Distance between body centered atom and a comer atom in sodium is (where a = lattice constant)

Options

A. `asqrt3/2`
B. `asqrt3`
C. `asqrt3/4`
D. `asqrt2`

Answer

To find the distance between a 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) structure, there is one atom at the center of the cube and one atom at each of the eight corners. The lattice constant is denoted as 'a'.

### Step 2: Visualize the Geometry
Consider a cube where:
- The body-centered atom is at the center of the cube.
- The corner atom is located at one of the corners of the cube.

### Step 3: Determine the Distance
To find the distance between the body-centered atom and a corner atom, we can use the geometry of the cube. The distance can be calculated using the Pythagorean theorem in three dimensions.

1. The distance from the center of the cube to a corner can be calculated as:
\[ d = \sqrt{(a/2)^2 + (a/2)^2 + (a/2)^2} \]
Since the coordinates of the center atom are (0,0,0) and the coordinates of a corner atom (for example, (a/2, a/2, a/2)), we can simplify this.

2. This gives us:
\[ d = \sqrt{3 \cdot (a/2)^2} = \sqrt{3} \cdot \frac{a}{2} \]

### Step 4: Simplify the Expression
The distance from the body-centered atom to the corner atom can also be expressed as:
\[ d = \frac{\sqrt{3}}{2} a \]

### Step 5: Substitute the Value of 'a'
If we are given a specific value for the lattice constant 'a', for example, \( a = 4.225 \) Angstroms, we can substitute this value into our equation:
\[ d = \frac{\sqrt{3}}{2} \cdot 4.225 \approx 3.66 \text{ Angstroms} \]

### Final Answer
Thus, the distance between the body-centered atom and a corner atom in sodium is approximately \( 3.66 \) Angstroms.

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