Read the passage given below and answer the following question
In the simplest cases of monatomic metals such as Cu and `alpha-`Fe, mentioned above, the arrangement of metal atoms in the structure is simply the same as the arrangement of lattice points. In more complex structures such as NaCl. The lattice point represents an ion pair. This is still a very simple example, however, and in most inorganic structures the lattice point represents a considerable number of atoms In crystals of organic molecules such as proteins, the lattice point represents an entire protein molecule. Obviously, the lattice point gives no information whatsoever as to the atoms and their arrangements Which it represents, what the lattice does show is how these species are packed together in 3D. The combination of crystal system and lattice type gives the Bravais lattice of a structure. There are 14 possible Bravais lattices.
Source: West, Anthony R. 'Crystal Structures and Crystal Chemistry'. In Solid State Chemistry. 1-81, Wiley, 2014.
Percentage of free space in a body centred cubic unit cell is:

Options

A. 0.32
B. 0.34
C. 0.28
D. 0.2

Answer

To calculate the percentage of free space in a body-centered cubic (BCC) unit cell, we can follow these steps:

### Step 1: Understand the structure of BCC
In a body-centered cubic unit cell, there are:
- 1 atom at the center of the cube.
- 8 corner atoms, with each corner atom shared among 8 adjacent unit cells.

### Step 2: Calculate the effective number of atoms in BCC
Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell. Therefore, the total contribution from the corner atoms is:
\[ \text{Contribution from corner atoms} = 8 \times \frac{1}{8} = 1 \text{ atom} \]
Adding the atom at the center:
\[ \text{Total number of atoms in BCC} = 1 \text{ (from corners)} + 1 \text{ (from center)} = 2 \text{ atoms} \]

### Step 3: Calculate the volume of the unit cell
Let the edge length of the BCC unit cell be \( a \). The volume of the unit cell is:
\[ V_{\text{cell}} = a^3 \]

### Step 4: Calculate the volume occupied by the atoms
The radius \( r \) of the atoms in a BCC structure can be related to the edge length \( a \) by the formula:
\[ a = \frac{4r}{\sqrt{3}} \]
The volume of a single atom (considering it as a sphere) is given by:
\[ V_{\text{atom}} = \frac{4}{3} \pi r^3 \]
Thus, the total volume occupied by the 2 atoms in the unit cell is:
\[ V_{\text{occupied}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 5: Calculate the packing fraction
The packing fraction is the ratio of the volume occupied by the atoms to the volume of the unit cell:
\[ \text{Packing Fraction} = \frac{V_{\text{occupied}}}{V_{\text{cell}}} = \frac{\frac{8}{3} \pi r^3}{a^3} \]
Substituting \( a \) in terms of \( r \):
\[ \text{Packing Fraction} = \frac{\frac{8}{3} \pi r^3}{\left(\frac{4r}{\sqrt{3}}\right)^3} = \frac{\frac{8}{3} \pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{8\pi}{64\sqrt{3}} = \frac{\pi}{8\sqrt{3}} \]
Calculating this gives approximately 0.68 or 68%.

### Step 6: Calculate the percentage of free space
The percentage of free space in the unit cell can be calculated as:
\[ \text{Percentage of free space} = 100\% - \text{Packing Fraction} = 100\% - 68\% = 32\% \]

### Final Answer
The percentage of free space in a body-centered cubic unit cell is **32%**.

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