Element X crystallizes in 12 co - ordination fcc lattice. On applyng high temperature it changes to bcc lattice. Find the ratio of the density of the crystal lattice before and and after applying high temperature

Options

A. `1:1`
B. `3:2`
C. `sqrt2:sqrt3`
D. `2(sqrt2)^(3):(sqrt3)^(3)`

Answer

To solve the problem of finding the ratio of the density of the crystal lattice before and after applying high temperature, we will follow these steps:

### Step 1: Understand the Density Formula
The density (\( \rho \)) of a crystal lattice is given by the formula:
\[ \rho = \frac{Z \cdot M}{A^3 \cdot N_A} \]
where:
- \( Z \) = number of effective atoms in the unit cell
- \( M \) = molar mass (atomic weight) of the element
- \( A \) = edge length of the unit cell
- \( N_A \) = Avogadro's number

### Step 2: Determine the Values for FCC and BCC
For the FCC (Face-Centered Cubic) lattice:
- \( Z_{FCC} = 4 \)
- The edge length \( A_{FCC} = \frac{4}{\sqrt{2}} R \)

For the BCC (Body-Centered Cubic) lattice:
- \( Z_{BCC} = 2 \)
- The edge length \( A_{BCC} = \frac{4}{\sqrt{3}} R \)

### Step 3: Calculate the Density for FCC
Substituting the values into the density formula for FCC:
\[ \rho_{FCC} = \frac{4 \cdot M}{\left(\frac{4}{\sqrt{2}} R\right)^3 \cdot N_A} \]
Calculating \( A_{FCC}^3 \):
\[ A_{FCC}^3 = \left(\frac{4}{\sqrt{2}} R\right)^3 = \frac{64}{2\sqrt{2}} R^3 = 32\sqrt{2} R^3 \]
Thus,
\[ \rho_{FCC} = \frac{4M}{32\sqrt{2} R^3 \cdot N_A} = \frac{M}{8\sqrt{2} R^3 \cdot N_A} \]

### Step 4: Calculate the Density for BCC
Substituting the values into the density formula for BCC:
\[ \rho_{BCC} = \frac{2 \cdot M}{\left(\frac{4}{\sqrt{3}} R\right)^3 \cdot N_A} \]
Calculating \( A_{BCC}^3 \):
\[ A_{BCC}^3 = \left(\frac{4}{\sqrt{3}} R\right)^3 = \frac{64}{3\sqrt{3}} R^3 \]
Thus,
\[ \rho_{BCC} = \frac{2M}{\frac{64}{3\sqrt{3}} R^3 \cdot N_A} = \frac{6M\sqrt{3}}{64 R^3 \cdot N_A} = \frac{3M\sqrt{3}}{32 R^3 \cdot N_A} \]

### Step 5: Find the Ratio of Densities
Now we can find the ratio of the densities:
\[ \frac{\rho_{FCC}}{\rho_{BCC}} = \frac{\frac{M}{8\sqrt{2} R^3 \cdot N_A}}{\frac{3M\sqrt{3}}{32 R^3 \cdot N_A}} = \frac{M \cdot 32}{8\sqrt{2} \cdot 3M} = \frac{32}{24\sqrt{2}} = \frac{4}{3\sqrt{2}} \]

### Step 6: Simplify the Ratio
To express the ratio in a more standard form, we can multiply the numerator and denominator by \( \sqrt{2} \):
\[ \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \]

### Final Answer
Thus, the ratio of the density of the crystal lattice before and after applying high temperature is:
\[ \frac{\rho_{FCC}}{\rho_{BCC}} = \frac{2\sqrt{2}}{3} \]