If `a` is the length of unit cell, then which one is correct relationship ?

Options

A. For simple cubic lattice. Radius of metal atoms `= (a)/(2)`
B. For bcc lattice. Radius of metal atom `= (sqrt(3)a)/(4)`
C. For fcc lattice, Radius of metal atom `= (a)/(2sqrt(2))`
D. All of these

Answer

To solve the problem regarding the relationships between the length of the unit cell (denoted as `a`) and the radius of metal ions in different types of crystal lattices, we will analyze each type of lattice step by step.

### Step 1: Simple Cubic Lattice
1. In a simple cubic lattice, atoms are located at the corners of the cube.
2. The atoms at the corners touch each other along the edge of the cube.
3. If `R` is the radius of the metal ion, then the distance between the centers of two adjacent corner atoms is equal to the edge length `a`.
4. Therefore, we can write the equation:
\[ 2R = a \]
5. Rearranging gives us:
\[ R = \frac{a}{2} \]
6. This confirms that the relationship for the simple cubic lattice is correct.

### Step 2: Body-Centered Cubic (BCC) Lattice
1. In a BCC lattice, there are atoms at the corners and one atom at the center of the cube.
2. The atoms at the corners touch the atom at the body center along the body diagonal of the cube.
3. The length of the body diagonal can be expressed in terms of the edge length `a`:
\[ \text{Body Diagonal} = \sqrt{3}a \]
4. The body diagonal consists of two corner atoms (each contributing a radius `R`) and the body-centered atom (also contributing a radius `R`):
\[ 4R = \sqrt{3}a \]
5. Rearranging gives us:
\[ R = \frac{\sqrt{3}}{4}a \]
6. This confirms that the relationship for the BCC lattice is correct.

### Step 3: Face-Centered Cubic (FCC) Lattice
1. In an FCC lattice, atoms are located at the corners and at the centers of each face of the cube.
2. The atoms at the corners touch the atoms at the face centers along the face diagonal.
3. The length of the face diagonal can be expressed in terms of the edge length `a`:
\[ \text{Face Diagonal} = \sqrt{2}a \]
4. The face diagonal consists of four radii (two from corner atoms and two from face-centered atoms):
\[ 4R = \sqrt{2}a \]
5. Rearranging gives us:
\[ R = \frac{\sqrt{2}}{4}a = \frac{a}{2\sqrt{2}} \]
6. This confirms that the relationship for the FCC lattice is correct.

### Conclusion
All three relationships derived above are correct:
1. For Simple Cubic: \( R = \frac{a}{2} \)
2. For BCC: \( R = \frac{\sqrt{3}}{4}a \)
3. For FCC: \( R = \frac{a}{2\sqrt{2}} \)

Thus, the final answer is that all of the given relationships are correct.