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

Options

A. `407 "pm"`
B. `407/sqrt(1) ` pm
C. `(407)/(sqrt(2)` pm
D. `407sqrt(3)/(2)`pm

Answer

To find the closest distance between gold atoms in a face-centered cubic (FCC) lattice, 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 corner of the cube and at the center of each face.

2. **Identify the Edge Length**: The edge length of the unit cell is given as \( a = 4.07 \, \text{Å} \).

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

4. **Relate the Face Diagonal to Atomic Radii**: In the FCC structure, the face diagonal is equal to the sum of the diameters of the atoms along that diagonal. Since there are two atomic radii at the corners and one at the face center:
\[ \text{Face Diagonal} = 4R \]
where \( R \) is the radius of the gold atom.

5. **Set Up the Equation**: From the above relationships, we can equate the two expressions for the face diagonal:
\[ 4R = a\sqrt{2} \]

6. **Solve for \( R \)**: Rearranging gives:
\[ R = \frac{a\sqrt{2}}{4} \]

7. **Calculate the Closest Distance**: The closest distance between two gold atoms is twice the atomic radius:
\[ 2R = 2 \times \frac{a\sqrt{2}}{4} = \frac{a\sqrt{2}}{2} \]

8. **Substitute the Edge Length**: Now substitute \( a = 4.07 \, \text{Å} \):
\[ 2R = \frac{4.07 \sqrt{2}}{2} = 2.035 \sqrt{2} \, \text{Å} \]

9. **Calculate the Numerical Value**: Now calculate \( 2.035 \sqrt{2} \):
\[ 2.035 \times 1.414 \approx 2.88 \, \text{Å} \]

10. **Convert to Picometers**: Since \( 1 \, \text{Å} = 100 \, \text{pm} \):
\[ 2.88 \, \text{Å} = 288 \, \text{pm} \]

### Final Answer:
The closest distance between gold atoms is approximately \( 288 \, \text{pm} \).