A metal crystallizes in foc lattice and the edge of the unit cell is 620 pm. The radius of the metal atom is:

Options

A. 265.5 pm
B. 310 pm
C. 219.2 pm
D. 438.6 pm

Explanation

To find the radius of a metal atom that crystallizes in a face-centered cubic (FCC) lattice with a given edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Lattice Structure**:
In a face-centered cubic lattice, atoms are located at each corner of the cube and at the center of each face. The relationship between the edge length (A) of the unit cell and the atomic radius (r) is given by the formula:
\[ r = \frac{A}{2\sqrt{2}} \]

2. **Identify the Given Data**:
The edge length (A) of the unit cell is provided as 620 pm (picometers).

3. **Substitute the Given Value into the Formula**:
We can substitute the value of A into the formula to find the radius (r):
\[ r = \frac{620 \text{ pm}}{2\sqrt{2}} \]

4. **Calculate the Denominator**:
First, calculate \(2\sqrt{2}\):
\[ 2\sqrt{2} \approx 2 \times 1.414 = 2.828 \]

5. **Perform the Division**:
Now, divide 620 pm by 2.828:
\[ r = \frac{620 \text{ pm}}{2.828} \approx 219.2 \text{ pm} \]

6. **Final Result**:
The radius of the metal atom is approximately 219.2 pm.

### Conclusion:
Thus, the radius of the metal atom that crystallizes in an FCC lattice with an edge length of 620 pm is **219.2 pm**.