In a diamond, carbon atom occupy fcc lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356pm, then diameter of carbon atom is:

Options

A. 77.07pm
B. 154.14pm
C. 251.7pm
D. 89pm

Answer

To find the diameter of a carbon atom in a diamond structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**: In a diamond, carbon atoms occupy the face-centered cubic (FCC) lattice points and alternate tetrahedral voids.

2. **Identify the Relationship**: The relationship between the edge length (A) of the unit cell and the radius (R) of the carbon atom is given by:
\[ \frac{\sqrt{3}}{4} A = 2R \]
Here, \(2R\) represents the diameter of the carbon atom.

3. **Substitute the Given Edge Length**: The edge length of the unit cell is given as \(A = 356 \text{ pm}\). We will substitute this value into the equation.
\[ \frac{\sqrt{3}}{4} \times 356 = 2R \]

4. **Calculate \(R\)**: First, calculate \(\frac{\sqrt{3}}{4} \times 356\):
\[ R = \frac{\sqrt{3}}{8} \times 356 \]
Now, calculate \(\sqrt{3} \approx 1.732\):
\[ R = \frac{1.732}{8} \times 356 \approx 77.07 \text{ pm} \]

5. **Calculate the Diameter**: Since we need the diameter \(D\) of the carbon atom, we use:
\[ D = 2R = 2 \times 77.07 \text{ pm} \approx 154.14 \text{ pm} \]

6. **Final Answer**: The diameter of the carbon atom is approximately \(154.14 \text{ pm}\).