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

Comment: Correct.
5 Marks

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

Comment: Correct.
5 Marks

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.

Comment: You are correct that the intensity is zero, meaning in effect that the reflection does not exist. The reason for this is the symmetry of the unit cell with cancelling of contributions from the various ions; this is explained fully in the model answer.
3 Marks

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

Comment: The logic is correct as far as putting h=2 or h=-2; however being cubic, the other two axes are identical so 020, 0-20, 002, 00-2 are also permissable giving a final multiplicity of 6.
1 Mark

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%

Polarisation produces a 86.2% reduction in intensity for the 200 reflection.

Comment: Correct.
4 Marks

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

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

Comment: Correct.
3 Marks

5. What is the intensity ratio I₁₁₁/I₂₀₀?

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

Comment: The method is correct but of course you are carrying a wrong value (2) for the multiplicity (6) so the actual numerical result obtained (0.288) is too large by a factor of 3.
3 Marks

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 -.....

Comment: You have correctly identified several prime suspects for erroneous intensity ratios (your numbers i, ii, iii, xi). The others are also plausible though I might make the following caveats: v) temperature changes might upset the ratios provided the changes are such that one peak is measured consitently at a different temperature to other peak(s); vii) absorption can indeed cause problems in principle but note this is an NaCl sample, not an organic; xii) the detector similarly might upset ratios if the detector is unstable with time (and therefore operating differently for different reflections) or in the case of a position sensitive detector it is non-linear with respect to position. In the case of the monochromator/filter (ix, x) this is less likely to cause a ratio upset since any such fault is likely to affect all peaks equally, and therefore the ratio will be unaffected.
8 Marks overall for this question