Body centred cubic and face centred cubic unit cells have `n_1 and n_2` effective number of atms. Which one the following `(n_1 , n_2)` combination is correct ?

Options

A. (4,1)
B. (4,2)
C. (1,4)
D. (2,4)

Answer

To solve the question regarding the effective number of atoms in Body Centred Cubic (BCC) and Face Centred Cubic (FCC) unit cells, we will calculate the values of \( n_1 \) and \( n_2 \) step by step.

### Step 1: Calculate \( n_1 \) for Body Centred Cubic (BCC)
- In a BCC unit cell, there are:
- 8 atoms located at the 8 corners of the cube.
- 1 atom located at the center of the cube.

- Contribution from corner atoms:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell because it is shared among 8 unit cells.
- Total contribution from 8 corner atoms:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]

- Contribution from the body-centered atom:
- The body-centered atom contributes fully to the unit cell.

- Therefore, the total effective number of atoms \( n_1 \) in BCC is:
\[ n_1 = 1 \text{ (from corners)} + 1 \text{ (from body center)} = 2 \]

### Step 2: Calculate \( n_2 \) for Face Centred Cubic (FCC)
- In an FCC unit cell, there are:
- 8 atoms located at the 8 corners of the cube.
- 6 atoms located at the center of each of the 6 faces of the cube.

- Contribution from corner atoms:
- Each corner atom contributes \( \frac{1}{8} \) of an atom.
- Total contribution from 8 corner atoms:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]

- Contribution from face-centered atoms:
- Each face-centered atom contributes \( \frac{1}{2} \) of an atom because it is shared between 2 unit cells.
- Total contribution from 6 face-centered atoms:
\[ \text{Contribution from faces} = 6 \times \frac{1}{2} = 3 \]

- Therefore, the total effective number of atoms \( n_2 \) in FCC is:
\[ n_2 = 1 \text{ (from corners)} + 3 \text{ (from faces)} = 4 \]

### Conclusion
- The effective number of atoms in BCC is \( n_1 = 2 \).
- The effective number of atoms in FCC is \( n_2 = 4 \).
- Thus, the correct combination of \( (n_1, n_2) \) is \( (2, 4) \).

### Final Answer
The correct combination of \( (n_1, n_2) \) is \( (2, 4) \).

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