Sodium metal crystallizes in body centred cubic lattice with the cell edge, `a=4.29Å`. What is the radius of sodium atom?

Options

A. ` 1.8574` Å
B. `2.8574` Å
C. `3.8574` Å
D. `4.8574` Å

Answer

To find the radius of a sodium atom in a body-centered cubic (BCC) lattice, we can use the relationship between the edge length of the cube (a) and the radius (r) of the atom.

### Step-by-Step Solution:

1. **Understand the BCC Structure**: In a body-centered cubic lattice, there are atoms at each corner of the cube and one atom at the center of the cube. The relationship between the edge length (a) and the radius (r) of the atom can be derived from the geometry of the cube.

2. **Use the Formula for BCC**: The formula that relates the radius of the atom to the edge length in a BCC structure is:
\[ r = \frac{\sqrt{3}}{4} a \]
where \( r \) is the radius of the atom and \( a \) is the edge length of the cube.

3. **Substitute the Given Value**: We are given that the edge length \( a = 4.29 \, \text{Å} \). Substitute this value into the formula:
\[ r = \frac{\sqrt{3}}{4} \times 4.29 \]

4. **Calculate \( \sqrt{3} \)**: The value of \( \sqrt{3} \) is approximately 1.732. Now substitute this value into the equation:
\[ r = \frac{1.732}{4} \times 4.29 \]

5. **Perform the Multiplication**: Calculate the multiplication:
\[ r = 0.433 \times 4.29 \approx 1.8574 \, \text{Å} \]

6. **Final Result**: Therefore, the radius of the sodium atom is approximately:
\[ r \approx 1.8574 \, \text{Å} \]

### Conclusion:
The radius of the sodium atom in a body-centered cubic lattice is approximately **1.8574 Å**.