Silver crystallizes in face-centred cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms).

Explanation

To calculate the radius of the silver atom in a face-centered cubic (FCC) unit cell, follow these steps:

### Step 1: Understand the FCC structure
In a face-centered cubic unit cell, atoms are located at each of the corners of the cube and at the center of each face. The atoms at the corners and the face centers touch each other along the face diagonal.

### Step 2: Determine the relationship between edge length and atomic radius
For an FCC unit cell, the relationship between the edge length (a) and the atomic radius (r) is given by the formula:
\[ 4r = \sqrt{2}a \]
This is because the face diagonal of the cube can be expressed in terms of the atomic radius.

### Step 3: Rearrange the formula to find the radius
To find the radius (r), rearrange the formula:
\[ r = \frac{a}{4\sqrt{2}} \]

### Step 4: Substitute the given edge length
The edge length (a) is given as 400 pm (picometers). Substitute this value into the equation:
\[ r = \frac{400 \, \text{pm}}{4\sqrt{2}} \]

### Step 5: Calculate the value of r
First, calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]
Now substitute this value:
\[ r = \frac{400 \, \text{pm}}{4 \times 1.414} \]
\[ r = \frac{400 \, \text{pm}}{5.656} \]
\[ r \approx 70.71 \, \text{pm} \]

### Step 6: Final calculation
Now, simplify the calculation:
\[ r \approx 70.71 \, \text{pm} \]

### Conclusion
The radius of the silver atom is approximately **70.71 pm**.

---