# Planar density

1. Oct 28, 2006

### asdf1

HOw od you calculate the planar density for {100}, {110}, {111} for FCC?

2. Oct 29, 2006

### Swapnil

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.

Last edited: Oct 29, 2006
3. Oct 29, 2006

### asdf1

^_^
Thank you very much for explaning that very clearly!!!

4. Jun 4, 2008

### rrc83

FCC for (2 0 0)

how do you find the planar density for the (2 0 0) FCC unit cell

5. Jun 4, 2008

### Defennder

First of all, can you picture the (200) plane in the FCC unit cell? Secondly, across how many atoms does it cut?

6. Jan 24, 2010

### davidalbertos

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).

7. Dec 25, 2011

### nour halawani

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 )