Powder Diffraction Assignments

Answers to Problem Set 2 by Stewart F. Parker

1. Use the structure factor equation to calculate the F² (intensity) values for (i) the 200-reflection, and (ii) the 100-reflection from sodium chloride. You have already been given a sample calculation for the 111-reflection,Calculating the Intensity of Diffraction Using the Structure Factor Equation, and all the information that you need is contained within this example. In the case of the 100-reflection there is something odd about the answer and you should try to explain this.

Sodium chloride (NaCl) is a face centred cubic structure with a unit cell a-spacing of 5.638 Å. The structure and fractional coordinates for the 4 Na and 4 Cl ions are given in the figure below.



Answer:

(i) Calculation of the intensity of the 200 reflection

The trigonometric form of the Structure Factor equation is:

F(S) = Sn fn cos {360 (hx + ky + lz)} + iSn fn sin {360 (hx + ky + lz)} (in degrees)

(1) Using Crystallography and Bragg's law

The d-spacing for the 200 reflection is required. The formula for d for cubic crystals is:

dhkl = a/Ö(h2 + k2 + l2 )

For h,k,l = 2,0,0 we have

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

From Bragg's law

l = 2d sinq

hence

sin q/l = 1/2d = 1/(2 ´ 2.819) = 0.177 Å-1

(2) Working out the atomic scattering factors (f_n) needed:

From the International Crystallographic Tables, the atomic scattering factors for Na⁺ and Cl^- in the vicinity of 0.177 Å are:

sin q/l	0.15	0.20
fNa+	9.035	8.39
fCl-	14.12	12.20

The form factor is a smooth function, so it is reasonable to interplolate between the values given. This results in:

sin q/l	0.15	0.177	0.20
fNa+	9.035	8.69	8.39
fCl-	14.12	13.08	12.20


Comment: Perfect; exactly the same answer as mine.
5 Marks

(3) Working out the cosine part of F₁₁₁

Since k = l = 0, only the h term is required in the structure factor equation. Inserting the appropriate f values, coordinates and retaining only h = 2 gives:

Sn fn cos{360 (hx + ky + lz)}	=		8.69 cos {360 (2 ´ 0)}
		+	8.69 cos {360 (2 ´ 0)}
		+	8.69 cos {360 (2 ´ 1/2)}
		+	8.69 cos {360 (2 ´ 1/2)}
		+	13.08 cos {360 (2 ´ 1/2)}
		+	13.08 cos {360 (2 ´ 0)}
		+	13.08 cos {360 (2 ´ 0)}
		+	13.08 cos {360 (2 ´ 1/2)}
	=		8.69 cos 0 + 8.69 cos 0 + 8.69 cos 360 + 8.69 cos 360
		+	13.08 cos 360 + 13.08 cos 0 + 13.08 cos 0 + 13.08 cos 360
	=		8.69 + 8.69 + 8.69 + 8.69 + 13.08 + 13.08 + 13.08 + 13.08
	=		87.08

(4) Working out the sine part of F₁₁₁

Similarly, only the h term needs to be considered:

Sn fn sin {2p (hx + ky + lz)}	=		8.69 sin {360(2 ´ 0)}
		+	8.69 sin {360(2 ´ 0)}
		+	8.69 sin {360(2 ´ 1/2)}
		+	8.69 sin {360(2 ´ 1/2)}
		+	13.08 sin {360(2 ´ 1/2)}
		+	13.08 sin {360(2 ´ 0)}
		+	13.08 sin {360(2 ´ 0)}
		+	13.08 sin {360(2 ´ 1/2)}
	=		8.69 sin 0 + 8.69 sin 0 + 8.69 sin 360 + 8.69 sin 360
		+	13.08 sin 360 + 13.97 sin 0 + 13.97 sin 0 + 13.97 sin 360
	=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
	=		0

(5) Combining the cosine and sine parts of F₂₀₀ gives:

F200 = 87.08 + i.0

(6) The intensity I₂₀₀

The intensity I is the square of the amplitude F:

I200 = F2002 = 7583 units


Comment: Exactly the same answer as mine.
5 Marks

(ii) Calculation of the intensity of the 100 reflection

To calculate the intensity the procedure above can be followed. This is very tedious! Is there a shortcut? YES! For the intensity to be non-zero the sum of the cosine and sine terms must be non-zero. Since it is the 100 reflection that is required, again only the first term is needed. In the expressions given above, the 2 is replaced by 1, thus terms like:
cos {360 (2 ´ 0)}
become:
cos {360 (1 ´ 0)}
For the cosine terms the values will consist of either cos 0 or cos 180 i.e. 1 and -1 respectively. Since there are equal numbers of each term the sum will be identically zero. Similarly for the sine terms, the terms will be either sin 0 or sin 180, both of which are zero Hence the sine terms are also zero. Thus the structure factor and hence the intensity are both zero. Further since cos 180n = -1 (n odd) and +1 (n=0 or even), the same situation will arise for all the reflections of the type h00 (h odd) and these relections will be absent. Exactly the same arguments apply to reflections of the type 0k0 and 00l, thus reflections with k odd will be absent for 0k0 and reflections with l odd will be absent for 0l0. Physically, this corresponds to destructive interference by neighbouring ions (of the same type).




