In a body centred cubic structure, the space occupied is about

Options

A. 0.74
B. 0.2
C. 0.68
D. 0.524

Answer

To determine the space occupied by atoms in a body-centered cubic (BCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Body-Centered Cubic (BCC) Structure**:
- In a BCC unit cell, there are atoms located at the eight corners of the cube and one atom at the center of the cube.

2. **Calculating the Number of Atoms in a BCC Unit Cell**:
- Each corner atom is shared by 8 unit cells, so the contribution of one corner atom is \( \frac{1}{8} \).
- There are 8 corner atoms, contributing a total of:
\[ 8 \times \frac{1}{8} = 1 \text{ atom} \]
- The atom at the center contributes fully, so we have:
\[ 1 \text{ (from corners)} + 1 \text{ (from center)} = 2 \text{ atoms in total} \]

3. **Volume of Atoms in the Unit Cell**:
- Assuming each atom is a sphere, the volume \( V \) of one atom is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume occupied by 2 atoms is:
\[ V_{\text{total}} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

4. **Volume of the Unit Cell**:
- The volume of the cubic unit cell with edge length \( a \) is:
\[ V_{\text{unit cell}} = a^3 \]

5. **Finding the Relationship Between Edge Length \( a \) and Atomic Radius \( r \)**:
- The body diagonal of the cube can be expressed as:
\[ \text{Body diagonal} = a\sqrt{3} \]
- In a BCC structure, the body diagonal is also equal to the diameter of the atom at the center plus two radii from the corner atoms:
\[ \text{Body diagonal} = 4r \]
- Equating the two expressions gives:
\[ a\sqrt{3} = 4r \implies a = \frac{4r}{\sqrt{3}} \]

6. **Calculating the Packing Fraction**:
- The packing fraction \( PF \) is defined as the ratio of the volume occupied by the atoms to the volume of the unit cell:
\[ PF = \frac{V_{\text{total}}}{V_{\text{unit cell}}} = \frac{\frac{8}{3} \pi r^3}{a^3} \]
- Substituting \( a = \frac{4r}{\sqrt{3}} \):
\[ a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \]
- Therefore, the packing fraction becomes:
\[ PF = \frac{\frac{8}{3} \pi r^3}{\frac{64r^3}{3\sqrt{3}}} = \frac{8\pi \sqrt{3}}{64} = \frac{\pi \sqrt{3}}{8} \]

7. **Calculating the Numerical Value**:
- Using \( \pi \approx 3.14 \) and \( \sqrt{3} \approx 1.732 \):
\[ PF \approx \frac{3.14 \times 1.732}{8} \approx \frac{5.441}{8} \approx 0.6801 \]
- Converting to percentage:
\[ PF \approx 0.6801 \times 100 \approx 68.01\% \]

### Final Answer:
The space occupied by atoms in a body-centered cubic structure is approximately **68.01%**.