Question

Find the distance (in pm) between the body-centered atom and one corner atom in an element ($a=2.32$ pm).

Answer 1

The relation between the radius and the side is $a=\frac{4r}{\sqrt{3}}$.
In compact BCC structure, diagonal is equivalent to $4\times r$.

So distance between centre and corner is $2\times r$.
Now,
$2.32=\frac{4r}{\sqrt{3}}$
$\sqrt{3}\times 2.32=4r$ $⟹\frac{\sqrt{3}\times 2.32}{2}=2r=2$

Hence, the distance between the body-centered atom and one corner atom in an element is $2$ pm.