Calculate the density `("in kg m"^(-3))` of potassium having a bcc structure with nearest neighbour distance of `4.52Å`. (Given atomic weight of potassium is 39)

Answer

To calculate the density of potassium with a body-centered cubic (BCC) structure given the nearest neighbor distance, we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC structure, there are 2 atoms per unit cell. This is because there are 8 corner atoms (each contributing 1/8th of an atom) and 1 atom in the center.

### Step 2: Identify the Given Data
- Nearest neighbor distance (d) = 4.52 Å
- Atomic weight of potassium (M) = 39 g/mol

### Step 3: Convert Units
Convert the nearest neighbor distance from angstroms to meters:
\[ d = 4.52 \, \text{Å} = 4.52 \times 10^{-10} \, \text{m} \]

### Step 4: Relate Nearest Neighbor Distance to Edge Length
In a BCC structure, the nearest neighbor distance is related to the edge length (a) of the unit cell by the following relationship:
\[ d = \sqrt{3} \cdot \frac{a}{2} \]
Thus, we can rearrange this to find the edge length (a):
\[ a = \frac{2d}{\sqrt{3}} \]

### Step 5: Calculate the Edge Length
Substituting the value of d:
\[ a = \frac{2 \times 4.52 \times 10^{-10}}{\sqrt{3}} \]
\[ a \approx \frac{9.04 \times 10^{-10}}{1.732} \approx 5.22 \times 10^{-10} \, \text{m} \]

### Step 6: Calculate the Volume of the Unit Cell
The volume (V) of the unit cell can be calculated using the edge length:
\[ V = a^3 = (5.22 \times 10^{-10})^3 \approx 1.42 \times 10^{-28} \, \text{m}^3 \]

### Step 7: Convert Atomic Weight to kg
Convert the atomic weight of potassium from grams to kilograms:
\[ M = 39 \, \text{g/mol} = 0.039 \, \text{kg/mol} \]

### Step 8: Use the Density Formula
The density (ρ) can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{V \cdot N_A} \]
where:
- Z = number of atoms per unit cell = 2 (for BCC)
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)

### Step 9: Substitute the Values
Substituting the values into the density formula:
\[ \rho = \frac{2 \cdot 0.039}{1.42 \times 10^{-28} \cdot 6.022 \times 10^{23}} \]
\[ \rho = \frac{0.078}{8.55 \times 10^{-6}} \approx 911.45 \, \text{kg/m}^3 \]

### Final Answer
The density of potassium is approximately:
\[ \rho \approx 911.45 \, \text{kg/m}^3 \]