The radius of an ion in a body-centred cube of edge a is:

Options

A. `a/2`
B. `(sqrt(2)a)/4`
C. `(sqrt(3)a)/4`
D. None of these

Answer

To find the radius of an ion in a body-centered cubic (BCC) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure of BCC:**
In a body-centered cubic (BCC) 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. **Identify the Relationship Between Edge Length and Atomic Radius:**
The edge length of the cube is denoted as \( a \). In a BCC structure, the body diagonal connects two corner atoms and the atom at the center. The body diagonal can be expressed in terms of the edge length \( a \).

3. **Calculate the Length of the Body Diagonal:**
The length of the body diagonal \( d \) of a cube can be calculated using the formula:
\[ d = \sqrt{3}a \]

4. **Relate the Body Diagonal to Atomic Radius:**
In a BCC unit cell, the body diagonal is equal to the sum of the diameters of the two corner atoms and the diameter of the center atom. Since the diameter is twice the radius, we can express this as:
\[ d = 4R \]
where \( R \) is the radius of the atom.

5. **Set Up the Equation:**
From the above relationships, we have:
\[ \sqrt{3}a = 4R \]

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

### Final Answer:
The radius of an ion in a body-centered cubic structure is:
\[ R = \frac{\sqrt{3}a}{4} \]