📚 25 Solid State Chemistry Numerical Problems

🟢 Easy (1-5)
🟡 Medium (6-15)
🔴 Hard (16-22)
🟣 Expert (23-25)
1
🧊 Unit Cell Easy
Calculate the number of atoms in a simple cubic unit cell.
📋 Given:
• Simple cubic structure
• Atoms located only at corners
🎯 Find: Number of atoms per unit cell
Step 1: Count corner atoms
Number of corners = 8
Step 2: Each corner atom is shared by 8 unit cells
Contribution per corner atom = 1/8
Step 3: Calculate total atoms
Total atoms = 8 × (1/8) = 1
Number of atoms = 1
2
📐 BCC Structure Easy
Iron crystallizes in body-centered cubic (BCC) structure. Calculate the number of atoms per unit cell.
📋 Given:
• BCC structure
• Atoms at corners and body center
🎯 Find: Number of atoms per unit cell
Step 1: Corner atoms contribution
Corner atoms = 8 × (1/8) = 1
Step 2: Body center atom contribution
Body center atoms = 1 × 1 = 1
Step 3: Total atoms
Total = 1 + 1 = 2
Number of atoms = 2
3
💎 FCC Structure Easy
Calculate the coordination number for atoms in a face-centered cubic (FCC) structure.
📋 Given:
• FCC structure
• Close-packed arrangement
🎯 Find: Coordination number
Step 1: Identify nearest neighbors
In FCC, atoms touch along face diagonals
Step 2: Count nearest neighbors
Each atom has 12 nearest neighbors
Step 3: This is maximum packing
FCC = Cubic close packing (CCP)
Coordination Number = 12
4
📊 Packing Efficiency Easy
What is the packing efficiency of simple cubic structure if atoms are touching along the edge?
📋 Given:
• Simple cubic structure
• Edge length a = 2r (where r is atomic radius)
🎯 Find: Packing efficiency (%)
Step 1: Volume of atoms in unit cell
Vatoms = 1 × (4/3)πr³
Step 2: Volume of unit cell
Vcell = a³ = (2r)³ = 8r³
Step 3: Packing efficiency
η = [(4/3)πr³ / 8r³] × 100% = π/6 × 100%
Packing Efficiency = 52.4%
5
🕳️ Voids Easy
Calculate the number of tetrahedral and octahedral voids in an FCC unit cell.
📋 Given:
• FCC structure with 4 atoms per unit cell
🎯 Find: Number of tetrahedral and octahedral voids
Step 1: Tetrahedral voids
Number of tetrahedral voids = 2N = 2 × 4 = 8
Step 2: Octahedral voids
Number of octahedral voids = N = 4
Step 3: Total voids
Total voids = 8 + 4 = 12
Tetrahedral = 8, Octahedral = 4
6
⚖️ Density Calculation Medium
Copper crystallizes in FCC structure with edge length 3.61 Å. Calculate the density of copper.
📋 Given:
• FCC structure (Z = 4)
• Edge length a = 3.61 Å = 3.61 × 10⁻⁸ cm
• Atomic mass of Cu = 63.5 g/mol
• NA = 6.022 × 10²³ mol⁻¹
🎯 Find: Density of spinel (g/cm³)
Step 1: Calculate unit cell volume
V = a³ = (8.08 × 10⁻⁸)³ = 5.27 × 10⁻²² cm³
Step 2: Apply density formula for complex structure
ρ = (Z × M) / (NA × V)
Step 3: Substitute values
ρ = (8 × 142.3) / (6.022 × 10²³ × 5.27 × 10⁻²²)
Step 4: Calculate numerator and denominator
Numerator = 1138.4 g/mol
Denominator = 3.17 × 10² cm³/mol
Step 5: Final calculation
ρ = 1138.4 / 317 = 3.59 g/cm³
Density of spinel = 3.59 g/cm³
24
🌊 Neutron Diffraction Expert
Calculate the de Broglie wavelength of thermal neutrons at 25°C and determine if they are suitable for crystal diffraction studies.
📋 Given:
• Temperature T = 25°C = 298 K
• Neutron mass mn = 1.675 × 10⁻²⁷ kg
• Boltzmann constant k = 1.38 × 10⁻²³ J/K
• Planck's constant h = 6.626 × 10⁻³⁴ J·s
• Typical d-spacing in crystals ~ 1-5 Å
🎯 Find: de Broglie wavelength and suitability assessment
Step 1: Calculate thermal energy
Ethermal = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 298
Step 2: Calculate thermal energy
Ethermal = 6.17 × 10⁻²¹ J
Step 3: Calculate neutron velocity
E = (1/2)mv², so v = √(2E/m)
v = √(2 × 6.17 × 10⁻²¹ / 1.675 × 10⁻²⁷) = 2.71 × 10³ m/s
Step 4: Calculate momentum
p = mv = 1.675 × 10⁻²⁷ × 2.71 × 10³ = 4.54 × 10⁻²⁴ kg·m/s
Step 5: Calculate de Broglie wavelength
λ = h/p = 6.626 × 10⁻³⁴ / 4.54 × 10⁻²⁴ = 1.46 × 10⁻¹⁰ m = 1.46 Å
Step 6: Assessment
λ ≈ 1.46 Å is comparable to crystal d-spacings (1-5 Å)
λ = 1.46 Å - Excellent for diffraction studies!
25
🧮 Advanced Crystallography Expert
A complex oxide Ln₂Ti₂O₇ (pyrochlore structure) has a cubic unit cell with 8 formula units. Given that the oxide ion radius is 1.40 Å and it adopts the ideal pyrochlore geometry, calculate the expected lattice parameter and predict the lanthanide ionic radius required for this structure.
📋 Given:
• Pyrochlore structure Ln₂Ti₂O₇ (Z = 8)
• Oxide ion radius rO²⁻ = 1.40 Å
• Ti⁴⁺ ionic radius = 0.605 Å
• Ideal pyrochlore geometry requirements
• Coordination: Ln³⁺ (8-fold), Ti⁴⁺ (6-fold)
🎯 Find: Expected lattice parameter and required Ln³⁺ radius
Step 1: Analyze Ti-O octahedral coordination
Ti-O distance = rTi⁴⁺ + rO²⁻ = 0.605 + 1.40 = 2.005 Å
Step 2: In pyrochlore, Ti-O distance relates to lattice parameter
Ti-O = a√2/4, therefore a = 4 × Ti-O / √2
Step 3: Calculate lattice parameter
a = 4 × 2.005 / √2 = 8.02 / 1.414 = 5.67 Å
Step 4: For ideal pyrochlore geometry (Ln-O coordination)
Ln-O distance = a√3/4 = 5.67 × 1.732 / 4 = 2.46 Å
Step 5: Calculate required Ln³⁺ radius
rLn³⁺ = Ln-O distance - rO²⁻
rLn³⁺ = 2.46 - 1.40 = 1.06 Å
Step 6: Identify suitable lanthanide
This radius corresponds to Nd³⁺ or Pr³⁺ (r ≈ 1.06-1.09 Å)
Lattice parameter = 5.67 Å, rLn³⁺ = 1.06 Å (Nd³⁺/Pr³⁺)

