The correct relation for radius of atom and edge length in case of FCC arragement is

Options

A. `r = a/2`
B. `r = (sqrt3a)/(4) `
C. `r = (a)/(2 sqrt 2)`
D. `r = (4a)/(sqrt3)`

Answer

To find the correct relation for the radius of an atom (R) and the edge length (A) in a face-centered cubic (FCC) arrangement, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identifying Atomic Touching Points**:
- In the FCC structure, the atoms at the corners and the face-centered atoms touch each other along the face diagonal of the cube.

3. **Calculating the Face Diagonal**:
- The length of the face diagonal (d) can be calculated using the Pythagorean theorem. Since the face diagonal forms a right triangle with two edges of the cube:
\[ d = \sqrt{A^2 + A^2} = \sqrt{2A^2} = A\sqrt{2} \]

4. **Relating the Face Diagonal to Atomic Radii**:
- Along the face diagonal, there are three atomic radii (R) involved: one from the corner atom, one from the face-centered atom, and another from the corner atom on the opposite side. Therefore, we can express this as:
\[ d = R + 2R + R = 4R \]

5. **Setting Up the Equation**:
- Now, we can equate the two expressions for the face diagonal:
\[ A\sqrt{2} = 4R \]

6. **Solving for Radius (R)**:
- Rearranging the equation to solve for R gives:
\[ R = \frac{A\sqrt{2}}{4} \]
- This can be simplified to:
\[ R = \frac{A}{2\sqrt{2}} \]

### Final Relation:
Thus, the correct relation for the radius of an atom (R) and the edge length (A) in the case of a face-centered cubic arrangement is:
\[ R = \frac{A}{2\sqrt{2}} \]