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. (a) 77.07pm
B. (b) 154.14pm
C. (c) 251.7pm
D. (d) 89pm

Answer

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

### Step 1: Understand the Structure
In a diamond, carbon atoms occupy the face-centered cubic (FCC) lattice points and alternate tetrahedral voids. The FCC unit cell has atoms at the corners and the centers of the faces.

### Step 2: Calculate the Number of Atoms in the FCC Unit Cell
In an FCC unit cell:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom, totaling \( 8 \times \frac{1}{8} = 1 \) atom.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom, totaling \( 6 \times \frac{1}{2} = 3 \) atoms.

Thus, the total number of atoms in the FCC unit cell is \( 1 + 3 = 4 \) atoms.

### Step 3: Identify the Tetrahedral Voids
In addition to the atoms at the FCC lattice points, carbon atoms also occupy alternate tetrahedral voids. In an FCC unit cell, there are 8 tetrahedral voids, but since carbon occupies alternate ones, effectively there are 4 carbon atoms in the tetrahedral voids.

### Step 4: Relate the Edge Length to the Radius of Carbon Atom
The relationship between the edge length \( a \) of the unit cell and the radius \( r \) of the carbon atom in tetrahedral voids is given by the formula:

\[ \frac{\sqrt{3}}{4} a = 2r \]

Where \( r \) is the radius of the carbon atom.

### Step 5: Solve for the Radius
Given the edge length \( a = 356 \) pm, we can substitute this value into the equation:

\[ \frac{\sqrt{3}}{4} \times 356 = 2r \]

Calculating the left side:

\[ \frac{\sqrt{3}}{4} \times 356 \approx 154.14 \text{ pm} \]

Thus,

\[ 2r = 154.14 \text{ pm} \]

### Step 6: Calculate the Diameter
The diameter \( d \) of the carbon atom is twice the radius:

\[ d = 2r = 154.14 \text{ pm} \]

### Final Answer
The diameter of the carbon atom is approximately **154.14 pm**.

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