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Scattering of Neutrons from 2d Crystal Lattice

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A two-dimensional rectangular crystal has a unit cell with sides a 6.28Å and
    b 3.14Å. A beam of monochromatic neutrons of wavelength 5.0 Å is used to
    examine the crystal.

    Using either the Laue condition for diffraction or Bragg's Law, determine
    whether it would be possible to observe the following reflections: (11), (20)
    and (02).

    2. Relevant equations

    Laue conditions: K[itex]\cdot[/itex]a=2πh
    K[itex]\cdot[/itex]b=2πk
    K[itex]\cdot[/itex]c=2πl

    for wavevector K, lattice vectors a,b,c and integers h,k,l.

    Bragg's law: 2dsinθ=nλ

    for lattice plane spacing d, Bragg angle θ, scattering order n and wavelength λ.

    3. The attempt at a solution

    Reciprocal space lattice is rectangular with lengths a* = 2π/a, b*=2π/b. Incoming wavevector k has magnitude 2π/5 = 1.26Å-1.

    Laue conditions:
    Direction (11) → K=a*+b*
    K[itex]\cdot[/itex]a=a*[itex]\cdot[/itex]a=2π
    K[itex]\cdot[/itex]b=b*[itex]\cdot[/itex]b=2π

    both are integer multiples of 2π and so (11) reflections are allowed.

    Direction (20) → K=2a*
    K[itex]\cdot[/itex]a=4π
    K[itex]\cdot[/itex]b=0

    allowed again for same reasons.

    Direction (02) → K=2b*
    K[itex]\cdot[/itex]a=0
    K[itex]\cdot[/itex]b=4π

    allowed.

    Bragg's law:

    (11) → d = ((h/a)2+(k/b)2)-1/2
    = ((1/6.28)2+(1/3.14)2)-1/2
    ≈ 2π/√5 (assuming a = 2π and b = π)

    (11) reflection has highest common factor 1 so it is an n=1 reflection:
    Bragg's law gives

    1*5=2*(2π/√5)sinθ

    rearrangement gives sinθ = 0.88970... which is certainly an allowed value for sinθ so I conclude that it is an allowed reflection.

    (20) is an n=2 reflection, d = a/h = a/2 ≈ π
    Bragg's law gives sinθ = 5/π > 1 which is not valid so the reflection cannot happen.

    (02) is also n=2, d = b/k ≈ π/2
    Bragg's law gives sinθ=3.18 and so it is not an allowed reflection.

    I'm not too confident in the way I've gone about solving this problem by either method and the contradictory conclusions definitely show that at least one of them is wrong. Could anyone give some hints as to where I'm going wrong? Thanks!
     
  2. jcsd
  3. Apr 6, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    If k is the wave-vector of the incoming wave and k’ for the diffracted wave, then the Laue condition is that k’- k = K where K is a reciprocal lattice vector.

    The dot product of a reciprocal lattice vector K with a direct lattice vector (a or b) will always yield an integer multiple of 2##\pi##. So, all you did was verify that fact for the particular reciprocal lattice vectors corresponding to (1,1), (2, 0) and (0,2).

    Since |k| = |k’|, see if you can use the condition k’- k = K to find the minimum possible value of |k| for a given |K|. That will tell you the maximum wavelength that the incident waves can have and still satisfy the Laue condition.
     
  4. Apr 6, 2013 #3
    Thanks for the reply.

    The vectors k, k' and K make a triangle because of the vector addition relating them. To minimise k, surely the triangle would be flattened such that k and k' run parallel to K.

    This gives k=k'=K/2.
    and we know
    λ=2[itex]\pi[/itex]/k

    (11): K=(([itex]\frac{2\pi}{a}[/itex])2+([itex]\frac{2\pi}{b}[/itex])2)[itex]\frac{1}{2}[/itex]
    = [itex]\sqrt{5}[/itex]Å-1

    corresponding to a maximum λ of 4[itex]\pi[/itex]/[itex]\sqrt{5}[/itex]Å

    Similarly I get

    (20): λ = 2[itex]\pi[/itex]Å (maximum)
    (02): λ = [itex]\pi[/itex]Å (maximum)

    As it is given that the incoming beam has wavelength 5Å then (11), (20) are allowed and (02) is excluded.

    Is this reasoning correct? Is it generally true that all wavelengths below a maximum are reflected by a crystal?
     
  5. Apr 6, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes, that all looks right to me.

    If λ is below the maximum for a set of lattice planes, then you will get reflection if the crystal is oriented such that the lattice planes make the proper angle with respect to the direction of the the incident wavevector. If you change λ then you would need to adjust the orientation. If you use a powder of little crystals oriented in all directions, then there will always be some crystals oriented to give reflection as long as λ is below the maximum.
     
  6. Apr 7, 2013 #5
    Thanks very much. You've pulled me out of that horrible mess of confusion! I had briefly thought of the condition you gave but I dismissed them before I figured out the consequences because I didn't think it was either the Laue condition or Bragg's law. Thanks again.
     
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