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Crystal Structure Lattice Problem

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Using the powder XRD data below, show that the substance has a face centred cubic structure. (xray lamda = 0.154056 nm)

    Peak No.------2(theta)
    1 -------------38.06
    2 -------------44.24
    3 -------------64.34
    4 -------------68.77
    5 -------------73.07

    2. Relevant equations

    [tex] 2dsin\theta = n\lambda[/tex]

    [tex] d = \frac{a}{\sqrt{N}}[/tex]

    [tex] \Delta sin\theta = \left(\frac{\lambda}{4a^{2}}\right)N_{2} - N_{1}[/tex]

    [tex] N= h^{2}+k^{2}+l^{2} [/tex]

    3. The attempt at a solution

    I've worked out sin theta for each sin theta squared and delta sin theta:

    Peak------2(theta)------sin theta----sin squared theta---delta sin squared theta

    The only example we've covered is with a primitive cubic structure which I almost knew what I was doing(!) and the only advice that the lecturer gave was to "look for the highest common factor of values in the list delta sin squared theta to find

    I obviously noted that the difference between peak 2 and 3 was the same value as Peak 2 but what I'm meant to do with that information I'm not so sure about!!?

    I know that a fcc structure only has N values of 3,4,8,11 etc but really could do with some advice as where to go from here!!?

    PS sorry that this is repeated from adv
    anced physics - i felt it is more relevant here and didn't know how to move it!!
    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 15, 2009 #2
    Any help at all would be appreciated!!! I've spent ages searching for answers and the people that i've spolen to at uni have no clue either so i'd love to be able to pass on the knowledge!!!
  4. Nov 16, 2009 #3
    [tex] d = {{n\lambda}\over{2sin\theta}}=\frac{a}{\sqrt{N}}[/tex]

    [tex] {{n^{2}\lambda^{2}}\over{4sin^{2}\theta}}=\frac{a^2}{N}[/tex]

    [tex] {{n^{2}\lambda^{2}}\over{4a^{2}}}=\frac{sin^{2}\theta}{N}[/tex]

    [tex] \frac{sin^{2}\theta_1}{N_1}=\frac{sin^{2}\theta_i}{N_i}[/tex]

    [tex] \frac{N_i}{N_1}=\frac{sin^{2}\theta_i}{sin^{2}\theta_1}[/tex] This might be more useful?
    Last edited: Nov 16, 2009
  5. Nov 16, 2009 #4
    You have left out the most important part of identifying the lattice type, which is the selection rules. fcc lattices have non-zero intensity where the planes have h,k,l all even or all odd. so you can use trial and error and this equation:
    [tex]\sin^{2}\theta = \left(\frac{\lambda}{4a^{2}}\right)[h^{2}+k^{2}+l^{2}][/tex]
    to get the h,k and l for your planes, and compare to the selection rule and... you're done!
  6. Nov 17, 2009 #5
    Thanks so far!! I have since been told that the values for peak 4 and 5 are wrong so to ignore them.

    Therefore I got N values for the first three peaks of 3,4,8 which I know are right for a fcc cube so now I just have the second part which says I need to confirm my findings with structure factor calculations so I'll see how they go and let you know when it's solved!!
  7. Nov 18, 2009 #6
    thanks everyone!! I'm handing in tomorrow - still not sure what structure factor calculations are - I thought that's what I'd been doing all along!!!
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