What is the critical radius ratio for the CsCl structure?

Answer

To find the critical radius ratio for the CsCl structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- In the CsCl structure, Cs⁺ ions are located at the body center of the cubic unit cell, while Cl⁻ ions are located at the corners of the cube.

2. **Determine the Geometry**:
- The distance between the body-centered Cs⁺ ion and a corner Cl⁻ ion can be expressed in terms of the lattice parameter (edge length) \( A \). The distance from the center of the cube to a corner is given by \( \frac{\sqrt{3}A}{2} \).

3. **Set Up the Equation**:
- The sum of the radii of the Cs⁺ and Cl⁻ ions must equal the distance between them:
\[ R_{Cs^+} + R_{Cl^-} = \frac{\sqrt{3}A}{2} \]

4. **Relate the Corner Ions**:
- Since there are 8 corners in the cubic unit cell and the distance between two corner Cl⁻ ions is equal to the edge length \( A \):
\[ 2R_{Cl^-} = A \implies R_{Cl^-} = \frac{A}{2} \]

5. **Substitute \( R_{Cl^-} \) into the Equation**:
- Substitute \( R_{Cl^-} \) into the distance equation:
\[ R_{Cs^+} + \frac{A}{2} = \frac{\sqrt{3}A}{2} \]

6. **Solve for \( R_{Cs^+} \)**:
- Rearranging gives:
\[ R_{Cs^+} = \frac{\sqrt{3}A}{2} - \frac{A}{2} = \frac{(\sqrt{3} - 1)A}{2} \]

7. **Calculate the Radius Ratio**:
- The critical radius ratio \( \frac{R_{Cs^+}}{R_{Cl^-}} \) can now be calculated:
\[ \frac{R_{Cs^+}}{R_{Cl^-}} = \frac{\frac{(\sqrt{3} - 1)A}{2}}{\frac{A}{2}} = \sqrt{3} - 1 \]

8. **Numerical Value**:
- Evaluating \( \sqrt{3} - 1 \) gives approximately \( 0.732 \).

### Final Answer:
The critical radius ratio for the CsCl structure is approximately **0.732**.

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