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 and BCC unit cells, we can follow these steps:

### Step 1: Understand the formula for density
The density (\(d\)) of a unit cell can be expressed as:
\[ d = \frac{z \cdot m}{A^3 \cdot N_A} \]
where:
- \(z\) = 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: Calculate the density for BCC
For a BCC unit cell:
- Edge length \(A_{BCC} = 2 \, \text{pm}\)
- Number of atoms per unit cell \(z_{BCC} = 2\)

Substituting these values into the density formula:
\[ d_{BCC} = \frac{z_{BCC} \cdot m}{A_{BCC}^3 \cdot N_A} = \frac{2 \cdot m}{(2 \, \text{pm})^3 \cdot N_A} \]
Calculating \(A_{BCC}^3\):
\[ (2 \, \text{pm})^3 = 8 \, \text{pm}^3 \]
Thus, the density for BCC becomes:
\[ d_{BCC} = \frac{2 \cdot m}{8 \cdot N_A} = \frac{m}{4 \cdot N_A} \]

### Step 3: Calculate the density for FCC
For an FCC unit cell:
- Edge length \(A_{FCC} = 4 \, \text{pm}\)
- Number of atoms per unit cell \(z_{FCC} = 4\)

Substituting these values into the density formula:
\[ d_{FCC} = \frac{z_{FCC} \cdot m}{A_{FCC}^3 \cdot N_A} = \frac{4 \cdot m}{(4 \, \text{pm})^3 \cdot N_A} \]
Calculating \(A_{FCC}^3\):
\[ (4 \, \text{pm})^3 = 64 \, \text{pm}^3 \]
Thus, the density for FCC becomes:
\[ d_{FCC} = \frac{4 \cdot m}{64 \cdot N_A} = \frac{m}{16 \cdot N_A} \]

### Step 4: Find the ratio of densities
Now, we can find the ratio of the densities of FCC to BCC:
\[ \frac{d_{FCC}}{d_{BCC}} = \frac{\frac{m}{16 \cdot N_A}}{\frac{m}{4 \cdot N_A}} = \frac{1/16}{1/4} = \frac{1}{16} \cdot \frac{4}{1} = \frac{4}{16} = \frac{1}{4} \]

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