In face -centered cubic unit cell, edge length is

Options

A. (a) `4r/sqrt 3`
B. (b) `r/sqrt(2)`
C. (c) `2 sqrt(2) r`
D. (d) ` 3sqrt(2)r/4`

Answer

To determine the relationship between the edge length (a) of a face-centered cubic (FCC) unit cell and the atomic radius (r), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the FCC Structure**:
- In a face-centered cubic unit cell, atoms are located at each of the eight corners and at the center of each of the six faces of the cube.

2. **Visualizing the Unit Cell**:
- Draw a cube to represent the unit cell. Mark the corners and the centers of the faces where the atoms are located.

3. **Identifying the Arrangement of Atoms**:
- Each corner atom contributes 1/8th of its volume to the unit cell (since it is shared among 8 unit cells), and each face-centered atom contributes 1/2 (since it is shared between 2 unit cells).

4. **Finding the Diagonal of the Face**:
- The face diagonal of the cube can be calculated using the Pythagorean theorem. For a cube with edge length 'a', the face diagonal (d) can be expressed as:
\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

5. **Relating the Face Diagonal to Atomic Radius**:
- In the FCC structure, along the face diagonal, there are 4 atomic radii (2 radii from one atom at the corner and 2 from the face-centered atom). Therefore, we can write:
\[ d = 4r \]
- Setting the two expressions for the face diagonal equal gives:
\[ a\sqrt{2} = 4r \]

6. **Solving for Edge Length (a)**:
- To find the edge length 'a', rearrange the equation:
\[ a = \frac{4r}{\sqrt{2}} = \frac{4r}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}r}{2} = 2\sqrt{2}r \]

7. **Final Result**:
- The relationship between the edge length 'a' and the atomic radius 'r' in a face-centered cubic unit cell is:
\[ a = 2\sqrt{2}r \]