The lattice parameter (edgelength) for a body-centred cubic cell of atomic radius r ? is : 

Options

A. `sqrt8r`
B. `(sqrt3)/(4)r`
C. `(1)/(2sqrt2)r`
D. `(4r)/(sqrt3)`

Answer

To find the lattice parameter (edge length) for a body-centered cubic (BCC) cell in terms of the atomic radius \( r \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the BCC Structure**:
- In a body-centered cubic unit cell, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Identifying Nearest Neighbor Distance**:
- The nearest neighbor distance \( D \) in a BCC structure is the distance between the atom at the center and one of the corner atoms. This distance is equal to twice the atomic radius \( r \):
\[ D = 2r \]

3. **Using Geometry**:
- In the BCC unit cell, we can visualize a right triangle formed by the center atom and two corner atoms. The edge length of the cube is denoted by \( A \), and the distance \( D \) (the diagonal across the face of the cube) can be expressed using the Pythagorean theorem.

4. **Applying Pythagorean Theorem**:
- The diagonal \( AC \) (hypotenuse) of the triangle formed by the edge length \( A \) can be calculated as:
\[ AC = \sqrt{A^2 + A^2} = \sqrt{2}A \]
- The distance \( AB \) (the diagonal across the body of the cube) can be calculated as:
\[ AB = \sqrt{AC^2 + A^2} = \sqrt{(\sqrt{2}A)^2 + A^2} = \sqrt{2A^2 + A^2} = \sqrt{3}A \]

5. **Relating \( D \) and \( A \)**:
- The distance \( D \) is half of the body diagonal \( AB \):
\[ D = \frac{1}{2}AB = \frac{1}{2}(\sqrt{3}A) = \frac{\sqrt{3}}{2}A \]

6. **Expressing \( r \) in terms of \( A \)**:
- Since \( D = 2r \), we can set up the equation:
\[ 2r = \frac{\sqrt{3}}{2}A \]
- Rearranging gives:
\[ A = \frac{4r}{\sqrt{3}} \]

### Final Result
Thus, the lattice parameter (edge length) \( A \) for a body-centered cubic cell in terms of atomic radius \( r \) is:
\[ A = \frac{4r}{\sqrt{3}} \]