Chemistry - Solid State Question with Solution | TestHub

As for bcc structure$\mathrm{r}=\frac{\sqrt{3}}{4}\mathrm{a}$
So$2\mathrm{r}=\frac{\sqrt{3}}{2}\mathrm{a}$
As 2r represents the distance between the bcc atom and one of the corner atom so in a cube
the distance between the body centred atom and one corner atom is$\frac{\sqrt{3}}{2}\mathrm{a}$.