If r is the radius of an atom in face-centred cubic unit cell of edge length a then

Options

A. `r=(a)/(2sqrt(2))`
B. `r=(4)/(sqrt(3))a`
C. `r=2sqrt(2)a`
D. `r=sqrt((3)/(4))a`

Answer

To solve the problem, we need to establish the relationship between the radius \( r \) of an atom in a face-centered cubic (FCC) unit cell and the edge length \( a \) of the unit cell.

### Step-by-Step Solution:

1. **Understanding the FCC Structure**:
In a face-centered cubic unit cell, there are atoms located at each corner of the cube and one atom at the center of each face.

2. **Identifying the Arrangement**:
The atoms at the corners and the face centers touch each other along the face diagonal of the cube.

3. **Calculating the Face Diagonal**:
The length of the face diagonal in terms of the edge length \( a \) can be calculated using the Pythagorean theorem. For a square face of the cube:
\[ \text{Face Diagonal} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

4. **Relating the Face Diagonal to Atomic Radius**:
In the FCC structure, the face diagonal is equal to four times the atomic radius \( r \) (since there are two radii from the corner atom and two from the face-centered atom):
\[ \text{Face Diagonal} = 4r \]

5. **Setting Up the Equation**:
Now we can set the two expressions for the face diagonal equal to each other:
\[ a\sqrt{2} = 4r \]

6. **Solving for \( r \)**:
To find \( r \), we rearrange the equation:
\[ r = \frac{a\sqrt{2}}{4} \]
Simplifying further gives:
\[ r = \frac{a}{2\sqrt{2}} \]

### Final Answer:
Thus, the relationship between the radius \( r \) of an atom in a face-centered cubic unit cell and the edge length \( a \) is:
\[ r = \frac{a}{2\sqrt{2}} \]