Let's go through the step-by-step solution to calculate the planar density (\(\rho_A\)) for iron atoms in a face-centered cubic (FCC) (100) plane. --- # Step 1: Understand the problem - **Atomic radius of iron, \( r \) = 139 pm = \( 139 \times 10^{-12} \) m** - **Structure: FCC (100) plane** - **Goal:** Calculate the planar density, \(\rho_A\), in atoms/m\(^2\) --- # Step 2: Determine the number of atoms per unit area in the (100) plane ## FCC (100) plane atom arrangement: - In FCC, the (100) plane contains **atoms at the corners** and **atoms at the face centers**. - Each corner atom is shared among 4 adjacent unit cells. - Each face-centered atom is shared between 2 unit cells. ### Number of atoms per (100) plane: - **Corners:** 4 corners, each contributing \(\frac{1}{4}\) of an atom: \[ 4 \times \frac{1}{4} = 1 \text{ atom} \] - **Face centers:** 1 face center on the (100) face, contributing \(\frac{1}{2}\) atom: \[ 1 \times \frac{1}{2} = .5 \text{ atom} \] - **Total atoms per (100) plane:** \[ 1 + .5 = 1.5 \text{ atoms} \] --- # Step 3: Find the area of the (100) plane In FCC, the **lattice parameter \(a\)** (edge length of the cube) relates to the atomic radius \(r\): \[ a = 2 \sqrt{2} r \] ### Calculation of \(a\): \[ a = 2 \sqrt{2} \times 139 \text{ pm} \] \[ a = 2 \times 1.2 \times 139 \text{ pm} \approx 2.8284 \times 139 \text{ pm} \] \[ a \approx 393.2 \text{ pm} = 393.2 \times 10^{-12} \text{ m} \] ### Area of the (100) face: \[ A_{(100)} = a^2 = (393.2 \times 10^{-12} \text{ m})^2 \] \[ A_{(100)} \approx (393.2)^2 \times 10^{-24} \text{ m}^2 \] \[ A_{(100)} \approx 154,626 \times 10^{-24} \text{ m}^2 \] \[ A_{(100)} \approx 1.54626 \times 10^{-19} \text{ m}^2 \] --- # Step 4: Calculate planar density \(\rho_A\) \[ \rho_A = \frac{\text{Number of atoms in the plane}}{\text{Area of the plane}} \] \[ \rho_A = \frac{1.5 \text{ atoms}}{1.54626 \times 10^{-19} \text{ m}^2} \] \[ \rho_A \approx 9.7 \times 10^{18} \text{ atoms/m}^2 \] --- # **Final answer:** ### **Planar density, \(\rho_A \approx 9.70 \times 10^{18}\) atoms/m\(^2\)** --- # **Rounded to 3 significant figures:** \[ \boxed{ \rho_A \approx 9.70 \times 10^{18} \text{ atoms/m}^2 } \]