Distance between two layers in FCC unit cell

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SUMMARY

The distance between any two layers in a Face-Centered Cubic (FCC) unit cell is definitively calculated as a/√3, where 'a' represents the edge length of the unit cell. This conclusion arises from analyzing the geometric arrangement of atoms and the density of planes within the crystal structure. Specifically, the distance pertains to planes parallel to the (111) Miller indices, which are of significant interest due to their high atomic density. The discussion clarifies that the distance a/2 pertains to the distance from a face-centered atom to the edges surrounding it, not the interlayer distance.

PREREQUISITES
  • Understanding of FCC unit cell geometry
  • Knowledge of Miller indices and their significance in crystallography
  • Familiarity with concepts of atomic density in crystal structures
  • Basic principles of geometric calculations in three-dimensional space
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  • Study the derivation of distances between planes in FCC structures
  • Learn about Miller indices and their application in crystallography
  • Explore the concept of atomic density and its implications in material science
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Students and professionals in materials science, crystallography, and solid-state physics, particularly those focusing on the geometric and density-related properties of crystal structures.

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How is the distance between any two layers in a FCC unit cell equal to a/√3, where a is the edge length? I think it should be a/2 because the distance of a face centered atom from any 4 edges surrounding it is a/2. Can someone explain it geometrically?
 
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Abdul Quadeer said:
How is the distance between any two layers in a FCC unit cell equal to a/√3, where a is the edge length? I think it should be a/2 because the distance of a face centered atom from any 4 edges surrounding it is a/2. Can someone explain it geometrically?

Divide the area density of atoms within a plane by the volume density of atoms in the crystal. This gives the distance between the chosen plane and the nearest parallel plane.

Of course, this still means you need to decide which planes you're measuring the distance between.
 
PhaseShifter said:
Divide the area density of atoms within a plane by the volume density of atoms in the crystal. This gives the distance between the chosen plane and the nearest parallel plane.

Going that way I got a/2.

PhaseShifter said:
Of course, this still means you need to decide which planes you're measuring the distance between.

a/√3 is the distance between which two planes in a fcc unit cell? Not the nearest ones?
 
Abdul Quadeer said:
a/√3 is the distance between which two planes in a fcc unit cell? Not the nearest ones?

Generally you'll be interested in the planes of greatest density rather than the closest planes. In fact, you can find planes arbitrarily close together, if you pick planes with a very low density.
 
That does not answer my question. a/√3 is the distance between which two planes in a fcc unit cell?
(btw I found this distance in the derivation of the height of a hexagonal unit cell)
 
It's the distance between planes parallel to the (111) plane.
 
(111) = ? :confused:
 
I learned something new there :smile:
Thanks!
 

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