Question

If edge length of a bcc crystallized $Fe$ is $8\sqrt{3}\mathring{A}$, the atomic radius (in $\mathring{A}$) is :

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

The correct option is A $6$

We have to use relation between a (edge length) and r (radius) of lattice.
As this is a BCC lattice, we know $r=\frac{A\times \sqrt{3}}{4}$
By putting the value, we get $r=\frac{8\times \sqrt{3}\times \sqrt{3}}{4}=6$
Hence, the atomic radius is $6\mathring{A}$.