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How to Calculate Distance Between Planes of FCC

The acronym FCC stands for "Face Centered Cubic." It is a specific subset of crystalline lattice configurations that describes the geometry of the atomic constituents of a solid whose primitive unit cell takes the form of a cube. The FCC structure describes an arrangement in which atoms are fixed at all eight corners of a cube, with an additional atom fixed at the center of each of the six cube faces. The planes of this structure take the shape of an equilateral triangle with three corner atoms fixed at each vertex. The distance between planes is simply the perpendicular length that connects two planes.

Things You'll Need

Pencil
Paper

Instructions

- 1
  Draw a cube oriented in Cartesian coordinates with the (0, 0, 0) point located on the far (back), bottom, left corner of the cube. Dimension the cube so that each line segment spans a length of "a." The length variable is a generalized length for which the distance between atoms can be substituted for any given compound. The diagram should display a cube with corners at the following Cartesian coordinates: (0, 0, 0), (a, 0, 0), (a, a, 0), (0, a, 0), (0, 0, a), (a, 0, a), (a, a, a), and (0, a, a).
- 2
  Draw the FCC planes in the cube diagram. They will appear as oppositely-oriented triangles. Sketch the first plane P1 by drawing the line segment that runs from (a, 0, 0) to (0, a, 0), the segment that runs from (0, a, 0) to (0, 0, a), and the segment that runs from (0, 0, a) to (a, 0, 0). The second plane P2 is formed from the line segments that run (a, 0, a) to (0, a, a), (0, a, a) to (a, a, 0), and (a, a, 0) to (a, 0, a).
- 3
  Write the equations of the planes. Recall that a plane equation takes the form of Ax + By + Cz - D = 0 where the coefficients A, B, C and are the components of the plane's normal vector N. D is the plane's constant which can be determined algebraically by substituting any point that lies on the plane into the equation and solving for D. The equation for P1 appears as P1 = x + y + z - a = 0. The equation for P2 appears as P2 = x + y + z - 2a = 0.
- 4
  Write the equation d = |P2 - P1|/|N|. This is the equation for the distance between the two planes, which is simply the difference between P1 and P2 scaled by the length of the normal vector N: |N| = sqrt(A^2 + B^2 + C^2).
- 5
  Substitute the known values into the distance equation to get d. Here we have d = |P2 - P1|/|N| = |x + y + z - 2a - (x + y + z - a)|/sqrt(1^2 + 1^2 + 1^2) = a/sqrt(3).

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