Let's go through the step-by-step solution to calculate the planar density (\(\rho_A\)) for iron atoms in an FCC (100) plane. --- # Step 1: Understand the problem - **Atomic radius of iron, \(r\):** 139 pm = \(139 \times 10^{-12}\) m - **Crystal structure:** FCC (face-centered cubic) - **Plane of interest:** (100) plane - **Goal:** Calculate the planar density \(\rho_A\) in atoms/m\(^2\) --- # Step 2: Find the atomic arrangement on the (100) plane In FCC: - Each unit cell has 4 atoms. - The (100) plane cuts through the cube, intersecting atoms at specific positions. **Atoms per (100) plane:** - Corner atoms: 4 atoms, each shared among 4 adjacent unit cells, so contribution per cell: \(4 \times \frac{1}{4} = 1\) - Face atoms: 2 atoms (on the face), each shared between 2 cells, so contribution per cell: \(2 \times \frac{1}{2} = 1\) **Total atoms in the (100) plane per unit cell:** \[ \text{Number of atoms} = 1 + 1 = 2 \] --- # Step 3: Determine the area of the (100) plane In FCC, the lattice parameter \(a\) relates to atomic radius \(r\): \[ a = 2 \sqrt{2} r \] - For FCC: \[ a = 2 \sqrt{2} r \] Calculate \(a\): \[ a = 2 \times \sqrt{2} \times 139 \text{ pm} \] \[ a \approx 2 \times 1.4142 \times 139 \text{ pm} \approx 2.8284 \times 139 \text{ pm} \] \[ a \approx 393.1 \text{ pm} = 393.1 \times 10^{-12} \text{ m} \] **Area of the (100) plane:** \[ A = a^2 \] \[ A = (393.1 \times 10^{-12})^2 \approx 1.546 \times 10^{-19} \text{ m}^2 \] --- # Step 4: Calculate the number of atoms per unit area (planar density) Number of atoms in the (100) plane: \[ N = 2 \] Planar density: \[ \rho_A = \frac{\text{Number of atoms}}{\text{Area}} = \frac{2}{A} \] \[ \rho_A = \frac{2}{1.546 \times 10^{-19}} \approx 1.294 \times 10^{19} \text{ atoms/m}^2 \] --- # **Final answer:** \[ \boxed{ \rho_A \approx 1.29 \times 10^{19} \text{ atoms/m}^2 } \] **Rounded to 3 significant figures.**