At room temperature, Polonium cryatsillises in Cubic primitive cell. If edge length is `3.0Å`, calculate the theoretical density of Po. (Atomic wt of Po = 207g)

Options

A. 25/3 amu/`Å^(3)`
B. 23/3 amu/`Å^(3)`
C. 21/3 amu/`Å^(3)`
D. 27/3 amu/`Å^(3)`

Answer

To calculate the theoretical density of Polonium (Po) crystallizing in a cubic primitive cell, we will follow these steps:

### Step 1: Understand the structure
Polonium crystallizes in a simple cubic unit cell. In a simple cubic structure, there is one atom per unit cell, as each of the 8 corners of the cube contributes 1/8 of an atom (8 corners x 1/8 = 1 atom).

### Step 2: Identify given values
- Edge length (A) = 3.0 Å (angstrom)
- Atomic weight (M) of Polonium = 207 g/mol

### Step 3: Convert atomic weight to atomic mass units (amu)
1 amu = 1 g/mol, so:
- M = 207 g/mol = 207 amu

### Step 4: Calculate the volume of the unit cell
The volume (V) of the cubic unit cell can be calculated using the formula:
\[ V = A^3 \]
Substituting the value of A:
\[ V = (3.0 \, \text{Å})^3 = 27.0 \, \text{Å}^3 \]

### Step 5: Calculate the density using the formula
The theoretical density (D) can be calculated using the formula:
\[ D = \frac{Z \times M}{N_A \times V} \]
Where:
- \( Z \) = number of atoms per unit cell = 1 (for simple cubic)
- \( M \) = mass of the atom in amu = 207 amu
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( V \) = volume of the unit cell in ų = 27.0 ų

Substituting the values:
\[ D = \frac{1 \times 207 \, \text{amu}}{6.022 \times 10^{23} \, \text{mol}^{-1} \times 27.0 \, \text{Å}^3} \]

### Step 6: Simplify the equation
Since we want the density in amu/ų, we can ignore \( N_A \) in this case:
\[ D = \frac{207}{27.0} \]
Calculating this gives:
\[ D = 7.67 \, \text{amu/Å}^3 \]

### Final Result
The theoretical density of Polonium is approximately \( 7.67 \, \text{amu/Å}^3 \).

---