Powder Diffraction Assignments

Answers to Problem Set 2 by Stewart F. Parker

  1. Use the structure factor equation to calculate the F2 (intensity) values for (i) the 200-reflection, and (ii) the 100-reflection from sodium chloride. You have already been given a sample calculation for the 111-reflection,Calculating the Intensity of Diffraction Using the Structure Factor Equation, and all the information that you need is contained within this example. In the case of the 100-reflection there is something odd about the answer and you should try to explain this.

    2. Sodium chloride (NaCl) is a face centred cubic structure with a unit cell a-spacing of 5.638 Å. The structure and fractional coordinates for the 4 Na and 4 Cl ions are given in the figure below.

    Answer:

    (i) Calculation of the intensity of the 200 reflection

    The trigonometric form of the Structure Factor equation is:

                      F(S) = Sn fn cos {360 (hx + ky + lz)} + iSn fn sin {360 (hx + ky + lz)} (in degrees)

    (1) Using Crystallography and Bragg's law

    The d-spacing for the 200 reflection is required. The formula for d for cubic crystals is:
                      dhkl = a/Ö(h2 + k2 + l2 )
    For h,k,l = 2,0,0 we have
                      d200 = 5.638/Ö(22 + 02 + 02 ) = 5.638/Ö4 = 2.819 Å
    From Bragg's law
                         l = 2d sinq
    hence
                    sin q/l = 1/2d = 1/(2 ´ 2.819) = 0.177 Å-1

    (2) Working out the atomic scattering factors (fn) needed:

    From the International Crystallographic Tables, the atomic scattering factors for Na+ and Cl- in the vicinity of 0.177 Å are:

    sin q/l 0.150.20
    fNa+ 9.035 8.39
    fCl- 14.12 12.20

    The form factor is a smooth function, so it is reasonable to interplolate between the values given. This results in:

    sin q/l 0.15 0.177 0.20
    fNa+9.035 8.698.39
    fCl- 14.12 13.0812.20

    (3) Working out the cosine part of F111

    Since k = l = 0, only the h term is required in the structure factor equation. Inserting the appropriate f values, coordinates and retaining only h = 2 gives:

    Sn fn cos{360 (hx + ky + lz)} =   8.69 cos {360 (2 ´ 0)}
        + 8.69 cos {360 (2 ´ 0)}
        + 8.69 cos {360 (2 ´ 1/2)}
        + 8.69 cos {360 (2 ´ 1/2)}
        + 13.08 cos {360 (2 ´ 1/2)}
        + 13.08 cos {360 (2 ´ 0)}
        + 13.08 cos {360 (2 ´ 0)}
        + 13.08 cos {360 (2 ´ 1/2)}
      =   8.69 cos 0 + 8.69 cos 0 + 8.69 cos 360 + 8.69 cos 360
        + 13.08 cos 360 + 13.08 cos 0 + 13.08 cos 0 + 13.08 cos 360
      =   8.69 + 8.69 + 8.69 + 8.69 + 13.08 + 13.08 + 13.08 + 13.08
      = 87.08

    (4) Working out the sine part of F111

    Similarly, only the h term needs to be considered:

    Sn fn sin {2p (hx + ky + lz)} =   8.69 sin {360(2 ´ 0)}
        + 8.69 sin {360(2 ´ 0)}
        + 8.69 sin {360(2 ´ 1/2)}
        + 8.69 sin {360(2 ´ 1/2)}
        + 13.08 sin {360(2 ´ 1/2)}
        +13.08 sin {360(2 ´ 0)}
        + 13.08 sin {360(2 ´ 0)}
        + 13.08 sin {360(2 ´ 1/2)}
      =   8.69 sin 0 + 8.69 sin 0 + 8.69 sin 360 + 8.69 sin 360
        + 13.08 sin 360 + 13.97 sin 0 + 13.97 sin 0 + 13.97 sin 360
      =  0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
      =   0

    (5) Combining the cosine and sine parts of F200 gives:

                      F200 = 87.08 + i.0

    (6) The intensity I200

    The intensity I is the square of the amplitude F:

