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Homework Help: Bragg Diffraction

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A beam of 3.55keV X-rays is directed at a crystal. As the angle of incidence is increased from zero, a first strong interferece maximum is found when the beam makes an angle of 18.0o with the planes of the crystal.

    Calculated d=5.67x10-10 from this (distance between adjacent planes)

    (c) Find the longest wavelength for which two interference maxima would be produced.
    2. Relevant equations

    ## 2d\sin \theta = n\lambda ##

    3. The attempt at a solution

    I set n=2, since we're looking for the second interference maxima, so then ## d\sin\theta = \lambda ##
    I was confused by where theta came from here, but to find the maximum wavelength I thought I would set ## \theta ## = 90, but this doesn't make physical sense to me.
  2. jcsd
  3. Nov 3, 2014 #2


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    Look up a picture. And wonder what to do with the 18o in the problem description.
    Randomly picking ##\theta=\pi/4## indeed doesn't make sense :)
  4. Nov 3, 2014 #3
    Where does ## \theta = \pi /4 ## come into it?
    I chose ## \theta = \pi /2 ## to maximise ## d\sin\theta##
    And isn't the 18o only applicable for the earlier part of the question, when you're effectively given ## \lambda ## as ##\lambda = hc/E ##? As when the wavelength changes, so will the angle for the first maxima.
    Part (c) tells us we're calculating a new value for ##\lambda##

    Not sure how well I explained myself there. Sorry if it's incomprehensible!
  5. Nov 3, 2014 #4


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    It happens when the incident ray falls perpendicularly at the crystal plane and reflects exactly backwards. ( In the picture, the rays are shifted for clarity.)

  6. Nov 4, 2014 #5


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    My mistake. Still 90 degrees was a random choice: apparently to maximize. But the thing to do is to make it 'fit'.
    Last edited: Nov 5, 2014
  7. Nov 5, 2014 #6
    90 degrees is the maximum range of angles that can be measured by a diffractometer. This corresponds to back-reflection (see ehild drawing), or 2θ=180 degree. In this case the diffraction condition is 2d=nλ. If you want to have the second order at this maximum angle, then you have
    d=λ. The first order will be at 2dsinθ=d or sinθ=1/2.
    If λ is larger than d, you get the first order but not the second order. It will require an angle with sin>1.
    If λ is larger than 2d you don't get any peak.

    Of course, in practice the range of angles is less tan 0-180, with restrictions at both ends.
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