An element crystallises in fcc structure. If the atomic radius is `2.2 Å` , what will be the edge length of the unit cell ?

Explanation

To find the edge length of a unit cell for an element that crystallizes in a face-centered cubic (FCC) structure with a given atomic radius, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The atomic radius (R) is given as \(2.2 \, \text{Å}\).
- We need to convert this into centimeters for consistency in units.
- \(1 \, \text{Å} = 10^{-8} \, \text{cm}\), so:
\[ R = 2.2 \, \text{Å} = 2.2 \times 10^{-8} \, \text{cm} \]

2. **Understand the Relationship in FCC Structure**:
- In a face-centered cubic (FCC) structure, the relationship between the edge length (A) and the atomic radius (R) is given by the formula:
\[ A = 2\sqrt{2}R \]
- Here, \(A\) is the edge length of the unit cell.

3. **Substitute the Value of R into the Formula**:
- Now, substitute \(R = 2.2 \times 10^{-8} \, \text{cm}\) into the formula:
\[ A = 2\sqrt{2} \times (2.2 \times 10^{-8} \, \text{cm}) \]

4. **Calculate \(2\sqrt{2}\)**:
- First, calculate \(\sqrt{2}\):
\[ \sqrt{2} \approx 1.414 \]
- Therefore:
\[ 2\sqrt{2} \approx 2 \times 1.414 = 2.828 \]

5. **Complete the Calculation**:
- Now, substitute this value back into the equation for A:
\[ A \approx 2.828 \times (2.2 \times 10^{-8} \, \text{cm}) \]
- Perform the multiplication:
\[ A \approx 6.22 \times 10^{-8} \, \text{cm} \]

6. **Final Result**:
- The edge length of the unit cell is:
\[ A \approx 6.22 \times 10^{-8} \, \text{cm} \]

### Summary:
The edge length of the unit cell for the element crystallizing in an FCC structure with an atomic radius of \(2.2 \, \text{Å}\) is approximately \(6.22 \times 10^{-8} \, \text{cm}\).