In diamond structure ,carbon atoms form fcc lattic and `50%` tetrahedral voids occupied by acrbon atoms . Evergy carbon atoms is surrounded tetrachedral by four carbon atom with bond length 154 pm . Germanium , silicon and tin also crystallise in same way as diamond `(N_(A)=6xx 10^(23))`
The side length of diamond unit cell is `("in pm")`:

Options

A. 154
B. 1422.63
C. 711.32
D. 355.66

Answer

To find the side length (edge length) of the diamond unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- In a diamond structure, carbon atoms form a face-centered cubic (FCC) lattice.
- 50% of the tetrahedral voids in the FCC lattice are occupied by carbon atoms.

2. **Bond Length Information**:
- The bond length between two carbon atoms in diamond is given as 154 pm.
- This bond length represents the distance between the centers of two adjacent carbon atoms.

3. **Relationship Between Bond Length and Atomic Radius**:
- Since the bond length (distance between two atoms) is 2r, we can express the atomic radius (r) as:
\[ 2r = 154 \text{ pm} \implies r = \frac{154}{2} = 77 \text{ pm} \]

4. **Determine the Relationship Between Edge Length and Atomic Radius**:
- For diamond, the relationship between the edge length (a) of the unit cell and the atomic radius (r) is given by:
\[ \frac{\sqrt{3}a}{4} = 2r \]
- Rearranging this gives:
\[ a = \frac{8r}{\sqrt{3}} \]

5. **Substituting the Radius**:
- Now substitute the value of r into the equation:
\[ a = \frac{8 \times 77 \text{ pm}}{\sqrt{3}} \]

6. **Calculate the Edge Length**:
- First, calculate \(8 \times 77\):
\[ 8 \times 77 = 616 \text{ pm} \]
- Next, divide by \(\sqrt{3} \approx 1.732\):
\[ a = \frac{616 \text{ pm}}{1.732} \approx 355.66 \text{ pm} \]

7. **Final Result**:
- The edge length of the diamond unit cell is approximately:
\[ a \approx 355.66 \text{ pm} \]

### Conclusion:
The side length of the diamond unit cell is **355.66 pm**.

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