Structure Factor Calculation for 200 Reflection of NaCl |
Using Crystallography and Bragg's law we can calculate the d spacing and other parameters for 200 reflection:
The next step is to work out the atomic scattering factors (fn) that will be needed. The course material already provided values for the atomic scattering factors for Na+ and Cl- ions (obtained from the International Tables for Crystallography) in the vicinity (0.15 and 0.20) of sinθ / λ = 0.177. So interpolating between these values you can obtain the required values given in bold in the middle column of the following table:
sinθ / λ | 0.15 | 0.177 | 0.20 |
fNa+ | 9.035 | 8.69 | 8.39 |
fCl− | 14.12 | 13.08 | 12.20 |
Note that the bold values given above were obtained by interpolating between the side-values at a ratio of 23:27 and are therefore slightly closer to the right-hand side values, but you will probably have noticed that 0.177 is almost exactly the mean (0.175) of 0.15 and 0.20 and therefore a simple mean, (9.035 + 8.39)/2 = 8.71 and (14.12 + 12.20)/2 = 13.16, would have been good enough.
We now put these atomic scattering factors into the structure factor equation given previously:
F(S) = | Σ n |
fn exp{2π i (hx + ky + lz)} |
It is easier to deal with this equation in two parts, first taking the cosine part which has 8 terms corresponding to the 8 atoms involved. Inserting the appropriate values (hkl = 200; and coordinates) and expressing in radians leads to:
Σ n |
fn cos{2π (hx + ky + lz)} | = | 8.69 cos {2π (2 × 0 + 0 × 0 + 0 × 0)} | |
+ | 8.69 cos {2π (2 × 0 + 0 × 1/2 + 0 × 1/2)} | |||
+ | 8.69 cos {2π (2 × 1/2 + 0 × 0 + 0 × 1/2 )} | |||
+ | 8.69 cos {2π (2 × 1/2 + 0 × 1/2 + 0 × 0)} | |||
+ | 13.08 cos {2π (2 × 1/2 + 0 × 0 + 0 × 0)} | |||
+ | 13.08 cos {2π (2 × 0 + 0 × 1/2 + 0 × 0)} | |||
+ | 13.08 cos {2π (2 × 0 + 0 × 0 + 0 × 1/2)} | |||
+ | 13.08 cos {2π (2 × 1/2 + 0 × 1/2 + 0 × 1/2)} | |||
= | 8.69 cos 0 + 8.69 cos 0 + 8.69 cos 2π + 8.69 cos 2π | |||
+ | 13.08 cos 2π + 13.08 cos 0 + 13.08 cos 0 + 13.08 cos 2π | |||
= | 8.69 + 8.69 + 8.69 + 8.69 + 13.08 + 13.08 + 13.08 + 13.08 | |||
= | + | 87.08 |
Similarly, working out the sine part of F200 gives:
Σ n |
fn sin {2π (hx + ky + lz)} | = | 8.69 sin {2π (2 × 0 + 0 × 0 + 0 × 0)} | |
+ | 8.69 sin {2π (2 × 0 + 0 × 1/2 + 0 × 1/2)} | |||
+ | 8.69 sin {2π (2 × 1/2 + 0 × 0 + 0 × 1/2)} | |||
+ | 8.69 sin {2π (2 × 1/2 + 0 × 1/2 + 0 × 0)} | |||
+ | 13.08 sin {2π (2 × 1/2 + 1 × 0 + 1 × 0)} | |||
+ | 13.08 sin {2π (2 × 0 + 0 × 1/2 + 0 × 0)} | |||
+ | 13.08 sin {2π (2 × 0 + 0 × 0 + 0 × 1/2)} | |||
+ | 13.08 sin {2π (2 × 1/2 + 0 × 1/2 + 0 × 1/2)} | |||
= | 8.69 sin 0 + 8.69 sin 0 + 8.69 sin 2π + 8.69 sin 2π | |||
+ | 13.08 sin 2π + 13.08 sin 0 + 13.08 sin 0 + 13.08 sin 2π | |||
= | 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 | |||
= | 0 |
Combining the cosine and sine parts of F200 into the complex number form gives:
© Copyright 1999-2006. Birkbeck College, University of London. | Author(s): Paul Barnes |