Powder Diffraction Assignments

Answers to Problem Set 2

  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.

    Answer:

    200-reflection:

    d200 = 5.638 Å / (22+02+02)1/2
    = 5.638 Å / 2
      =>   sin q / l = 1/(2d)
    = (5.638 Å)-1
    = 0.177 Å-1.

    Similarly, for 100-reflection:

    d100 = 5.638 Å / (12+02+02)1/2
    = 5.638 Å
      =>   sin q / l = (2 · 5.638 Å)-1
    = 0.0887 Å-1.

    From the table and using proportion, we find for 200-reflection:

    fNa+ = 8.39 + (9.035 - 8.39) · 0.027 / 0.05 = 8.74
    fCl- = 12.20 + (14.12 - 12.20) · 0.027 / 0.05 = 13.24.


    Comment: The idea is right but in fact you weighted the values the wrong way round so that they were closer to the 0.15 rather than the 0.20 columns; the resulting values are therefore slightly wrong, but this is only a small error.
    4 Marks

    From the equation F(S) = nS fn e2p i (hx + ky + lz) we have for 200-reflection:

    F200(S) = fNa+ { 1 + 1 + e 2p i ( 2 · 1/2 ) + e 2p i ( 2 · 1/2 ) } +
    fCl- { e 2p i ( 2 · 1/2 ) + 1 + 1 + e 2p i ( 2 · 1/2 ) }
    = 2 · ( fNa+ + fCl- ) ( 1 + e 2p i )
    = 2 · ( fNa+ + fCl- ) ( 1 + cos 2p + i sin 2p )
    = 87.90,

    and for 100-reflection:

    F100(S) = fNa+ { 1 + 1 + e 2p i ( 1 · 1/2 ) + e 2p i ( 1 · 1/2 ) } +
    fCl- { e 2p i ( 1 · 1/2 ) + 1 + 1 + e 2p i ( 1 · 1/2 ) }
    = 2 · ( fNa+ + fCl- ) ( 1 + e p i )
    = 2 · ( fNa+ + fCl- ) ( 1 + cos p + i sin p )
    = 0.

    Obviously, 100-reflection is an example of destructive interference ("extinction").


    Comment: This is the right idea. You might also point out that the 100-reflection is one of many examples of destructive interference brought about by crystal symmetry.
    5 Marks

    Finally, the intensity for the first reflection will be:

    F2002 = 87.902 = 7726.41.


    Comment: This is right. The small difference from my value is due only to my using slightly different f values for Na and Cl.
    5 Marks

  2. Work out the multiplicity allowance factor needed for the 200-reflection. The 111-reflection example has already been given in Multiplicity, j.

    Answer:

    There are five more planes that have the same d as 200-reflection: -200, 020, 0-20, 002, 00-2. Therefore, the multiplicity for 200-reflection is: 6.


    Comment: This is correct.
    3 Marks

  3. Calculate the Polarisation allowance factor needed for the 200-reflection. The equivalent calculation for the 111-reflection example has already been given in Polarisation, P.

    Answer:

    d200 = 5.638 Å / (22+02+02)1/2
    = 5.638 Å / 2
      =>   sin q = l /(2d200)
    = 1.54 Å / 5.638 Å
    = 0.27315

    We deal with case (3) (unpolarised X-rays), and formula P = (1 + cos2 2q )/2. The Polarisation allowance factor for this case will be:

    P = 0.5 · [ 1 + ( cos2 q - sin2 q )2 ]
    = 0.5 · [ 1 + (1 - 2 sin2 q )2 ]
    = 0.5 · [ 1 + (1 - 2 · 0.273152 ) 2 ]
    = 0.86.


    Comment: This is right. I like the neat way of calculating the answer without having to use the actual value of the angle.
    4 Marks

  4. Calculate the Lorentz allowance factor needed for the 200-reflection. The equivalent calculation for the 111-reflection example has already been given in Lorentz Factor, L.

    Answer:

    L = (sin2q cos q)-1
    = [sin2q ( 1 - sin2q )1/2 ]-1
    = [ 0.273152 · ( 1 - 0.273152 )1/2 ]-1
    = 13.93.

    [The value for sin q calculated in the previous answer was used here.]


    Comment: This is right. Again a neat avoidance of finding the angle.
    3 Marks

  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 . The way to do this has already been explained in Agreement?.

    Answer:

    I111/ I200  = ( j111 P111 L111 F1112 ) / ( j200 P200 L200 F2002 )
    = ( 8 · 0.89 ·18.39 · 398 ) / ( 6 · 0.86 ·13.93 · 7726 )
    = 0.0938.


    Comment: This is correct.
    3 Marks

  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.

    Two possible examples might be:

    X-ray absorption: the intensities might require an absorption corrections, since you do not know whether the measurement was made in transmission geometry or reflection geometry.

    A badly stabilised power supply: with a badly stabilised power supply the X-ray flux on the sample could vary by several percent over the time between measuring two reflections such as 111 and 200.

    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.

    Answer:

    Generally, here belong all the factors that favorize some reflections.

    • Preferred orientation.
    • Crystallities are not small enough - poor statistics may favorize some reflections.
    • Impurities may have different influences on different reflections (through overlapping, for example).
    • Anisotropy of the sample.
    • Anisotropy due to temperature effects.
    • Strain introduced into sample after grinding/milling.
    • Sample too transparent in combination with preferred orientation.


    Comment: You have many good answers though very slightly more explanation might have been given for example to explain why one reflection is affected more than another (see my model answer comments). I have given 1 mark for each reason totalling...
    7 Marks


    Overall Comment: This is an excellent and well-thought out set of answers.

Student: Siniûa Prokic