Let's go through the step-by-step solution calculate the planar density of atoms (\(\rho_A\)) for iron atom in face-centered cubic (FCC) (100) plane. --- ## Step 1: Understand the problem - **Material:** Iron (Fe) - **Crystal structure:** FCC - **Plane:** (100) - **Atomic radius, \(r\):** 139 pm = \(139 \times 10^{-12}\) m - **Goal:** Find the planar density \(\rho_A\) in atoms/m\(^2\) --- ## Step : Determine the number of atoms per (100) plane In FCC structure: - The (100) plane: - Contains **2 atoms per unit cell** ( FCC 4 atoms per unit cell, and the (100) cuts through 1/2 of the atoms the corners 1/2 of the atoms at the face). **In FCC:** - Each corner is shared 4 unit cells. - Each face atom is shared between 2 unit cells. - For the (100 plane: - **Corner atoms:** 4 corners, each shared among 4 unit cells, so each contributes \(1/4\) atom per unit cell. - **Face atoms:** 2 face atoms on the (100) plane, each shared between 2 unit cells, so each contributes \(1/2\) atom per unit cell. Total atoms per (100) plane in one unit cell: \[ \text{Atoms from corners} = 4 \times \frac{1}{4} = 1 \] \[ \text{Atoms from faces} = 2 \times \frac{1}{2} = 1 \] \[ \text{Total} = 1 + 1 = 2 \text{ atoms} \] --- ## Step 3: Find the interplanar spacing, \(d_{100}\) For FCC: - Lattice parameter \(a\): \[ a = 2 \sqrt{2} r \] Calculate \(a\): \[ a = 2 \sqrt{2} \times 139\, \text{pm} = 2 \times 1.4142 \times 139\, \text{pm} \approx 2.8284 \times 139\, \text{pm} \approx 393.09\, \text{pm} \] - The interplanar spacing \(d_{hkl}\): \[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \] For (100): \[ d_{100} = \frac{a}{\sqrt{1^2 + + }} = a = 393.09\, \text{pm} = 393.09 \times 10^{-12}\, \text{m} \] --- ## Step 4: Calculate the planar density \(\rho_A\) - **Planar density**: number of atoms per unit area on the (100) plane \[ \rho_A = \frac{\text{Number of atoms in the plane}}{\text{Area of the plane}} \] - The **area of the (100) plane**: \[ A_{plane} = a^2 \] - Total atoms per plane: \[ N_{atoms} = 2 \] - Therefore, \[ \rho_A = \frac{2}{a^2} \] Calculate \(a^2\): \[ a^2 = (393.09 \times 10^{-12})^2 \approx 1.546 \times 10^{-19}\, \text{m}^2 \] Finally, \[ \rho_A = \frac{2}{1.546 \times 10^{-19}} \approx 1.293 \times 10^{19}\, \text{atoms/m}^2 \] --- ## **Final answer (rounded to 3 significant figures):** \[ \boxed{ \rho_A \approx 1.29 \times 10^{19}\, \text{atoms/m}^2 } \] --- ## **Summary:** - The planar density of atoms on the (100) plane in FCC iron is approximately **\(1.29 \times 10^{19}\) atoms/m\(^2\)**.