An element crystallizes in face centred unit cell. What is the length of the side of the unit cell, if atomic radius of the element is 0.144 nm ?

Explanation

To find the length of the side of the unit cell (denoted as 'a') for an element that crystallizes in a face-centered cubic (FCC) structure with a given atomic radius (r) of 0.144 nm, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the FCC Structure**:
- In a face-centered cubic (FCC) unit cell, atoms are located at each of the eight corners and at the center of each of the six faces of the cube.

2. **Identifying the Relationship Between Atomic Radius and Edge Length**:
- The relationship between the atomic radius (r) and the edge length (a) in an FCC unit cell can be derived from the geometry of the cube.
- The face diagonal of the cube can be expressed in terms of the edge length 'a'. The face diagonal (AC) can be calculated using the Pythagorean theorem:
\[ AC = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

3. **Relating Atomic Radius to Face Diagonal**:
- The face diagonal (AC) in an FCC unit cell is also equal to four times the atomic radius (r) because it spans across two atomic radii from one corner atom to the face-centered atom and then to the opposite corner atom:
\[ AC = 4r \]

4. **Setting Up the Equation**:
- We can set the two expressions for the face diagonal equal to each other:
\[ 4r = a\sqrt{2} \]

5. **Solving for Edge Length (a)**:
- Rearranging the equation to solve for 'a':
\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

6. **Substituting the Given Radius**:
- Now, substitute the given atomic radius (r = 0.144 nm) into the equation:
\[ a = 2\sqrt{2} \times 0.144 \text{ nm} \]
- Calculate \( \sqrt{2} \) (approximately 1.414):
\[ a = 2 \times 1.414 \times 0.144 \text{ nm} \]

7. **Final Calculation**:
- Performing the multiplication:
\[ a \approx 2 \times 1.414 \times 0.144 \approx 0.407 \text{ nm} \]

### Conclusion:
The length of the side of the unit cell (a) is approximately **0.407 nm**.