## Step-by-Step Solution Given data: - Density of thorium (\(\rho\)) = 11.72 g/cm³ - Atomic weight of thorium (M) = 232 g/mol - Structure: Face Centered Cubic (FCC) - Avogadro's number (\(N_A\)) = \(6.022 \times 10^{23}\) atoms/mol For an FCC structure, the number of atoms per unit cell (\(n\)) = 4. The density formula for a cubic unit cell is: \[ \rho = \frac{n \cdot M}{a^3 \cdot N_A} \] Where: - \(\rho\) = density (g/cm³) - \(n\) = number of atoms per unit cell (4 for FCC) - \(M\) = atomic weight (g/mol) - \(a\) = lattice parameter (cm) - \(N_A\) = Avogadro's number Rearrange to solve for \(a\): \[ a^3 = \frac{n \cdot M}{\rho \cdot N_A} \] \[ a = \sqrt[3]{\frac{n \cdot M}{\rho \cdot N_A}} \] Substitute the values: \[ a = \sqrt[3]{\frac{4 \times 232}{11.72 \times 6.022 \times 10^{23}}} \] Calculate the denominator: \[ 11.72 \times 6.022 \times 10^{23} = 70.59584 \times 10^{23} \] Calculate the numerator: \[ 4 \times 232 = 928 \] Now, \[ a = \sqrt[3]{\frac{928}{70.59584 \times 10^{23}}} \] \[ a = \sqrt[3]{\frac{928}{7.059584 \times 10^{24}}} \] Calculate the ratio: \[ \frac{928}{7.059584 \times 10^{24}} = 1.315 \times 10^{-22} \] Take the cube root: \[ a = (1.315 \times 10^{-22})^{1/3} \] Calculate the cube root: \[ a \approx (1.315)^{1/3} \times (10^{-22})^{1/3} \] \[ a \approx 1.09 \times 10^{-7.33} \] \[ a \approx 1.09 \times 4.68 \times 10^{-8} \text{ cm} \] \[ a \approx 5.1 \times 10^{-8} \text{ cm} \] Convert to Ångstroms (\(1 \text{ Å} = 10^{-8} \text{ cm}\)): \[ a \approx 5.1 \text{ Å} \] ## Final Answer The lattice parameter \(a\) for thorium with FCC structure is **\(5.1 \ \text{Å}\) (angstroms)**.