The atomic radius for a body - centred cubic cell of lattice is 

Options

A. `(sqrt2)/(4)a`
B. `(sqrt3)/(4)a`
C. `(a)/(2)`
D. `(a)/(sqrt8)`

Answer

To find the atomic radius for a body-centered cubic (BCC) cell of lattice, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the BCC Structure**:
In a body-centered cubic lattice, there is one atom at each corner of the cube and one atom at the center of the cube. The atoms at the corners touch the atom at the center.

2. **Identify the Body Diagonal**:
The body diagonal of the cube connects two opposite corners of the cube and passes through the center atom. The length of the body diagonal can be expressed in terms of the atomic radius (r). In a BCC structure, the body diagonal is equal to four times the atomic radius:
\[ \text{Body Diagonal} = 4r \]

3. **Calculate the Length of the Body Diagonal**:
The body diagonal can also be calculated using the side length (a) of the cube. The formula for the body diagonal (d) in terms of the cube's side length is:
\[ d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

4. **Set the Two Expressions for the Body Diagonal Equal**:
Since both expressions represent the same body diagonal, we can set them equal to each other:
\[ 4r = a\sqrt{3} \]

5. **Solve for the Atomic Radius (r)**:
Rearranging the equation to solve for r gives:
\[ r = \frac{a\sqrt{3}}{4} \]

### Final Result:
The atomic radius for a body-centered cubic cell of lattice is:
\[ r = \frac{a\sqrt{3}}{4} \]