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Interplanar spacing d using braggs law

  1. Dec 5, 2009 #1
    hi, ive been bashing my head against the wall for the past few days trying to work this one out!!!
    ive worked out a de broglie wavelength of electrons accelerated from rest by a voltage of 100v to be around (1.23 x 10 ^-10) m.
    now using a KCl diffraction , these electrons cause a first diffraction peak at angle of 11.3 degrees.......so therefore what is the interplanar spacing of the KCl???????

    im pretty sure the braggs formula has got to be implemented but unsure how??it couldnt be as simple as rearranging the bragg formula to get the d vaule could it?

    2 d sin theta = n lambda

    so basic info is that electrons are accelrated by 100v from rest
    de broglie wavelength is (1.23 x 10 ^-10) m
    and the first diffraction peak is at 11.3 degree angle

    please, would apreciate ur ideas and help!!!
  2. jcsd
  3. Dec 5, 2009 #2
    Bragg's formula will provide the value of d, the inter-planar spacing corresponding to this angle.
    If you actually need the lattice constant of the sample, this d may or may not be it.
    But it seems that you only need the inter-planar spacing here.
  4. Dec 6, 2009 #3
    thanks for that, but can the lattice constant be worked out from this information???and how can i use the bragg formula to ge this???
  5. Dec 6, 2009 #4


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    Hello strugglin, welcome to PF :)

    A lot of threads here discuss the same question, https://www.physicsforums.com/showthread.php?t=346987". You can follow the same procedure in order to get the needed values of both lattice constants and interplanar spacing d [Taking into consideration the crystal structure of KCl].
    Last edited by a moderator: Apr 24, 2017
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