The regular three dimensional arrangement of points in a crystal is known as crystal lattice and the smallest repeating pattern in the lattice is called unit cell. The unit cells are characterised by the edge lengths a, b, c and the angles between them `alpha, beta and gamma` respectively. Based on this, there are seven crystal systems. In a cubic unit cell:
`a=b=c and alpha = beta=gamma=90^(@)` The number of points in simple, body centred and face centred cubic cells are 1,2 and 4 respectively In both the hcp and ccp of spheres, the number of tetrahedral voids per sphere is two while the octahedral voids is one.
In a face centred cubic cell, an atom at the face contributes to the unit cell

Options

A. 1 part
B. `1//2` part
C. `1//4` part
D. `1//8` part

Explanation

To solve the question regarding the contribution of an atom located at the face of a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Face-Centered Cubic Structure**:
- In a face-centered cubic (FCC) unit cell, atoms are located at each of the eight corners of the cube and at the center of each of the six faces.

2. **Counting the Atoms at the Corners**:
- Each corner atom is shared by 8 adjacent unit cells. Therefore, the contribution of one corner atom to a single unit cell is \( \frac{1}{8} \).
- Since there are 8 corners, the total contribution from the corner atoms is:
\[ 8 \times \frac{1}{8} = 1 \text{ atom} \]

3. **Counting the Atoms at the Faces**:
- Each face-centered atom is shared between 2 adjacent unit cells. Therefore, the contribution of one face-centered atom to a single unit cell is \( \frac{1}{2} \).
- Since there are 6 faces, the total contribution from the face-centered atoms is:
\[ 6 \times \frac{1}{2} = 3 \text{ atoms} \]

4. **Calculating Total Contribution**:
- The total number of atoms in a face-centered cubic unit cell can be calculated by adding the contributions from the corner and face-centered atoms:
\[ \text{Total atoms} = \text{Contribution from corners} + \text{Contribution from faces} = 1 + 3 = 4 \text{ atoms} \]

5. **Conclusion**:
- Therefore, the contribution of an atom located at the face of a face-centered cubic cell to the unit cell is \( \frac{1}{2} \).

### Final Answer:
The contribution of an atom at the face of a face-centered cubic cell to the unit cell is \( \frac{1}{2} \).

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