Question

Percentage of free space in a BCC unit cell is ?

• A. $32\%$
• B. $34\%$
• C. $28\%$
• D. $30\%$

Answer 1

The correct option is A

$32\%$

Explanation for the correct option:

Option(A): $32\%$

Step 1: Known data about BCC unit cell:

In the BCC unit cell, the atoms are present at each corner of a cube that is a total of 8 and one atom in the center of the cube.

Body-centered unit cells contain two atoms per unit cell. That is Z = 2;

Step 2: Formula for the calculation of packing efficiency:

We know the formula for packing efficiency is given as :
$\mathrm{packing}\mathrm{efficiency}=\frac{\mathrm{Z}\times \mathrm{volume}\mathrm{of}\mathrm{sphere}}{\mathrm{volume}\mathrm{of}\mathrm{cube}}$

Here, the Volume of the sphere = $\frac{4}{3}\pi r^3$
The volume of the cube = $a^3$

Step 3: Calculation of packing efficiency:

Hence, we can calculate the packing efficiency of the unit cell as :
$\mathrm{packing}\mathrm{efficiency}=\frac{2\times \frac{4}{3}\pi r^3}{a^3}$

(Also, we know that there is a relation between r value and edge length a , which is given by:
$$r=\frac{\sqrt{3}}{4}a$$
$$=\frac{2\times \frac{4}{3}\pi r^3}{a^3}$$)

Hence,
$\begin{array}{rl}& \mathrm{P}.\mathrm{E}={\frac{2\times \frac{4}{3}\times \pi \times \left[\frac{\sqrt{3}}{4}a\right]}{a^3}}^3\\ & =2\times \frac{4}{3}\times \frac{22}{7}\times \frac{\sqrt{3}\times \sqrt{3}\times \sqrt{3}}{4\times 4\times 4}\\ & =0.68\end{array}$
$\begin{array}{rl}& =0.68\times 100\\ & =68\text{\%}\end{array}$

Step 4: Calculation of percentage of free space :

The percentage of free space can be calculated with the formula,

$$\begin{gathered} \mathrm{Percentage}\mathrm{of}\mathrm{Free}\mathrm{space}=100-\mathrm{packing}\mathrm{efficiency} \\ =100-68 \\ =32\% \end{gathered}$$

Hence, the Percentage of free space in a BCC unit cell is $32\%$.