1 a) Calculated intensity from the structure factor equation, for the 200 reflection:

• Atomic scattering factor of Na⁺ and Cl^-:

            d_hkl = a/Ö(h² + k² + l² ); a = 5.638 Å

            d₂₀₀ = 5.638/Ö(2² + 0² + 0²) = 5.638/Ö4 = 2.819 Å

        From Bragg´s Law:

           sin q/l = 1/2d = 1/(2 * 2.819Å) = 0.1774

        Finally from ICT by interpolation, assuming that there is a linear decrease, the calculated factors are:

            D (f_Na+) = 0.645 and D (f_Cl-) = 1.92 for sin q/l between 0.15 and 0.20.

             f_Na+ = 9.035 - [D (f_Na+)*(0.1774-0.15)/0.05] = 8.682

            f_Cl- = 14.12 - [D (f_Cl-)*(0.1774-0.15)/0.05] = 13.068

• 
  Comment: Correct.
  5 Marks

  Calculation of the sine and cosine part of the structure factor equation

        S f_n cos{360 (h_x + k_y + l_z)}=

        =   8.682 cos{360(2*0 + 0*0 + 0*0)}      +   8.682 cos{360(2*0 + 0*0.5 + 0*0.5)}+
        +   8.682 cos{360(2*0.5 + 0*0 + 0*0.5)}+   8.682 cos{360(2*0.5 + 0*0.5 + 0*0)}+
        + 13.068 cos{360(2*0.5 + 0*0 + 0*0)}   + 13.068 cos{360(2*0 + 0*0.5 + 0*0)}   +
        + 13.068 cos{360(2*0 + 0*0 + 0*0.5)}   + 13.068 cos{360(2*0.5 + 0*0.5 + 0*0.5)} =

        = 8.682 (cos 0 + cos 0 + cos 360 + cos 360) + 13.068 (cos 360 + cos 0 + cos 0 + cos 360) = 4 * 8.682 + 4 * 13.068 =
        = 87

        sine part of F₂₀₀:

        S f_n sin{360 (h_x + k_y + l_z)}=

         =   8.682 sin{360(2*0 + 0*0 + 0*0)}     +   8.682 sin{360(2*0 + 0*0.5 + 0*0.5)}+
        +   8.682 sin{360(2*0.5 + 0*0 + 0*0.5)}+   8.682 sin{360(2*0.5 + 0*0.5 + 0*0)}+
        + 13.068 sin{360(2*0.5 + 0*0 + 0*0)}   + 13.068 sin{360(2*0 + 0*0.5 + 0*0)}   +
        + 13.068 sin{360(2*0 + 0*0 + 0*0.5)}   + 13.068 sin{360(2*0.5 + 0*0.5 + 0*0.5)} =

        = 8.682 (sin 0 + sin 0 + sin 360 + sin 360) + 13.068 (sin 360 + sin 0 + sin 0 + sin 360) = 8.682 * 0 + 13.068 * 0 =
        = 0

After combining the two parts one gets:

        F₂₀₀ = 87 + i * 0, with F₂₀₀² = I₂₀₀ =  87² + 0² = 7569 units

Comment: Correct (my value was 7583).
5 Marks

1 b) Calculated intensity from the structure factor equation, for the 100 reflection:

• Atomic scattering factor of Na⁺ and Cl^-:

            d_hkl = a/Ö(h² + k² + l² ); a = 5.638 Å

            d1₀₀ = 5.638/Ö(1² + 0² + 0²) = 5.638/Ö1 = 5.638 Å

        From Bragg´s Law:

            sin q/l = 1/2d = 1/(2 * 5.638Å) = 0.0886

        As in 1 a) one would have to search the values above and below sin q/l for the 100 reflection and calculate the correct value by linear interpolation.
        Knowing that there is something strange about this "reflection" I did the calculation for the structure factor equation first with the result that it is 0!
        Hence I didn´t look up the scattering factors for Na⁺ and Cl^-.

• Calculation of the sine and cosine part of the structure factor equation

        S f_n cos{360 (h_x + k_y + l_z)}=

        =   8.682 cos{360(1*0 + 0*0 + 0*0)}      +   8.682 cos{360(1*0 + 0*0.5 + 0*0.5)}+
        +   8.682 cos{360(1*0.5 + 0*0 + 0*0.5)}+   8.682 cos{360(1*0.5 + 0*0.5 + 0*0)}+
        + 13.068 cos{360(1*0.5 + 0*0 + 0*0)}   + 13.068 cos{360(1*0 + 0*0.5 + 0*0)}   +
        + 13.068 cos{360(1*0 + 0*0 + 0*0.5)}   + 13.068 cos{360(1*0.5 + 0*0.5 + 0*0.5)} =

        = 8.682 (cos 0 + cos 0 + cos 180 + cos 180) + 13.068 (cos 180 + cos 0 + cos 0 + cos 180) = 8.682 * (1+1-1-1) + XXX * (-1+1+1-1) =
        = 0

        sine part of F₂₀₀:

        S f_n sin{360 (h_x + k_y + l_z)}=

         =   8.682 sin{360(1*0 + 0*0 + 0*0)}     +   8.682 sin{360(1*0 + 0*0.5 + 0*0.5)}+
        +   8.682 sin{360(1*0.5 + 0*0 + 0*0.5)}+   8.682 sin{360(1*0.5 + 0*0.5 + 0*0)}+
        + 13.068 sin{360(1*0.5 + 0*0 + 0*0)}   + 13.068 sin{360(1*0 + 0*0.5 + 0*0)}   +
        + 13.068 sin{360(1*0 + 0*0 + 0*0.5)}   + 13.068 sin{360(1*0.5 + 0*0.5 + 0*0.5)} =

