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 mass of diamond unit cell is:

Options

A. ` 96 ` amu
B. `96 g`
C. `144 ` amu
D. `144 g`

Answer

To find the mass of the diamond unit cell, we will follow these steps:

### Step 1: Determine the structure of diamond
Diamond has a face-centered cubic (FCC) lattice structure. In this structure, carbon atoms are located at the corners and the face centers of the cube.

### Step 2: Calculate the number of carbon atoms in the FCC lattice
In an FCC lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating the total number of carbon atoms:
\[ \text{Number of atoms from corners} = 8 \times \frac{1}{8} = 1 \]
\[ \text{Number of atoms from face centers} = 6 \times \frac{1}{2} = 3 \]
\[ \text{Total number of carbon atoms in FCC} = 1 + 3 = 4 \]

### Step 3: Calculate the number of tetrahedral voids
In an FCC lattice, the number of tetrahedral voids is given by:
\[ \text{Number of tetrahedral voids} = 2 \times \text{Number of atoms in FCC} \]
Thus,
\[ \text{Number of tetrahedral voids} = 2 \times 4 = 8 \]

### Step 4: Determine how many tetrahedral voids are occupied
It is given that 50% of the tetrahedral voids are occupied by carbon atoms:
\[ \text{Number of carbon atoms in tetrahedral voids} = 50\% \times 8 = 4 \]

### Step 5: Calculate the total number of carbon atoms in the unit cell
The total number of carbon atoms in the diamond unit cell is the sum of the carbon atoms in the FCC lattice and those in the tetrahedral voids:
\[ \text{Total number of carbon atoms} = 4 + 4 = 8 \]

### Step 6: Calculate the mass of the unit cell
The mass of one carbon atom is 12 amu. Therefore, the mass of the diamond unit cell can be calculated as:
\[ \text{Mass of unit cell} = \text{Number of carbon atoms} \times \text{Mass of one carbon atom} \]
\[ \text{Mass of unit cell} = 8 \times 12 \text{ amu} = 96 \text{ amu} \]

### Final Answer:
The mass of the diamond unit cell is **96 amu**.

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