Calculate the radius of Xe atom , If the edge of the unit cell (FCC) is 620 pm.

Options

A. 189.37 pm
B. 209.87 pm
C. 219.25 pm
D. 235.16 pm

Answer

To calculate the radius of a xenon (Xe) atom in a face-centered cubic (FCC) unit cell with an edge length of 620 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**: In a face-centered cubic (FCC) structure, atoms are located at each of the corners of the cube and at the center of each face. Each corner atom is shared among eight adjacent unit cells, and each face-centered atom is shared between two unit cells.

2. **Identify the Relationship Between Edge Length and Atomic Radius**: In an FCC unit cell, the atoms touch along the face diagonal. The relationship can be expressed as:
\[ \text{Face diagonal} = 4r \]
where \( r \) is the radius of the atom.

3. **Calculate the Face Diagonal**: The face diagonal \( d \) of a cube can be calculated using the edge length \( a \):
\[ d = \sqrt{2}a \]
Here, \( a = 620 \, \text{pm} \).

4. **Substituting the Edge Length**:
\[ d = \sqrt{2} \times 620 \, \text{pm} = 1.414 \times 620 \, \text{pm} \approx 874.6 \, \text{pm} \]

5. **Relate the Face Diagonal to the Atomic Radius**:
Since the face diagonal is equal to \( 4r \):
\[ 4r = 874.6 \, \text{pm} \]

6. **Solve for the Radius \( r \)**:
\[ r = \frac{874.6 \, \text{pm}}{4} \approx 218.65 \, \text{pm} \]

7. **Final Result**: Rounding to three significant figures, the radius of the xenon atom is approximately:
\[ r \approx 219 \, \text{pm} \]

### Summary:
The radius of the xenon atom in an FCC unit cell with an edge length of 620 pm is approximately **219 pm**.