# Calculating the Planar Density (PA) for Iron in FCC (100) Plane ## Step 1: Understand the problem - **Structure:** Face-Centered Cubic (FCC) - **Plane:** (100- **Atomic radius (r): 139 pm - **Goal Find **planar density (atoms²)** for the (100) plane. --- ## Step 2: Determine the atomic arrangement on the (100) plane ### FCC (100) plane: - The FCC unit cell has atoms at: - Corners: 8 atoms (each shared by 8 cells) - Face centers: 6 atoms (each shared by 2 cells) ### Atoms in the (100) plane: - The (100) plane cuts through the FCC cell parallel to the x-y plane. - In the (100) plane: - **Corner atoms**: 4 corners visible in the cross-section - **Face atoms**: 1 atom at the face center (on the top face) ### Atomic contribution: - **Corner atoms:** Each contributes \( \frac{1}{4} \) atom (since each corner is shared among 4 adjacent cells) - **Face atom:** Fully on the plane (since it's a face-centered atom on that face) **Total atoms on the (100) plane per unit cell:** \[ N_{atoms} = 4 \times \frac{1}{4} + 1 = 1 + 1 = 2 \] --- ## Step 3: Determine the area of the (100) plane ### Lattice parameter \(a\): In FCC, the relation between atomic radius \(r\) and lattice parameter \(a\): \[ a = 2 \sqrt{2} r \] Substitute \(r = 139\, \text{pm}\): \[ a = 2 \sqrt{2} \times 139\, \text{pm} \approx 2 \times 1.414 \times 139 \approx 2.828 \times 139 \approx 392.8\, \text{pm} \] ### Area of the (100) plane: \[ A_{plane} = a^2 \] Convert \(a\) to meters: \[ a = 392.8\, \text{pm} = 392.8 \times 10^{-12}\, \text{m} \] \[ A_{plane} = (392.8 \times 10^{-12})^2 \approx 1.543 \times 10^{-19}\, \text{m}^2 \] --- ## Step 4: Calculate the planar density \[ PA = \frac{\text{Number of atoms on the plane}}{\text{Area of the plane}} \] \[ PA = \frac{2}{A_{plane}} = \frac{2}{1.543 \times 10^{-19}} \approx 1.297 \times 10^{19}\, \text{atoms/m}^2 \] --- ## **Final Answer:** \[ \boxed{ \textbf{Planar density } PA \approx 1.30 \times 10^{19}\, \text{atoms/m}^2 } \]