Powder Diffraction Assignments

Answers to Problem Set 1 by Stewart F. Parker

  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?

    The equation for X-ray absorption is:
    I(l) / Io(l) = exp{- (m / r)l rx}
    where I(l) and Io(l) are the transmitted and incident X-ray intensities, respectively.
    (m/r)l is the mass absorption coefficient at the wavelength l,
    r is the density of the material.
    For lead r = 11.3 g/cm3, and m/r = 230 cm2/gm at the Cu Ka wavelength of 1.54 Å.
    It is required to reduce a Cu Ka X-ray beam from Io(l) = 108 photons/s to I(l) = 10 photons/s
    Substituting the numerical values in the equation gives:
    10/108 = exp{-230.11.3.x}
    Þ10-7 = exp{-2599x}
    taking natural logs and re-arranging gives:
    x = -ln(10-7)/2599 = 16.1/2599 = 0.0062 cm = 0.062 mm.

    The output from a non-monochromated Cu X-ray source is a mixture of continuum white radiation and the characteristic X-ray lines. The white radiation extends up to much shorter wavelengths (ie higher energies) than the characteristic X-ray lines and is consequently both more harmful and more penetrating. To remove the hazard associated with this the beamstop needs to be thicker. In addition, the larger the mass, the easier it is to cope with the heat deposited in the beamstop.


    2. 5 Marks

  2. Powder diffraction patterns of organic compounds are often broadened due to sample transparency. Briefly discuss how narrow peaks might be obtained (a) from a laboratory diffractometer configured for Bragg-Brentano geometry, and (b) from an angle-dispersive synchrotron powder diffractometer.

    3. Sample transparency arises because X-rays are scattered by electrons, thus they penetrate more in materials of low atomic number. The effect is that X-rays diffracted from the surface and from deep in the bulk are separated in space. Thus a detector will register the two contributions at different scattering angles. Since there is more "bulk" than "surface" the "bulk" X-rays will be the major contribution and the diffracted beam will appear at lower angle. The solution is to reduce the sample size such that the difference in surface and bulk beams is negligible.
    (a) For Bragg-Brentano geometry this can be done by dispersing the sample as a fine powder on a suitable sample holder. The disadvantage is that this may introduce texture into the sample.
    Comment: Specify more details about the sample holder
    2 Marks
    (b) For an angle-dispersive synchrotron powder diffractometer, the incident flux is so much greater by that the sample size can be reduced to a very small quantity.
    Comment: This is not the major reason why narrow peaks can be obtained
    0 Marks

  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 0.70 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.

    Re-arranging Bragg's law in terms of q gives:
    q = sin-1(l/2d)
    Using Si(111) with d = 3.136 Å to select:
    l = 0.70 Å requires q = 6.408°
    l = 0.65 Å requires q = 5.948°
    Thus a rotation of (6.408 - 5.948) = 0.46° is required.
    5 Marks

  4. In energy dispersive diffraction, the equation relating an X-ray photon energy (E in keV) to its wavelength (l in Å) is:
    E = 12.398 / l
    What range of d-spacings can be measured for a detector designed to record X-ray photons with energies in the range 20-100 keV when the detector angle (2q) is set to 3.5°. Show your workings clearly in your answer.

    We are given that:
    E = 12.398 / l
    Re-arranging in terms of l and substituting for l in Bragg's law gives:
    12.938/E = 2dsinq
    For 2q = 3.5° sinq = 0.061, thus:
    Error: 0.061 = sin 2q
    d = 106.05/E
    So for E = 20 keV, d = 5.302 Å
    and for
    E = 100 keV, d = 1.061 Å
    thus d spacings in the range 1.061 to 5.302 Å may be measured with X-rays in the 20 – 100 keV range when the detector angle is set to 3.5°.
    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, 1/2+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. 1/2-x, 1/2-y, z
    10. x, 1/2-y, z
(You should note that none of the answers involves high-order, i.e. n > 2, symmetry axes.)

