An element crystallizes both in fcc and bcc lattice. If the density of the element in the two forms is the same, the ratio of unit cell length of fcc to that of bcc lattice is

Options

A. `(2)^(1//3)`
B. `((1)/(2))^(1//3)`
C. `(4)^(1//3)`
D. `((1)/(4))^(1//3)`

Answer

To solve the problem of finding the ratio of unit cell lengths of FCC (Face-Centered Cubic) to BCC (Body-Centered Cubic) lattices given that their densities are the same, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Density Formula**:
The density (ρ) of a crystal lattice can be expressed using the formula:
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]
where:
- \( Z \) = number of atoms per unit cell,
- \( M \) = molar mass of the element,
- \( N_A \) = Avogadro's number,
- \( a \) = edge length of the unit cell.

2. **Identify the Number of Atoms in Each Lattice**:
- For FCC, \( Z = 4 \) (4 atoms per unit cell).
- For BCC, \( Z = 2 \) (2 atoms per unit cell).

3. **Set Up the Density Equations**:
Since the densities of FCC and BCC are the same, we can set their density equations equal to each other:
\[ \frac{4M}{N_A \cdot a_{fcc}^3} = \frac{2M}{N_A \cdot a_{bcc}^3} \]

4. **Cancel Common Terms**:
The molar mass \( M \) and Avogadro's number \( N_A \) appear in both equations, so they can be canceled out:
\[ \frac{4}{a_{fcc}^3} = \frac{2}{a_{bcc}^3} \]

5. **Cross-Multiply to Solve for the Ratio**:
Cross-multiplying gives:
\[ 4 \cdot a_{bcc}^3 = 2 \cdot a_{fcc}^3 \]
Rearranging this leads to:
\[ \frac{a_{fcc}^3}{a_{bcc}^3} = \frac{4}{2} = 2 \]

6. **Take the Cube Root**:
To find the ratio of the edge lengths, take the cube root of both sides:
\[ \frac{a_{fcc}}{a_{bcc}} = 2^{1/3} \]

### Final Result:
The ratio of the unit cell length of FCC to that of BCC is:
\[ \frac{a_{fcc}}{a_{bcc}} = 2^{1/3} \]