Calculating Planar Density for FCC {100}, {110}, {111}

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SUMMARY

The discussion focuses on calculating planar density for face-centered cubic (FCC) structures specifically for the {100}, {110}, and {111} planes. For the {100} plane, the surface density is determined to be 2/a² atoms per unit area. The {110} plane has a surface density of 1/(a√2) atoms per unit area, while the {111} plane yields a density of 4/(a²√3) atoms per unit area. Additionally, the conversation touches on the (200) plane and its relation to the FCC unit cell, emphasizing the importance of visualizing atomic arrangements in these calculations.

PREREQUISITES
  • Understanding of face-centered cubic (FCC) crystal structures
  • Familiarity with lattice constants and atomic radii
  • Knowledge of geometric calculations for areas of squares and triangles
  • Basic principles of planar density in crystallography
NEXT STEPS
  • Research the calculation of planar density for other crystal structures, such as body-centered cubic (BCC)
  • Explore the concept of atomic packing factor (APF) in FCC and BCC lattices
  • Learn about the significance of Miller indices in crystallography
  • Investigate the effects of temperature and pressure on planar density in materials science
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Materials scientists, crystallographers, and students studying solid-state physics or materials engineering will benefit from this discussion, particularly those interested in the properties and calculations related to FCC crystal structures.

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HOw od you calculate the planar density for {100}, {110}, {111} for FCC?
 
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Lets start with [100] plane which is a plane parallel to a face of the unit cell and it looks like a square. There is one atom at the center of the square and a total of 4*1/4 atoms on the coners of the plane. Hence there are a net total of 2 atoms inside the square. Now the area is just the area of the square which is a^2 (where a is the lattice constant). So the suface density is 2/a^2 atoms per unit area.

The [110] plane is the plane which cuts the unit cell diagonally in half and it looks like a square. There are just 4*1/4 atoms on the corners of the square - a net total of 1 atom inside the square. The length of one of the sides of the plane is a*sqrt(2). Hence the surface density is 1/(a*sqrt(2)) atoms per unit area.

The [111] plane is a plane that touches the three far corners of the unit cell and it looks like a triangle. There are then a total 1/6*3 atoms that make up the vertices of the triangle and there are a total of 1/2*3 atoms that make up the three edges of the triangle. So you have a net total of 2 atoms inside the triangle. The triangle is an equlilateral triangle with a leg of length a*sqrt(2). The area of an equilateral triangle is s^2*sqrt(3)/4 which then gives us a^2*sqrt(3)/2 as the area of that trinagle. Hence the density is 2/(a^2*sqrt(3)/2) atoms per unit area.
 
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Thank you very much for explaning that very clearly!
 
FCC for (2 0 0)

how do you find the planar density for the (2 0 0) FCC unit cell
 
First of all, can you picture the (200) plane in the FCC unit cell? Secondly, across how many atoms does it cut?
 
Just in case, in Si diamond structure, the plane (110) includes a total of 4 atoms, which increases the density to 9.6*10^(14).
 
I need to know the number at surface atoms in a cube of a FCC lattice of gold atoms knowing that R= 144.2 pm with respect to L( length of the cube) and a ( length of a unit cell )
 

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