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 when the densities of the element in both forms are the same, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the number of atoms in each unit cell:**
- For FCC, the number of atoms (Z) is 4.
- For BCC, the number of atoms (Z) is 2.

2. **Write the formula for density:**
- The density (ρ) of a crystal can be expressed as:
\[ \rho = \frac{Z \times M}{A^3 \times N_A} \]
where:
- \( Z \) = number of atoms in the unit cell,
- \( M \) = molar mass of the element,
- \( A \) = edge length of the unit cell,
- \( N_A \) = Avogadro's number.

3. **Set up the density equations for FCC and BCC:**
- For FCC:
\[ \rho_{FCC} = \frac{4M}{A_{FCC}^3 \times N_A} \]
- For BCC:
\[ \rho_{BCC} = \frac{2M}{A_{BCC}^3 \times N_A} \]

4. **Since the densities are equal, set the two density equations equal to each other:**
\[ \frac{4M}{A_{FCC}^3 \times N_A} = \frac{2M}{A_{BCC}^3 \times N_A} \]

5. **Cancel out the common terms (M and \( N_A \)):**
\[ \frac{4}{A_{FCC}^3} = \frac{2}{A_{BCC}^3} \]

6. **Cross-multiply to solve for the relationship between \( A_{FCC} \) and \( A_{BCC} \):**
\[ 4 \times A_{BCC}^3 = 2 \times A_{FCC}^3 \]

7. **Rearranging gives:**
\[ \frac{A_{FCC}^3}{A_{BCC}^3} = \frac{4}{2} = 2 \]

8. **Taking the cube root of both sides:**
\[ \frac{A_{FCC}}{A_{BCC}} = 2^{1/3} \]

### Final Answer:
Thus, the ratio of the unit cell length of FCC to that of BCC is:
\[ \frac{A_{FCC}}{A_{BCC}} = 2^{1/3} \]