Copper crystallises in fcc with a unit cell length of 330 pm. What is the radius of copper atom?

Options

A. 117
B. 181
C. 108
D. 128

Explanation

To find the radius of a copper atom in a face-centered cubic (FCC) structure with a unit cell length of 330 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
- In an FCC unit cell, atoms are located at each of the eight corners of the cube and at the center of each of the six faces.

2. **Identify the Edge Length**:
- The edge length (a) of the FCC unit cell is given as 330 pm.

3. **Calculate the Face Diagonal**:
- The face diagonal (d) of the cube can be calculated using the formula:
\[ d = a \sqrt{2} \]
- Substituting the value of a:
\[ d = 330 \, \text{pm} \times \sqrt{2} \approx 330 \, \text{pm} \times 1.414 \approx 466.5 \, \text{pm} \]

4. **Relate the Face Diagonal to Atomic Radius**:
- In an FCC structure, the face diagonal is equal to the sum of the diameters of the three atoms that lie along that diagonal. Since the diameter of an atom is twice its radius (d = 2R), we can express this as:
\[ d = 4R \]
- Therefore, we can set up the equation:
\[ 4R = a \sqrt{2} \]

5. **Solve for the Radius (R)**:
- Rearranging the equation gives:
\[ R = \frac{a \sqrt{2}}{4} \]
- Substituting the value of a:
\[ R = \frac{330 \, \text{pm} \times \sqrt{2}}{4} \approx \frac{330 \, \text{pm} \times 1.414}{4} \approx \frac{466.5 \, \text{pm}}{4} \approx 116.625 \, \text{pm} \]

6. **Round the Result**:
- Rounding 116.625 pm gives approximately 117 pm.

### Final Answer:
The radius of the copper atom is approximately **117 pm**.