# Distance between two layers in FCC unit cell

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?

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.

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.

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?

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) = ? I learned something new there Thanks!!