Logo Structure Factor Calculation for 200 Reflection of NaCl

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Using Crystallography and Bragg's law we can calculate the d spacing and other parameters for 200 reflection:

dhkl = a/√(h2 + k2 + l2 )
For hkl = 200 we have
d200 = 5.638/√(22 + 02 + 02 ) = 5.638/√4 = 2.819 Å
From Bragg's law:
λ = 2d sinθ
so
sinθ / λ = 1 / 2d = 1 / (2 × 2.819) = 0.177 Å-1

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 8.698.39
fCl14.12 13.0812.20

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:

F200 = +87.08 + 0i
which on squaring for the intensity leads to:

I200 = F2002 = 7583 units


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© Copyright 1999-2006.  Birkbeck College, University of London. Author(s): Paul Barnes