        = 8.682 (sin 0 + sin 0 + sin 180 + sin 180) + 13.068 (sin 180 + sin 0 + sin 0 + sin 180) = 8.682 * 0 + 13.068 * 0 =
        = 0

After combining the two parts one gets:

        F₁₀₀ = 0 + i * 0, with F₁₀₀² = I₁₀₀ =  0² + 0² = 0 units

This result is not astonishing, because in cubic systems there is high symmetry which leads to certain (symmetry-specific) exclusions. 100 is such an exclusion in cubic systems. Also 010 and 001 would give no reflection. The unit cell defines all atom places which are introduced later into the structure factor equation. Due to the fact that the sine and cosine part are made up of sums it is possible under certain circumstances that both parts of the structure factor equation can become zero. Result of this is that such a specific reflection shows no intensity and hence cannot be measured..
In opposite of this an elongation of one axis would allow to observe I₁₀₀. In a triclinic or orthorhombic unit cell all three possible reflections 100, 010 and 001 could be measured. Each of them then belongs to one of the three axises.

Comment: Basically correct logic; note that your comment is not exactly right ("100 is excluded in cubic systems"): it is excluded in this case of face centred cubic systems, but not in all cubic systems.
6 Marks

2 ) Multiplicity for 200 reflection:

• Applying the same permutations as with the 111 reflection one ends up with the following possibilities:

        Knowing that -0 and +0 are the same there remain only two senseful and logical permutations: 200 and -200!
        Hence j = 2 (in opposite of j = 8 for the 111 reflection).

        For 100 j would be the same as for 200, j = 2

Comment: The argument is correct as far as 100 is concerned, but this case is more special in that the other 2 axis are identical (see model answer), so the correct answer is 6.
1 Mark

3 ) Polarisation for 200 reflection:

• First q is calculated:

• Now P_100gf = (1+cos² 2q)/2 = (1+cos² 31.7 °)/2 = 0.862

Comment: Correct.
4 Marks

4 ) Lorentz Factor for 200 reflection:

        q = 15.85 °; sin q = 0.273; sin² q = 0.0746 and cos q = 0.962.

        L = 1/sin² q cos q = 13.936

Comment: Correct.
3 Marks

5 ) Intensity ratio I₁₁₁/I₂₀₀:

        I = j * P * L * A * F², with a neglible value for A if it is a flat sample and Bragg Brentano geometry

        I₁₁₁ = 8 * 0.894 * 18.39 * 398 = 52347 c units
        I₂₀₀ = 2 * 0.862 * 13.936 * 7563 = 181706 c units

         I₁₁₁ /I₂₀₀ = 0.288; I₂₀₀ is approximately 3,5 times stronger than I₁₁₁

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

6 ) Reasons for a different intensity ratio in comparison to the experiment:
 

Sample treatment:
How is the sample prepared? Is it sieved, or is it just ground in a mortar or mill?
The preparation might lead to preferred orientation or texture effects in the sample. If one of the two cases occurs the intensity information can be seriously influenced! A comparison of different samples of the same material might give different intensities!

Comment: Yes, you have really given two causes - crystallite size (re texture) and preferred orientation.
3 Marks

Spinning or not?
If the sample is measured without spinning it it migh lead to a preferred orientation, which increases the intensity for reflections of the preferred orientation.

Comment: Yes in combination with the previous factor, this could make a difference.
1 Mark

Absorption correction:
Depending on how the sample is measured a specific absorption correction for transmission or reflection geometry has to be made!

Comment: Correct, but of course this was given as an example in the original question.

Temperature effects:
Temperature has an influence on the excursions of atoms in a crystal. Hence intensity information can be altered by measurements at different temperatures. A cooled sample would have other intensities as a heated one.

Comment: Yes, but note that in addition there could in principle even be an effect at constant temperature since the temperature effect varies with angle and therefore with hkl.
1.5 Marks

Badly stabilised power supply for the X-ray tube:
This might lead to a change in intensity of the original emitted radiation of the X-ray tube.

Comment: Correct, but again this was given as an example in the original question.

Badly stabilised power supply for the detector:
If the high voltage supply of a scintillation detector is not stable, the multiplication of an incoming signal would change and lead to different measured intensities for the same signal! A PSD detector loses the linearity with high count rates. Therefore high intensities might not be as exact as low ones.

Comment: This is a good one.
2 Marks

Polarisation of the monochromator:
A monochromator polarises the X-ray beam and hence the intensity of the diffraction pattern. The absorption correction should be small and mostly neglible, but two different monochromators (same type) might give slighlty different intensity informations in the same measurement setup.

Comment: I think it highly unlikely that an experimenter would change the monochromator to another one during a scan! Hope I have interpreted this correctly.

Scattering factors:
It is assumed that there is a linear change in the scattering factor in the ICT. How good is this assumption? High intensities are more influenced by a slight change in the scattering factors due to the product which is formed for the intensity.

Comment: The scattering factors in the ITC are very precise, though differences would be achieved depending on whether ionic or neutral atomic scattering factors are used.

Total: 7.5 Marks for this part

Comment: Overall a very good assignment; note the multiplicity mistake which propagated an error into the later value.