CsBr has bcc lattice with the edge length `9.4` Å, the shortest inter- ionic distance in between `Cs^(+)` & `Br^(-)` is:

Options

A. `8.14` Å
B. `7.14` Å
C. `7.50` Å
D. `4.07`Å

Answer

To find the shortest inter-ionic distance between \( Cs^+ \) and \( Br^- \) in a body-centered cubic (BCC) lattice of CsBr with an edge length of \( 9.4 \) Å, follow these steps:

### Step 1: Understand the BCC Structure
In a BCC lattice, there are atoms located at the corners of the cube and one atom at the center of the cube. Each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell, and there are 8 corners, leading to a total contribution of 1 atom from the corners. The atom at the center contributes 1 full atom. Thus, there are 2 atoms per unit cell.

### Step 2: Identify the Body Diagonal
The shortest inter-ionic distance in a BCC lattice can be found by calculating half of the body diagonal of the cube. The body diagonal connects two opposite corners of the cube through the center atom.

### Step 3: Calculate the Body Diagonal
The formula for the body diagonal \( BD \) of a cube with edge length \( a \) is given by:
\[ BD = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]
Here, \( a \) is the edge length of the cube.

### Step 4: Substitute the Edge Length
Given that the edge length \( a = 9.4 \) Å, we can substitute this value into the body diagonal formula:
\[ BD = 9.4 \sqrt{3} \, \text{Å} \]

### Step 5: Calculate the Value
Now, calculate \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
Thus,
\[ BD = 9.4 \times 1.732 \approx 16.27 \, \text{Å} \]

### Step 6: Find the Shortest Inter-Ionic Distance
The shortest inter-ionic distance is half of the body diagonal:
\[ \text{Shortest Inter-Ionic Distance} = \frac{BD}{2} = \frac{16.27}{2} \approx 8.14 \, \text{Å} \]

### Final Answer
The shortest inter-ionic distance between \( Cs^+ \) and \( Br^- \) in CsBr is approximately \( 8.14 \) Å.

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