0
$\begingroup$

So I have a couple of crystals I am analysing (LiF RbCl NaCl...) and I have the x ray diffraction patterns for them. How do I determine the different structures of these crystals as I need to calculate the respective radii for the ions and thought I would need the cubic structure in order to obtain a relationship between the lattice constant and the radius of each of the ions?

$\endgroup$
  • $\begingroup$ Most alkali metal halide crystals are face centred cubic; caesium (and I guess francium) can be body centred. Are you asking how you determine that from your diffraction data? $\endgroup$ – PM 2Ring Sep 30 '18 at 0:57
0
$\begingroup$

If the lattice structure is cubic, one can decide whether it is bcc, fcc or simple cubic from the diffraction data. The Bragg's Law gives the relation $ \lambda = 2d sin(\theta) $. For a cubic structure, $$ d = \frac{a}{(h^{2}+k^{2}+l^{2})^\frac{1}{2}} $$ $$h^{2}+k^{2}+l^{2} = \frac{4a^{2}}{\lambda^{2}} sin^{2}(\theta)$$

Since h,k,l are integers, $h^{2}+k^{2}+l^{2}$ is also restricted to some integer values, for instance 7 can never be a sum of squares of 3 integers. Pick the smallest angle for which there is a diffraction peak and make a guess for the value of $h^{2}+k^{2}+l^{2}$, say start with 1, which corresponds to the [100] family. Find the value of $\frac{4a^{2}}{\lambda^{2}}$ using this guess and use this value to find $h^{2}+k^{2}+l^{2}$ for the other diffraction peaks. If the values do not come out to be integers or very close to an integer or you obtain the value 7 for a peak, then the initial guess is wrong. Then one can move onto the 2nd guess, say $h^{2}+k^{2}+l^{2} = 2$ corresponding to the [110] family. Repeat the procedure until you have identified $h^{2}+k^{2}+l^{2}$ and hence the [hkl] family for all the diffraction peaks.

Now remember that the structure factor for the bcc lattice is propotional to $(1 + e^{i(h+k+l)}) $ and hence for the diffraction peaks for a bcc lattice, $h+k+l$ must necessarily add up to an even integer. If that is the case, one can safely conclude that the lattice is bcc. Similarly for fcc, the structure factor $(1 + e^{i(h+k)}+ e^{i(k+l)}+e^{i(l+h)})$ imposes the condition that h,k,l muat be all odd or all even. If that is the case, then the lattice is most probably fcc. If it is neither bcc nor fcc,it will be a simple cubic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.