Reading Kittel, I understand that the structure factor of an fcc lattice is such that there is only scattering from those planes with all even or all odd indices. He then shows XRD for KCl and KBr powders and says that KCl, while fcc, simulates a sc lattice with lattice constant a/2 and KBr is fcc.
However, in the XRD plots, KCl has FEWER peaks than KBr. I would have thought the opposite: that all indices would show peaks for the sc, while only the all even/all odd indices would show peaks for KBr. What am I missing?
Thanks!
Best Answer
You're missing the fact that you're really not making an apples-to-apples comparison when you categorize KCl as a (almost) simple cubic structure and KBr as an fcc structure. KBr has an fcc Bravais lattice with lattice constant $a$, whereas in order for for KCl to look like an (almost) simple cubic structure, not only do you have to consider the K and Cl atoms to be the same (not a bad assumption since they have similar atomic numbers) but - as you yourself noted - you also have to view the lattice as having a reduced lattice constant $a/2$. So you're not really comparing fcc to sc. You're comparing a fcc structure with lattice constant $a$ to a simple cubic (sc) structure with lattice constant $a/2$.
It's actually not surprising that KCl has fewer strong diffraction peaks than KBr. Both of these ionic crystals have the NaCl crystal structure, but KCl extinguishes more diffraction peaks because K and Cl have very similar atomic numbers of 19 and 17, respectively. Br has an atomic number of 35 which is quite a bit different from the atomic number of K (atomic number 19) so, if one considers both KCl and KBr to have the same NaCl structure with about the same lattice constant $a$, fewer diffraction peaks will be extinguished for KBr than for KCl.