Solutions: Problem Set 2

Christy De Meyer

  1. Thickness of the lead foil = 6,20165 mm.

    Formula used:
    I(l)/ I0(l)= exp{-(m/r)l rx}
    or:
    10/108 = exp{-(230.11,3).x}

    Usually the beam stop is thicker because the X-ray beam is unfiltered and non-monochromated. As a result, X-rays with another energy (higher as well as lower) have to be captured too.

  2. Organic materials consist mainly of light elements (C, H, O, N,...). The X-rays penetrate deeper into the sample and therefore they are also scattered at atoms below the surface. The resulting detection angle for these scattered X-rays is always a little lower then for the X-rays scattered at the surface. So, transparency leads to broader peaks shifted towards lower angles.

    This can be overcome by using a silicon single crystal sample holder covered by a very thin "coating" of the organic powder (dusting the powder on the holder.)
    Result: the peaks will be much sharper, but the intensities will be much less reliable and smaller.

    The characteristic feature of synchrotron radiation that is particularly beneficial for powder diffraction studies of organic powder samples,is the possibility to select the incident X-ray wavelength from the highly intense continuous synchrotron radiation source. Another wavelength then the usual Cu Ka1 radiation can be used as monochromatic indicent radiation. So the wavelength for organic materials can be chosen slightly above or well below any of the absorption edges of the elements in the sample.

    The two design features of synchrotron angle-despersive powder diffractometers used: Debye Scherrer and Flat plate.

  3. Formula's used:

    E = 12.398 / l
    and
    n.l = 2.d.sinq
    results in:
    C = E.d.sinq (E in keV, d in Å, the C = 6,1993keVÅ)

    So, d = C / (E.sinq)
    for E = 20keV and 2q = 3.5° or q = 1.75° is d = 10.15Å
    for E = 100keV and 2q = 3.5° or q = 1.75° is d = 2.03Å

  4. Given:

    Si(111) - d = 3.3136Å
    l = 2.d.sin(qM/2)
    wavelength l0 = 0.75Å
    wavelength l1 = 0.65Å

    Problem: change of the angle 2qM

    Answer:
    sin(qM0/2) = 0.75Å/(2x3.3136Å)
    qM0 = 0.2268 rad
    sin(qM1/2) = 0.65Å/(2x3.3136Å)
    qM0 = 0.1965 rad
    So the change in 2q = - 0.0303 rad
    or the change in 2q = - 1° 44'

    1. -x,y,-z : a twofold rotation axis parallel to the y-direction and passing through the origin, (0,0,0), symmetry element: 2.
    2. 1/2+x,1/2+y,1/2+z : translation in the [1,1,1] direction by 1/2a, 1/2b and 1/2c. (??)
    3. -x,-y,-z : centre of inversion at the origin, symmetry element: -1 (minus sign above the digit)
    4. -x,-y,1/2+z : a two-one screw axis with a twofold rotation about the c-axis, i.e. about the line 0,0,z , followed by a translation of 1/2c along it, symmetry element: 21
    5. x,y,-z : mirror plane perpendicular to the c-axis and passing through the origin (plane x,y,0), symmetry element: m
    6. 1/2+x,-y,z : mirror plane perpendicular to the b-axis and passing through the origin (plane x,0,z) followed by a translation of 1/2a along a, symmetry element: a (because translation in "a" direction)
    7. 1/2+x,-y,-z : a twofold rotation about the x-axis, followed by a 1/2a translation along it, element: 21
    8. 1/2+x,1/2+y,z : a translation 1/2a + 1/2b (??)
    9. 1/2-x,1/2-y,z : a twofold rotation about z-axis followed by a translation along 1/2a + 1/2b, symmetry element: 21
    10. x,1/2-y,z : mirror plane perpendicular to b-axis passing through the origin (plane x,0,z) followed by a translation of 1/2b along b, symmetry element: m