(i). 200 reflection.

d₂₀₀ = 5.638 / Ö 2²+0²+0² = 2.819

½ (2.819) = sin q / l = 0.177Å^-1

Comment: You have calculated this value correctly. However you should note that it is different to the value for 111 (which was 0.154). This means however that the values for f_n will also be different and their (new) values should have also been calculated at this point - see model answer.
2 Marks

Cosine part.

å _{n ¦ n} cos {360 (h_x+k_y+l_z)} = 4 (8.39 cos {360 (0) }) + 4(8.39 cos {360 (1) })

=33.56+33.56 = 67.12

å _{n¦ n}sin {2p (h_x+k_y+l_z)}= 4 (8.39 {sin{2p (0)}) + 4 (8.39 {sin{2p (1)})

= 0 +3.67 = 3.67

F₂₀₀² = (67.12)² +(3.67)² = 4518 (to 0dp).

I₂₀₀ = F₂₀₀² = 4518 units.

Comment: As pointed out the f-values haven't been recalculated. Further you seem to use the same value for both Na and Cl. Also note that the cos() and sin(hx+ky+lz) terms need to be calculated for each of the 4 Na's and 4 Cl's as they are all different - you cannot simply multiply one value by 4. However this is all made clear in the model answer.
0.5 Mark

(ii). d₁₀₀ = 5.638/Ö 1²+0²+0² = 5.638.

½ (5.638) = sin q /l = 0.0887Å^-1

å _{n¦ n}cos {360(h_x+k_y+l_z)} = 4 (9.05 cos {360(0)}) + 4(9.035 cos {360(1/2)})

=36.14 –35.14 = 0.

å _{n¦ n}sin {2p (h_x+k_y+l_z)}= 4 (9.035 {sin{2p (0)}) + 4 (9.035 {sin{2p (1/2)})

= 0 +1.9806.

F₁₀₀² = 0²+(1.9806)² = 3.923.

I₁₀₀² = F₂₀₀² = 3.923 units.

J₂₀₀ = 2.

2 sets of planes (200 and -200).

Comment: The two situations you point out (200 and -200) are correct, but because it is a cubic crystal, the other two axes are identical and so the 020, 0-20, 002 and 00-2 will also be the same giving a final multiplicity values of 6.
1 Mark

Question 3.

d₂₀₀ = 5.638/Ö 2²+0²+0² = 5.638/2 = 2.819Å^-1.

1.54 = 2 x 2.819 x sinq

sinq = 0.273

q = sin^-1 (0.273) = 15.843

2q = 31.69o (to 2dp).

Cos (2q ) = 0.724

1+cos²(2q ) = 1.724

P₂₀₀ = (1+cos²(2q ))/2 = 0.862.

Polarization produces a reduction in intensity of 86% in the case of the 200 reflection.

Comment: correct.
4 Marks

Question 4.

q = 15.84^o

Sin q = 0.273

Sin² q = 0.075

Cos q = 0.962

Sin^2q cosq = 0.0722

L₂₀₀ = 1/sin^2q cosq = 13.85 (200 reflection).

Comment: correct.
3 Marks

Question 5.

I₂₀₀ = j₂₀₀xP₂₀₀xL₂₀₀xF₂₀₀²

I₂₀₀ = 2 x 0.862 x 13.85 x4518

I₂₀₀ = 107, 878 c units (to 0dp).

I₁₁₁/I₂₀₀ = 52347 / 107878 = 0.485 (to 3dp).

Constant (c) = 0.485 (to 3dp).

Question 6.

Plausible factors (additional to j, p and l) which might be the cause of the disagreement between the measured and predicated I₁₁₁ / I₂₀₀ ratio:

(i). Sample preparation - if crystals within sample are not randomly orientated, then preferred orientation will give rise to a higher intensity value than the predicted value.

2 Marks

(ii). Texture - Presence of large crystals that are therefore not numerous enough, to provide a true powder average.

2 Marks

(iii). Spinning - if the spinning rate of the sample is too low, then not enough crystallites will satisfy the Bragg condition and diffraction intensities will be lower than predicted.

1 Mark

(iv). Transparency - If the sample is transparent, then diffraction peaks will be shifted to lower scattering angles, therefore intensities at predicted peaks will be different in the measured spectrum.

Comment: yes but not the shift in angle will be very small and so the effect will be very small.
0.5 Mark

(v). Signal to noise ratio - the predicted value does not take into account the noise that appears in the measured spectrum.

Comment: There will be similar noise in both peaks the effect of which will be part (but I guess not complete) cancellation.
0.5 Mark

6 marks overall for this question

Overall Comments: The second half of the assignment was essentially correct, but you made some mistakes in the first part (the f-values and F-calculations); assure yourself, using the model answer, that you do now understand these points.