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Bragg Scattering

  1. Oct 8, 2008 #1
    A beam of thermal neutrons emerges from a nuclear reactor and is incident on a crystal. The beam is Bragg scattered from crystal whose scattering planes are separated by 0.247 nm. From the continuous energy spectrum of the beam we wish to select neutrons of energy 0.0105 eV. Find the Bragg scattering angle that results in a scattered beam of this energy. Will other energies also be present in the scattered beams?
  2. jcsd
  3. Oct 8, 2008 #2
    so what i did was i have d =0.247 nm and K = 0.0105 eV. I used the equation 2dsin(theta) = hc/ sqrt( 2(mc^2) K). and then I got theta = 34.4 degrees. So, it asks if other energies will also be present in the scattered beams and I am not quite sure about that but my guess would be yes. would anyone please help me understand this?
  4. Oct 8, 2008 #3
    Bragg's formula is

    Keeping the angle fixed, it will be satisfied by various wavelengths, corresponding to various values of n.
    Your value is for n=1 but you can have higher order reflections of other wavelengths (that will be different energies for neutrons)
  5. Oct 8, 2008 #4
    thanks a lot.
    i was thinking the same but didn't know how to express it.
  6. Oct 8, 2008 #5
    I have another question. So, an electron bounces elastically back and forth in one dimension between two walls that are L apart. Assuming that the electron is represented by a de Broglie standing wave with a node at each wall, show that the permitted de Broglie wavelengths are lambda = 2L/n where n =1,2,3...
    What I don't understand here is why is lambda = 2L/n but not L/n ?
    I would really appreciate if anyone would help me with this.
  7. Oct 8, 2008 #6
    If you have nodes at the walls, you need to fit an integer number of half-wavelengths between the walls. The distance between two nodes is wavelength/2.
    L=n (lambda/d)
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