If the lattice parameter of `Si = 5.43 Å` and the mass of `Si` atom is `28.08 xx 1.66 xx10^(-27) kg`, the density of silicon in `kg m^(-3)` is (Given: Silicon has diamondcubic structure)

Options

A. `2330`
B. `1115`
C. `3445`
D. `1673`

Explanation

To find the density of silicon, we can follow these steps:

### Step 1: Identify the given values
- Lattice parameter (a) of silicon = 5.43 Å
- Mass of silicon atom = 28.08 × 1.66 × 10^(-27) kg

### Step 2: Convert the lattice parameter from Ångströms to meters
1 Å = 10^(-10) m, thus:
\[ a = 5.43 \, \text{Å} = 5.43 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the cubic unit cell is given by:
\[ V = a^3 \]
Substituting the value of \(a\):
\[ V = (5.43 \times 10^{-10})^3 \]
Calculating this gives:
\[ V = 1.6 \times 10^{-28} \, \text{m}^3 \]

### Step 4: Calculate the mass of the unit cell
In a diamond cubic structure, there are 8 atoms per unit cell. Therefore, the mass of the unit cell (M) is:
\[ M = 8 \times \text{mass of one silicon atom} \]
Substituting the mass of one silicon atom:
\[ M = 8 \times (28.08 \times 1.66 \times 10^{-27}) \]
Calculating this gives:
\[ M = 3.73 \times 10^{-25} \, \text{kg} \]

### Step 5: Calculate the density of silicon
Density (\(\rho\)) is given by the formula:
\[ \rho = \frac{\text{mass}}{\text{volume}} \]
Substituting the values we calculated:
\[ \rho = \frac{3.73 \times 10^{-25}}{1.6 \times 10^{-28}} \]
Calculating this gives:
\[ \rho \approx 2330 \, \text{kg/m}^3 \]

### Final Answer
The density of silicon is approximately **2330 kg/m³**.

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