🎯 Problem Difficulty Summary

🟢 Easy (1-5)

Basic unit cell calculations, coordination numbers, simple packing efficiency

🟡 Medium (6-15)

Density calculations, Miller indices, Bragg's law, radius ratios, d-spacing

🔴 Hard (16-22)

Lattice energy, thermal expansion, defect concentration, diffusion, phase transitions

🟣 Expert (23-25)

Complex structures, neutron diffraction, advanced crystallography applications

© 2025 Advanced Solid State Chemistry Problem Set

25 Comprehensive numerical problems covering all aspects of solid state chemistry

From basic crystal structures to advanced characterization techniques and complex oxide systems

freetestmaker.com
📚 freetestmaker.com | Anonymous | 2025-10-31 21:56:46
10²³ mol⁻¹
🎯 Find: Density (g/cm³)
Step 1: Calculate volume of unit cell
V = a³ = (3.61 × 10⁻⁸)³ = 4.70 × 10⁻²³ cm³
Step 2: Apply density formula
ρ = (Z × M) / (NA × V)
Step 3: Substitute values
ρ = (4 × 63.5) / (6.022 × 10²³ × 4.70 × 10⁻²³)
Step 4: Calculate
ρ = 254 / 28.3 = 8.98 g/cm³
Density = 8.98 g/cm³
7
📏 Atomic Radius Medium
Calculate the atomic radius of gold if it crystallizes in FCC structure with density 19.3 g/cm³.
📋 Given:
• FCC structure (Z = 4)
• Density ρ = 19.3 g/cm³
• Atomic mass of Au = 197 g/mol
• NA = 6.022 × 10²³ mol⁻¹
🎯 Find: Atomic radius (Å)
Step 1: Calculate unit cell volume
V = (Z × M) / (ρ × NA) = (4 × 197) / (19.3 × 6.022 × 10²³)
Step 2: Calculate volume
V = 788 / (1.162 × 10²⁶) = 6.78 × 10⁻²³ cm³
Step 3: Find edge length
a = ∛V = ∛(6.78 × 10⁻²³) = 4.08 × 10⁻⁸ cm
Step 4: For FCC, a = 2√2r
r = a/(2√2) = 4.08 × 10⁻⁸/(2√2) = 1.44 × 10⁻⁸ cm = 1.44 Å
Atomic Radius = 1.44 Å
8
🔍 Miller Indices Medium
A crystallographic plane intercepts the coordinate axes at (3a, 2a, 6a). Determine the Miller indices.
📋 Given:
• Intercepts on axes: x = 3a, y = 2a, z = 6a
🎯 Find: Miller indices (hkl)
Step 1: Write intercepts as coefficients
Intercepts: (3, 2, 6)
Step 2: Take reciprocals
Reciprocals: (1/3, 1/2, 1/6)
Step 3: Clear fractions (multiply by LCM = 6)
(1/3 × 6, 1/2 × 6, 1/6 × 6) = (2, 3, 1)
Miller Indices = (2 3 1)
9
📡 Bragg's Law Medium
X-rays of wavelength 1.54 Å are diffracted by a crystal at an angle of 30°. Calculate the d-spacing for first-order diffraction.
📋 Given:
• Wavelength λ = 1.54 Å
• Diffraction angle θ = 30°
• First-order diffraction (n = 1)
🎯 Find: d-spacing (Å)
Step 1: Apply Bragg's law
nλ = 2d sin θ
Step 2: Rearrange for d
d = nλ / (2 sin θ)
Step 3: Substitute values
d = (1 × 1.54) / (2 × sin 30°)
Step 4: Calculate
d = 1.54 / (2 × 0.5) = 1.54 Å
