Noreen Hanif
ASSIGNMENT 3

NaCl
a = 5.638 Å

Na (0,0,0) (0,1/2,1/2) (1/2,0,1/2) (1/2,1/2,0)
Cl (1/2,0,0) (0,1/2,0) (0,0,1/2) (1/2,1/2,1/2)

Structure factor equation :

F(S)  =

å n

fn e2p i (hx + ky + lz)

NaCl cubic hence
d_hkl = a/(h²+k²+l²)^1/2

(i) 200 reflection :

d₂₀₀ = 5.638/(2²+0²+l⁰)^1/2
d₂₀₀ = 2.819 Å

Braggs Law : l = 2dsinq
sinq/l = 1/2d = 1/2(2.819) = 0.177 Å^-1

From International Crystallographic Tables :

sinq/l	0.15	0.2
fNa+	9.035	8.39
fCl-	14.12	12.2

Use intermediate values weighted towards the 0.2 value by 0.027:0.023, hence

sinq/l	0.15	0.177	0.2
fNa+	9.035	8.69	8.39
fCl-	14.12	13.08	12.2

Comment: Correct.
5 Marks

cosine of F₂₀₀ :
f₈cos{360(hx+ky+lz)}
=8.69cos{360(2x0 + 0x0 + 0x0)}
+8.69cos{360(2x0 + 0x1/2 + 0x1/2)}
+8.69cos{360(2x1/2 + 0x0 + 0x1/2)}
+8.69cos{360(2x1/2 + 0x1/2 + 0x1/2)}
+13.08cos{360(2x1/2 + 0x0 + 0x0)}
+13.08cos{360(2x0 + 0x1/2 + 0x1/2)}
+13.08cos{360(2x0 + 0x0 + 0x1/2)}
+13.08cos{360(2x1/2 + 0x1/2 + 0x1/2)}
=8.69cos0 + 8.69cos0 + 8.69cos360 +8.69cos360 + 13.08cos360 + 13.08cos0 + 13.08cos0 + 13.08cos360
=8.69+8.69+8.69+8.69+13.0+813.08+813.08+813.08
= 87.08

sine part of F₂₀₀ :
f₈sin{360(hx+ky+lz)}
=8.69sin{360(2x0 + 0x0 + 0x0)}
+8.69sin{360(2x0 + 0x1/2 + 0x1/2)}
+8.69sin{360(2x1/2 + 0x0 + 0x1/2)}
+8.69sin{360(2x1/2 + 0x1/2 + 0x1/2)}
+13.08sin{360(2x1/2 + 0x0 + 0x0)}
+13.08sin{360(2x0 + 0x1/2 + 0x1/2)}
+13.08sin{360(2x0 + 0x0 + 0x1/2)}
+13.08sin{360(2x1/2 + 0x1/2 + 0x1/2)}
=8.69sin0 + 8.69sin0 + 8.69sin360 + 8.69sin360 + 13.08sin360 + 13.08sin0 + 13.08sin0 + 13.08sin360
= 0

F₂₀₀ = 87.08 + i.0

Intensity I₂₀₀ = F₂₀₀² = 7583 units

Comment: Correct.
5 Marks

(ii) 100 reflection

d₁₀₀ = 5.638/(1²+0²+l⁰)^1/2
d₁₀₀ = 5.638 Å

Braggs Law : l = 2dsinq
sinq/l = 1/2d = 1/2(5.638) = 0.089 Å^-1

cosine of F₁₀₀ :
f_Na+cos{360(hx+ky+lz)}
=f_Na+cos{360(1x0 + 0x0 + 0x0)}
+f_Na+cos{360(1x0 + 0x1/2 + 0x1/2)}
+f_Na+cos{360(2x1/2 + 0x0 + 0x1/2)}
+f_Na+cos{360(1x1/2 + 0x1/2 + 0x1/2)}
+f_Cl-cos{360(1x1/2 + 0x0 + 0x0)}
+f_Cl-cos{360(1x0 + 0x1/2 + 0x1/2)}
+f_Cl-cos{360(1x0 + 0x0 + 0x1/2)}
+f_Cl-cos{360(1x1/2 + 0x1/2 + 0x1/2)}
=f_Na+cos0 + f_Na+cos0 + f_Na+cos180 + f_Na+cos180 + f_Cl-cos180 + f_Cl-cos0 + f_Cl-cos0 + f_Cl-cos180
=f_Na++f_Na++(-f_Na+)+(-f_Na+)+(-f_Cl-)+f_Cl-+f_Cl-+(-f_Cl-)
= 0

sine part of F₁₀₀ :
f_nsin{360(hx+ky+lz)}
=f_Na+sin{360(1x0 + 0x0 + 0x0)}
+f_Na+sin{360(1x0 + 0x1/2 + 0x1/2)}
+f_Na+sin{360(1x1/2 + 0x0 + 0x1/2)}
+f_Na+sin{360(1x1/2 + 0x1/2 + 0x1/2)}
+f_Cl-sin{360(1x1/2 + 0x0 + 0x0)}
+f_Cl-sin{360(1x0 + 0x1/2 + 0x1/2)}
+f_Cl-sin{360(1x0 + 0x0 + 0x1/2)}
+f_Cl-sin{360(1x1/2 + 0x1/2 + 0x1/2)}
=f_Na+sin0 + f_Na+sin0 + f_Na+sin180 + f_Na+sin180 + f_Cl-sin180 + f_Cl-sin0 + f_Cl-sin0 + f_Cl-sin180
= 0

No intensity for reflection 100 due to systematic absence. Certain reflections for fcc structures are always forbidden such as the 100.

