# Calculation of Planar Density (PA) for Iron (Fe) in an (100) Plane ## Step 1: Understand the problem - ** (Face-Centered Cubic)** structure - **(100) plane** in FCC - Atomic radius \( r = 139\, \text{pm} = 139 \times 10^{-12}\, \text{m} \) - Find the **planar atomic density (atoms/m²)**: number of atoms per unit area on (100) plane. --- ## Step 2: lattice parameter \(a\) In FCC, the relationship between radius \(r\) and lattice parameter \(a): [ a = 2 \sqrt{2} r ] Calculate \(a\): \[ a 2 \sqrt{2} \times 139 \times 10^{-12}\,text{m} \] \[ a = 2 \times 1.4142 \times 139 \times 10^{-12} \] \[ a \approx 2.8284 \times 139 \times 10^{-12} \] \[ a \approx 393.4 \times 10^{-12}\, \text{m} = 393.4\, \text{pm} \] --- ## Step 3: Identify atoms in the (100) plane In FCC: - The (100) plane cuts through the cube parallel to the x, y, z axes. - Atoms in the FCC unit cell: - **Corner atoms:** 8 corners, each shared among 8 unit cells. - **Face atoms:** 6 faces, each shared between 2 unit cells. - In the (100) plane: - The **atoms lying exactly in the plane** are: - 4 corner atoms (each shared among 4 unit cells in the plane) - 1 face atom at the center of the face (not present in the (100) plane itself but in the same plane if it were a face) But, more straightforwardly for the (100) plane: - The **(100) plane** intersects the cube at a face. - **Atoms in the (100) plane:** - **Corner atoms:** 4 corners in the plane, each shared among 4 in-plane unit cells. - **Face atom:** The atom at the face center, fully within the plane. However, in FCC, the (100) plane contains: - 1 atom at the face center (full atom) - 4 corner atoms, each shared among 4 neighboring unit cells, contributing \( \frac{1}{4} \) each to the plane. **Total atoms per (100) plane:** \[ N_{atoms} = 1 + 4 \times \frac{1}{4} = 1 + 1 = 2 \] --- ## Step 4: Determine the area of the (100) plane The (100) plane's area corresponds to the face of the cube: \[ A = a^2 \] Calculate: \[ A = (393.4 \times 10^{-12})^2 = 393.4^2 \times 10^{-24} \] \[ A \approx 154,084 \times 10^{-24} = 1.54084 \times 10^{-19}\, \text{m}^2 \] --- ## Step 5: Calculate the planar density \(PA\) \[ PA = \frac{\text{Number of atoms in the plane}}{\text{Area of the plane}} \] \[ PA = \frac{2}{A} = \frac{2}{1.54084 \times 10^{-19}} \approx 1.298 \times 10^{19}\, \text{atoms/m}^2 \] --- ## **Final Answer:** \[ \boxed{ \textbf{Planar density } PA \approx 1.30 \times 10^{19}\, \text{atoms/m}^2 } \] *(rounded to 3 significant figures)*