  1. -x, y, -z

    2. In the section of the course notes on Rotational Symmetry II Symmetry Operators, we are shown that for a twofold rotation about the z-axis, the point (x,y,z) maps to (-x,-y,z). Applying the same arguments shows that:
    for a twofold rotation about the x-axis, the point (x,y,z) maps to (x,-y,-z)
    and
    for a twofold rotation about the y-axis, the point (x,y,z) maps to (-x,y,-z).
    Thus the symmetry element in this case is a twofold rotation about the y-axis.
    1 Mark

  2. 1/2+x, 1/2+y, 1/2+z

    3. A translation is the movement of a point to another point by addition of a vector. Thus a co-ordinate cannot undergo a sign change. (This is true even for a translation by a negative amount, note that in general 1/2 - x is NOT equal to x - 1/2 since subtraction is not commutative). Hence the required symmetry element is a translation of (1/2, 1/2, 1/2).
    1 Mark

  3. -x, -y, -z

    4. In the course notes on Rotary-Inversion Symmetry I. Inversion Symmetry we are given that inversion at the origin has the symmetry operator (-x,-y,-z), hence the symmetry element in the present case is inversion at the origin.
    1 Mark

  4. -x, -y, 1/2+z

    5. In the section of the course notes on Screw Symmetry II. Two-One Screw Axis we are shown that for a two-one screw axis parallel to the a-axis, the point (x, y, z) maps to (1/2 + x, -y,-z). Applying the same argument to a two-one screw axis parallel to the b-axis shows that the point (x,y,z) maps to (-x, 1/2 + y,-z). Similarly for a two-one screw axis parallel to the
    c-axis the point (x, y, z) maps to (-x, -y, 1/2 + z). Hence the symmetry element required is a 21 screw axis parallel to the
    c-axis.
    1 Mark

  5. x, y, -z

    6. In the course notes on Rotary-Inversion Symmetry II. Mirror Planes we are shown that for a mirror plane perpendicular to the x-axis (ie the yz-plane) the point (x,y,z) maps to (-x, y, z). Similarly, for a plane perpendicular to the y-axis (ie the xz-plane) the point (x,y,z) maps to (x, -y, z) and for a plane perpendicular to the z-axis (ie the xy-plane) the point (x,y,z) maps to (x, y, -z). Hence the required symmetry element is reflection in a mirror plane perpendicular to the z-axis.
    1 Mark

  6. 1/2+x, -y, z

    7. In the course notes on Glide Symmetry we are shown that an a-glide with the reflection plane being the xz-plane, the point (x,y,z) maps to (1/2 + x, -y, z). Hence this is the required symmetry element.
    1 Mark

  7. 1/2+x, -y, -z

    8. In the section of the course notes on Screw Symmetry II. Two-One Screw Axis we are shown that for a two-one screw axis parallel to the a-axis, the point (x,y,z) maps to (1/2 + x, -y,-z). Hence this is the required symmetry element.
    1 Mark

  8. 1/2+x, 1/2+y, z

    9. From part ii) above it can be seen that a translation is the only symmetry operation that does not involve a sign change of at least one co-ordinate. For the present case a translation of (1/2, 1/2, 0) is the required symmetry element.
    1 Mark

  9. 1/2-x, 1/2-y, z

    10. In the section of the course notes on Rotational Symmetry II Symmetry Operators, we are shown that for a twofold rotation parallel to the z-axis about an arbitrary point (Tx, Ty, z), the point (x,y,z) maps to (2Tx-x, 2Ty-y, z). Applying the same arguments shows that: for a twofold rotation parallel to the y-axis about an arbitrary point (Tx, y, Tz), the point (x,y,z) maps to (2Tx-x, y, 2Tz-z) and
    for a twofold rotation parallel to the x-axis, the point (x,y,z) maps to (x, 2Ty-y, 2Tz-z).
    Thus to map to the point (1/2-x, 1/2-y, z), rotation parallel to the z-axis is required and since the displacement is twice the co-ordinate, it follows that the symmetry element in this case is a twofold rotation parallel to the z-axis passing through the point (1/4, 1/4, z).
    1 Mark

  10. x, 1/2-y, z

    11. In the present example, the x and z co-ordinates are unchanged, this implies reflection in a plane parallel to the xz-plane (ie perpendicular to the y-axis). However, the change in y is not simply a mapping to –y as would be generated by reflection in the y = 0 plane, but there is an additional displacement. Thus the plane is located parallel to the y = 0 plane. Consider a plane parallel to the xz plane but at an arbitrary value Ty. For the y co-ordinate, the distance from the plane is y – Ty, after reflection, by symmetry, the distance is still y – Ty. Thus the new y co-ordinate is given by:
    y – (y – Ty) – (y – Ty) = 2Ty – y
    The symmetry operator is (x, 1/2-y, z), comparison with the general result, shows that the required symmetry element is reflection in the plane (x, 1/4, z).
    Minor comment: beware use of non-standard 8-bit characters in answer to (x) above
    1 Mark