| Q. No. | Topic/Concept | Key Formula/Method | Answer | Unit/Format |
|---|---|---|---|---|
| 1 | Packing Efficiency in FCC |
For FCC: a = 2√2 × r Packing efficiency = (4 × (4/3)πr³) / a³ × 100 4 atoms per FCC unit cell |
74 | Percentage (%) |
| 2 | BCC Density Calculation |
Density = (Z × M) / (NA × a³) M = (ρ × NA × a³) / Z Z = 2 for BCC structure |
64 | g/mol |
| 3 | Voids in CCP Structure |
In CCP: Tetrahedral voids = 2N Original close-packed spheres = 875 Removed spheres were from voids, not positions |
1750 | Number of voids |
| 4 | Rock Salt Structure |
Space occupied = (Vions / Vunit cell) × 100 4 cations + 4 anions per unit cell a = 2(r+ + r-) |
65 | Percentage (%) |
| 5 | Fluorite Structure |
M = (ρ × NA × a³) / Z Z = 4 for fluorite (CaF2 type) 4 formula units per unit cell |
80 | g/mol |
| 6 | HCP Unit Cell Volume |
a = 2r, c = 1.633a V = (3√3/2) × a² × c Answer as X in X × 105 pm³ |
647 | × 105 pm³ |
| 7 | Frenkel Defects |
Defect % = (1015 / 6.022×1020) × 100 = 0.000166% = 0.166 × 10-3 Rounded to nearest integer × 10-3 |
0 | × 10-3 % |
| 8 | Spinel Structure |
A²⁺: (1/8) × 64 tetrahedral = 8 B³⁺: (1/2) × 32 octahedral = 16 Total cations = 8 + 16 = 24 |
24 | Number of cations |
| 9 | Semiconductor Doping |
Si atoms/cm³ = (ρ × NA) / M B atoms = Si atoms × 10-6 Answer as X in X × 1016 |
5 | × 1016 atoms/cm³ |
| 10 | Anti-fluorite Structure |
Distance = (√3/4) × a = (√3/4) × 640 pm K⁺ to nearest O²⁻ distance |
277 | pm |