Assignment Problem Set 3
by Nicki Agron-Olshina

Problem 1

200-reflection

Given Bragg's Law, l = 2dsinq
sinq/l=1/2d
= 1/(2´2.819)
= 1/5.638
= 0.177Å^-1

100-reflection

Given Bragg's Law, l = 2dsinq
sinq/l = 1/2d
= 1/(2´5.638)
= 1/11.276
= 0.089Å^-1

Atomic Scattering factors for Na⁺ and Cl^-. Bold type indicates calculated data.

Comment: Correct.
5 Marks

Using the Structure Factor Equation, F(S) = S ¦_n e^{2pi(hx+ky+lz)}
and the fractional coordinates for the 4 Na⁺ and 4 Cl^- as follows:

(i) Working out the cosine part of F₂₀₀
S¦_n cos{360(hx+ky+lz)}
= 8.69cos{360(2´0 +0´0 +0´0)}
+8.69cos{360(2´0 +0´1/2 +0´1/2)}
+8.69cos{360(2´1/2 +0´0 +0´1/2)}
+8.69cos{360(2´1/2 +0´1/2 +0´0)}
+13.08cos{360(2´1/2 +0´0 +0´0)}
+13.08cos{360(2´0 +0´1/2 +0´0)}
+13.08cos{360(2´0 +0´0 +0´1/2)}
+13.08cos{360(2´1/2 +0´1/2 +0´1/2)}

= 8.69cos0+8.69cos0 +8.69cos360+8.69cos360+13.08cos360 +13.08cos0+13.08cos0 +13.08cos360
= (8.609)4 + (13.08)4
= 87.08 cos part

Working out the sine part of F₂₀₀
S¦_n sin{360(hx+ky+lz)}
= 8.69sin{360(2´0 +0´0 +0´0)}
+8.69sin{360(2´0 +0´1/2 +0´1/2)}
+8.69sin{360(2´1/2 +0´0 +0´1/2)}
+8.69sin{360(2´1/2 +0´1/2 +0´0)}
+13.08sin{360(2´1/2 +0´0 +0´0)}
+13.08sin{360(2´0 +0´1/2 +0´0)}
+13.08sin{360(2´0 +0´0 +0´1/2)}
+13.08sin{360(2´1/2 +0´1/2 +0´1/2)}

= 8.69sin0+8.69sin0 +8.69sin360+8.69sin360+13.08sin360 +13.08sin0+13.08sin0 +13.08sin360
= (0.0)4+(0.0)4
= 0 sin part
Combining cosine and sine parts
F₂₀₀²= {cosine part}²+ {sine part}²
F₂₀₀² = (87.08)²+0² = 7583 units

Comment: Correct.
5 Marks

(ii) Working out the cosine part of F₁₀₀
S¦_n cos{360(hx+ky+lz)}
= 9.82cos{2p (1´0 +0´0 +0´0)}
+9.82cos{2p (1´0 +0´1/2 +0´1/2)}
+9.82cos{2p (1´1/2 +0´0 +0´1/2)}
+9.82cos{2p (1´1/2 +0´1/2 +0´0)}
+16.46cos{2p (1´1/2 +0´0 +0´0)}
+16.46cos{2p (1´0 +0´1/2 +0´0)}
+16.46cos{2p (1´0 +0´0 +0´1/2)}
+16.46cos{2p (1´1/2 +0´1/2 +0´1/2)}

= 9.82cos0+9.82cos0 +9.82cosp+9.82cosp +16.46cosp +16.46cos0+13.08cos0 +16.46cosp
= (9.82)4+(-9.82)4+(13.08)4+(-13.08)4
= 0 cos part

Working out the sine part of F₁₀₀
S¦_n sin{360(hx+ky+lz)}
= 9.82sin{2p (1´0 +0´0 +0´0)}
+9.82sin{2p (1´0 +0´1/2 +0´1/2)}
+9.82sin{2p (1´1/2 +0´0 +0´1/2)}
+9.82sin{2p (1´1/2 +0´1/2 +0´0)}
+16.46sin{2p (1´1/2 +0´0 +0´0)}
+16.46sin{2p (1´0 +0´1/2 +0´0)}
+16.46sin{2p (1´0 +0´0 +0´1/2)}
+ 16.46sin{2p (1´1/2 + 0´1/2 + 0´1/2)}

