Problem 8 - Entrance Test

Let's calculate the density for BCC (α-Fe) and FCC (γ-Fe) structures. The molar mass (M) and Avogadro's number (NA) are the same for both. Density ρ = (Z × M) / (a³ × NA) For BCC (α-Fe): Number of atoms per unit cell (Z_BCC) = 2 Relation between edge length (a_BCC) and atomic radius (r): 4r = a_BCC√3 => a_BCC = 4r/√3 So, a_BCC³ = (4r/√3)³ = 64r³ / (3√3) ρ_BCC = (2 × M) / ((64r³ / (3√3)) × NA) = (6√3 × M) / (64r³ × NA) For FCC (γ-Fe): Number of atoms per unit cell (Z_FCC) = 4 Relation between edge length (a_FCC) and atomic radius (r): 4r = a_FCC√2 => a_FCC = 4r/√2 = 2√2 r So, a_FCC³ = (2√2 r)³ = (8 × 2√2)r³ = 16√2 r³ ρ_FCC = (4 × M) / ((16√2 r³) × NA) = (4M) / (16√2 r³ × NA) = M / (4√2 r³ × NA) Now, calculate the ratio ρ_BCC / ρ_FCC: ρ_BCC / ρ_FCC = [ (6√3 × M) / (64r³ × NA) ] / [ M / (4√2 r³ × NA) ] ρ_BCC / ρ_FCC = (6√3 / 64) × (4√2 / 1) ρ_BCC / ρ_FCC = (24√6) / 64 = (3√6) / 8 Now, substitute the value of √6 ≈ 2.449: ρ_BCC / ρ_FCC = (3 × 2.449) / 8 = 7.347 / 8 ≈ 0.918375 Let's recheck the calculation of a³. For BCC: a_BCC = 4r/√3 => a_BCC³ = (4r/√3)³ = 64r³ / (3√3) For FCC: a_FCC = 2√2 r => a_FCC³ = (2√2 r)³ = 16√2 r³ ρ_BCC / ρ_FCC = [ (Z_BCC * M) / (a_BCC³ * NA) ] / [ (Z_FCC * M) / (a_FCC³ * NA) ] ρ_BCC / ρ_FCC = (Z_BCC / a_BCC³) / (Z_FCC / a_FCC³) ρ_BCC / ρ_FCC = (Z_BCC / Z_FCC) * (a_FCC³ / a_BCC³) ρ_BCC / ρ_FCC = (2 / 4) * (16√2 r³ / (64r³ / 3√3)) ρ_BCC / ρ_FCC = (1/2) * (16√2 / (64 / 3√3)) ρ_BCC / ρ_FCC = (1/2) * (16√2 * 3√3 / 64) ρ_BCC / ρ_FCC = (1/2) * (48√6 / 64) ρ_BCC / ρ_FCC = (1/2) * (3√6 / 4) ρ_BCC / ρ_FCC = 3√6 / 8 (3 × 2.4494897) / 8 = 7.3484691 / 8 = 0.9185586 This is closest to 0.925. Let me check for rounding differences in roots or common values used. √3 ≈ 1.732 √2 ≈ 1.414 √6 ≈ 2.449 The calculation looks correct. The value 0.918 is closest to 0.925. This ratio is less than 1, meaning that FCC has a higher density than BCC for the same atomic radius. This is expected as FCC has a higher packing efficiency (74%) compared to BCC (68%). The ratio is 0.9186. Let's see if 0.925 can be obtained by different rounding or approximation. (3 * 2.45) / 8 = 7.35 / 8 = 0.91875. Still close to 0.918. Let's re-calculate using the density of α-Fe and γ-Fe with exact values. Atomic radius r = 124 pm = 124 × 10⁻¹² m For BCC: a_BCC = 4r/√3 = 4 * 124 pm / √3 = 496 / 1.732 = 286.37 pm For FCC: a_FCC = 4r/√2 = 4 * 124 pm / √2 = 496 / 1.414 = 350.78 pm Now, we need the ratio of densities. Since M and NA cancel out, the ratio is (Z_BCC / a_BCC³) / (Z_FCC / a_FCC³) Ratio = (2 / (286.37)³) / (4 / (350.78)³) = (2 / 23485750) / (4 / 43100000) Ratio = (2 / 2.348 × 10⁷) / (4 / 4.31 × 10⁷) Ratio = (0.8517 × 10⁻⁷) / (0.928 × 10⁻⁷) Ratio = 0.8517 / 0.928 = 0.9177. This is consistently giving 0.918 approximately. Option A is 0.925. This is the closest. Final check: 3√6 / 8 = 0.91855... Option A (0.925) is the closest option. It's a difference of about 0.006. This might be a rounding difference in the given choices for NEET exams. Therefore, the ratio of densities ρ_BCC / ρ_FCC = 3√6 / 8 ≈ 0.9186. Among the given options, 0.925 is the closest value. This means the density of BCC α-Fe is slightly less than that of FCC γ-Fe, which is consistent with BCC having lower packing efficiency (68%) than FCC (74%).