When trying to calculate the structur factor for diamond I came across the calculation on the wikipedia. In the calculation they just add up the atomic form factors of alle the atoms in the unit cell to calculate the structure factor.

However I thought, that the Atomic structure factor is the Fouriertranform of the electron density distribution. As not all atom points in the Diamond lattice are the same (the (1/4,1/4,1/4) Atom electron bounds are rotated 90 Degrees in relation to the (0,0,0) Atom electron bounds), I would expect that due two different electron density funcitons there would be two structure factors in the calculations. Am I missing something?

  • $\begingroup$ "bounds are rotated 90 Degrees in relation to the (0,0,0)" In a diamond? Now, I don't do solid state or crystals, but I vaguely recall that the bonds of a single carbon in diamond point at the points (heh!) of a regular tetrahedron, right? So certainly not 90 degrees. $\endgroup$ – dmckee --- ex-moderator kitten Jan 29 '13 at 16:19

We need to get some terminology straight first: "Diamond structure" in solid state refers to a way of arranging atoms in a crystal. That does not necessarily mean that the compound we are talking about is diamond, the carbon compound. For example, silicon also crystallizes in the diamond structure.

When atomic form factors are calculated (and tabulated), it is usually assumed that the electron density distribution of the atom in question has spherical symmetry. If we make this assumption, the form factor calculation on wikipedia you are referring to is correct and exact.

Moving on to actual diamond (the carbon compound), you are quite correctly pointing out that the assumption of spherical symmetry is not perfect in this case, where the covalent bonds between the carbon atoms are highly directional in nature, breaking the spherical symmetry of the electron density distribution. You should therefore consider the result an approximation for diamond (the carbon compound). I don't know exactly how good or bad this approximation is. However, I would expect the approximation to be better for silicon that diamond-carbon, since silicon has more core electrons not involved in bonding for which the spherical symmetry approximation is probably very good.

  • $\begingroup$ Thanks for answering this old question, I already figured it would be like this. $\endgroup$ – miceterminator May 21 '13 at 9:43

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