An element (at. mass = 60) having face centred cubic unit cell has a density of 6.23g `cm^(-3)`. What is the edge length of the unit cell? (Avogadro's constant = `6.023 xx 10^(23) mol^(-1)`)

Explanation

To find the edge length of a face-centered cubic (FCC) unit cell given the atomic mass, density, and Avogadro's number, we can follow these steps:

### Step 1: Identify the known values
- Atomic mass (m) = 60 g/mol
- Density (D) = 6.23 g/cm³
- Avogadro's number (Nₐ) = 6.023 x 10²³ mol⁻¹
- For FCC, the number of atoms per unit cell (Z) = 4

### Step 2: Use the formula for density
The formula relating density (D), the number of atoms per unit cell (Z), atomic mass (m), and edge length (a) is given by:

\[ D = \frac{Z \cdot m}{a^3 \cdot N_a} \]

### Step 3: Rearrange the formula to find \(a^3\)
Rearranging the formula to solve for \(a^3\):

\[ a^3 = \frac{Z \cdot m}{D \cdot N_a} \]

### Step 4: Substitute the known values into the equation
Substituting the values into the equation:

\[ a^3 = \frac{4 \cdot 60 \, \text{g/mol}}{6.23 \, \text{g/cm}^3 \cdot 6.023 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 5: Calculate \(a^3\)
Calculating the numerator:

\[ 4 \cdot 60 = 240 \, \text{g/mol} \]

Now calculating the denominator:

\[ 6.23 \cdot 6.023 \times 10^{23} \approx 3.75 \times 10^{24} \, \text{g/cm}^3 \]

Now substituting back into the equation:

\[ a^3 = \frac{240}{3.75 \times 10^{24}} \approx 6.4 \times 10^{-23} \, \text{cm}^3 \]

### Step 6: Calculate the edge length \(a\)
Now, take the cube root to find \(a\):

\[ a = (6.4 \times 10^{-23})^{1/3} \approx 4.0 \times 10^{-8} \, \text{cm} \]

### Final Result
Thus, the edge length of the unit cell is approximately:

\[ a \approx 4.0 \times 10^{-8} \, \text{cm} \]

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