Silver crystallises in a face centred cubic unit cell. Each side of the unit cell has a length of 400 pm. Calculate radius of the silver atoms (Assume that the atoms just touch each other on the diagonal across the face of the unit cell i.e. each face atom is touching four atoms)

Explanation

To solve the problem of calculating the radius of silver atoms in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step 1: Understand the FCC Structure
In a face-centered cubic unit cell, atoms are located at each corner of the cube and at the center of each face. The atoms at the corners are shared among adjacent unit cells, while the face-centered atoms belong entirely to the unit cell.

### Step 2: Identify the Relationship Between Side Length and Atomic Radius
In an FCC structure, the relationship between the edge length (a) of the unit cell and the atomic radius (r) can be expressed as:
\[ r = \frac{a}{2\sqrt{2}} \]

### Step 3: Substitute the Given Value
We are given the side length of the unit cell:
\[ a = 400 \text{ pm} \]
Now, we can substitute this value into the formula:
\[ r = \frac{400 \text{ pm}}{2\sqrt{2}} \]

### Step 4: Calculate the Value of \( \sqrt{2} \)
The value of \( \sqrt{2} \) is approximately 1.414. Therefore, we can rewrite the equation as:
\[ r = \frac{400 \text{ pm}}{2 \times 1.414} \]

### Step 5: Perform the Calculation
Calculating the denominator:
\[ 2 \times 1.414 \approx 2.828 \]
Now, substituting back into the equation:
\[ r = \frac{400 \text{ pm}}{2.828} \]

### Step 6: Final Calculation
Now we perform the division:
\[ r \approx 141.4 \text{ pm} \]

### Conclusion
The radius of the silver atoms is approximately:
\[ r \approx 141.4 \text{ pm} \]