Distance between two layers in FCC unit cell

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Discussion Overview

The discussion centers on the geometric interpretation of the distance between layers in a face-centered cubic (FCC) unit cell, specifically addressing the claim that this distance is equal to a/√3, where a is the edge length of the unit cell. Participants explore different perspectives on the calculation and significance of this distance, including comparisons to other distances such as a/2.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question the assertion that the distance between layers in an FCC unit cell is a/√3, suggesting it should be a/2 based on the geometry of face-centered atoms.
  • One participant proposes dividing the area density of atoms within a plane by the volume density of atoms in the crystal to find the distance between chosen planes.
  • Another participant notes that the distance a/√3 may not correspond to the nearest planes but rather to planes of greater density.
  • Clarification is provided that a/√3 is the distance between planes parallel to the (111) plane.
  • Participants express uncertainty about the meaning of Miller indices and their relevance to the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the correct distance between layers in the FCC unit cell, with no consensus reached on whether it is a/√3 or a/2. The discussion remains unresolved regarding which planes are being referenced in the calculations.

Contextual Notes

Participants highlight the need to specify which planes are being measured when discussing distances, indicating that the choice of planes affects the calculated distance.

Who May Find This Useful

This discussion may be of interest to students and professionals studying crystallography, solid-state physics, or materials science, particularly those exploring the geometric properties of crystal structures.

zorro
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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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