An element crystallizes in fcc lattice having edge length `400 pm` Calculate the maximum diameter of an atom which can be place in interstitial site without distorting the structure.

Answer

To solve the problem of finding the maximum diameter of an atom that can be placed in an interstitial site of a face-centered cubic (FCC) lattice without distorting the structure, follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**:
- In an FCC lattice, atoms are located at the corners and the centers of each face of the cube.

2. **Identify the Edge Length**:
- Given the edge length \( a = 400 \, \text{pm} \).

3. **Determine the Radius of Atoms in FCC**:
- Let \( r_2 \) be the radius of the atoms in the FCC lattice.
- The relationship between the edge length \( a \) and the radius \( r_2 \) can be derived from the geometry of the FCC structure. The face diagonal \( d \) of the cube can be expressed as:
\[ d = 4r_2 \]
- The face diagonal can also be expressed in terms of the edge length \( a \) using the Pythagorean theorem:
\[ d = \sqrt{a^2 + a^2} = a\sqrt{2} \]

4. **Set the Two Expressions for the Face Diagonal Equal**:
- Equating the two expressions for the face diagonal:
\[ 4r_2 = a\sqrt{2} \]

5. **Solve for \( r_2 \)**:
- Rearranging gives:
\[ r_2 = \frac{a\sqrt{2}}{4} \]
- Substitute \( a = 400 \, \text{pm} \):
\[ r_2 = \frac{400 \times \sqrt{2}}{4} = 100\sqrt{2} \, \text{pm} \]

6. **Calculate \( r_2 \)**:
- Calculate \( r_2 \):
\[ r_2 \approx 100 \times 1.414 = 141.4 \, \text{pm} \]

7. **Determine the Radius of the Interstitial Atom**:
- For an octahedral void in an FCC structure, the relationship between the radius \( r_1 \) of the interstitial atom and the radius \( r_2 \) of the atoms in the lattice is given by:
\[ \frac{r_1}{r_2} = 0.414 \]
- Therefore, we can express \( r_1 \) as:
\[ r_1 = 0.414 r_2 \]

8. **Substitute \( r_2 \) into the Equation for \( r_1 \)**:
- Substitute \( r_2 \):
\[ r_1 = 0.414 \times 141.4 \, \text{pm} \approx 58.6 \, \text{pm} \]

9. **Calculate the Diameter of the Interstitial Atom**:
- The diameter \( d \) of the interstitial atom is:
\[ d = 2r_1 = 2 \times 58.6 \approx 117.2 \, \text{pm} \]

### Final Answer:
The maximum diameter of an atom that can be placed in an interstitial site without distorting the FCC structure is approximately **117.2 pm**.