Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately :

Options

A. 3.22 Å
B. 5.72 Å
C. 0.93 Å
D. 1.86 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 unit cell and the radius of the atom.

### Step-by-Step Solution:

1. **Identify the Structure**:
Sodium crystallizes in a body-centered cubic (BCC) lattice. In a BCC lattice, there is one atom at the center of the cube and 8 atoms at the corners (each corner atom contributes 1/8 of an atom to the unit cell).

2. **Understand the Geometry**:
In a BCC unit cell, the atoms along the body diagonal touch each other. The body diagonal can be calculated using the edge length (A) of the unit cell.

3. **Calculate the Body Diagonal**:
The body diagonal (d) of a cube with edge length A is given by the formula:
\[ d = \sqrt{3} \times A \]

4. **Relate the Body Diagonal to Atomic Radius**:
In a BCC structure, the body diagonal is equal to four times the radius (R) of the sodium atom (since there are two radii from the corner atoms and one radius from the center atom):
\[ d = 4R \]

5. **Combine the Equations**:
Setting the two expressions for the body diagonal equal gives:
\[ \sqrt{3} \times A = 4R \]

6. **Solve for the Radius (R)**:
Rearranging the equation to solve for R:
\[ R = \frac{\sqrt{3} \times A}{4} \]

7. **Substitute the Given Edge Length**:
Given that the edge length (A) is 4.29 Å:
\[ R = \frac{\sqrt{3} \times 4.29}{4} \]

8. **Calculate the Radius**:
First, calculate \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
Now substitute this value in:
\[ R = \frac{1.732 \times 4.29}{4} \approx \frac{7.42668}{4} \approx 1.857 \]

9. **Round the Result**:
Rounding off gives:
\[ R \approx 1.86 \, \text{Å} \]

### Final Answer:
The radius of the sodium atom is approximately **1.86 Å**.