Question

In FCC how many atoms are at a distance of $2\sqrt{3}r$ (where r is radius of atom) from a corner atom?

Answer 1

For fcc lattice, $a=2\sqrt{2}r$ (Where a is the side length)
$\therefore 2\sqrt{3}r=\sqrt{\frac{3}{2}}a$

If we want to find the distance between a corner atom and the face center atoms of the faces which are not touching that corner atom (let's say, XT) then,
$X{T}^2=T{Z}^2+X{Z}^2$
Again, $X{Z}^2=X{Y}^2+Y{Z}^2=(\frac{a}{2}{)}^2+a^2=\frac{5a^2}{4}$
So, $X{Z}^2=(\frac{a}{2}{)}^2+\frac{5a^2}{4}=\frac{6a^2}{4}$
$\Rightarrow XZ=\sqrt{\frac{3}{2}}a$
So there will be 3 face center atoms which are at a distance $\sqrt{\frac{3}{2}}a$ from a particular corner atom and again that corner atom will be shared by 8 cubes.
So, the total number of atoms that are at a distance $2\sqrt{3}r$ (where r is radius of atom) from a corner atom will be $8\times 3=24$