I am trying to find the (a,b,c) vs (h,k,l) relation to determine the interplanar spacing of a crystal I believe to be face-centered orthorhombic (FCO). The basic equation available everywhere is valid for the primitive orthorhombic, but I cannot agree that it would work for a FCO system, with obviously lower symmetry. After playing with VESTA, I simulated two different diffraction pattern made from three cubic cells, a primitive, a body-centered, and a face-centered, with identical cell parameters. It seems that the BCC has 100 and 111 extinct, while the FCC has 100 and 110 extinct. How can you demonstrate that algebraically?
Cheers
The diamond lattice consist of a face centered cubic Bravais point lattice which contains two identical atoms per lattice point. The distance between the two atoms equals one quarter of the body diagonal of the cube. The diamond lattice represents the crystal structure of diamond, germanium and silicon.
For all orthogonal crystal systems (orthorhombic, tetragonal, cubic) the interplanar distance of a lattice plane with Miller indices is:
d=1./sqrt(h^2/a^2 + k^2/b^2 + l^2/c^2). This holds for each of the possible Bravais lattices in these systems. Thus all reflections h,k,l exist. If the lattice is primitive, all reflections may have non-zero intensity, if the lattice is body centered only reflections for which h+k+l=2n i.e. even may have non-zero intensity and for a face-centered lattice the reflection conditions are h+k and h+l and k+l must all be even. This means for a face centered lattice that Bragg reflections must have all even or all odd Miller indices.
I would not say that a face centered lattice has lower symmetry than a primitive one. As a matter of fact there are more symmetry operators in the group (the additional translations)!
In a body centered lattice there are always pairs of atoms in the structure. For each atom at a position x,y,z there is an identical atom at x, y, z + [1/2 , 1/2, 1/2]. You need to calculate the common contribution for both of these atoms to the scattered wave, the contribution to the structure factor F(hkl). This exercise will show you that the two contributions cancel each other (= add up to zero) for all Bragg reflections with h+k+l = 2n+1, while they add up to the sum of the two atomic form factors for h+k+l=2n.
Similar calculations can be done for all other Bravais lattices, and likewise for space groups that contain gliede planes and screw axes.
To understand these fundamental concepts of diffraction I strongly suggest to read the introduction to X-ray / neutron diffraction in any good text book, for example
Giacovazzo et al. "Fundamentals of Crystallography" ; Als-Nielsen "Elements of modern X-ray physics" etc etc etc. For a quick summary see slides 10 to 12 in the attached lecture notes.
At Merga, one should not use the wrong term "diamond lattice"; this is the diamond structure. A lattice is a mathematical object, where each point can be described as : p=n*a + m*b + p*c, with abc the base vectors, and nmp all whole integer numbers. The atom positions in the diamond structure do not fulfill this condition.
Dear Prof. Neder,
Thank you for your thorough reply and for your time. Deeply appreciated.
Please check the fully-ready preprint article DOI: 10.13140/RG.2.2.27720.65287/3, Title: Qualitative Analyses of Thin Film-Based Materials Validating New Structures of Atoms
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