Question

In a BCC unit cell, if half of the atoms per unit cell are removed, then percentage void is:

• A. $68$
• B. $32$
• C. $34$
• D. $66$

Answer 1

The correct option is D $66$

Number of atoms per unit cell in BCC lattice = 2

If half of the atoms are removed, number of remaining atoms per unit cell = 1

Let a be the side of cube.

As we know that,

Packing efficiency of BCC unit cell = $\frac{\text{no. of atoms per unit cell}\times \text{volume of 1 atom}}{\text{volume of unit cell}}\times 100$

No of atoms per unit cell = 1

volume of 1 atom = $\frac{\sqrt{3}}{16}\pi a^3$

Volume of unit cell = $a^3$

$\therefore$ Packing efficiency = $\frac{1\times \frac{\sqrt{3}}{16}\pi a^3}{a^3}\times 100=\frac{\sqrt{3}}{16}\pi \times 100=33.99\%\approx 34\%$

$\therefore$ percentage void space = 100 - packing efficiency = $(100-34)\%=66\%$