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 a volume of unit cell is $A\times \frac{\pi \sqrt{3}}{64}$. Find the value of A.

Answer 1

$r_A=\frac{r_B}{2}$

Volume of $B=\frac{4\pi }{3}\times r_B^3$

Volume of $A=\frac{4\pi }{3}\times \frac{r_B^3}{8}$

Unoccupied volume $=\frac{4\pi }{3}(r_B^3-\frac{r_B^3}{8})$

In BCC $4r_B=a\sqrt{3}$

$⟹r_B=\frac{a\sqrt{3}}{4}=\frac{4\pi }{3}\times \frac{7}{8}\times \frac{3\sqrt{3}{a}^3}{64}=\frac{7\pi \sqrt{3}{a}^3}{128}$ (For one atom)

Effective no. of atoms of $B=\frac{1}{8}\times 8+1=2$

Volume of unit cell $=a^3$

Ratio $=\frac{2\times 7\pi \sqrt{3}{a}^3}{128\times 93}=\frac{7\pi \sqrt{3}}{64}$

$⟹A=7$