Fraser Rowan - Assignment II

Question 1 - Absorption of X-rays using Pb foil.

Workings:-

		ln(I/Io) = -(µr)rx

		ln(10/10E+8) = -(230)(11.3x)

		-18.42 = -230 (11.3x)

		0.08 = 11.3x

		x = 0.007cm

		x = 0.7mm 

A 0.7mm Pb foil would be required to reduce the intensity of a CuKa X-ray from 10E+8 p/s to 10p/s. If the incident beam was both unfiltered and non-monochromated, then it would contain a host of other emissions of higher energy than the CuKa X-ray. Such emissions, commonly referred to as "White radiation" would require a foil of much greater thickness than 0.7mm to fully stop all X-radiation passing through.
2 Marks

Question 2 - Line broadening

Elements such as H, C, O and N are poor X-ray scatterers (or, as they are more commonly referred to, are transparent to X-rays). The diffraction peaks of organic compounds are therefore usually broad and suffer slight shifts (unless samples are thin films) because they consist mostly of light elements. One method of minimising the effect of line broadening, using Bragg-Brentano geometry is to prepare the sample as a fine coating on a Si single crystal sample holder. This method is not, however, without its draw backs which include (i) preferred orientation and (ii) low intensities.

(i) Preferred orientation occurs when differences in lattice parameters exist. The greater the magnitude of difference between lattice parameters, the larger the extent of preferred orientation. As a consequence of preferred orientation, some reflections falsely have greater intensities than expected and other may be completely absent.  An extreme analogy would be a compact disc box; if one was to empty a box containing 20 or so compact disc boxes onto the floor, very few if any would come to rest on their side. As a consequence, planes perpendicular to the side of the boxes would be absent from a diffraction pattern.

(ii) Less sample would be used as a fine coating on an Si substrate, hence yielding lower intensities. This would be problematic as smaller peaks would be lost in the background. Signal to noise ratio would also be decreased, resulting in the introduction of larger errors than would normally be tolerated.

More generally, the cause of line broadening in Bragg-Brentano design diffractomers is due to axial divergence which results in a beam which is fanned out as opposed to collimated. This effect is minimised using parallel plate collimators. Parallel plate collimators (or soller slits) can be placed after the anode in the path of the incident beam, or prior to the detector in the path of the diffracted beam. A more disruptive solution to the problem would be to use a Cr anode. (This unfortunately involves replacing the X-ray tube and realigning the goniometer.) As Cr K X-rays have greater wavelength than the more commonly used Cu K (2.29 and 1.54, respectively) they will give better resolution at low angles, where the majority of the peaks exist for organic samples, thus giving narrower peaks.
Minor comment: Excellent answer, but note that under examination conditions, a much shorter answer would be expected. You will not have time for long answers when it comes to the written examination paper
3 Marks

Synchrotron radiation has several advantages over ordinary laboratory X-radiation (LXR). Two of the most significant advantages are (i) the intensity of the X-ray beam arriving at the analyte is much greater, and (ii) the beam is significantly more parallel than that of  LXR. As the intensity of the beam is much greater, the resultant peaks of the diffraction pattern are bigger giving a higher signal to noise ratio and better counting statistics. The focussing geometry used in conventional diffractometers is not required with synchrotron radiation as the beam is already almost parallel. A third advantage of synchrotrons over LXR is that the wavelength of the beam is tuneable. (In order to obtain variable wavelengths in conventional diffractometers one would have to change the X-ray tube. Even then the wavelengths are essentially fixed at six values for those of Cr, Fe, Co, Cu, Mo and Ag.)

Synchrotron radiation is particularly useful for analysing organic samples as high count rates are obtained and the wavelength can be long enough to allow good resolution of low angle peaks. This is further enhanced by the omission of focussing geometry which is inaccurate at low angles. Along with these attributes, the fact that the beam is highly collimated means that the data obtained is of much higher resolution than that obtained from a conventional laboratory diffractometer.
Comment: You explained the why, but not the how, i.e. how the instrument is designed to exploit the parallel nature of the incoming beam.
1 Mark

Question 3 - Rotation angle of monochromator

Workings:-

Bragg Equation,	 l = 2dsinq

Angle (q) at 0.75Å,

		0.75 = (2 x 3.136)sinq

		sinq = 0.120

		q = 6.89

Angle (q) at 0.65Å,

		0.65 = (2 x 3.136)sinq

		sinq = 0.104

		q = 5.95

Changing from 0.75 to 0.65Å , the angle (q) would have to be reduced by 0.94 q from 6.89 to 5.95 q or 1.88 2q.
Minor comment: Be aware that although the axis is referred as 2q, you only change the angle by q. You answer is a little ambiguous!
1 Mark

Question 4 - Range of d-spacings available over the range 20-100keV

Workings:-

		E = 12.398/l

		or l = 12.398/E

Where E=20keV,		l = 0.6199Å

Where E=100keV,	l = 0.1240Å

When 2q = 2.7, then q = 1.35.

Bragg Equation,    l = 2dsinq

		2d = l/sinq = l/sin1.35 = l/0.0236

For l = 0.6199, 2d=26.267 \ d = 13.134

For l = 0.1240, 2d = 5.254 \ d = 2.627

When a detector, designed to record X-ray photons with energies in the range 20-100keV,  is set at an angle of 2.7q, d-spacings within the range 2.627-13.134Å can be measured.
1 Mark

Question 5 - Symmetry operators.

1. 1 Mark


2. 0 Mark : See model answer


3. 1 Mark


4. 1 Mark


5. 1 Mark


6. 0 Mark : See model answer - You need to specify type of glide plane too


7. 1 Mark


8. 0 Mark : See model answer


9. 1 Mark : Minor comment - note that (0,1/4,z) is a line and not a point


10. 1 Mark