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XRD structure factor

  1. May 1, 2012 #1
    Hi all,

    structure factor is 2πi(hx+ky+lz),
    (hkl)is plane index, and (xyz) is index within a unit cell of real lattice.

    For example, we are now calculating FCC's structure factor. In general, textbook
    taught us using convention unit cell of FCC( that is FCC in form of cubic) to index (xyz).
    so we got (xyz)=(000), (1/2 1/2 0), (1/2 0 1/2),.....

    But what if we use FCC's primitive unit cell instead of convention unit cell?
    Then (xyz) will be (000) only, and systematic absence will not happen.

    So what is wrong in my idea?

    Thanks!
     
  2. jcsd
  3. May 1, 2012 #2
    The structure factor in general is defined as

    [itex]
    f(Q) = \sum\limits_{i} f_i(Q) e^{i Q r_i} [/itex]

    where f_i(Q) is the form factor of atom i at position r_i, and Q = a* H + b*K+c*L is the scattering vector.

    If you use the conventional unit cell, then you have the identical atom at the positions you list.

    If you use the primitive unit cell, then you change the basis vectors (a,b,c), and with that the reciprocal space basis vectors (a* b*, c*). Also, you change the indexing of reflections.
    (HKL)-(200) in the conventional unit cell will have a different (HKL) in the primitive unit cell (I'm too drunk to work that out now :-p).

    In the full formula for the scattered intensity you normalize to the unit cell volume somewhere, so in the end the result does not depend on the choice of the unit cell.
     
  4. May 1, 2012 #3
    The short answer is:

    The indexing of reflections change, (HKL)_conventional is not the same as (HKL)_primitive.
     
  5. May 1, 2012 #4
    Hi,

    I am not quite sure I understand your answer. But I already checked the general form
    f(Q)=∑ifi(Q)eiQri before I post this thread.

    so FCC convention unit cell with lattice constant A.
    And the lattice vectors are a=Ax, b=Ay, c=Az. x,y,z are unit vectors in these directions.
    And the reciprocal lattice vectors(a* b* c*) are also forming simple cubic,
    bearing a "dot" a* = 2pi.

    And then as I wanted, change it into FCC primitive unit cell vectors:
    a=A/2(x+y), b=A/2(x+z), c=A/2(y+z).
    Therefore the reciprocal lattice vectors change to BCC primitive uc vectors.

    Well, no matter how I change the real lattice vectors, these vectors "dot" its reciprocal vectors = 2pi. (I neglect 2pi delta).

    For eiQri, Q "dot" ri is still the same: 2π(hx+ky+lz).

    Since I changed (a b c) into FCC primitive cell form, there is no (xyz) point within unit cell
    any more. Every point is just right at lattice point, so every point become (000).
    And you put (000) into 2π(hx+ky+lz), it got zreo.
     
  6. May 2, 2012 #5
    Yes, but calculate the primitive UC reciprocal space coordinates (Q_x, Q_y and Q_z) to those of allowed and forbidden reflections of the conventional UC.

    Can you find any (HKL)_primitive that falls onto a forbidden (HKL)_conventional?
     
  7. May 2, 2012 #6
    Ok, sober now :-) so here we go

    [itex]
    a_p = \frac{A}{2}(x+y)\\
    b_p = \frac{A}{2}(y+z)\\
    c_p = \frac{A}{2}(z+x)\\

    V_p = a_p \cdot (b_p \times c_p) = \frac{A^3}{4}
    [/itex]
    as it should be, given that there are 4 atoms in the conventional unit cell but only one in the primitive one.

    Next up are the reciprocal basis vectors, plug-and-play with the textbook formula.
    [itex]
    a^\star_p = \frac{2\pi}{V}(b \times c) = \frac{2\pi}{A}(x+y-z) \\
    b^\star_p = \frac{2\pi}{V}(c \times a) = \frac{2\pi}{A}(-x+y+z) \\
    c^\star_p = \frac{2\pi}{V}(a \times b) = \frac{2\pi}{A}(x-y+z) \\
    [/itex]

    Note that these are different from the conventional reciprocal basis vectors
    [itex]
    a^\star_c = \frac{2\pi}{A} x
    [/itex] etc.

    Now try a few combinations of (HKL)_p:

    [itex]
    (100)_p = \frac{2\pi}{A} (x+y-z) = (1 1 \bar{1})_c \\
    (110)_p = \frac{2\pi}{A} (2y) = (0 2 0)_c \\
    (111)_p = \frac{2\pi}{A} (x+y+z) = (1 1 1)_c \\
    (211)_p = \frac{2\pi}{A}(2x + 2y) = (2 2 0)_c
    [/itex]
    all of these are allowed. In the primitive basis there are no special selection rules, in the conventional one you need (H+K)=even and so on.

    Note also that some forbidden reflections in the conventional basis end up with half-integer indices in the primitive basis and are therefore forbidden:
    [itex]
    (010)_c = \frac{2\pi}{A}(y) = (\frac{1}{2} \frac{1}{2} 0)_p \\
    (110)_c = \frac{2\pi}{A}(x+y) = (1 \frac{1}{2} \frac{1}{2})_p
    [/itex]

    BTW, this is a nice example of why the conventional unit cell is used: (100)_p and (111)_p look very different, but describe reflections that are related by symmetry. In the conventional basis this is much easier to see.
     
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