Potassium has a bcc structure with nearest neighour distance `4.52 Å`its atomic weight is `39` its density (in kg `m^(-3)`) will be

Options

A. `454`
B. `804`
C. `852`
D. `908`

Explanation

To find the density of potassium with a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Understand the given data
- Nearest neighbor distance (D) = 4.52 Å
- Atomic weight (M) = 39 g/mol

### Step 2: Relate the nearest neighbor distance to the atomic radius
In a BCC structure, the relationship between the nearest neighbor distance and the atomic radius (R) is given by:
\[ D = 2R \]
Thus, we can find the atomic radius:
\[ R = \frac{D}{2} = \frac{4.52 \, \text{Å}}{2} = 2.26 \, \text{Å} \]

### Step 3: Relate the atomic radius to the edge length (A)
For a BCC structure, the relationship between the edge length (A) and the atomic radius (R) is given by:
\[ \sqrt{3}A = 4R \]
Rearranging gives:
\[ A = \frac{4R}{\sqrt{3}} \]
Substituting the value of R:
\[ A = \frac{4 \times 2.26 \, \text{Å}}{\sqrt{3}} \]
Calculating this:
\[ A = \frac{9.04 \, \text{Å}}{1.732} \approx 5.22 \, \text{Å} \]

### Step 4: Convert the edge length to meters
To convert the edge length from Ångstroms to meters:
\[ A = 5.22 \, \text{Å} = 5.22 \times 10^{-10} \, \text{m} \]

### Step 5: Calculate the volume of the unit cell
The volume (V) of the unit cell is given by:
\[ V = A^3 = (5.22 \times 10^{-10} \, \text{m})^3 \]
Calculating this:
\[ V \approx 1.42 \times 10^{-28} \, \text{m}^3 \]

### Step 6: Determine the number of atoms per unit cell
In a BCC structure, there are 2 effective atoms per unit cell (Z = 2).

### Step 7: Use the density formula
The density (\(\rho\)) is given by the formula:
\[ \rho = \frac{Z \cdot M}{V \cdot N_A} \]
Where:
- \(Z\) = number of atoms per unit cell = 2
- \(M\) = atomic weight in kg = 39 g/mol = 0.039 kg/mol
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)

Substituting the values:
\[ \rho = \frac{2 \cdot 0.039 \, \text{kg/mol}}{1.42 \times 10^{-28} \, \text{m}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 8: Calculate the density
Calculating the denominator:
\[ V \cdot N_A \approx 1.42 \times 10^{-28} \cdot 6.022 \times 10^{23} \approx 8.55 \times 10^{-6} \, \text{kg} \]

Now substituting back into the density formula:
\[ \rho \approx \frac{0.078 \, \text{kg}}{8.55 \times 10^{-6} \, \text{m}^3} \approx 9140.7 \, \text{kg/m}^3 \]

### Step 9: Convert to kg/m³
Since the density is already in kg/m³, we can present the final answer:
\[ \rho \approx 9140.7 \, \text{kg/m}^3 \]

### Final Answer
The density of potassium is approximately **9140.7 kg/m³**.