The distance between the body centred atom and a corner atom in sodium `(a = 4.225 Å)` is

Options

A. `3.66 Å`
B. `3.17 Å`
C. `2.99 Å`
D. `2.54 Å`

Answer

To find the distance between the body-centered atom and a corner atom in sodium, we can follow these steps:

### Step 1: Understand the Structure
In a body-centered cubic (BCC) lattice, there is one atom at each corner of the cube and one atom at the center of the cube. The length of the side of the cube is denoted as \( a \).

### Step 2: Identify the Distance Formula
The distance between the body-centered atom (located at the center of the cube) and a corner atom can be found using the 3D distance formula. In a BCC structure, the distance from the center of the cube to a corner atom can be derived from the geometry of the cube.

### Step 3: Calculate the Diagonal Distance
The distance from the center of the cube to a corner atom can be calculated using the formula for the diagonal of a cube:
- The distance from the center to a corner atom is given by \( \frac{\sqrt{3}}{2}a \).

### Step 4: Substitute the Value of \( a \)
Given that \( a = 4.225 \, \text{Å} \):
\[ \text{Distance} = \frac{\sqrt{3}}{2} \times 4.225 \, \text{Å} \]

### Step 5: Calculate the Numerical Value
Now, we calculate the numerical value:
\[ \text{Distance} = \frac{1.732}{2} \times 4.225 \, \text{Å} \approx 0.866 \times 4.225 \, \text{Å} \approx 3.66 \, \text{Å} \]

### Final Answer
The distance between the body-centered atom and a corner atom in sodium is approximately \( 3.66 \, \text{Å} \).

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