Problem Set 2 - Wolfgang Seidl

1. The lead foil must have a thickness of 0,062 mm.

    X-ray absorption equation: I(l) / I0 (l) = exp [(-m / p)l px]

The equation above is solved to the thickness x after applying ln,

    x = (ln I0 - ln I) p / m*p = (ln 108 - ln 10) * 1/230 * 1/11,3 = 6,2 * 10-3 cm = 0,062 mm

Unfiltered incoming X-ray radiation from a X-ray tube is "white" radiation, with a wide range of different wavelengths. The equation above is wavelength specific so every single wavelength has its own absoption coefficient which has to be taken into account. In the normal case one would look at the wavelength with the highest intensity and the shortest wavelength and calculate x from the absorption equation. For secureness the main beam stop would be slightly thicker than calculated.
The main beam stopper for a monochromated beam could be thinner than with the non monochromated beam, because the monochromator reduces the intensity and mostly absorbs a large part of the incoming radiation with wavelengths away from the maxima.


2. To obtain narrow diffraction peaks from organic compounds they have to be applied as a very thin layer onto the flat sample holder to avoid diffraction from different areas in the sample which cause the peak broadening. In doing this the diffracted intensity mostly is significantly reduced and the peak heights become unreliable. A very thin layer causes another unwanted fact. The sample holder itself might start to diffract, if it isn´t made of a completely amorphous material such as glass. With single crystal silicon one would get additional peaks from the sample holder. A much better way would be to use another geometry, which allows transmission measurements.

The ultimate way is to use X-rays from a synchrotron source in an angle dispersive geometry. The intensitiy of the synchrotron radiation itself is much higher than that of a laboratory X-ray tube which is the basic advantage. This allows to have a much higher absorption or exclusion of X-rays within the measurement and still get sufficient intensity for high quality diffractograms.
Narrow peaks can be measured with the use of an analyser crystal between the sample holder and the detector. The analyser crystal is like a filter which passes on only a small and to the analyser crystal specific angular range to the detector. This brings a loss of intensity with it, but the gained peaks are narrow and still intense enough to have reliable peak heights. The measurement can be improved by longer collection times, which improves the signal/noise ratio.
Another bur similar way is to exchange the analyser crystal against a set of foils (Hart-Perish), that are built out of a set of parallel thin metal foils with distinct length. As above this allows only a narrow angular range of the diffracted beam to pass on to the detector. Again there occurs loss of intensity, but as above there is enough intensity for a good diffractogram.
An important advantage of these to ways is, that the organic sample can be applied onto the flat holder in a really "thick" layer up to a few mm. By adjusting the detector and the foils/analyser crystal in the right way the measured area in the sample is defined. This prevents diffraction of the sample holder with use of only thin layers. Also movements of the sample don´t disturb the measurement.


3. The rotating angle for the monochromator can be calculated from the Bragg-equation:

   l = 2d sin Q

d spacing of Si 111 = 3.136 Å, and l changing from 0.65 to 0.75 Å. The equation is solved to sin Q:

    sin Q1 = l1 / 2d = 0.65 / 2*3.136 and sin Q2 = 0.75 / 2*3.136

   Q1 = 5.94 ° and Q2 = 6.87 °, D Q = 0.92 °

hence the rotating angle for the monochromator is twice the difference: D 2Q = 1.84 °


4. Again the Bragg-equation has to be used, but changed to calcualte with keV for l.

   E = 12.398 / l [keV/Å]; l = 12.398 / E

This inserted into the Bragg-equation yields:

    12.398 / E = 2d sin Q

After solving this equation to d the given values are inserted:

d1 (20 keV) = 12.398 / (2*E*sin Q) = 12.398 / (2*20*sin 1.35°) = 13.155 Å
d2 (100 keV) = 12.398 / (2*100*sin 1.35°) = 2.631 Å

d values between these to calculated values can be measured with the given photon energy.


5. The symmetry elements for the given symmetry operators are:
 

  1. -x, y, -z is a twofold rotation axis about the y-axis. y always remains as it was and x, z come back to the starting values when using the operator twice.
  2. 1/2+x, 1/2+y,z is a translation along the diagonal of the x and y-axis and perpendicular to the z-axis.
  3. -x, -y, -z is an inversion through the origin, which is combined out of a twofold rotation for example -x, y, -z and a mirror plane x, -y, z perpendicular to the y-axis.
  4. -x, -y, 1/2+z gives a 21 screw axis with a translation in z and a twofold rotation about the z-axis. Twice applied the result is x, y, 1+z, which is equal to the starting except a translation along z.
  5. x, y, -z is a mirror plane perpendicular to the cartesian z-axis, points with z = 0 remain where they are!
  6. 1/2+x, -y, z is a glide symmetry combined out of a mirror plane perpendicular to the cartesian y-axis and a translation along the x-axis.
  7. 1/2+x, -y, -z is again a 21 screw axis but this time with a translation and a twofold rotation about the x-axis. Twice applied the operator yields: 1+x, y, z which is overall a translation along x.
  8. 1/2-x, 1/2-y, -z is an inversion through the point (1/4, 1/4, 0). This is an inversion because all of the coordinates bear a minus and the 1/2 shows, that the inversion doesn´t occur through the origin. By calculating Tx and Ty the inversion center can be defined.
  9. -x, 1/2-y, z is a twofold rotation about (0, 1/4, z). -x, -y, z is a twofold rotation similar to case i. except along z. 1/2-y indicates a rotation not located at the origin.
  10. x, 1/2-y, z finally is a shifted mirror plane. x, -y, z would be the mirror plane perpendicular to the y-axis. 1/2-y shifts it to the point (x, 1/4, z).
I hope this is sufficiently funded as you requested giving reasons for my answers. The optimum would be to make some drawings, but I guess this is far beyond the scope of this question.