Question

A bcc lattice is made up of hollow spheres of B. Spheres of solids A are present in hollow spheres of B. The radius of A is half of the radius of B. The ratio of total volume of spheres of B unoccupied by A in a unit cell and volume of unit cell is A$\times \frac{\pi \sqrt{3}}{64}$. Find the value of A.

Answer 1

Let $r$ be the radius of hollow sphere B and $0.5r$ be the radius of the solid sphere A.

Volume of hollow sphere B $=\frac{4\pi r^3}{3}$

Volume of hollow solid sphere A $=\frac{4\pi (0.5r{)}^3}{3}$

There are two hollow spheres B and two solid spheres A in bcc unit cell.

The total volume of spheres of B unoccupied by A in a unit cell $=\frac{8\pi (r{)}^3}{3}(1-(0.5{)}^3)$ $...\left(1\right)$

Total volume of unit cell $=a^3=\frac{1}{(0.433{)}^3}{r}^3$ $...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we can say that

the ratio of the total volume of spheres of B unoccupied by A in a unit cell and volume of a unit cell is

$\frac{\frac{8\pi (r{)}^3}{3}(1-(0.5{)}^3)}{\frac{1}{(0.433{)}^3}{r}^3}$

But this is equal to $A\times \frac{\pi \sqrt{3}}{64}$.

Hence, $A=7$