Answer:

1.
Structure Factor equation:

F(S)  =

å n

fn e2p i (hx + ky + lz)

F(S)  =

å n

fn cos {2p(hx + ky + lz)} + i

å n

fn sin {2p(hx + ky + lz)} (in radians)

or

F(S)  =

å n

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

å n

fn sin {360 (hx + ky + lz)} (in degrees),

(I)
(1). Using crystallography and Bragg's law
The formula for d for cubic crystals is:

dhkl = a/Ö(h2 + k2 + l2 )

For hkl = 200 we have

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

From Bragg's law

l = 2d sinq

so

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

(2) Working out what atomic scattering factors (f_n):

From the International Crystallographic Tables we get the following f values:

Using an intermediate value weighted produces the following result(needed values in bold):

Comment: Correct.
5 Marks

(3) Working out the cosine part of F₂₀₀:

å n	fn cos{360 (hx + ky + lz)}	=		8.69 cos {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	8.69 cos {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	8.69 cos {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	8.69 cos {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.08 cos {360 (2 ´ 1/2 + 0 ´ 0 + 0 ´0)}
			+	13.08 cos {360 (2 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.08 cos {360 (2 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	13.08 cos {360 (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 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 F₂₀₀

å n	fn sin {2p (hx + ky + lz)}	=		8.69 sin {2p (2 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	8.69 sin {2p (2 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	8.69 sin {2p (2 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	8.69 sin {2p (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.08 sin {2p (2 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	13.08 sin {2p (2 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	13.08 sin {2p (2 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	13.08 sin {2p (2 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		8.69 sin 0 + 8.69 sin 0 + 8.69 sin 2p + 8.69 sin 2p
			+	13.08 sin 2p + 13.08 sin 0 + 13.08 sin 0 + 13.08 sin 2p
		=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
		=		0

(5) Combining the cosine and sine parts of F₂₀₀

F200 = 87.08 + i ´0

so

Fhkl2 = {cosine part}2 + {sine part}2

F2002 = (87.08)2 + 02 = 7583

Comment: Correct answer. However you are following the original lecture case (111) style too closely. For example you worked out the cosine part in degrees but then the sine part in radians; the reason I used both styles in the original was for the benefit of students' individual preferences. However I expect students to choose one style then stick with it in their own answers.
4 Marks

(II)
(1). Using crystallography and Bragg's law
The formula for d for cubic crystals is:

dhkl = a/Ö(h2 +k2 + l2 )

For hkl = 100 we have

d100 = 5.638/Ö(12 + 02 + 02 ) = 5.638/Ö1 = 5.638 Å

From Bragg's law

l = 2d sinq

so

sin q/l = 1/2d = 1/(2 ´ 5.638) = 0.0887 Å-1

(2) Working out what atomic scattering factors (f_n):

From the International Crystallographic Tables we get the following f values:

Using an intermediate value weighted produces the following result(needed values in bold):

(3) Working out the cosine part of F₁₀₀:

å n	fn cos{360 (hx + ky + lz)}	=		9.83 cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	9.83 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	9.83 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	9.83 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	16.47 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´0)}
			+	16.47 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	16.47 cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	16.47 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		9.38 cos 0 + 9.38 cos 0 + 9.38 cos 180 + 9.38 cos 180
			+	16.47 cos 180 + 16.47 cos 0 + 16.47 cos 0 + 16.47 cos 180
		=		9.38 + 9.38 + (-9.38) + (-9.38) + (-16.47) + 16.47 + 16.47 + (-16.47)
		=		0

(4) Working out the sine part of F₁₀₀

å n	fn sin {2p (hx + ky + lz)}	=		9.83 sin {2p (1 ´ 0 + 0 ´ 0 + 0 ´ 0)}
			+	9.83 sin {2p (1 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
			+	9.83 sin {2p(1 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
			+	9.83 sin {2p (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
			+	16.47 sin {2p (1 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
			+	16.47 sin {2p (1 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
			+	16.47 sin {2p (1 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
			+	16.47 sin {2p (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
		=		9.83 sin 0 + 9.83 sin 0 + 9.83 sin p + 9.83 sin p
			+	16.47 sin p + 16.47 sin 0 + 16.47 sin 0 + 16.47 sin p
		=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
		=		0

(5) Combining the cosine and sine parts of F₁₀₀

F100 = 0 + i 0

so

Fhkl2 = {cosine part}2 + {sine part}2

F1002 = (0)2 + 02 = 0

(III)

When two X-rays start off in phase, they will reach the detector and have different distance to travel. So we would say that the two X-ray photons are out of phase by the time they reach the detector. It is more usual to express this phase difference in terms of an angle (degrees or radians) whereby one angular revolution corresponds to a phase difference of one wavelength. If the phase difference is half a wavelength in distance then destructive interference occurs. In above case, the two waves end up being about half wavelengths, i.e. the phase difference is l/2 in distance. So the intensities of the two X-rays are cancelled each other and the detector eventually gets nothing. Th is is the reason why the value of intensity for 100-reflection of NaCl cubic is 0.

