Question

The relation between edge length (a) and radius of atom (r) for FCC lattice is $\sqrt{2a}=4r$.
If true enter 1, else enter 0.

Answer 1

Remember that a face-centered unit cell has an atom in the middle of each face of the cube. The square represents one face of a face-centered cube:

Applying Pythagoras theorem,

$a^2+a^2={(r+2r+r)}^2$

$\Rightarrow \sqrt{2}a=4r$