How Does Bragg's Diffraction Apply to FCC Lattices?

Yair Galili · May 14, 2017

Handwritten derivation on grid paper: reciprocal lattice vectors A, B, C given as 2π/a times (1,1,-1), (-1,1,1), (1,-1,1); G = 2π/a (h+l-k, h+k-l, k-l-h); and d = a / sqrt((h+l-k)^2 + (h+k-l)^2 + (k-l-h)^2).

I tried to treat FCC lattice with a lattice constant a, as a monoatomic lattice.
The reciprocal lattice is a BCC with a lattice constant 2pi/a.
And the structure factor is 1.
But I can treat FCC lattice with a lattice constant a, as a simple cubic lattice,
with 4 atoms in each unit cell. As one can see in Wikipedia:

Text excerpt deriving the FCC structure factor: F_hkl = f times the sum over four basis atoms of exp(-2 pi i (hx+ky+lz)), giving f[1+(-1)^(h+k)+(-1)^(k+l)+(-1)^(h+l)], resulting in 4f for all even or all odd indices and 0 for mixed parity.

Now the distance between adjacent planes in the reciprocal lattice is
as in a simple cubic: a/sqrt(h^2+k^2+l^2),and I cannot see
how I can get the same wavelengths,according to Bragg's diffraction.

nasu · May 15, 2017

You d

Yair Galili said:

View attachment 203502 I tried to treat FCC lattice with a lattice constant a, as a monoatomic lattice.
The reciprocal lattice is a BCC with a lattice constant 2pi/a.
And the structure factor is 1.
But I can treat FCC lattice with a lattice constant a, as a simple cubic lattice,
with 4 atoms in each unit cell. As one can see in Wikipedia:
View attachment 203503
Now the distance between adjacent planes in the reciprocal lattice is
as in a simple cubic: a/sqrt(h^2+k^2+l^2),and I cannot see
how I can get the same wavelengths,according to Bragg's diffraction.

You did not calculate any specific values for the lattice spacings if you do you'll see they are the same. Try to calculate the first 5 peaks, for example.

How Does Bragg's Diffraction Apply to FCC Lattices?

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