                      I200 = F2002 = 7583 units

    (ii) Calculation of the intensity of the 100 reflection

    To calculate the intensity the procedure above can be followed. This is very tedious! Is there a shortcut? YES! For the intensity to be non-zero the sum of the cosine and sine terms must be non-zero. Since it is the 100 reflection that is required, again only the first term is needed. In the expressions given above, the 2 is replaced by 1, thus terms like:
    cos {360 (2 ´ 0)}
    become:
    cos {360 (1 ´ 0)}
    For the cosine terms the values will consist of either cos 0 or cos 180 i.e. 1 and -1 respectively. Since there are equal numbers of each term the sum will be identically zero. Similarly for the sine terms, the terms will be either sin 0 or sin 180, both of which are zero Hence the sine terms are also zero. Thus the structure factor and hence the intensity are both zero. Further since cos 180n = -1 (n odd) and +1 (n=0 or even), the same situation will arise for all the reflections of the type h00 (h odd) and these relections will be absent. Exactly the same arguments apply to reflections of the type 0k0 and 00l, thus reflections with k odd will be absent for 0k0 and reflections with l odd will be absent for 0l0. Physically, this corresponds to destructive interference by neighbouring ions (of the same type).

  2. Work out the multiplicity allowance factor needed for the 200-reflection.

    3. For the multiplicity of the 200 reflection, the equivalent planes are:

    200 and -200

    Since a "bar zero" index is the same as a "zero" index, these are the only two planes. Hence j= 2 for 200.

  3. Calculate the Polarisation allowance factor needed for the 200-reflection.

    4. (1) Working out the angles q, 2q :

    From above:
                      d200 = 2.819 Å

    Assuming that Copper K-alpha radiation with a wavelength of 1.54 Å is used and inserting into Bragg's law gives:
                      1.54 = 2 ´ 2.819 ´ sinq
                      sinq = 0.273
                      q = sin-1(0.273) = 15.85°
                      2q = 31.70°

    (2) Calculating P200 :

    For the laboratory powder diffractometer we use the formula of Case (3):
    2q = 31.70°
    cos (2q) = 0.851
    cos2 (2q) = 0.724
    1 + cos2 (2q) = 1.724
    P200 = (1 + cos2 (2q ))/2 = 0.862
    So polarisation produces a reduction in intensity to 86% in the case of the 200 reflection.

  4. Calculate the Lorentz allowance factor needed for the 200-reflection.

    For the NaCl-200 reflection, and using the value for theta already obtained in the polarisation calculation:

                q = 15.85°
                sin q = 0.273
                sin2q = 0.0746
                cos q = 0.962
                sin2q cos q = 0.0718
                L = 1/sin2q cos q = 13.94

  5. The next step is to bring together the F2 value and 3 allowance factors determined above for a prediction of the intensity ratio I111/ I200 .

    6. Now:

    I111   j111 P111 L111 F1112   (8) (0.894) (18.39) (398)   52347
    ¾¾  =  ¾¾¾¾¾¾¾¾¾  =  ¾¾¾¾¾¾¾¾¾  =  ¾¾  =    0.287
    I200   j200 P200 L200 F2002   (2) (0.862) (13.94) (7583)   182239

  6. In the model answer that you will eventually receive for this assignment, an actual experimentally measured ratio, I111/ I200 , will be presented to you. However, for the moment I would like to present you with a hypothetical situation for the purposes of this question. Imagine someone informs you that s/he has measured the ratio I111/ I200 for sodium chloride but you have no details of the experiment and you suspect that the quality of the measurement is far from good. True to your suspicions the measured ratio differs significantly from your prediction. I would like you to think creatively as to what plausible factors (additional to j, P, L) might be the cause of this disagreement. I only want brief answers, such as a one-phrase/sentence explanation for each factor you dream up. List them in order of severity.

    It is hoped that, with the combination of your imagination and all the course material you have received so far, you should have no difficulty devising some 5-10 plausible factors; perhaps you might even come up with more than 10.

    1. Human error The calculations for the intensities are long and intricate, a slip of the calculator will result in an incorrect answer. (This applies to both me and the other experimentalist!) There are all sorts of other possibilities in here including measuring the wrong reflection and getting one or other of the correction factors wrong.

    2. Texture For NaCL it is possible to grow large single crystals quite easily, thus it would be possible to produce a sample with preferred orientation.

    3. Impure sample If the sample is not single phase, then the impurity/ies could have diffraction lines at the same position as the NaCL lines, so distorting the measured intensities.

    4. Temperature If the measurements are made at different temperatures then the FT will be different for the two measurements

    5. Use of synchrotron radiation If the measurement was carried out on a synchrotron, then the polarisation correction is different (ie = 1) for both reflections and the ratio will be different from the laboratory measurement.

    6. Detector non-linearity The 200 reflection is more intense than the 111, if this is sufficeient to saturate the detector, then the measured intensity ratio will be incorrect.