Problem 11 - Entrance Test

Let's analyze the properties for a simple cubic (SC) lattice: 1. **Atom Location**: In a simple cubic unit cell, atoms are located only at the 8 corners of the cube. Each corner atom is shared by 8 adjacent unit cells, so the effective number of atoms per unit cell (Z) is 8 × (1/8) = 1. 2. **Relation between edge length (a) and atomic radius (r)**: In a simple cubic lattice, the atoms at the corners touch along the edge. So, a = 2r. 3. **Volume of the unit cell**: V_cell = a³ = (2r)³ = 8r³. 4. **Volume occupied by atoms**: Since there is 1 effective atom per unit cell, the volume occupied by atoms (V_atoms) = 1 × (4/3)πr³. 5. **Packing Efficiency (PE)**: PE = (Volume occupied by atoms / Volume of unit cell) × 100% PE = [(4/3)πr³ / 8r³] × 100% PE = [ (4/3)π / 8 ] × 100% PE = [ (4π) / 24 ] × 100% PE = [ π / 6 ] × 100% PE = [ 3.14159 / 6 ] × 100% PE = 0.52359 × 100% ≈ 52.36% Rounding to one decimal place, it's 52.4%. Now, let's compare with the options: - Option A (74%, atoms at corners and face centers) describes an FCC lattice. - Option B (68%, atoms at corners and body center) describes a BCC lattice. - Option C (52.4%, atoms only at corners) correctly describes the packing efficiency and atom location for a simple cubic lattice. - Option D (50%, atoms only at corners) has the correct atom location but incorrect packing efficiency. Therefore, option C is the correct description for a simple cubic lattice.