A metal crystallise into two cubic phases FCC and BCC whose unit cell lengths are 2.5 and 2.0 `A^0` respectively, the ratio of densities of BCC and FCC

Options

A. 0.977
B. 1.024
C. 0.5
D. 2

Answer

To find the ratio of densities of BCC (Body-Centered Cubic) and FCC (Face-Centered Cubic) structures, we can use the formula for density of a unit cell:

\[ \text{Density} (d) = \frac{Z \cdot M}{N_A \cdot a^3} \]

Where:
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass of the substance
- \( N_A \) = Avogadro's number
- \( a \) = edge length of the unit cell

### Step-by-Step Solution:

1. **Identify the values for FCC and BCC:**
- For BCC:
- \( Z_{BCC} = 2 \)
- Edge length \( a_{BCC} = 2.0 \, \text{Å} \)
- For FCC:
- \( Z_{FCC} = 4 \)
- Edge length \( a_{FCC} = 2.5 \, \text{Å} \)

2. **Write the density formulas for both structures:**
- Density of BCC:
\[ d_{BCC} = \frac{Z_{BCC} \cdot M}{N_A \cdot a_{BCC}^3} = \frac{2M}{N_A \cdot (2.0)^3} \]
- Density of FCC:
\[ d_{FCC} = \frac{Z_{FCC} \cdot M}{N_A \cdot a_{FCC}^3} = \frac{4M}{N_A \cdot (2.5)^3} \]

3. **Calculate the ratio of densities:**
\[ \frac{d_{BCC}}{d_{FCC}} = \frac{\frac{2M}{N_A \cdot (2.0)^3}}{\frac{4M}{N_A \cdot (2.5)^3}} \]

4. **Simplify the ratio:**
- The \( M \) and \( N_A \) cancel out:
\[ \frac{d_{BCC}}{d_{FCC}} = \frac{2}{4} \cdot \frac{(2.5)^3}{(2.0)^3} \]
- This simplifies to:
\[ \frac{d_{BCC}}{d_{FCC}} = \frac{1}{2} \cdot \frac{(2.5)^3}{(2.0)^3} \]

5. **Calculate \( (2.5)^3 \) and \( (2.0)^3 \):**
- \( (2.5)^3 = 15.625 \)
- \( (2.0)^3 = 8 \)

6. **Substitute the values back into the ratio:**
\[ \frac{d_{BCC}}{d_{FCC}} = \frac{1}{2} \cdot \frac{15.625}{8} = \frac{15.625}{16} \]

7. **Calculate the final ratio:**
\[ \frac{15.625}{16} = 0.9765625 \approx 0.977 \]

### Final Result:
The ratio of densities of BCC to FCC is approximately \( 0.977 \).