Ar crystallizes in a F.C.C lattice with one atom at each lattice point. If the edge length is `5.311A^(0)` at OK, the distance between nearest neighbouring atoms in Ar at 'O'K is 

Options

A. `3.755A^(@)`
B. `7.355A^(@)`
C. `5.735A^(@)`
D. `1.877^(@)`

Answer

To find the distance between nearest neighboring atoms in Argon (Ar) which crystallizes in a Face-Centered Cubic (FCC) lattice, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Lattice Structure**:
- In an FCC lattice, atoms are located at each corner of the cube and at the center of each face. This means there are atoms at 8 corners and 6 face centers.

2. **Identify the Edge Length**:
- The edge length (a) of the FCC unit cell is given as 5.311 Å (angstrom).

3. **Calculate the Face Diagonal**:
- The formula for the face diagonal (d) in an FCC lattice is:
\[ d = \sqrt{2} \times a \]
- Substituting the value of edge length:
\[ d = \sqrt{2} \times 5.311 \, \text{Å} \]

4. **Calculate the Distance Between Nearest Neighbors**:
- The nearest neighbors in an FCC lattice are located along the face diagonal. The distance between nearest neighboring atoms is half of the face diagonal:
\[ \text{Distance} = \frac{d}{2} = \frac{\sqrt{2} \times a}{2} \]
- This simplifies to:
\[ \text{Distance} = \frac{a}{\sqrt{2}} \]
- Now substituting the edge length:
\[ \text{Distance} = \frac{5.311 \, \text{Å}}{\sqrt{2}} \]

5. **Final Calculation**:
- Calculate the numerical value:
\[ \text{Distance} \approx \frac{5.311}{1.414} \approx 3.755 \, \text{Å} \]

### Conclusion:
The distance between nearest neighboring atoms in Argon at 0 K is approximately **3.755 Å**.