Question

Lithium forms body centred cubic structure. The length of the side of its unit cell is $360\text{pm}$. Atomic radius (in pm) of the lithium will be
(Given $\sqrt{3}=1.7$)

Answer 1

For BCC structure, $\sqrt{3}a=4r$
Where, $a=\text{edge length}$
$r=\text{radius of the atom}$
So,
$r=\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{4}\times 360=153\text{pm}$