Silver metal crysatllises with a face centred cubic lattice. The length of the unit cell is found to be 450 pm. Calulate the atomic radius.

Explanation

To solve the problem of calculating the atomic radius of silver metal, which crystallizes in a face-centered cubic (FCC) lattice with a unit cell length of 450 pm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Type of Lattice**:
- The problem states that silver crystallizes in a face-centered cubic (FCC) lattice.

2. **Recall 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:
\[ R = \frac{A}{2\sqrt{2}} \]
- Here, \( \sqrt{2} \) is approximately 1.414.

3. **Substitute the Given Edge Length**:
- The edge length (A) is given as 450 pm. We can substitute this value into the formula:
\[ R = \frac{450 \, \text{pm}}{2\sqrt{2}} \]

4. **Calculate the Value of \( \sqrt{2} \)**:
- Calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]

5. **Perform the Calculation**:
- Now substitute \( \sqrt{2} \) back into the equation:
\[ R = \frac{450 \, \text{pm}}{2 \times 1.414} \]
- Calculate the denominator:
\[ 2 \times 1.414 \approx 2.828 \]
- Now, divide:
\[ R = \frac{450 \, \text{pm}}{2.828} \approx 159.12 \, \text{pm} \]

6. **Final Result**:
- Therefore, the atomic radius \( R \) of silver is approximately:
\[ R \approx 159.12 \, \text{pm} \]

### Final Answer:
The atomic radius of silver is approximately **159.12 pm**.