A metal crystallizes into two cubic phases, face-centred cubic and body-centred cubic, which have unit cell lengths `3.5` and `3.0 A`, respectively. Calculate the ration of densities of fcc and bcc.

Options

A. `1.259:1`
B. `1:1.259`
C. `3:2`
D. `1.142:1`

Explanation

To calculate the ratio of densities of face-centered cubic (FCC) and body-centered cubic (BCC) structures, we can follow these steps:

### Step 1: Understand the formula for density
The density (\( \rho \)) of a crystal structure can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{V \cdot N_A} \]
where:
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass of the substance (in g/mol)
- \( V \) = volume of the unit cell (in cm³)
- \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))

### Step 2: Calculate the volume of the unit cells
For FCC:
- The unit cell length \( a_{fcc} = 3.5 \, \text{Å} = 3.5 \times 10^{-8} \, \text{cm} \)
- The volume \( V_{fcc} = a_{fcc}^3 = (3.5 \times 10^{-8})^3 \, \text{cm}^3 \)

For BCC:
- The unit cell length \( a_{bcc} = 3.0 \, \text{Å} = 3.0 \times 10^{-8} \, \text{cm} \)
- The volume \( V_{bcc} = a_{bcc}^3 = (3.0 \times 10^{-8})^3 \, \text{cm}^3 \)

### Step 3: Determine the number of atoms per unit cell
- For FCC, \( Z_{fcc} = 4 \) (4 atoms per unit cell)
- For BCC, \( Z_{bcc} = 2 \) (2 atoms per unit cell)

### Step 4: Calculate the densities
Now we can express the densities for FCC and BCC:
- Density of FCC:
\[ \rho_{fcc} = \frac{Z_{fcc} \cdot M}{V_{fcc} \cdot N_A} = \frac{4M}{(3.5 \times 10^{-8})^3 \cdot N_A} \]

- Density of BCC:
\[ \rho_{bcc} = \frac{Z_{bcc} \cdot M}{V_{bcc} \cdot N_A} = \frac{2M}{(3.0 \times 10^{-8})^3 \cdot N_A} \]

### Step 5: Calculate the ratio of densities
Now, we can find the ratio of densities:
\[ \frac{\rho_{fcc}}{\rho_{bcc}} = \frac{\frac{4M}{(3.5 \times 10^{-8})^3 \cdot N_A}}{\frac{2M}{(3.0 \times 10^{-8})^3 \cdot N_A}} = \frac{4}{2} \cdot \frac{(3.0 \times 10^{-8})^3}{(3.5 \times 10^{-8})^3} \]
This simplifies to:
\[ \frac{\rho_{fcc}}{\rho_{bcc}} = 2 \cdot \left(\frac{3.0}{3.5}\right)^3 \]

### Step 6: Calculate the numerical value
Calculating the numerical value:
\[ \frac{3.0}{3.5} = \frac{30}{35} = \frac{6}{7} \]
Now, calculating \( \left(\frac{6}{7}\right)^3 \):
\[ \left(\frac{6}{7}\right)^3 = \frac{216}{343} \]
Thus,
\[ \frac{\rho_{fcc}}{\rho_{bcc}} = 2 \cdot \frac{216}{343} = \frac{432}{343} \approx 1.259 \]

### Final Answer
The ratio of densities of FCC to BCC is approximately:
\[ \frac{\rho_{fcc}}{\rho_{bcc}} \approx 1.259 : 1 \]