Question

In body−centered cubic cell (BCC) of lattice parameter a, the atomic radius is −

• A. $\frac{\sqrt{3}a}{4}$
• B. $\frac{a}{\sqrt{3}}$
• C. $2\sqrt{2a}$
• D. $\frac{a}{4}$

Answer 1

The correct option is A $\frac{\sqrt{3}a}{4}$


In body-centered cubic lattice, the atom at the body centre touches the spheres at the corners,
$\therefore$ Body diagonal, PS = 4r
Face diagonal $=PR=\sqrt{\left(PQ{)}^2\right(QR{)}^2}=\sqrt{2}a$
Body diagonal $=PR=\sqrt{\left(PR{)}^2\right(RS{)}^2}=\sqrt{2a^2+a^2}=\sqrt{3}a$
Now,$\sqrt{3}a=4r$
$\Rightarrow r=\frac{\sqrt{3}a}{4}$