Number of atoms in FCC gold cube with only the radius given

[nour halawani]

number of atoms in FCC gold cube with only the radius given!

1)it is given that the radius of a gold atom sphere is 144.2 pm, gold is packed in an FCC manner

Give the ratio Ns/Nv as a function of L (gold cube side) and a parameter

of FCC latice, knowing that Ns is the number of surface atoms and Nv is the number of atoms per volume.
2) and 3)
what i am thinking is:
in a unit cell we have 1/8 x 8 corner atoms = 1 atom and in the center of each phase we have 1/2 x 6 = 3 then the total number of atoms will be 4
atoms per unit = 4
volume of the cube is L^3 and volume of the unit is a^3 so if we want to know how many units are there in this cube we can divide the volume of the cube over the volume of the unit cell and we will have L^3/a^3
and if we multiply them by 4 which is the number of atoms per unit we will have the number of atoms in a cubic volume Nv= 4L^3/a^3

now talking about the surface is the thing confusing me
counting the atoms in unit cell is it 2 coming from 4 corners shared by 4 cells which is 1/4 x 4 and 1 center atom and by this we will have 2 atoms or my thinking is not reasonable?

 

[nour halawani]

won't i have an answer :(

 

[ehild]

1)it is given that the radius of a gold atom sphere is 144.2 pm, gold is packed in an FCC manner

Give the ratio Ns/Nv as a function of L (gold cube side) and a parameter

of FCC latice, knowing that Ns is the number of surface atoms and Nv is the number of atoms per volume.
2) and 3)
what i am thinking is:
in a unit cell we have 1/8 x 8 corner atoms = 1 atom and in the center of each phase we have 1/2 x 6 = 3 then the total number of atoms will be 4
atoms per unit = 4
volume of the cube is L^3 and volume of the unit is a^3 so if we want to know how many units are there in this cube we can divide the volume of the cube over the volume of the unit cell and we will have L^3/a^3

Correct so far...

and if we multiply them by 4 which is the number of atoms per unit we will have the number of atoms in a cubic volume Nv= 4L^3/a^3

now talking about the surface is the thing confusing me
counting the atoms in unit cell is it 2 coming from 4 corners shared by 4 cells which is 1/4 x 4 and 1 center atom and by this we will have 2 atoms or my thinking is not reasonable?

You do it well, and it will work well for large number of cells, but it is worth to count the atoms more carefully.
The gold cube has a sheet of surface cells on each side, 6 altogether. So the number of the inner cells is (L/a-2)^3.

Try to draw a cube of a few units. The cube in the picture is of 8x8x8 units, with 6x6x6 inner cells.

On one surface sheet, there are (L/a-2)^2 inner squares, which share the atoms at the corners with 3 other squares, and have their own atoms in the middle.

You have 12 edges of the gold cube. From each edge, (L/a-2) units are inner ones. The units have two unshared atoms at the centre of the outer sides, and common atoms at the outer corners.
There are eight corner cells with unshared atoms on the outer sides and corner.

Count all surface atoms, and take care so you do not count any of them twice.

ehild