Let's analyze and solve the problem step-by-step. --- ## Step 1: Understand the problem - **Objective:** Calculate the **planar density** \( \rho_A \) of atoms for Iron (Fe) in a **FCC (face-centered cubic)** structure. - **Given Data:** - Atomic radius of Iron, \( r = 139 \text{ pm} \) - Structure: FCC - Atomic number density units: atoms/m\(^2\) --- ## Step 2: Recall relevant formulas ### 2.1. Atomic radius in FCC structure In FCC, the atoms touch along the face diagonal: \[ \text{Face diagonal} = 4r \] and the face diagonal relates to the lattice parameter \( a \): \[ \text{Face diagonal} = a \sqrt{2} \] so, \[ a \sqrt{2} = 4r \Rightarrow a = \frac{4r}{\sqrt{2}} = 2 \sqrt{2} r \] ### 2.2. Number of atoms per unit cell in FCC \[ \text{Atoms per FCC unit cell} = 4 \] ### 2.3. Planar density \( \rho_A \) Planar density is defined as: \[ \rho_A = \frac{\text{Number of atoms centered on a particular plane}}{\text{Area of that plane}} \] For the FCC structure, the atoms are located at: - Corners (each shared by 8 unit cells) - Face centers (each shared by 2 unit cells) --- ## Step 3: Determine the relevant plane The plane with the highest planar density in FCC is usually the (111) plane: - The (111) plane passes through 3 atoms per unit cell face. --- ## Step 4: Find the number of atoms per (111) plane In FCC, the (111) plane contains: - 3 atoms per plane, each shared between 2 unit cells (since face atoms are shared). Total atoms per (111) plane per unit cell: \[ \text{Atoms per (111) plane} = 3 \times \frac{1}{2} = 1.5 \] --- ## Step 5: Calculate the area of the (111) plane - The (111) plane forms an equilateral triangle with edge length \( a \). - The area of the (111) plane per unit cell: \[ \text{Area of the (111) plane} = \frac{\sqrt{3}}{4} a^2 \] --- ## Step 6: Compute \( a \) (lattice parameter) \[ a = 2 \sqrt{2} r = 2 \sqrt{2} \times 139 \text{ pm} \] Calculate: \[ a = 2 \times 1.4142 \times 139 \text{ pm} \approx 2.8284 \times 139 \text{ pm} \approx 393.3 \text{ pm} \] --- ## Step 7: Calculate the area of the (111) plane \[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \] \[ = \frac{1.732}{4} \times (393.3)^2 \text{ pm}^2 \] \[ = .433 \times 154,701 \text{ pm}^2 \] \[ \approx 66,944 \text{ pm}^2 \] --- ## Step 8: Convert units to m\(^2\) \[ 1 \text{ pm} = 10^{-12} \text{ m} \] \[ \text{Area} = 66,944 \times (10^{-12})^2 \text{ m}^2 = 66,944 \times 10^{-24} \text{ m}^2 = 6.6944 \times 10^{-20} \text{ m}^2 \] --- ## Step 9: Calculate planar density \( \rho_A \) Number of atoms per (111) plane: \[ \text{Atoms} = 1.5 \] Area: \[ A = 6.6944 \times 10^{-20}\, \text{m}^2 \] Planar density: \[ \rho_A = \frac{1.5}{A} = \frac{1.5}{6.6944 \times 10^{-20}} \approx 2.24 \times 10^{19} \text{ atoms/m}^2 \] --- ## **Final answer:** ### **Planar density of atoms in the (111) plane:** \[ \boxed{ \rho_A \approx 2.24 \times 10^{19} \text{ atoms/m}^2 } \] --- ## **Rounded to 3 significant figures:** \[ \boxed{ \rho_A \approx 2.24 \times 10^{19} \text{ atoms/m}^2 } \]