Using the data given below, find the type of cubic lattice to which the crystal belongs.
`{:(,Fe,V,Pd),("a in pm",286,301,388),(rho" in gm cm"^(-3),7.86,5.96,12.16):}`

Answer

To determine the type of cubic lattice for the given crystals (Fe, V, Pd), we will use the formula relating density, molar mass, edge length, and the number of atoms per unit cell (Z). The steps to solve the problem are as follows:

### Step 1: Convert Edge Length from Picometers to Centimeters
The edge lengths are given in picometers (pm). We need to convert them to centimeters (cm) for consistency with the density units.

- For Iron (Fe):
\[ a = 286 \, \text{pm} = 286 \times 10^{-10} \, \text{cm} \]

- For Vanadium (V):
\[ a = 301 \, \text{pm} = 301 \times 10^{-10} \, \text{cm} \]

- For Palladium (Pd):
\[ a = 388 \, \text{pm} = 388 \times 10^{-10} \, \text{cm} \]

### Step 2: Use the Density Formula to Find Z
The formula for density (ρ) is given by:
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \]
Where:
- \( \rho \) = density in g/cm³
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass in g/mol
- \( a \) = edge length in cm
- \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))

We can rearrange this formula to solve for \( Z \):
\[ Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \]

### Step 3: Calculate Z for Each Element

#### For Iron (Fe):
- Molar mass (M) = 56 g/mol
- Density (ρ) = 7.86 g/cm³
- Edge length (a) = \(286 \times 10^{-10} \, \text{cm}\)

Calculating \( a^3 \):
\[ a^3 = (286 \times 10^{-10})^3 = 2.33 \times 10^{-29} \, \text{cm}^3 \]

Now substituting into the formula for Z:
\[ Z = \frac{7.86 \cdot 2.33 \times 10^{-29} \cdot 6.022 \times 10^{23}}{56} \approx 2 \]

#### For Vanadium (V):
- Molar mass (M) = 51 g/mol
- Density (ρ) = 5.96 g/cm³
- Edge length (a) = \(301 \times 10^{-10} \, \text{cm}\)

Calculating \( a^3 \):
\[ a^3 = (301 \times 10^{-10})^3 = 2.74 \times 10^{-29} \, \text{cm}^3 \]

Now substituting into the formula for Z:
\[ Z = \frac{5.96 \cdot 2.74 \times 10^{-29} \cdot 6.022 \times 10^{23}}{51} \approx 2 \]

#### For Palladium (Pd):
- Molar mass (M) = 106.4 g/mol
- Density (ρ) = 12.16 g/cm³
- Edge length (a) = \(388 \times 10^{-10} \, \text{cm}\)

Calculating \( a^3 \):
\[ a^3 = (388 \times 10^{-10})^3 = 5.85 \times 10^{-29} \, \text{cm}^3 \]

Now substituting into the formula for Z:
\[ Z = \frac{12.16 \cdot 5.85 \times 10^{-29} \cdot 6.022 \times 10^{23}}{106.4} \approx 4 \]

### Step 4: Determine the Type of Lattice
- If \( Z = 1 \): Simple Cubic
- If \( Z = 2 \): Body-Centered Cubic (BCC)
- If \( Z = 4 \): Face-Centered Cubic (FCC)

From our calculations:
- Iron (Fe): \( Z \approx 2 \) → BCC
- Vanadium (V): \( Z \approx 2 \) → BCC
- Palladium (Pd): \( Z \approx 4 \) → FCC

### Final Answer
- Iron (Fe) and Vanadium (V) belong to the Body-Centered Cubic (BCC) lattice.
- Palladium (Pd) belongs to the Face-Centered Cubic (FCC) lattice.