Question 1. Calculating the Intensity of Diffraction Using the Structure Calculation

Crystallography and Bragg

d/(hkl) = a/ Ö(h²+k²+l²)

d200 = 5.638/Ö(2²+0²+0²) d100 = 5.638/Ö(1²+0²+0²)

d200 = 2.819Å d100 = 5.638Å

l = 2dSinq

(200) Sinq/l = 0.177Å-1

(100) Sinq/l = 0.089Å-1

Scattering Factors (100) (200)

Comment: Correct.
5 Marks

From here on each reflection will be considered separately F200 then F100

Cosine part of F200

8.687 Cos{360(2x0)} = 8.687 Cos0 = 8.687

+ 8.687 Cos{360(2x½)} = 8.687 Cos360 = 8.687

+ 8.687 Cos{360(2x½)} = 8.687 Cos360 = 8.687

+ 8.687 Cos{360(2x½)} = 8.687 Cos360 = 8.687

+ 13.083 Cos{360(2x½)} = 13.083 Cos360 = 13.083

+ 13.083 Cos{360(2x0)} = 13.083 Cos0 = 13.083

+ 13.083 Cos{360(2x0)} = 13.083 Cos0 = 13.083

+ 13.083 Cos{360(2x½)} = 13.083 Cos360 = 13.083

= 87.08

Sine part of F200

8.687 Sin{360(2x0)} = 8.687 Sin0 = 0.000

+ 8.687 Sin{360(2x½)} = 8.687 Sin360 = 0.000

+ 8.687 Sin{360(2x½)} = 8.687 Sin360 = 0.000

+ 8.687 Sin{360(2x½)} = 8.687 Sin360 = 0.000

+ 13.083 Sin{360(2x½)} = 13.083 Sin360 = 0.000

+ 13.083 Sin{360(2x0)} = 13.083 Sin0 = 0.000

+ 13.083 Sin{360(2x0)} = 13.083 Sin0 = 0.000

+ 13.083 Sin{360(2x½)} = 13.083 Sin360 = 0.000

= 0.000

Combining Cosine and Sine parts of F200

Fhkl² = {Cosine}² + {Sine}²

= 87.08² + 0²

= 7582.93

Intensity

I200 = Fhkl² = 7582.93

The intensity of the 200 reflection is 7582.93

Comment: Correct.
5 Marks

Cosine part of F100

9.64 Cos{360(1x0)} = 9.64 Cos0 = 9.64

+ 9.64 Cos{360(1x½)} = 9.64 Cos360 = 9.64

+ 9.64 Cos{360(1x½)} = 9.64 Cos180 = -9.64

+ 9.64 Cos{360(1x½)} = 9.64 Cos180 = -9.64

+ 16.337 Cos{360(1x½)} = 16.337 Cos180 = -16.337

+ 16.337 Cos{360(1x0)} = 16.337 Cos360 = 16.337

+ 16.337 Cos{360(1x0)} = 16.337 Cos360 = 16.337

+ 16.337 Cos{360(1x½)} = 16.337 Cos180 = -16.337

= 87.08

Sine part of F200

9.64 Sin{360(1x0)} = 9.64 Sin0 = 0.000

+ 9.64 Sin{360(1x½)} = 9.64 Sin180 = 0.000

+9.64 Sin{360(1x½)} = 9.64 Sin180 = 0.000

+ 9.64 Sin{360(1x½)} = 9.64 Sin180 = 0.000

+ 16.337 Sin{360(1x½)} = 16.337 Sin180 = 0.000

+ 16.337 Sin{360(1x0)} = 16.337 Sin360 = 0.000

+ 16.337 Sin{360(1x0)} = 16.337 Sin360 = 0.000

+ 16.337 Sin{360(1x½)} = 16.337 Sin180 = 0.000

= 0.000

Combining Cosine and Sine parts of F100

Fhkl² = {Cosine}² + {Sine}²

= 0² + 0²

= 0

Intensity

I100 = Fhkl² = 0

The intensity of the 100 reflection is 0

A maximum number of possible lattice planes and d-spacings can be calculated for a given structure. Theoretically each of these lattice planes will reflect/diffract but for some the resultant intensity will be zero. This is due to the phase of neighbouring reflections interfering destructively. These zero instensity reflections can be categorised either as systematic absences or a function of a peculiarity in the structure. Systematic absences can be expected if the lattice type is non-primitive or if elements of space symmetry are present.  In a face centred close packed cubic array such as that of NaCl, any reflections where h, k, l are not all  odd or all even have a zero intensity.

