
#1
Oct2806, 09:55 AM

P: 741

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




#2
Oct2906, 01:42 AM

P: 460

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. 



#3
Oct2906, 08:59 AM

P: 741

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



#4
Jun408, 03:05 PM

P: 1

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




#5
Jun408, 10:37 PM

HW Helper
P: 2,618

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




#6
Jan2410, 09:25 PM

P: 1

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
Dec2511, 07:49 AM

P: 5

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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