An element with density 10 g `cm^(-3)` forms a cubic unit cell with edge length `3xx10^(-8)cm`. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g `mol^(-1)`.

Explanation

To determine the nature of the cubic unit cell for the given element, we will follow these steps:

### Step 1: Understand the given data
- Density (D) = 10 g/cm³
- Edge length (a) = 3 × 10^(-8) cm
- Atomic mass (M) = 81 g/mol

### Step 2: Use the formula for density
The formula for density (D) in terms of the number of atoms per unit cell (Z), atomic mass (M), Avogadro's number (N_A), and edge length (a) is given by:

\[ D = \frac{Z \cdot M}{N_A \cdot a^3} \]

Where:
- \(D\) = Density
- \(Z\) = Number of atoms per unit cell
- \(M\) = Molar mass of the element
- \(N_A\) = Avogadro's number (approximately \(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(a\) = Edge length of the unit cell

### Step 3: Rearrange the formula to find Z
We can rearrange the formula to solve for Z:

\[ Z = \frac{D \cdot N_A \cdot a^3}{M} \]

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

- \(D = 10 \, \text{g/cm}^3\)
- \(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)
- \(a = 3 \times 10^{-8} \, \text{cm}\)
- \(M = 81 \, \text{g/mol}\)

Calculating \(a^3\):

\[ a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} \, \text{cm}^3 \]

Now substituting into the equation for Z:

\[ Z = \frac{10 \cdot (6.022 \times 10^{23}) \cdot (27 \times 10^{-24})}{81} \]

### Step 5: Calculate Z
Calculating the numerator:

\[ 10 \cdot (6.022 \times 10^{23}) \cdot (27 \times 10^{-24}) = 10 \cdot 6.022 \cdot 27 \times 10^{-1} = 1625.94 \, \text{(approx)} \]

Now dividing by 81:

\[ Z = \frac{1625.94}{81} \approx 20.1 \]

### Step 6: Determine the nature of the unit cell
Since Z is approximately 2, we can conclude that the nature of the cubic unit cell is:

- **Body-Centered Cubic (BCC)**: A BCC unit cell has Z = 2.

### Final Answer
The nature of the cubic unit cell is **Body-Centered Cubic (BCC)**.

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