Sodium metal crystallises in body centred cubic lattic with cell edge `5.20Å` .What is the radius of sodium atom ?

Explanation

To find the radius of a sodium atom in a body-centered cubic (BCC) lattice with a given cell edge length of 5.20 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- Sodium metal crystallizes in a body-centered cubic (BCC) lattice. In a BCC lattice, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Understand the Body Diagonal**:
- In a BCC lattice, the body diagonal of the cube contains three atomic radii (from the corner atom to the center atom and back to the opposite corner atom). Thus, the body diagonal can be expressed in terms of the radius \( r \) of the sodium atom.

3. **Calculate the Length of the Body Diagonal**:
- The length of the body diagonal \( d \) of a cube with edge length \( a \) is given by the formula:
\[ d = \sqrt{3}a \]
- Here, \( a = 5.20 \, \text{Å} \).

4. **Set Up the Equation**:
- Since the body diagonal consists of three radii plus one radius at the center, we can express this as:
\[ d = 4r \]
- Therefore, we have:
\[ \sqrt{3}a = 4r \]

5. **Solve for the Radius \( r \)**:
- Rearranging the equation gives:
\[ r = \frac{\sqrt{3}a}{4} \]

6. **Substitute the Value of \( a \)**:
- Substitute \( a = 5.20 \, \text{Å} \) into the equation:
\[ r = \frac{\sqrt{3} \times 5.20}{4} \]

7. **Calculate the Radius**:
- First, calculate \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
- Now substitute and calculate:
\[ r = \frac{1.732 \times 5.20}{4} \approx \frac{8.9856}{4} \approx 2.2464 \, \text{Å} \]
- Rounding off gives:
\[ r \approx 2.25 \, \text{Å} \]

### Final Answer:
The radius of the sodium atom is approximately **2.25 Å**.