In a CsCl structure, if edge length is a, then distance between one Cs atom and one Cl atoms is

Options

A. `(a sqrt(3))/(2)`
B. `(a sqrt(3))/(4)`
C. `a sqrt(2)`
D. `(a)/(sqrt(2))`

Answer

To find the distance between one Cs atom and one Cl atom in a CsCl structure where the edge length is \( a \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the CsCl Structure**:
- In the CsCl crystal structure, cesium ions (Cs\(^+\)) are located at the body center of the cubic unit cell, while chloride ions (Cl\(^-\)) are located at the corners of the cube.

2. **Identifying the Positions**:
- The body center position of the Cs atom can be represented as \( (a/2, a/2, a/2) \).
- The corner positions of the Cl atoms can be represented as \( (0, 0, 0) \), \( (0, 0, a) \), \( (0, a, 0) \), \( (0, a, a) \), \( (a, 0, 0) \), \( (a, 0, a) \), \( (a, a, 0) \), and \( (a, a, a) \).

3. **Calculating the Distance**:
- To find the distance between the Cs atom at the body center and one of the Cl atoms at the corner, we can use the distance formula.
- The distance \( d \) between the Cs atom at \( (a/2, a/2, a/2) \) and a Cl atom at \( (0, 0, 0) \) is calculated as follows:
\[ d = \sqrt{(a/2 - 0)^2 + (a/2 - 0)^2 + (a/2 - 0)^2} \]
\[ d = \sqrt{(a/2)^2 + (a/2)^2 + (a/2)^2} \]
\[ d = \sqrt{3 \times (a/2)^2} \]
\[ d = \sqrt{3} \cdot \frac{a}{2} \]
\[ d = \frac{\sqrt{3}a}{2} \]

4. **Final Result**:
- Therefore, the distance between one Cs atom and one Cl atom in a CsCl structure is \( \frac{\sqrt{3}a}{2} \).