Copper has face centred cubic `(fc c)` lattice with interatomic spacing equal to `2.54 Å`. The value of the lattice constant for this lattice is

Options

A. `1.27 Å`
B. `5.08 Å`
C. `2.54 Å`
D. `3.59 Å`

Answer

To find the lattice constant \( A \) for copper, which has a face-centered cubic (FCC) lattice with an interatomic spacing of \( 2.54 \, \text{Å} \), 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 corner of the cube and at the center of each face. Each unit cell contains 4 atoms.

2. **Identify the Relationship Between Atomic Radius and Lattice Constant**:
In an FCC lattice, the relationship between the atomic radius \( R \) and the lattice constant \( A \) can be derived from the diagonal of the face of the cube:
\[ \text{Diagonal} = \sqrt{2} A \]
The diagonal of the face of the cube also equals \( 4R \) (since there are two atomic radii from one corner to the center of the face and two from the center to the opposite corner).

3. **Set Up the Equation**:
From the above relationships, we can equate:
\[ 4R = \sqrt{2} A \]
Rearranging gives:
\[ A = \frac{4R}{\sqrt{2}} = 2\sqrt{2} R \]

4. **Relate Interatomic Spacing to Atomic Radius**:
The interatomic spacing given is the distance between the centers of two adjacent atoms, which is equal to \( 2R \). Thus, we have:
\[ 2R = 2.54 \, \text{Å} \]
From this, we can find \( R \):
\[ R = \frac{2.54}{2} = 1.27 \, \text{Å} \]

5. **Substitute \( R \) into the Equation for \( A \)**:
Now substitute \( R \) back into the equation for \( A \):
\[ A = 2\sqrt{2} \times 1.27 \, \text{Å} \]

6. **Calculate the Value of \( A \)**:
Calculate \( A \):
\[ A = 2 \times 1.414 \times 1.27 \approx 3.59 \, \text{Å} \]

### Final Answer:
The value of the lattice constant \( A \) for copper is approximately \( 3.59 \, \text{Å} \).