behind X-ray crystallography is the diffraction of X-rays by atoms in a crystalline structure. The theory was developed by Braggs. Essentially, we consider the crystal to be a diffraction grating for an X-ray.
  2. What is XRD used for?
    X-ray Diffraction is used to determine the phase/structure of a crystalline structure.
  3. What is Bragg’s law?
    Bragg’s law gives the condition for constructive interference to occur for X-rays scattered from different planes.
    The condition is:
    2d\sin{\theta}=n\lambda
    Here, \lambda is the wavelength of the X-ray beam, d is the interplanar spacing and \theta is the incident angle.. Therefore the path difference between the X-rays scattered from two planes should be an integer multiple of the wavelength.
  4. Why are X-rays used as the source beam?
    Since the interplanar distances in crystals are of the order of Angstroms, therefore X-rays are the suitable choice, as the wavelength of the wave should be of the order of the slit size for the phenomenon of diffraction to take place.
  5. How are X-rays produced?
    Whenever charged particles (electrons or ions) of sufficient energy hit a material, X-rays are produced. X-rays are generated by using a vacuum tube that accelerates the electrons released by a hot cathode to high velocities using high voltages. When these electrons collide with the anode(a metal target), X-rays are produced. There are two types of X-ray generated: characteristic radiation and bremsstrahlung radiation.
  6. What do the x and y-axis in an XRD pattern represent?
    An XRD pattern is a plot of Intensity versus 2\theta . The intensity is plotted on the y-axis and the angles are plotted on the x-axis. The intensity of the highest peak is set to 100 and the rest of the peaks are scaled with respect to this. Then the peaks are mapped to a particular set of lattice planes given by the corresponding miller indices.
  7. What information does the XRD pattern of a crystal provide?
    The XRD pattern of a material can help us determine the structure and phase of our material. One can compare the obtained XRD pattern to some database like JCPDS database. And thus determine the structure/phase.
  8. Why do we observe peaks of different heights in the XRD pattern?
    The amount of scattered radiation received at the detector corresponds to the intensity. This intensity is recorded using some counter like the Geiger counter or a scintillation counter. The amount of radiation scattered from a plane depends mainly upon the no. of atoms per unit area. Since, the X-rays scatter by striking with atoms, therefore, more the no. of atoms, more is the intensity. Secondly, the intensity also depends upon interplanar spacing and the no. of planes corresponding to a particular miller index.
  9. How can you determine the crystal plane corresponding to a particular peak in the XRD pattern?
    To determine the Miller indices of the plane that corresponds to a particular peak in the XRD pattern, we first find the interplanar spacing d, by using the relation,
    2d\sin \theta=\lambda
    You can see that in the above relation we have set n=1, as we are only interested in the first order peaks.
    Once we have ‘d’, we can calculate the corresponding Miller indices by using a relation between d and the (hkl) indices for different lattice types:
    Cubic:
    \frac{1}{d^2}=\frac{h^2+k^2+l^2}{a^2}
    Tetragonal:
    \frac{1}{d^2}=\frac{h^2+k^2}{a^2}+\frac{l^2}{c^2}
    Hexagonal:
    \frac{1}{d^2}=\frac{4}{3}\left(\frac{h^2+hk+k^2}{a^2}\right)+\frac{l^2}{c^2}
    Rhombohedral:
    \frac{1}{d^{2}}=\frac{(h^{2}+k^{2}+l^{2})\sin ^{2}\alpha +2(hk+kl+hl)(\cos ^{2}\alpha -\cos \alpha )}{a^{2}(1-3\cos ^{2}\alpha +2\cos ^{3}\alpha )}
    Orthorhombic:
    \frac{1}{d^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}
    Monoclinic:
    \frac{1}{d^2}=\left(\frac{h^2}{a^2}+\frac{k^2\sin^2{\beta}}{b^2}+\frac{l^2}{c^2}-\frac{2hl\cos{\beta}}{ac}\right)\csc^2{\beta}
    Triclinic:
    \frac{1}{d^2}=\frac{\frac{h^2}{a^2}\sin^{2}\alpha +\frac{k^2}{b^2}\sin^2\beta +\frac{l^2}{c^2}\sin^2\gamma +\frac{2kl}{bc}(\cos\beta \cos\gamma-\cos \alpha) +\frac {2hl}{ac}(\cos\alpha \cos\gamma-\cos \beta) +\frac{2hk}{ab}(\cos\alpha \cos\beta-\cos \gamma) }{1-\cos^2\alpha -\cos^2\beta -\cos^2\gamma +2\cos \alpha \cos \beta \cos \gamma }
  10. How do the different planes contribute to the formation of peaks in the XRD pattern?
    Generally, a crystal would consist of many planes oriented in different directions. When X-rays strike the crystal at an angle \theta to a particular plane, then if this angle, satisfies the Bragg’s law, then we would get an intensity peak corresponding to that particular plane in the XRD pattern. All the parallel planes contribute to a particular peak. If the incident angle of X-rays is changed to different values, then we may or may not get intensity peaks in the XRD pattern corresponding to the different planes depending on whether they satisfy the Bragg’s law or not. A particular peak in the XRD pattern is usually, due to the first order(n=1) diffraction by a particular set of planes. Other planes may also satisfy the Bragg’s condition at the same angle but for different orders. But the intensities due to the secondary orders is very less as compared to the first order.
  11. How do you determine the phase/crystal structure/lattice type from the  XRD pattern?
  12. Why are peaks corresponding to some planes missing or not observed in the XRD pattern?
    Sometimes some the scattered X-rays from some set of planes may cancel out due to the scattered rays from another plane. For example consider the BCC lattce type. Here the X-rays scattered from the corner atoms are completely out of phase with those scattered from the central atoms. And since the number of atoms at the center is equal to the number of atoms at the corners, therefore the peaks corresponding to the (100) planes are absent.
  13. What are the relations between interplanar spacing and miller indices of the planes for different lattice types?
    Cubic:
    \frac{1}{d^2}=\frac{h^2+k^2+l^2}{a^2}
    Tetragonal:
    \frac{1}{d^2}=\frac{h^2+k^2}{a^2}+\frac{l^2}{c^2}
    Hexagonal:
    \frac{1}{d^2}=\frac{4}{3}\left(\frac{h^2+hk+k^2}{a^2}\right)+\frac{l^2}{c^2}
    Rhombohedral:
    \frac{1}{d^{2}}=\frac{(h^{2}+k^{2}+l^{2})\sin ^{2}\alpha +2(hk+kl+hl)(\cos ^{2}\alpha -\cos \alpha )}{a^{2}(1-3\cos ^{2}\alpha +2\cos ^{3}\alpha )}
    Orthorhombic:
    \frac{1}{d^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}
    Monoclinic:
    \frac{1}{d^2}=\left(\frac{h^2}{a^2}+\frac{k^2\sin^2{\beta}}{b^2}+\frac{l^2}{c^2}-\frac{2hl\cos{\beta}}{ac}\right)\csc^2{\beta}
    Triclinic:
    \frac{1}{d^2}=\frac{\frac{h^2}{a^2}\sin^{2}\alpha +\frac{k^2}{b^2}\sin^2\beta +\frac{l^2}{c^2}\sin^2\gamma +\frac{2kl}{bc}(\cos\beta \cos\gamma-\cos \alpha) +\frac {2hl}{ac}(\cos\alpha \cos\gamma-\cos \beta) +\frac{2hk}{ab}(\cos\alpha \cos\beta-\cos \gamma) }{1-\cos^2\alpha -\cos^2\beta -\cos^2\gamma +2\cos \alpha \cos \beta \cos \gamma }
  14. What is diffraction?
    Diffraction is the bending of light around the corners of an obstacle or slit that is comparable in size to it’s wavelength.
  15. What are Miller indices?
    Miller Indices are a method of describing the orientation of a plane or set of planes within a lattice in relation to the unit cell. They were developed by William Hallowes Miller. These indices are useful in understanding many phenomena in materials science, such as explaining the shapes of single crystals, the form of some materials’ microstructure, the interpretation of X-ray diffraction patterns, and the movement of a dislocation , which may determine the mechanical properties of the material.
  16. Explain an experimental procedure for finding the XRD pattern.
    A simple X-ray diffraction experimental setup requires the following: a radiation source, a sample crystal, and a detector. These are arranged as shown in the given figure.