Comment: Correct answer and basically correct explanation.
6 Marks

2. Multiplicity allowance factor for 200 reflection:
2,0,0; -2,0,0
multiplicity j₂₀₀ is 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. Polarisation allowance factor needed for 200 reflection :
Angles q, 2q
d₂₀₀ = a/a/(h²+k²+l²)^1/2
= 2.819 Å
Cu K_a wavelength is 1.54 Å
l = 2dsinq
1.54=2x2.819xsinq
q=15.85°, 2q=31.7°

P₂₀₀ :
2q=31.7°
cos2q=0.8508
cos²(2q)=0.7239
1+cos²(2q)=1.7239
P₂₀₀=1+cos²(2q)/2=0.862

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

Comment: Correct.
4 Marks

4. Lorentz allowance factor for 200 reflection :
From question 3, q=15.85°
sinq=0.273
sin²q=0.075
cosq=0.962
sin²qcosq=0.072

L₂₀₀=1/(sin²qcosq)=1/0.072=13.89

Comment: Correct answer; I'd advise keeping to at least 3 figure accuracy everywhere in this context.
2.5 Marks

5. I_hkl=cj_hklP_hklL_hklAF_hkl²

The constant c disappears due to division when finding the ratio between the intensities. Consider experimental measurements of I₁₁₁ and I₂₀₀ identical and ignore A. Values for 111 reflection taken from notes.

I₁₁₁/I₂₀₀=(j₁₁₁P₁₁₁L₁₁₁F₁₁₁²) / (j₂₀₀P₂₀₀L₂₀₀F₂₀₀²)
=(8x0.894x18.39x398) / (2x0.862x13.89x7583)
=52347/181585
=0.288

Prediction of intensity ratio I₁₁₁/I₂₀₀=0.288

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.
(i) Sample preparation : NaCl may not have been ground to a fine or even powder causing a poor powder average and differences in peak intensities.

2 Marks

(ii) Sample holder : if measurement made in transmission geometry and flat plate holders have been used then there will be a risk of preferred orientation causing differences in peak intensity. In capillary holders the cubic crystals may also align.

Comment: Correct; I would personally always put this as my No.1 suspicion, especially with flat plate as you point out, though in practice capillary with spinning is usually reliable.
2 Marks

(iii) Temperature effects : the temperature of the experiment needs to be considered which changes intensity.

Comment: This simple calculation of structure factor has ignored the atomic displacement parameters which are a function of Bragg angle and temperature. You need to be more specific here.
1 Mark

(iv) Position of X-ray beam : the beam should be falling on the same area of the sample so data obtained is from a constant area of the sample. This is important if the surface of the sample in a flat-plate holder is not smooth or well packed.

Comment: Yes since the beam footprint will change with angle.
1 Mark

(v) Damaged X-ray filter : beam non-monochromatic causing diffraction of different wavelengths which may interfere with the 111 and 200 reflections.

Comment: Interesting idea, but very unlikely.

(vi) Count time : not known if the instrument is a step-scanning diffractometer or has a stationary detector. Need to optimise the time over which the data is collected for either technique.

Comment: But the count time would have to be altered differently for the two peaks and its not obvious that this would happen from the circumstances you describe.

(vii) Sample spinning : if the instrument is a flat-plate Bragg Brentano diffractometer it may not have a facility for spinning the sample , therefore problems with sample averaging and texture may cause problems with peak intensity.

Comment: Correct, though I have bracketed this answer with your No.(i).
1 Mark

(viii) Background radiation : scattering of X-rays, e.g. air, other than Bragg reflection not accounted for.

Comment: I think this would be negligible since any such effect is small to begin with and is not going to vary dramatically with angle.

(ix) Crystallinity and phase purity : sample of NaCl not pure or crystalline will create difference in peak intensity.

Comment: Yes, NaCl could be impure. NaCl is always crystalline, though crystallinity can be a problem with some samples.
1 Mark

(x) Detector not calibrated : if the instrument is fitted with a position sensitive detector then it's calibration is required regularly so peak intensities can be reliably determined.

Comment: I agree that the detector status is important, whatever type of detector is used, but it requires a specific type of mal-function (such as a non-linearity or saturation) to cause a different type of change for the two peaks and thereby show up in the ratio.
0.5 Mark

Total: 8.5 Marks overall for this part

Comment: A very good assignment. Note the mistake in the multiplicity value propagated an error into the later value.