GENERAL COMMENT: This particular submission gave us much grief since it failed to display properly on non Microsoft-based computers. On this occasion JKC edited the HTML so that it strictly conformed to the HTML taught in the course material. For future submissions, you should not use products like MS-Word 2000. Note that the project work is partly judged on the quality of the HTML that you write. Therefore you should use the assignments (which do not give marks for HTML except for the first assignment set) as practice for writing HTML.

Please compare the HTML as modified by JKC to that produced by Word and look at the relative simplicity of the marked version (which could be made even simpler)

1. Lead foil is often used for the absorption of X-rays. Given (i) the equation for X-ray absorption, (ii) that m/r for lead at 1.54 Å is 230 cm²/g, and (iii) that the density of lead is 11.3 g/cm³, determine the thickness (in mm) of lead foil required to attenuate a Cu Ka X-ray beam of intensity 10⁸ photons/s down to 10 photons/s. Explain why the main beam-stop for an unfiltered non-monochromated copper X-ray source would be much thicker than this value?

Given parameters

     I(l) = 10 photons/sec

          I₀(l)= 10⁸ photons/sec

          (m/r)_1.54 = 230 cm²/g

            r = 11.3 g/cc

     Substituting these values in the equation,

I(l) / I_o(l) = exp{- (m / r)_l rx}

The final value obtained was 53 mm.

HTML comment: when viewed with my browser the value originally appeared to be 53 mm (i.e. millimetres!). Hence my comments about your source HTML
Comment: You took the log of 10^-8 by mistake, and you were asked to give the answer in mm and not microns. It would be better style to show your workings in more detail
1 Mark

The much higher thickness than the above value for a beam stop in an unfiltered non-monochromated copper X-ray source could be due to absorb the high energy background radiation.

Comment: correct but a bit too brief
1 Mark

2.     Powder diffraction patterns of organic compounds are often broadened due to sample transparency. Briefly discuss how narrow peaks might be obtained from a laboratory diffractometer configured for Bragg-Brentano geometry. Briefly explain how and why narrow peaks can be obtained with organic samples on synchrotron angle-dispersive powder diffractometers.

The best way to obtain the narrow peak in Bragg-Brentano geometry is to grind the sample into fine powder. Other alternative way, for organic compounds in particular, is to apply a fine coating of the powder on a Silicon single crystal.

Comment: only partially correct; see model answer
1 Mark

However, the diffraction pattern produced by synchroton radiation is much narrower due to the two important properties of such radiations. Firstly, the synchroton radiation is highly intense leading to better signal:noise ratio.   Secondly, the near parallel beam obtained with the synchroton radiations eliminates the low angle peak broadening found in conventional laboratory diffractometer.

Comment: only partially correct; see model answer
1 Mark

3.Crystal monochromators exploit the fact that only planes in the Bragg condition diffract X-rays in the required direction. On synchrotron powder diffractometers, double-bounce monochromators are rotated about a 2q axis so as to tune the instrument to a particular wavelength. If the wavelength on the instrument has to be changed from 0.75 to 0.65 Å (e.g. to avoid the high absorption of a zirconium compound), by what angle would the monochromator have to be rotated if Si (111) planes are used? (The d-spacing for Si (111) planes is 3.136 Å.) Show your workings clearly in your answer.

Given parameters

            l = 0.65 Å

          d = 3.136 Å

Substituting these values in the equation l = 2d sinq one can calculate q. From this the 2q can be calculated and found to be 11.9°. Thus the double bounce monochromator is to be rotated about an able 11.9° in order to use the Si (111) plane.

Comment: you have missed the point of the question since it is a change in angle that was to be calculated. See model answer for further details
1 Mark

4. energy dispersive diffraction, the equation relating an X-ray photon energy (E in keV) to its wavelength (l in Å) is:

            Given parameter,

                        Energy = 20 - 100 keV

                        2q = 2.7°

            Using the relation

Ed sin q = C

           Where C= 6.1993 keV Å. Substituting values of 20 keV and 100 keV in the above equation and 1.35 for q, the range of d-spacings have been found to be ranging from 2.631Å to 13.156 Å.

Comment: the question specifically asked you to show your workings clearly - you risk losing many marks in the event of a small numerical error
4 Marks

5. For each symmetry operator listed below, state the corresponding symmetry element giving reasons for your answer:
1. -x, y, -z
2. 1/2+x, 1/2+y, z
3. -x, -y, -z
4. -x, -y, 1/2+z
5. x, y, -z
6. 1/2+x, -y, z
7. 1/2+x, -y, -z
8. 1/2-x, 1/2-y, -z
9. -x, 1/2-y, z
10. x, 1/2-y, z

Ans:

1. two fold rotation along the y-axis

1 Mark
2. two fold rotation along the z-axis in a clock wise direction that does not pass through the origin

0 Marks
3. inversion

1 Mark
4. 2₁ screw axis along z-axis

1 Mark
5. reflection along xy plane

Minor comment: you should say "in" and not "along"
1 Mark
6. a glide plane along xz plane

Minor comment: you should say "in" and not "along"
1 Mark
7. 2₁ screw axis along X-axis

1 Mark
8.  

0 Marks
9. b glide plane along yz

0 Marks
10.  

0 Marks