= 9.82sin0 + 9.82sin0 + 9.82sinp+9.82cosp + 16.46sinp + 16.46sin0+13.08cos0 + 16.46sinp
= (9.82)4 + (-9.82)4 + (13.08)4 + (-13.08)4
= 0 sin part

Combining cosine and sine parts
F₁₀₀² = {cosine part}²+ {sine part}²
F₁₀₀²=0²+0² = 0 units

The amplitude of the 200-reflection from NaCl is 7583 times stronger than that from a single electron. The amplitude of the 100-reflection from NaCl is zero because none of the reciprocal lattice points of 100-reflection of NaCl intersect the Ewald Sphere.

Comment: You are correct in determining the 100 intensity as zero. However your explanation is not quite right. We can assume that with all possible orientations available in a powder of NaCl the reciprocal lattice ring (for a powder) would have intersected the Ewald sphere and therefore diffraction is possible on geometric grounds; however the position of the Na's and Cl's in the unit cell conspire to cancel out each other's contribution for this particular case - a full explanation is given in the model answer. ..
3 Marks

Problem 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

Problem 3

Given wavelength for CuKa = 1.54 Å and l = 2dsinq
Then 1.54 = 2(2.819)sinq
1.54 = 5.638sinq
0.273 = sinq
15.85° = q and 31.70° = 2q
Then to calculate P₂₀₀ given 31.70° = 2q
cos(2q) = 0.850
cos²(2q) = 0.724
and 1 + cos²(2q) = 1.724
P₂₀₀ = 1+cos²(2q)/2
P₂₀₀ = 1.724/2
P₂₀₀ = 0.862
So polarisation produces a reduction in intensity to approximately 86% in the case of the 200-reflection.

Comment: Correct.
4 Marks

Problem 4

Comment: Correct.
3 Marks

Problem 5

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

Problem 6

What plausible factors (additional to j,P,L) might be the cause of the discrepancy between the measured ratio I₁₁₁/I₂₀₀ and my predicted ratio?

1. Error in calculations: Even though the measured experimental values are suspect, there is always the possibility (some might say a strong probabiliby) that there is some careless, silly, should have been spotted, type of error in my calculations that resulted in my calculated allowance factors being more than slightly inaccurate. Perfection can only be approached asymptotically, but alas, never reached.
2. Particle size too coarse: If the particle size of the sample is too coarse some of the crystals might be diffracting intensly on certain reflections and not so intensly on others and this could result in erroneous measured peak intensities.
3. Preferred orientation of crystallites: This is probably not much concerne with a cubic crystal such as NaCl.
4. Poor allignment of instrument:
5. Sample not pure: Ionic substitution can result in changes in peak intensities. Peak overlap from impurity phases in a sample can result in errors is measured peak intensity.
6. Absorption of water during measurement: NaCl is hygroscopic and could absorb water, if conditions are humid enough, during measurement. This would alter peak intensities as the NaCl changed to a hydrated phase.
7. Thermal parameters: Na⁺ and Cl^- have large thermal parameters due to their single charge.
8. Temperature of specimen: Specimen temperature would probably have been at room temperature so this factor is probably not contributing much to peak intensity discrepinsies.
9. Anisotropic line broadening:
10. Inadequate time alloted to data collection: Poor counting statistics can result in less than optimal data from which to calculate intensity values.
11. X-ray tube intensity too low: Low tube intensity results in poor counting statistics and less than optimal data quality.
12. Malfunction of X-ray detector: If an X-ray detector is not functioning consistantly through out the measurement then the collected data will not reflect true beam intensities.

Comment: I agree particularly with your Nos. 2,3,5,7,12; your No.1 is equivalent to my "human error" (in model answer); No.4 is also valid; No.6 is a nice idea, even if unlikely (see model answer "Environment"), No.8 connects with No.7; No.10 connects with No.11 which is also valid. Altogether you have enough valid causes for:
10 Marks