To solve the problem, we need to find the radius of an aluminum atom and the nearest neighbor distance in a face-centered cubic (FCC) unit cell given that the edge length (a) is 400 picometers (pm).
### Step-by-Step Solution:
1. **Identify the structure of the FCC unit cell**:
- In a face-centered cubic unit cell, atoms are located at each of the corners and at the center of each face of the cube.
2. **Understand the relationship between the edge length and the face diagonal**:
- The face diagonal (d_f) of the cube can be calculated using the formula:
\[
d_f = a\sqrt{2}
\]
- Here, \(a\) is the edge length of the cube.
3. **Calculate the face diagonal**:
- Given \(a = 400 \, \text{pm}\),
\[
d_f = 400 \, \text{pm} \times \sqrt{2} \approx 400 \, \text{pm} \times 1.414 \approx 565.7 \, \text{pm}
\]
4. **Determine the nearest neighbor distance**:
- In an FCC structure, the nearest neighbor distance (d) is half of the face diagonal:
\[
d = \frac{d_f}{2} = \frac{a\sqrt{2}}{2} = \frac{400 \, \text{pm} \times \sqrt{2}}{2} = 200 \, \text{pm} \times \sqrt{2} \approx 282.8 \, \text{pm}
\]
5. **Calculate the radius of the aluminum atom**:
- In an FCC unit cell, the nearest neighbor distance (d) is equal to \(4r\), where \(r\) is the radius of the atom. Thus:
\[
d = 2r \quad \text{(since each atom contributes half its radius to the nearest neighbor distance)}
\]
- Therefore, we can express the radius as:
\[
r = \frac{d}{2} = \frac{282.8 \, \text{pm}}{2} \approx 141.4 \, \text{pm}
\]
### Final Answers:
- **Radius of aluminum atom**: \(141.4 \, \text{pm}\)
- **Nearest neighbor distance**: \(282.8 \, \text{pm}\)
