1. Calculation of intensity values for the 200 and 100 reflections of sodium chloride.

The calculation of structure factors (and then intensities) of 200 and 100 - reflections could be done according to the equation:

F(S)  =

å n

fn cos {360 (hx + ky + lz)} + i

å n

fn sin {360 (hx + ky + lz)}

1A. 200 reflection
a) using given coordinates of atoms in NaCl I calculate sin q/l (for cubic structure and given a) d₂₀₀ = 5.638/Ö(2² + 0² + 0² ) = 5.638/2 = 2.819 Å
From Bragg's law
sin q/l = 1/2d = 1/(2 ´ 2.819) = 0.1774 Å^-1

b) calculation of atomic scattering factors (f_Cl- and f_Na+) by interpolation from given values, resulted in

Comment: Correct.
5 Marks

c) calculation of the cosine part of F₂₀₀

å n	fn cos{360 (hx + ky + lz)}	=		8.681 cos {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	8.681 cos {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	8.681 cos {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	8.681 cos {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.245 cos {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	13.245 cos {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.245 cos {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	13.245 cos {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		8.681 cos 0 + 8.681 cos 0 + 8.681 cos 360 + 8.681 cos 360
			+	13.245 cos 360 + 13.245 cos 0 + 13.245 cos 0 + 13.245 cos 360
		=		8.681 + 8.681 + 8.681 + 8.681 + 13.245 + 13.245 + 13.245 + 13.245
		=	+	87.704

d) calculation of the sine part of F₂₀₀

å n	fn sin {360 (hx + ky + lz)}	=		8.681 sin {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	8.681 sin {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	8.681 sin {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	8.681 sin {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.245 sin {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	13.245 sin {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.245 sin {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	13.245 sin {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		8.681 sin 0 + 8.681 sin 0 + 8.681 sin 360 + 8.681 sin 360
			+	13.245 sin 360 + 13.245 sin 0 + 13.245 sin 0 + 13.245sin360
		=		0

e) Combining the cosine and sine parts of F₂₀₀ gives

F200 = 87.704 + i ´ 0

(Structure factor of (200) is a complex number with zero imaginary part.)

f) The intensity I (200)  is

I200 (without corrections) (F2002 ) = 7692.0 units

Comment: Correct.
5 Marks

1B. 100 reflection
a) using given coordinates of atoms in NaCl I calculated d₁₀₀ and sin q/l
d₁₀₀ = 5.638/Ö(1² + 0² + 0² ) = 5.638 Å
From Bragg's law
sin q/l = 1/2d = 1/(2 ´ 5.638) = 0.0887 Å^-1

b) I denoted the atomic scattering factors (f_Cl- and f_Na+) as V₂ and V₁ because it seemed to me that doing extrapolation has little sense here

c) calculation of the cosine part of F₁₀₀

å n	fn cos{360 (hx + ky + lz)}	=		V1cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	V1 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	V1 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	V1 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	V2 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	V2 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	V2 cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	V2 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		V1 cos 0 + V1 cos 0 + V1 cos 180 + V1 cos 180
			+	V2 cos 180 + V2 cos 0 + V2 cos 0 + V2 cos 180
		=		0

d) calculation of the sine part of F₁₀₀

å n	fn sin {360 (hx + ky + lz)}	=		V1 sin {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	V1 sin {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	V1 sin {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	V1 sin {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	V2 sin {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	V2 sin {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	V2 sin {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	V2 sin {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		V1 sin 0 + V1 sin 0 + V1 sin 180 + V1 sin 180
			+	V2 sin 180 + V2 sin 0 + V2 sin 0 + V2 sin 180
		=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
		=		0

e) Combining the cosine and sine parts of F₁₀₀ gives

F100 = 0 + i ´ 0 = 0

Structure factor F and intensity I of 100 reflection equals zero, diffraction will not be observable.

