Calculating Structure Factor for Sodium Chloride Crystal Structure

[captainjack2000]

Homework Statement

Sodium chloride has a face-centred cubic structure which can be regarded as a primitive cubic structure with a basis of sodium atoms at fractional coordinatese (0,0,0) (0.5,0.5,0) (0,0.5,0.5) and (0.5,0,0.5) and chlorine atoms at (0.5,0,0) (0,0.5,0) (0,0,0.5) and (0.5,0.5,0.5). The sodium ion has 10 electrons and the chlorine ion 18. Calculate the three different values of this structure factor. Show that the non zero values satify the relationship h+k=2n and k+l=2n where n is an integer.

Homework Equations

I know that the structure factor is given by S=Sigma[bjexp(iK.r)] and that Shkl=Sbasis * Sffc

The Attempt at a Solution

Sbasis = N(sodium) +N(chloride)exp(-i*PI*h)
=10+18 = 28 if h, k l are all even
=10-18 = -8 if h,k, l are all odd

I know this is wrong because the answers should be 112, -36 and 0
But my notes also say that If h,k,l are all odd the Shkl=4N and if they are all even Shkl=4N and if one or more is odd and the others are even Shkl=0. I can see that you get the required numbers by using these three cases with the numbers calculated above but I don't know how it all fits together and you can justify multiplying them by four.

Also, I have no Idea how to show the relationships h+k=2n and k+l=2n

 

[Hemmer]

Because you simply have list of atoms I don't think you need to break it down into S = S_{basis} \times S_{fcc} (although I suppose it is possible to do it that way?).

Using the sum and substituting in the list of vectors you should get:

\sum_j = n_j exp(iK.r_j) = n_{Na}( 1 + e^{i\pi(h + k)} + ... ) + n_{Cl} ( e^{i \pi h} + e^{i \pi k} + ...)

where n_{Na}, n_{Cl} are number of electrons in Sodium and Chlorine respectively. There should be 8 parts to the sum, and using e^{i \pi 2n} = 1 and e^{i \pi (2n+1)} = -1 where n is an integer, you should get the desired result.

To get you started, i think for h, k, l all even you should get S_{hkl} = 4(n_{Na} + n_{Cl}). There should be different three values S can take.