    Schematic of a diffractometer

    The source gives off X-rays that strike the crystal at some angle \theta , and the detector detects the scattered radiation, at angle of 2\theta from the source. The X-rays hit the atoms in the crystal plane and scatter in different directions. For diffraction to occur, the scattered radiation must be coherent(in phase), therefore the scattering must be elastic. The angular position of the source is varied over the range of interest and correspondingly the detector, and the intensities of the scattered radiations are recorded and plotted versus 2\theta . This plot of intensity of the scattered radiaiton vs 2\theta is known as an XRD pattern. (Alternatively, sample can be rotated.)The detector can be of several kinds. It can be a simple photographic plate or a sophisticated Geiger/Scintillation counter. The intensity of the highest peak is set to 100 and the rest of the peaks are scaled with respect to this. Then the peaks are mapped to a particular set of lattice planes given by the corresponding miller indices. This is done by calculating the interplanar spacing(d) corresponding to a particular peak. And then using a relation between d and the (hkl) indices for different lattice types.

  17. What are some selection rules for different lattice types for diffraction to occur?

     

    Unit Cell Type Allowed Reflections Forbidden Reflections
    Primitive Any H,K,L None
    Body-centered H+K+L = 2n (even) H+K+L = 2n+1 (odd)
    Face-centered H,K,L all odd or all even H,K,L mixed odd, even
    FCC Diamond H,K,L all odd
    H,K,L all even and H+K+L = 4n
    H,K,L mixed odd, even
    H,K,L all even and H+K+L not = 4n
    Hexagonal (HCP) L even
    H+2K not = 3n
    L odd and H+2K = 3n

     

References:

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