An element crystallizes in face centred cubic lattice. Calculate the length of the side of the unit cell if the radius of atom is 200 pm.

Explanation

To solve the problem of finding the length of the side of the unit cell for an element crystallizing in a face-centered cubic (FCC) lattice, given the radius of the atom is 200 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
In a face-centered cubic lattice, the atoms are located at each of the corners and the centers of all the faces of the cube. The relationship between the radius (r) of the atom and the side length (a) of the unit cell can be derived from the geometry of the FCC structure.

2. **Use the Formula for FCC**:
The relationship between the radius (r) of the atom and the edge length (a) of the unit cell in an FCC lattice is given by:
\[ r = \frac{a}{2\sqrt{2}} \]
where \( r \) is the atomic radius and \( a \) is the length of the side of the unit cell.

3. **Rearrange the Formula**:
To find the side length \( a \), we can rearrange the formula:
\[ a = 2r\sqrt{2} \]

4. **Substitute the Given Radius**:
Now, substitute the given radius \( r = 200 \) pm into the equation:
\[ a = 2 \times 200 \, \text{pm} \times \sqrt{2} \]

5. **Calculate \( \sqrt{2} \)**:
The value of \( \sqrt{2} \) is approximately 1.414.

6. **Perform the Calculation**:
Now, calculate \( a \):
\[ a = 2 \times 200 \, \text{pm} \times 1.414 \]
\[ a = 400 \, \text{pm} \times 1.414 \]
\[ a \approx 565.6 \, \text{pm} \]

7. **Final Result**:
Therefore, the length of the side of the unit cell is approximately:
\[ a \approx 565.7 \, \text{pm} \]

### Final Answer:
The length of the side of the unit cell is **565.7 pm**.