A metal can be crystallized in both BCC and FCC unit cells whose edge lengths are 2 pm and 4 pm respectively. Then ratio of densities of FCC and BCC unit cells is

Options

A. `(1)/(4)`
B. 4
C. `(1)/(16)`
D. 16

Answer

To find the ratio of densities of FCC (Face-Centered Cubic) and BCC (Body-Centered Cubic) unit cells, we can follow these steps:

### Step 1: Understand the formula for density
The density (D) of a unit cell can be calculated using the formula:
\[ D = \frac{Z \cdot M}{A^3 \cdot N_A} \]
where:
- \(D\) = density
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass of the substance
- \(A\) = edge length of the unit cell
- \(N_A\) = Avogadro's number

### Step 2: Determine the number of atoms per unit cell (Z)
- For BCC: \(Z = 2\) (1 atom at the body center + 8 corner atoms contributing \(\frac{1}{8}\) each)
- For FCC: \(Z = 4\) (6 face-centered atoms contributing \(\frac{1}{2}\) each + 8 corner atoms contributing \(\frac{1}{8}\) each)

### Step 3: Identify the edge lengths
- Edge length for BCC: \(A_{BCC} = 2 \, \text{pm}\)
- Edge length for FCC: \(A_{FCC} = 4 \, \text{pm}\)

### Step 4: Calculate \(A^3\) for both unit cells
- For BCC:
\[ A_{BCC}^3 = (2 \, \text{pm})^3 = 8 \, \text{pm}^3 \]
- For FCC:
\[ A_{FCC}^3 = (4 \, \text{pm})^3 = 64 \, \text{pm}^3 \]

### Step 5: Write the density expressions for both unit cells
- Density of BCC:
\[ D_{BCC} = \frac{Z_{BCC} \cdot M}{A_{BCC}^3 \cdot N_A} = \frac{2M}{8 \cdot N_A} \]
- Density of FCC:
\[ D_{FCC} = \frac{Z_{FCC} \cdot M}{A_{FCC}^3 \cdot N_A} = \frac{4M}{64 \cdot N_A} \]

### Step 6: Calculate the ratio of densities
Now, we can find the ratio of the densities:
\[ \frac{D_{FCC}}{D_{BCC}} = \frac{\frac{4M}{64 \cdot N_A}}{\frac{2M}{8 \cdot N_A}} = \frac{4M \cdot 8 \cdot N_A}{64 \cdot N_A \cdot 2M} \]
The \(M\) and \(N_A\) cancel out:
\[ = \frac{4 \cdot 8}{64 \cdot 2} = \frac{32}{128} = \frac{1}{4} \]

### Conclusion
The ratio of densities of FCC to BCC is:
\[ \frac{D_{FCC}}{D_{BCC}} = \frac{1}{4} \]