The number of atoms present per unit cell in simple, fcc and bcc are ………, ……….., and …………, respectively.

Answer

To find the number of atoms present per unit cell in simple cubic (SC), body-centered cubic (BCC), and face-centered cubic (FCC) structures, we can follow these steps:

### Step 1: Simple Cubic (SC)
1. In a simple cubic unit cell, atoms are located only at the corners.
2. There are 8 corners in a cubic unit cell.
3. Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell because it is shared among 8 adjacent unit cells.
4. Therefore, the total number of atoms in a simple cubic unit cell is:
\[ \text{Number of atoms} = 8 \times \frac{1}{8} = 1 \]

### Step 2: Body-Centered Cubic (BCC)
1. In a body-centered cubic unit cell, atoms are located at the corners and one atom is located at the center of the cube.
2. Again, there are 8 corner atoms, each contributing \( \frac{1}{8} \).
3. The body-centered atom contributes 1 full atom.
4. Therefore, the total number of atoms in a BCC unit cell is:
\[ \text{Number of atoms} = 8 \times \frac{1}{8} + 1 = 1 + 1 = 2 \]

### Step 3: Face-Centered Cubic (FCC)
1. In a face-centered cubic unit cell, atoms are located at the corners and at the centers of each face of the cube.
2. There are 8 corner atoms, each contributing \( \frac{1}{8} \).
3. There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) since each face-centered atom is shared between two adjacent unit cells.
4. Therefore, the total number of atoms in an FCC unit cell is:
\[ \text{Number of atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Final Answer
- The number of atoms per unit cell in:
- Simple Cubic (SC) = 1
- Body-Centered Cubic (BCC) = 2
- Face-Centered Cubic (FCC) = 4

Thus, the answer is:
- Simple Cubic: 1
- BCC: 2
- FCC: 4