The density of a face centred cubic element (atomic mass = 40 ) is 4.25 gm `cm^(-3)`, calculate the edge length of the unit cell.

Explanation

To solve the problem of calculating the edge length of a face-centered cubic (FCC) unit cell given its density and atomic mass, we can follow these steps:

### Step 1: Identify the given data
- Atomic mass (m) = 40 g/mol
- Density (ρ) = 4.25 g/cm³
- For a face-centered cubic unit cell, the number of atoms per unit cell (Z) = 4
- Avogadro's number (Na) = 6.022 × 10²³ atoms/mol

### Step 2: Write the formula for density
The formula for density in terms of the unit cell parameters is given by:
\[ \rho = \frac{Z \cdot m}{a^3 \cdot N_a} \]
Where:
- ρ = density
- Z = number of atoms per unit cell
- m = atomic mass
- a = edge length of the unit cell
- Na = Avogadro's number

### Step 3: Rearrange the formula to solve for \(a^3\)
From the density formula, we can rearrange it to find \(a^3\):
\[ a^3 = \frac{Z \cdot m}{\rho \cdot N_a} \]

### Step 4: Substitute the known values into the equation
Substituting the values into the equation:
- Z = 4
- m = 40 g/mol
- ρ = 4.25 g/cm³
- Na = 6.022 × 10²³ atoms/mol

So we have:
\[ a^3 = \frac{4 \cdot 40}{4.25 \cdot 6.022 \times 10^{23}} \]

### Step 5: Calculate \(a^3\)
Calculating the numerator:
\[ 4 \cdot 40 = 160 \]

Calculating the denominator:
\[ 4.25 \cdot 6.022 \times 10^{23} \approx 2.558 \times 10^{24} \]

Now substituting these values:
\[ a^3 = \frac{160}{2.558 \times 10^{24}} \approx 6.25 \times 10^{-23} \text{ cm}^3 \]

### Step 6: Calculate \(a\) by taking the cube root of \(a^3\)
To find \(a\), we take the cube root:
\[ a = \sqrt[3]{6.25 \times 10^{-23}} \approx 3.94 \times 10^{-8} \text{ cm} \]

### Step 7: Round the answer
Rounding the answer gives:
\[ a \approx 4.0 \times 10^{-8} \text{ cm} \]

### Final Answer
The edge length of the unit cell is approximately \(4.0 \times 10^{-8} \text{ cm}\).

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