Metallic gold crystallises in face centred cubic lattice with edge-length `4.07Å`. Closest distance between gold atoms is:

Options

A. 576.6pm
B. 287.8pm
C. 352.5pm
D. 704.9pm

Answer

To find the closest distance between gold atoms in a face-centered cubic (FCC) lattice with an edge length of 4.07 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
- In a face-centered cubic lattice, atoms are located at each of the corners and at the centers of each face of the cube.

2. **Identify the Edge Length**:
- The edge length (A) of the cube is given as 4.07 Å.

3. **Calculate the Face Diagonal**:
- The distance between two atoms located at the face centers can be calculated using the face diagonal of the cube. The face diagonal (d) can be calculated using the formula:
\[ d = \sqrt{2} \times A \]

4. **Substitute the Edge Length**:
- Substitute A = 4.07 Å into the formula:
\[ d = \sqrt{2} \times 4.07 \, \text{Å} \]

5. **Calculate the Face Diagonal**:
- Calculate the value:
\[ d = \sqrt{2} \times 4.07 \approx 5.76 \, \text{Å} \]

6. **Determine the Closest Distance**:
- The closest distance between two atoms in the FCC lattice is half of the face diagonal:
\[ \text{Closest Distance} = \frac{d}{2} = \frac{5.76 \, \text{Å}}{2} \approx 2.88 \, \text{Å} \]

7. **Convert to Meters**:
- To express the distance in meters, convert Ångströms to meters:
\[ 2.88 \, \text{Å} = 2.88 \times 10^{-10} \, \text{m} \]

8. **Convert to Picometers**:
- Since \(1 \, \text{Å} = 100 \, \text{pm}\), we can convert:
\[ 2.88 \, \text{Å} = 288 \, \text{pm} \]

### Final Answer:
The closest distance between gold atoms in the FCC lattice is approximately **288 pm**.

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