A metal crystallize in a body centered cubic lattice (bcc) with the edge of the unit cell `5.2Å`. The distance between the two nearest neighour is

Options

A. `10.4 Å`
B. `4.5Å`
C. `5.2Å`
D. `9.0Å`

Answer

To find the distance between two nearest neighbors in a body-centered cubic (BCC) lattice with a unit cell edge length of \(5.2 \, \text{Å}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the BCC Structure**:
In a BCC lattice, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Identify Nearest Neighbors**:
The nearest neighbors in a BCC lattice are the corner atom and the atom at the body center. The distance between these two atoms is what we need to calculate.

3. **Determine the Relationship**:
The distance between the nearest neighbors can be expressed in terms of the atomic radius \(R\). In the BCC structure, the distance between the nearest neighbors is \(2R\).

4. **Use the Body Diagonal**:
The body diagonal of the cube can be calculated using the edge length \(A\). The body diagonal \(d\) can be expressed as:
\[ d = \sqrt{A^2 + A^2 + A^2} = \sqrt{3A^2} = A\sqrt{3} \]

5. **Relate the Body Diagonal to Atomic Radii**:
The body diagonal in the BCC structure is also equal to the distance across the body center atom and two corner atoms, which is \(4R\). Therefore, we can set up the equation:
\[ A\sqrt{3} = 4R \]

6. **Solve for \(R\)**:
Rearranging the equation gives:
\[ R = \frac{A\sqrt{3}}{4} \]

7. **Substitute the Given Edge Length**:
Now, substitute \(A = 5.2 \, \text{Å}\):
\[ R = \frac{5.2 \sqrt{3}}{4} \]

8. **Calculate \(R\)**:
First, calculate \(\sqrt{3} \approx 1.732\):
\[ R = \frac{5.2 \times 1.732}{4} \approx \frac{9.0016}{4} \approx 2.2504 \, \text{Å} \]

9. **Find the Distance Between Nearest Neighbors**:
The distance between the two nearest neighbors is \(2R\):
\[ 2R = 2 \times 2.2504 \approx 4.5008 \, \text{Å} \]

10. **Final Answer**:
Rounding off gives us the distance between the two nearest neighbors as approximately \(4.5 \, \text{Å}\).

### Conclusion:
The distance between the two nearest neighbors in the BCC lattice is approximately \(4.5 \, \text{Å}\).

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