Comment: Yes, this is the right idea.
6 Marks

2. Work out the multiplicity allowance factor needed for the 200-reflection.

For the multiplicity of the 200 reflection, the equivalent planes are:

200 and -200

Since a "bar zero" index is the same as a "zero" index, these are the only two planes. Hence j= 2 for 200.




Comment: This is really the only part of your answers that were wrong. Although you correctly spotted 200 and -200, the equivalent along the other two axes, i.e. 020 and 0-20; 002 and 00-2, will also have the same d-spacings since for cubic crystals the three axes are exactly equivalent. So the multiplicity is, rather, 6.
1 Mark

3. Calculate the Polarisation allowance factor needed for the 200-reflection.

(1) Working out the angles q, 2q :

From above:

d200 = 2.819 Å

Assuming that Copper K-alpha radiation with a wavelength of 1.54 Å is used and inserting into Bragg's law gives:

1.54

= 2 ´ 2.819 ´ sinq

sinq

= 0.273

q

= sin-1(0.273) = 15.85°

2q

= 31.70°

(2) Calculating P₂₀₀ :

For the laboratory powder diffractometer we use the formula of Case (3):

2q	= 31.70°
cos (2q)	= 0.851
cos2 (2q)	= 0.724
1 + cos2 (2q)	= 1.724
P200 = (1 + cos2 (2q ))/2	= 0.862

So polarisation produces a reduction in intensity to 86% in the case of the 200 reflection.



Comment: Exactly right.
4 Marks

4. Calculate the Lorentz allowance factor needed for the 200-reflection.

   For the NaCl-200 reflection, and using the value for theta already obtained in the polarisation calculation:

   										q	= 15.85°
   										sin q	= 0.273
   										sin2q	= 0.0746
   										cos q	= 0.962
   										sin2q cos q	= 0.0718
   										L = 1/sin2q cos q	= 13.94

   
   Comment: Exactly right.
   3 Marks

5. The next step is to bring together the F² value and 3 allowance factors determined above for a prediction of the intensity ratio I₁₁₁/ I₂₀₀ .

Now:

I111		j111 P111 L111 F1112		(8) (0.894) (18.39) (398)		52347
¾¾	=	¾¾¾¾¾¾¾¾¾	=	¾¾¾¾¾¾¾¾¾	=	¾¾	=		0.287
I200		j200 P200 L200 F2002		(2) (0.862) (13.94) (7583)		182239


Comment: This is correct except of course that the wrong value (2) for the multiplicity (6) was used. If you divide your answer by 3 then you get 0.096 which is the correct value.
3 Marks
6. In the model answer that you will eventually receive for this assignment, an actual experimentally measured ratio, I₁₁₁/ I₂₀₀ , will be presented to you. However, for the moment I would like to present you with a hypothetical situation for the purposes of this question. Imagine someone informs you that s/he has measured the ratio I₁₁₁/ I₂₀₀ for sodium chloride but you have no details of the experiment and you suspect that the quality of the measurement is far from good. True to your suspicions the measured ratio differs significantly from your prediction. I would like you to think creatively as to what plausible factors (additional to j, P, L) might be the cause of this disagreement. I only want brief answers, such as a one-phrase/sentence explanation for each factor you dream up. List them in order of severity.

   It is hoped that, with the combination of your imagination and all the course material you have received so far, you should have no difficulty devising some 5-10 plausible factors; perhaps you might even come up with more than 10.

1. Human error The calculations for the intensities are long and intricate, a slip of the calculator will result in an incorrect answer. (This applies to both me and the other experimentalist!) There are all sorts of other possibilities in here including measuring the wrong reflection and getting one or other of the correction factors wrong.

2 Marks
2. Texture For NaCL it is possible to grow large single crystals quite easily, thus it would be possible to produce a sample with preferred orientation.

2 Marks
3. Impure sample If the sample is not single phase, then the impurity/ies could have diffraction lines at the same position as the NaCL lines, so distorting the measured intensities.

Comment: Yes, assuming that the impurities express themselves as a second (separate) phase from NaCl. I might point out that impurities might also be incorporated into the NaCl lattice (e.g. K for Na) in which case they would also change the ratio by their effect on the average atomic scattering factors from those of pure Na and Cl.
2 Marks
4. Temperature If the measurements are made at different temperatures then the F_T will be different for the two measurements

Comment: The effect will also vary with Bragg angle (even if the temperature doesn't change) though it would be very small in this case since the peaks are fairly close to each other.
1 Mark
5. Use of synchrotron radiation If the measurement was carried out on a synchrotron, then the polarisation correction is different (ie = 1) for both reflections and the ratio will be different from the laboratory measurement.

Comment: This is an interesting possibility that I hadn't thought of.
1 Mark
6. Detector non-linearity The 200 reflection is more intense than the 111, if this is sufficeient to saturate the detector, then the measured intensity ratio will be incorrect.

Comment: This is a good one as it will very clearly affect the two reflections differently as you point out.
2 Marks


Overall Mark of 10 for this question.


Overall Comment: A very good piece of work.