The relation between atomic radius (r ) and the edge (a) of the unit cell are given below. Which is correctly matched.

Options

A. Simple cubic `r=(a)/(2)`
B. Face centred `r=0.3535a`
C. Body centred r = 0.433a
D. All of the above

Answer

To determine the correct relationships between atomic radius (r) and the edge length (a) of the unit cell for different types of cubic structures, we will analyze the simple cubic, face-centered cubic (FCC), and body-centered cubic (BCC) unit cells step by step.

### Step 1: Simple Cubic Unit Cell
1. **Understanding the Structure**: In a simple cubic unit cell, atoms are located only at the corners of the cube.
2. **Edge Length and Atomic Radius**: The edge length (a) is equal to twice the atomic radius (r) because the atoms at the corners touch each other along the edge.
\[ a = 2r \implies r = \frac{a}{2} \]
3. **Conclusion for Simple Cubic**: The relationship for the simple cubic unit cell is correctly given as \( r = \frac{a}{2} \).

### Step 2: Face-Centered Cubic Unit Cell
1. **Understanding the Structure**: In a face-centered cubic unit cell, atoms are located at each corner and at the center of each face of the cube.
2. **Calculating the Face Diagonal**: The face diagonal can be calculated using Pythagoras' theorem. The length of the face diagonal (d) is given by:
\[ d = a\sqrt{2} \]
3. **Atoms Along the Face Diagonal**: Along the face diagonal, there are 4 atomic radii (2 from each corner atom and 1 from the face-centered atom):
\[ d = 4r \]
4. **Setting the Equations Equal**: Set the two expressions for the face diagonal equal:
\[ a\sqrt{2} = 4r \implies r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}} \approx 0.3535a \]
5. **Conclusion for Face-Centered Cubic**: The relationship for the face-centered cubic unit cell is correctly given as \( r = \frac{a\sqrt{2}}{4} \).

### Step 3: Body-Centered Cubic Unit Cell
1. **Understanding the Structure**: In a body-centered cubic unit cell, atoms are located at each corner and one atom at the center of the cube.
2. **Calculating the Body Diagonal**: The body diagonal can be calculated as:
\[ d = a\sqrt{3} \]
3. **Atoms Along the Body Diagonal**: Along the body diagonal, there are 4 atomic radii (1 from the center atom and 1 from each corner atom):
\[ d = 4r \]
4. **Setting the Equations Equal**: Set the two expressions for the body diagonal equal:
\[ a\sqrt{3} = 4r \implies r = \frac{a\sqrt{3}}{4} \approx 0.433a \]
5. **Conclusion for Body-Centered Cubic**: The relationship for the body-centered cubic unit cell is correctly given as \( r = \frac{a\sqrt{3}}{4} \).

### Final Summary of Relationships
- Simple Cubic: \( r = \frac{a}{2} \)
- Face-Centered Cubic: \( r = \frac{a\sqrt{2}}{4} \approx 0.3535a \)
- Body-Centered Cubic: \( r = \frac{a\sqrt{3}}{4} \approx 0.433a \)