The maximum proportion of available volume that can be filled by hard sphere in diamond is

Options

A. 0.52
B. 0.34
C. 0.32
D. 0.68

Answer

To find the maximum proportion of available volume that can be filled by hard spheres in diamond, we will follow these steps:

### Step 1: Determine the number of atoms in one unit cell of diamond
Diamond forms a face-centered cubic (FCC) lattice. In an FCC lattice:
- There are 8 atoms located at the corners, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 faces, each contributing \( \frac{1}{2} \) of an atom.

Calculating the contribution:
- Contribution from corners: \( 8 \times \frac{1}{8} = 1 \)
- Contribution from faces: \( 6 \times \frac{1}{2} = 3 \)

In addition, diamond has carbon atoms located in half of the tetrahedral voids. In an FCC lattice, there are 8 tetrahedral voids, so:
- Contribution from tetrahedral voids: \( \frac{8}{2} = 4 \)

Total number of atoms in one unit cell of diamond:
\[ 1 + 3 + 4 = 8 \text{ atoms} \]

### Step 2: Calculate the radius of each atom in diamond
The radius of each atom in diamond can be expressed as:
\[ r = \frac{\sqrt{3}a}{8} \]
where \( a \) is the edge length of the unit cell.

### Step 3: Calculate the volume of one atom
The volume \( V \) of one atom (considered as a sphere) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the expression for \( r \):
\[ V = \frac{4}{3} \pi \left(\frac{\sqrt{3}a}{8}\right)^3 \]
Calculating this:
\[ V = \frac{4}{3} \pi \cdot \frac{3\sqrt{3}a^3}{512} = \frac{4 \cdot 3\sqrt{3} \pi a^3}{1536} = \frac{12\sqrt{3} \pi a^3}{1536} = \frac{\sqrt{3} \pi a^3}{128} \]

### Step 4: Calculate the total volume of 8 atoms
Since there are 8 atoms in the unit cell, the total volume occupied by the atoms is:
\[ \text{Total Volume} = 8 \times \frac{\sqrt{3} \pi a^3}{128} = \frac{8\sqrt{3} \pi a^3}{128} = \frac{\sqrt{3} \pi a^3}{16} \]

### Step 5: Calculate the volume of the unit cell
The volume of the unit cell is given by:
\[ \text{Volume of unit cell} = a^3 \]

### Step 6: Calculate the fraction of volume occupied by the atoms
The fraction of the volume occupied by the atoms is:
\[ \text{Fraction} = \frac{\text{Total Volume of atoms}}{\text{Volume of unit cell}} = \frac{\frac{\sqrt{3} \pi a^3}{16}}{a^3} \]
This simplifies to:
\[ \text{Fraction} = \frac{\sqrt{3} \pi}{16} \]

### Step 7: Numerical approximation
Calculating the numerical value:
\[ \frac{\sqrt{3} \pi}{16} \approx 0.34 \]

Thus, the maximum proportion of available volume that can be filled by hard spheres in diamond is approximately **0.34**.

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