A metal crystallizes in fcc lattice and edge of the unit cell is 620 pm. The radius of metal atoms is

Options

A. 265.6 pm
B. 310 pm
C. 219.2 pm
D. 438.6 pm

Explanation

To find the radius of the metal atoms in a face-centered cubic (FCC) lattice with a unit cell edge length of 620 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
In an FCC lattice, atoms are located at each corner of the cube and at the center of each face. The atoms touch each other along the face diagonal.

2. **Identify the Edge Length**:
The edge length (a) of the unit cell is given as 620 pm.

3. **Calculate the Length of the Face Diagonal**:
The length of the face diagonal (d) can be calculated using the Pythagorean theorem. For a cube, the face diagonal can be expressed as:
\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]
Substituting the value of a:
\[ d = 620 \, \text{pm} \times \sqrt{2} \]

4. **Determine the Relationship Between the Face Diagonal and Atomic Radius**:
In the FCC structure, along the face diagonal, there are 4 atomic radii (r) from the corner atoms and the face-centered atom:
\[ d = 4r \]

5. **Set Up the Equation**:
From the above relationships, we can equate the two expressions:
\[ 4r = 620 \, \text{pm} \times \sqrt{2} \]

6. **Solve for the Radius (r)**:
Rearranging the equation to find r:
\[ r = \frac{620 \, \text{pm} \times \sqrt{2}}{4} \]

7. **Calculate the Value**:
First, calculate \(620 \times \sqrt{2}\):
\[ 620 \times \sqrt{2} \approx 620 \times 1.414 \approx 876.2 \, \text{pm} \]
Now divide by 4:
\[ r \approx \frac{876.2 \, \text{pm}}{4} \approx 219.05 \, \text{pm} \]

8. **Final Result**:
The radius of the metal atoms is approximately:
\[ r \approx 219.05 \, \text{pm} \]

### Summary:
The radius of the metal atoms in the FCC lattice with an edge length of 620 pm is approximately **219.05 pm**.