the edge length of the unit cells in terms of the radius of sphere constituting fcc ,bcc and simple cubic unit cells are respectively ………. .

Options

A. `2sqrt2r,"4r"/sqrt3 , 2r`
B. `(4r)/sqrt3, 2sqrt2r,2r`
C. `2r,2sqrt2r,(4r)/sqrt3`
D. `2r,(4r)/sqrt3,2sqrt2r`

Answer

To find the edge lengths of the unit cells in terms of the radius of the spheres constituting the face-centered cubic (FCC), body-centered cubic (BCC), and simple cubic unit cells, we can derive the relationships step by step.

### Step 1: Face-Centered Cubic (FCC) Unit Cell

In an FCC unit cell, the spheres are located at each corner and the centers of each face. The relationship between the edge length (a) and the atomic radius (R) can be derived from the diagonal of the face of the cube.

1. The face diagonal of the FCC unit cell can be expressed as:
\[ \text{Face diagonal} = a\sqrt{2} \]
2. The face diagonal also equals four times the radius of the atom (since there are two radii from each of the two atoms touching along the diagonal):
\[ \text{Face diagonal} = 4R \]
3. Setting these two expressions equal gives:
\[ a\sqrt{2} = 4R \]
4. Solving for a:
\[ a = \frac{4R}{\sqrt{2}} = 2\sqrt{2}R \]

### Step 2: Body-Centered Cubic (BCC) Unit Cell

In a BCC unit cell, there is one atom at each corner and one atom in the center of the cube. The relationship between the edge length (a) and the atomic radius (R) can be derived from the body diagonal.

1. The body diagonal of the BCC unit cell can be expressed as:
\[ \text{Body diagonal} = a\sqrt{3} \]
2. The body diagonal also equals four times the radius of the atom (since there are two radii from the corner atom and two from the center atom):
\[ \text{Body diagonal} = 4R \]
3. Setting these two expressions equal gives:
\[ a\sqrt{3} = 4R \]
4. Solving for a:
\[ a = \frac{4R}{\sqrt{3}} = \frac{4}{\sqrt{3}}R \]

### Step 3: Simple Cubic Unit Cell

In a simple cubic unit cell, there is one atom at each corner of the cube.

1. The edge length (a) is directly related to the atomic radius (R) since the edge length equals twice the radius:
\[ a = 2R \]

### Summary of Results

- For FCC: \( a = 2\sqrt{2}R \)
- For BCC: \( a = \frac{4}{\sqrt{3}}R \)
- For Simple Cubic: \( a = 2R \)