Comment: Correct answer, though I was also looking for some explanation as to how the cancellation to zero arises; see the model answer for this.
3 Marks

2. There are in my opinion 6 equivalent sets of planes to 200-reflection, all with the same d spacing. These 6 sets of planes carry the hkl indices:
 

Comment: Correct answer.
3 Marks

3. Calculation of the polarization allowance factor P for 200 reflection:

a) to calculate the d₂₀₀ spacing in cubic NaCl crystals:

d200 = 5.638/(22 + 02 + 02 )1/2 = 5.638/Ö4 = 2.819 Å

For copper Ka radiation the wavelength is about 1.542 Å, and so inserting into Bragg's law gives:

1.542

= 2 ´ 2.819 ´ sinq

sinq

= 0.2735

q

= sin-1(0.2735) = 15.87°

2q

= 31.75°

b) calculating P₂₀₀ :
For the laboratory powder diffractometer I use the formula P = (1 + cos² 2q )/2

2q	= 31.75°
cos (2q)	= 0.850
cos2 (2q)	= 0.723
1 + cos2 (2q)	= 1.723
P200 = (1 + cos2 (2q ))/2	= 0.862

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

Comment: Correct answer.
4 Marks

4. Calculation of the Lorentz allowance factor needed for the 200-reflection.

I used the formula L = 1/sin²q cos q.

Again, taking the NaCl 200 reflection into consideration, and using the value for theta already obtained in the calculation of P:

										q	= 15.875°
										sin q	= 0.274
										sin2q	= 0.0748
										cos q	= 0.962
										sin2q cos q	= 0.0719
										L = 1/sin2q cos q	= 13.90

Comment: Correct answer.
3 Marks

5.The ratio of  I₁₁₁/ I₂₀₀ ..

Multiplying the F value and correction values (and ignoring absorption) gives for 200 reflection:

When I use the I₁₁₁ value calculated by the tutor I obtain

The intensity of reflection 200 should be ten times stronger then that of 111 reflection.
 

Comment: Your evaluation of the 200 intensity and the ratio is correct, though you have expressed the ratio the other way around to that used in the model answer.
2.5 Marks

6. Possible causes of disagreement in intensities ratio could in my opinion be:

a) Sample of NaCl exhibits preferred orientation - the crystallites are of prolonged or platelet - like shape and tend to arrange themselves with the axe of the needle or plane of the plate to be parallel with the flat surface of the sample. Then in the measurred pattern  reflections do not correspond to the pattern gained on powder where crystallites are isotropically oriented.

b) Sample of NaCl exhibits preferred orientation - one can imagine that the pattern was collected on NaCl film epitaxially grown on a single crystal.

c) Sample of NaCl exhibits texture - it can be caused by crystallites that are much larger then the rest of the sample, and their influence on diffraction picture depens on their orientation in the sample. It is probable that they are not oriented randomly and then change the diffraction picture. In this case sieving may provide a solution by removing the largest crystallites.

d) This needed perfect misorientation of crystallites (all possible orientations of grains should be present in the sample) in powder can be spoiled also in case when only very small quantity of sample, poorly grinded Pparticles large then 10 microns) was used for the measurement.

e) The intensities might require an absorption corrections, since you do not know whether the measurement was made in transmission geometry or reflection geometry.

f) The intensities might require an temperature corrections. Even if difference in 2q position of 111 and 200 reflection are only about 5° this could play some role.

g) The intensity of 111 reflection is too small, and in combination with the short maesurement time the negative influence of the background on calculation of intensity is large.

h) There was used bad profile function (or were used two different functions) for the calculation of intensities.

ch) The value I_200/I₁₁₁ was by mistake calculated from two different patterns (measured on the same apparatus, at diferrent times and so under different conditions) - a rather quaint cause.

i) The value I_200/I₁₁₁ was by mistake calculated from patterns measured on completely different apparatus - with neutron diffraction or synchrotrone x-ray radiation ....

j) A badly-stabilised power supply: with a badly stabilised power supply the X-ray flux on the sample could vary by several percent over the time between measuring two reflections such as 111 and 200.
 
 

Comment: I heartily agree with your possible causes a, b (similar to a), c, d ("crystallite size" I think would have been a better overall description), f; g and h were new ones on me, but fair enough they could explain the differences; your ch was equivalent to my (model answer) "human error" category; e and j were, of course already given. There's enought here for:
8.5 Marks