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Calculation of the first two X-ray Bragg reflections from KCl and KBr

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

    HI, I am practicing some exam questions to get ready for a repeat, below is the question I am having trouble with.

    "The lattice constants of KCl and KBr are 6.29 angstroms, and 6.59 angstroms, respectively.
    What would be the first two observed Bragg diffraction angles in each case , if the wavelength of X-rays was 1.932 angstroms? (assume Debye-Scherrer method is used.)


    2. Relevant equations

    Here, Braggs law is relevant,

    nλ= 2dsinθ.


    3. The attempt at a solution

    Braggs law can be rearranged to give θ = sin^-1(nλ/2dhkl).

    the Questions asks for the first two angles for each case, so n=1 and n=2 for the first and second angle.

    my two problems, are, how can I use α, the lattice constant to get dhkl?

    Since the two Crystals are both FCC, the lattice parameters are a=b=c, and α=β=γ,

    dhkl =[1/a^2 + ( h^2 +k^2 + l^2)]^-1/2

    I haven't been given any miller indices for h,k &l, so do I assume some, or just use α as dhkl?

    Also, will the angles be in units of radians or degrees?

    Any help or push in the right direction will be much appreciated.

    Cheers
     
  2. jcsd
  3. Aug 1, 2012 #2

    TSny

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  4. Aug 1, 2012 #3
    Thanks TSny, that article is a great help. : ) I'm fairly sure that the question refers to θ, but i will try to verify.

    I realized one small (er.. big ) detail, I overlooked,

    the Question that I posted was a part (d) of a four part question, and the part before (c), gave a list of miller indices for planes observed from x-ray diffraction from KCl and KBr crystals, the point of question (c) was to explain why KCl was missing some planes that KBr had.

    It just dawned on me that these miller indices are probably the ones meant to be used for hkl in the following question to determine the lattice spacing.

    the list reads as follows,

    KBr;( 1 1 1),(2 0 0), (2 2 0), (3 1 1), (2 2 2), (4 0 0),
    KCl; (2 0 0), (2 2 0), (2 2 2), (4 0 0 ),

    So my question is now, since I use the first two respective miller indices for each crystal to calculate D-hkl, what do I use for n in Braggs formula?? do I use 1 for the first angle calculation, and then 2 for each second angle calculation.
     
  5. Aug 1, 2012 #4
    Nope, just use 1 throughout. If you are using Miller indices, the convention is that you always talk about, e.g., the (200) reflection rather than the “second-order (100) reflection”.

    Did you get part (c)? That's a nice question!
     
  6. Aug 2, 2012 #5
    Well, I think I got part (c), I'll give you my answer (it may not be the most concise, but i think im on the right track)

    The cause of the diffraction differences begin with the K+ and Cl- ions, they both have an argon electron shell structure, and hence scatter x-rays almost equally.

    whereas K+ and Br- have different scattering strengths. the scattering factor of a crystal is similar to that of a lone atom, but the electron density distribution is perturbed when a solid is formed. Reflection intensity falls off as the the angle of incidence increases, and back scattering becomes weaker. the electron cloud which scatters the X-ray is comparable in size to the wavelength of the x-ray, and interference effects become apparent. The X-rays that the KCl crystal diffracts, destructively interfere, to such an extent that they dissapear , whereas in KBr, even though interference is present, there are weak enough Bragg reflections to be measured. In the case of KCl, even though the Laue conditions are fulfilled, there will be no reflected beam, if k h & l are not either all even or all odd.
     
  7. Aug 2, 2012 #6
    also here are my calculations for part (d)

    for KBr the first two indices are ( 1 1 1 ), and (2 0 0 ), so h =1, k=1, l=1, and h=2, k=0, l=0, respectivley.

    for the fist case d-hkl is 3.805 angstroms, and 3.295angstroms for the second.
    plugging them into Braggs formula with the variable I gave above gives the first two angles for Kbr as θ = sin^-1(nλ/2dhkl) = 14.71 degrees, for the first angle, and 17.05 degrees for the second angle. I assume that (if my math is correct) for the KCl angles, I just skip the missing miller indices, and use (2 0 0), (2 2 0) to determin D-hkl and therefore the first two angles for KCl.

    Thanks in advance for any further advice or corrections. I would appreciate it if i could get help making my Answer for part (c) as concise as possible. :D
     
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