Lars-Erik Briggner, Assignment 3

1. Calculate the intensity values for 200- and 100 reflection for NaCl.

i) d_hkl = a/(h²+k²+l²)^½,, i.e. d₂₀₀ = 5.638/2 = 2.819 Å. l=2dsinq i.e. sinq/l=(2d)^-1 = (2x2.819)^-1= 0.177Å

From the Int. Cryst. Tables can scattering factors for 0.177 be interpolated: f_Na= 8.69, f_Cl= 13.08

Using the Structure Factor equation: F = Sf_ne^{2pi(hx+ky+lz)} gives F₂₀₀² = (cosine part)² + (sine part)²

Cosine part = Sf_ncos{cos(hx+ky+lz)}= 8.69cos{360(2x0+0x0+0x0)}+....according the example +

13.08cos{360(2x½+0x0+0x0)}+.....= 87.08, likewise the Sine part = 0

F₂₀₀²= I₂₀₀= 87.08²= 7583 units

ii) I₁₀₀=?. d₁₀₀=5.638/1=5.638 Å. sinq/l=(2x5.638)^-1=0.089. The scatter factors become irrelevant since

in the Structure Factor equation both the Cosine and the Sine part become zero which means that I₁₀₀=0 units

The reflection does not exist.

2. What is the multplicity for the 200 reflection?

2,0,0; -2,0,0; 2,-0,0; -2,-0,0; 2,0,-0; -2,0,-0; 2,-0,-0; -2,-0,-0 but since 0 and -0 is the same there are only

two differents sets of planes, thus J₂₀₀=2

3. What is the polarisation allowance factor the 200 reflection?

d₂₀₀ =  2.819 Å. Ka(Cu)=1.54 Å, i.e. 1.54=2x2.819sinq, thus q=15.85° and 2q=31.70°

P=(1+cos²2q)/2= 0.862, i.e. the intensity will be reduced to about 86%

4. What is the lorentz allowance factor neede for the 200 reflection?

L= 1/(sin²qcosq), q=15.85 i.e. L = 13.94

5. What is the intensity ratio I₁₁₁/I₂₀₀?

I₁₁₁/I₂₀₀= (J₁₁₁P₁₁₁L₁₁₁F₁₁₁²)/(J₂₀₀P₂₀₀L₂₀₀F₂₀₀²), I₂₀₀= 182239, I₁₁₁=52347, i.e. approx 0.29

6. List plausible factors that could influence measured I₁₁₁/I₂₀₀ ratio!

i) Preferred orientation - one reflection will have higher probability than the other due to orientation of crystallites

ii) Texture - To few and large particles to give a good statistical base

iii) Spinning - No spinning means that not the whole sample will be averaged, biased

iv) Sampling time - The sampling time should be adjusted following the wavelength in order to compensate.

v) Temperature - changes in temperature will affect different reflection to a different extent

vi) X-ray absorption - ....

vii) Transparency - different reflections will be affected to different extent especially with thick organic samples

viii) Primary slit - The primary slit should be adjusted following different 2q

ix) Monochromator - bad adjustment could affect the reflections to a different extent

x) Filter - same as above

xi) Alignment - The instrumet need optimum alignment in order to perform with accuracy

xii) Detector type - A bad choice of detector could produce erroneous results

xiii) A badly stabilised power supply -.....