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 grey tin also crystallise in same way as diamond `(N_(A)=6xx 10^(23))`
The side length of diamond unit cell (in pm):
a. 154
b. 1422.63
c. 711.32
d. 355.66

Answer

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

### Step 1: Understand the structure
The diamond structure is based on a face-centered cubic (FCC) lattice. In this structure, each carbon atom is tetrahedrally coordinated with four other carbon atoms.

### Step 2: Identify the relationship between edge length and atomic radius
In an FCC lattice, the relationship between the edge length (a) and the radius (r) of the atoms is given by the formula:
\[ \frac{\sqrt{3}}{4} a = 2r \]
This formula accounts for the tetrahedral arrangement of atoms in the diamond structure.

### Step 3: Rearrange the formula to find edge length
From the formula, we can rearrange it to solve for the edge length (a):
\[ a = \frac{2r \cdot 4}{\sqrt{3}} = \frac{8r}{\sqrt{3}} \]

### Step 4: Substitute the bond length
The bond length between carbon atoms is given as 154 pm. Since this bond length corresponds to the distance between two carbon atoms, we can consider this as twice the radius (r):
\[ r = \frac{154 \text{ pm}}{2} = 77 \text{ pm} \]

### Step 5: Calculate the edge length
Now, substituting the value of r into the equation for a:
\[ a = \frac{8 \times 77 \text{ pm}}{\sqrt{3}} = \frac{616 \text{ pm}}{\sqrt{3}} \approx 355.66 \text{ pm} \]

### Step 6: Conclusion
Thus, the side length of the diamond unit cell is approximately 355.66 pm.

### Final Answer
The side length of the diamond unit cell is **d. 355.66 pm**.

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