Comment: Your final answer is correct (i.e. zero intensity) though in the preceding explanation you appear to show that the cosine part is non-zero; this was obviously just a word processing error since you got the right answer. The explanation was good; the only part I didn't like was the sentence referring to "neighbouring reflections interfering" which is not quite right as a concept, but obviously you have got the general idea
5.5 Marks

Question 2. Calculating the Multiplicity Allowance Factor for the 200 Reflection

Two possibilities:- 200 and -200

\ The Multiplicity of the 200 reflection 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

Question 3. Calculating the Polarisation Allowance Factor for the 200 Reflection

d/(hkl) = a/Ö(h²+k²+l²)

d200 = 5.638/Ö(4)

d200 = 2.819Å

l = 2dSinq

1.54 = 2 x 2.819 Sinq

Sinq = 0.273

q = 15.8

2q = 31.69°

Cos (2q) = 0.85

Cos²2q = 0.724

1+Cos²2q = 1.724

(1+Cos²2q)/2 = 0.862

The polarisation factor for the 200 reflection is 0.862

Comment: Correct.
4 Marks

Question 4. Calculating the Lorentz Allowance Factor for the 200 Reflection

q = 15.8 from Polarization equation

Sinq = 0.272

Sin²q = 0.074

Cos q = 0.962

Sin²qCosq = 0.0712

L = 1/(Sin²qCosq) = 14.04

The Lorentz Factor for the 200 reflection is 14.04

Comment: Correct.
3 Marks

Question 5. Calculating the Intensity Ratio Between the 111 and 200 Reflections

I111/I200 = (J111 P111 L111 F111²)/(J200 P200 L200 F200²)

= (8x0.894x18.39x398)/(2x0.862x14.04x7582.93)

= 52347.06/183544.52 (654337)

As the 200 reflection is the most intense, it can be scaled to 100% which gives a I111/I200 ratio of 28.5/100

(On checking the JCPDF for NaCl, the I111/I200 ratio is closer to 8/100. I have rechecked my calculation and cannot identify any errors. I remain unable to explain this difference.)

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 (0.285) is too large by a factor of 3. Well done for checking out the ICDD/JCPDS value; but this was effectively wrong anyway for the kinds of reasons that follow in the next part of this assignment; some more information is contained in the model answer
3 Marks

Question 6. Possible explanations for differences in intensity ratio.

The majority of explanations for differences in measured intensity ratio can either be attributed to instrumental configuration or sample preparation:-

Instrumental Configuration

Soller slits - When no soller slits are in place, asymmetric broadning of peaks can occur at low 2q values. Axial divergence also introduces a decreasing negative 2q error up to 90° and an increasing positive error beyond 90°.

Divergence slit - When fixed divergence slits are in place the divergence of the beam decreases with an increase of 2q. At low 2q values a larger area of the sample is illuminated with the incident beam .. When an automatic slit is used a fixed area is illuminated regarless of 2q. Intensities obtained using fixed slits will therefore be expexted to vary from those obtained by automatic slits.

Receiving slit - The receiving slit assists in determining the resolution of the peaks. A very narrow (0.02mm)slit will allow separation of a1 and a2 peaks. Wide slits (0.5mm) will not show this, and addition peak distrotion can lead to ambiguous peak maximum and hence errors in 2q and intensity occur.

Instrumental alignment - Variations in intensity can occur if the goniometer is not aligned appropriately.

Configuration - In poorly configured goniometers in which the sample remains fixed and the anode/detector are mobile, excessive weighing on the gears can result in a variation in counting times for each step. This will result in significant errors in the intensity ratios measured.

Sample Preparation

Preferred orientation - Due to method of sample preparation, the orientation of the analyte may not be completely randomized. Significant variations in intensity ratio can occur as a consequence of preferred orientation.

Sample height displacement - Due to poor sample prep., areas of the analyte may lie on a different foccusing circle  (i.e a different height). This will affect the angle of the diffracted beam which will result in a shift of 2q and a variation of intensity due to the X-rays not being directed towards the detector.

Other possibilities

Extinction - If the sample was a near perfect crystal it may have reduced diffracting powers, hence affecting the intensity ratios.

Through lack of practical experience I am unable to prioritse each individual variable in the list. Preferred orientation and the precence/absence/automation of the three types of slits discussed above will have the greatest contribution to varying intensity ratio. With NaCl, Extinction is most likely to have the least contribution.