Question

Sodium metal crystallises in a body centred cubic lattice with unit cell edge of 4.29 $A^{\circ }$. The radius of sodium atom in $A^{\circ }$ is approximately

Answer 1

Let r be the radius of atom and a be the edge length of the lattice. So, for BCC lattice,
$\sqrt{3}a=4r$
r = $\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{4}\times 4.29A^o$
r = 1.86 $A^o$