How Does Bragg's Diffraction Apply to FCC Lattices?

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SUMMARY

The discussion centers on the application of Bragg's diffraction to face-centered cubic (FCC) lattices. It establishes that the reciprocal lattice of an FCC lattice is body-centered cubic (BCC) with a lattice constant of 2π/a. The structure factor for the FCC lattice is confirmed to be 1. The conversation emphasizes the importance of calculating specific lattice spacings to demonstrate that the wavelengths obtained through Bragg's diffraction yield consistent results across different lattice treatments.

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  • Understanding of FCC and BCC lattice structures
  • Familiarity with Bragg's diffraction principles
  • Knowledge of reciprocal lattice concepts
  • Ability to perform calculations involving lattice constants and structure factors
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  • Calculate the first five peaks of Bragg's diffraction for FCC lattices
  • Explore the relationship between lattice constants and reciprocal lattices
  • Study the implications of structure factors in diffraction patterns
  • Review examples of Bragg's law applications in crystallography
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Physicists, materials scientists, and students studying crystallography who are interested in the diffraction properties of FCC lattices and their applications in material analysis.

Yair Galili
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20170514_110831-1.jpg
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:
upload_2017-5-14_11-14-34.png

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.
 

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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.
 

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