Let's go step-by-step to calculate the **planar density** of atoms in an FCC (100) surface. --- ## Step 1: Understand the problem - **Structure:** FCC (face-centered cubic) - **Surface plane:** (100) - **Atomic radius, \( r \):** 139 pm = \( 139 \times 10^{-12} \) m - **Goal:** Calculate **planar density** \( \rho_A \) in atoms/m² --- ## Step 2: Find the surface atom arrangement for FCC (100) In an FCC crystal, the (100) plane: - Contains **4 atoms per unit cell** (due to the face atoms and corner atoms) - The **unit cell** has: \[ a = 2 \sqrt{2} r \] because in FCC, the relation between lattice parameter \( a \) and atomic radius \( r \) is: \[ a = 2 \sqrt{2} r \] --- ## Step 3: Determine the number of atoms per (100) surface area The FCC (100) surface: - Contains **1/4 of the corner atoms** (since each corner atom is shared among 4 cubes) - Contains **1 face atom** (which is fully on the surface) But for planar density, it's easier to consider the **area of the (100) surface** of the unit cell. --- ## Step 4: Calculate the surface area For the FCC unit cell: \[ a = 2 \sqrt{2} r \] Calculate \( a \): \[ a = 2 \sqrt{2} \times 139 \times 10^{-12} \text{ m} \] \[ a \approx 2 \times 1.4142 \times 139 \times 10^{-12} \] \[ a \approx 2.8284 \times 139 \times 10^{-12} \] \[ a \approx 393.1 \times 10^{-12} \text{ m} \] --- ## Step 5: Identify atoms per surface unit cell In the (100) plane: - Each face contains **1 atom** (face atom) - The surface unit cell is the same as the face of the cube with area: \[ A_{cell} = a^2 \] Number of atoms per unit cell on the (100) surface: - One atom from the face (since the face atom is fully on the surface) --- ## Step 6: Calculate the number of atoms per unit area \[ \rho_A = \frac{\text{Number of atoms on surface}}{\text{Surface area of the unit cell}} \] Number of atoms on the surface per unit cell = 1 Surface area: \[ A_{cell} = a^2 = (393.1 \times 10^{-12})^2 \] \[ A_{cell} \approx (393.1)^2 \times 10^{-24} \] \[ A_{cell} \approx 154,537 \times 10^{-24} = 1.545 \times 10^{-19} \text{ m}^2 \] --- ## Step 7: Calculate planar density \[ \rho_A = \frac{1 \text{ atom}}{1.545 \times 10^{-19} \text{ m}^2} \] \[ \rho_A \approx 6.47 \times 10^{18} \text{ atoms/m}^2 \] --- ## **Final Answer:** **Planar density of atoms on FCC (100) surface:** \[ \boxed{ \rho_A \approx 6.47 \times 10^{18} \text{ atoms/m}^2 } \] --- ### **Rounded to 3 significant figures:** \[ \boxed{ \rho_A \approx 6.47 \times 10^{18} \text{ atoms/m}^2 } \]