Comment: Correct answer re zero intensity. However again you are copying all the words in the original lecture regardless of their meaning; you must check the meaning of all words before using them; e.g. in your table, the values you determine are referred to as "intermediate value weighted..." whereas 0.0887 is clearly not intermediate between 0.15 and 0.20; but make no mistake the answer was correct. In your explanation part, you correctly connect the zero intensity to a destructive interference situation. However you do not make clear why it is destructive in this case (100) yet constructive in other cases (e.g.(200)). A full explanation can be found in the model answer.
5 Marks

2.

The number +0 is identical in every respect to the number -0, so there are in fact two equivalent sets of planes to 200 (i.e.200 and -200), all with the same d spacing. So we would say that the multiplicity j₂₀₀ is 2 in this NaCl cubic case.

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.

(1)Working out the angles q, 2q :

First we remind you how to calculate the d spacing in cubic crystals:

dhkl = a/(h2 + k2 + l2 )1/2

d200 = 5.638/(22 + 02 + 02 )1/2 = 5.638/Ö4 = 2.814 Å

For Copper Ka radiation the wavelength is about 1.54 Å, and so inserting into Bragg's law gives us:

1.54

= 2 ´ 2.814 ´ sinq

sinq

= 0.273

q

= sin-1(0.273) = 15.84°

2q

= 31.68°

(2) Calculating P₂₀₀ :

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

2q	= 31.68°
cos (2q)	= 0.849
cos2 (2q)	= 0.721
1 + cos2 (2q)	= 1.721
P200 = (1 + cos2 (2q ))/2	= 0.8605

So polarisation produces a reduction in intensity to 86.05% in the case of the 200-reflection.

Comment: Correct.
4 Marks

4.

Lorentz Factor, L

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

										q	= 15.84°
										sin q	= 0.273
										sin2q	= 0.0745
										cos q	= 0.962
										sin2qcos q	= 0.0717
										L = 1/sin2q cos q	= 13.95

Comment: Correct.
3 Marks

5.

The formula previously given, relating the measured intensity to the structure factor, I_hkl = c j_hkl P_hkl L_hkl A F_hkl², can now be used with respect to all parameters except A (absorption) which we will ignore for the purposes of this exercise. The various values that have been calculated in this section for the case of 200 diffraction are:

Multiplying these (and ignoring absorption) gives:

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.

(1). Temperature Effects:Temperature will have an effect on the intensities of the powder diffraction pattern. The equation of modified atomic scattering factor f_T is: f_T = f exp(-B sinq²/l)
(2). Preferred Orientation of the Sample:The intensities of sample in different mounting ways are quite different, such as in capillary, flat-plate.
(3). Sample Texture: The sample texture causes that in the sample there are insufficient crystallites to provide a true powder diffraction pattern.
(4). Transpareny of Sample:For the Bragg-Brentano geometry, poor data may be obtained when the sample is too transparent for X-rays.
(5). A bad X-ray filters: With a bad metal foil filter, the X-ray pattern has high energy background radiation.
(6).A bad X-ray source: Not pure monochromatic beam causes varied pattern.
(7).Magnetic effect:Strong magnetic field effects on the X-ray so as to bad X-ray pattern comes out.
(8).Orientation of X-ray instrument:With bad orientation of X-ray instrument, X-ray running in vibration condition causes bad result of X-ray pattern.
(9).The holder of sample:Different shape and quality of the holder of sample produce different background of X-ray pattern.
(10).Count Times:The setup of count times is important to obtain perfected X-ray pattern.
(11).Slit effect: The good slit gives a less asymmetric and narrower peak shape of X-ray pattern, especially at low scattering angles.
(12).Calibration of X-ray Detectors:The position sensitive detectors require careful calibration for both wire position and efficiency so that both scattering angles and intensities can be reliably determined.
(13).Choice of X-ray Target:With bad choice of X-ray target, a powder diffraction yields the data with a very high background and very little signal.
(14).Gaseous environment:The intensity is changed while the sample is under gaseous environment.
(15).Pressure of environment:The intensity of a sample under different pressure is changed.

Comment: I agree with your answer Nos. 1-6,8,10-13, possibly Nos.14,15, but note that a full answer needs to show that the effect differs for different peaks (i.e. different 2-theta angles); it is not enough to show that the intensity is changed; it must be changed by different amounts for different angles. You did give a hint of this in Nos 11,12. The model answer gives more details on this. Total marks for this question: 7

Overall Comment on Assignment: A good solid assignment. However, you must take care in writing answers that you do not just reproduce text (from other sources) that is not completely relevant.