Diamond has face-centred cubic lattice. There are two atoms at `(0, 0, 0)` and `((a)/(4), (a)/(4), (a)/(4))` coordinates. The ratio of the carbon-carbon bond distance to the edge of the unit cell is

Options

A. `sqrt((3)/(16))`
B. `sqrt((1)/(4))`
C. `(1)/(4)`
D. `(1)/(sqrt2)`

Explanation

To find the ratio of the carbon-carbon bond distance to the edge of the unit cell in diamond, we can follow these steps:

### Step 1: Understand the Lattice Structure
Diamond has a face-centered cubic (FCC) lattice structure. In this structure, there are two carbon atoms located at the coordinates:
- Atom A at (0, 0, 0)
- Atom B at \((\frac{a}{4}, \frac{a}{4}, \frac{a}{4})\)

### Step 2: Identify the Edge Length
Let the edge length of the unit cell be denoted as \(a\).

### Step 3: Calculate the Distance Between the Two Atoms
To find the distance between the two carbon atoms (A and B), we can use the distance formula in three-dimensional space:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of atoms A and B:
- \(x_1 = 0, y_1 = 0, z_1 = 0\) (for atom A)
- \(x_2 = \frac{a}{4}, y_2 = \frac{a}{4}, z_2 = \frac{a}{4}\) (for atom B)

The distance \(d\) becomes:
\[ d = \sqrt{\left(\frac{a}{4} - 0\right)^2 + \left(\frac{a}{4} - 0\right)^2 + \left(\frac{a}{4} - 0\right)^2} \]
\[ = \sqrt{\left(\frac{a}{4}\right)^2 + \left(\frac{a}{4}\right)^2 + \left(\frac{a}{4}\right)^2} \]
\[ = \sqrt{3 \left(\frac{a}{4}\right)^2} \]
\[ = \frac{a}{4} \sqrt{3} \]

### Step 4: Find the Ratio of Carbon-Carbon Bond Distance to Edge Length
Now, we need to find the ratio of the carbon-carbon bond distance to the edge length \(a\):
\[ \text{Ratio} = \frac{\text{Carbon-Carbon Bond Distance}}{\text{Edge Length}} = \frac{\frac{a}{4} \sqrt{3}}{a} \]
\[ = \frac{\sqrt{3}}{4} \]

### Step 5: Final Result
Thus, the ratio of the carbon-carbon bond distance to the edge of the unit cell is:
\[ \frac{\sqrt{3}}{4} \]