d-spacing = 1.54 Å
10
⚡ Ionic Radius Medium
Calculate the radius ratio for LiF and predict its structure type.
📋 Given:
• Ionic radius of Li⁺ = 0.76 Å
• Ionic radius of F⁻ = 1.33 Å
🎯 Find: Radius ratio and structure type
Step 1: Calculate radius ratio
r⁺/r⁻ = 0.76/1.33 = 0.571
Step 2: Compare with standard ranges
0.414 < 0.571 < 0.732
Step 3: Determine structure
This range corresponds to octahedral coordination
Radius ratio = 0.571, Structure = NaCl type (6:6)
11
🌊 Packing Efficiency Medium
Calculate the packing efficiency of body-centered cubic (BCC) structure.
📋 Given:
• BCC structure with 2 atoms per unit cell
• Atoms touch along body diagonal
🎯 Find: Packing efficiency (%)
Step 1: Relationship between a and r
Body diagonal = 4r = a√3, so r = a√3/4
Step 2: Volume of atoms
Vatoms = 2 × (4/3)πr³ = 2 × (4/3)π × (a√3/4)³
Step 3: Simplify
Vatoms = (π√3/8) × a³
Step 4: Packing efficiency
η = (π√3/8) × 100% = 68.0%
Packing Efficiency = 68.0%
12
📏 d-spacing Medium
Calculate the d-spacing for (100) planes in a simple cubic crystal with lattice parameter a = 4.0 Å.
📋 Given:
• Simple cubic structure
• Lattice parameter a = 4.0 Å
• Miller indices (100)
🎯 Find: d₁₀₀ spacing (Å)
Step 1: Apply d-spacing formula for cubic system
dhkl = a/√(h² + k² + l²)
Step 2: Substitute Miller indices (100)
h = 1, k = 0, l = 0
Step 3: Calculate
d₁₀₀ = 4.0/√(1² + 0² + 0²) = 4.0/√1 = 4.0 Å
d₁₀₀ = 4.0 Å
13
🔬 NaCl Structure Medium
NaCl has a face-centered cubic structure with edge length 5.64 Å. Calculate the density of NaCl.
📋 Given:
• FCC structure for NaCl (Z = 4 formula units)
• Edge length a = 5.64 Å = 5.64 × 10⁻⁸ cm
• Molecular mass of NaCl = 58.5 g/mol
🎯 Find: Density (g/cm³)
Step 1: Calculate unit cell volume
V = a³ = (5.64 × 10⁻⁸)³ = 1.79 × 10⁻²² cm³
Step 2: Apply density formula
ρ = (Z × M) / (NA × V)
Step 3: Substitute values
ρ = (4 × 58.5) / (6.022 × 10²³ × 1.79 × 10⁻²²)
Step 4: Calculate
ρ = 234 / 107.8 = 2.17 g/cm³
Density = 2.17 g/cm³
14
🎯 Interplanar Spacing Medium
Calculate the interplanar spacing for (111) planes in an FCC crystal with lattice parameter a = 3.6 Å.
📋 Given:
• FCC structure
• Lattice parameter a = 3.6 Å
• Miller indices (111)
🎯 Find: d₁₁₁ spacing (Å)
Step 1: Apply d-spacing formula
dhkl = a/√(h² + k² + l²)
Step 2: Substitute Miller indices (111)
h = 1, k = 1, l = 1
Step 3: Calculate
d₁₁₁ = 3.6/√(1² + 1² + 1²) = 3.6/√3
Step 4: Final answer
d₁₁₁ = 3.6/1.732 = 2.08 Å
d₁₁₁ = 2.08 Å
15
🧪 CsCl Structure Medium
CsCl crystallizes in a simple cubic structure where Cs⁺ and Cl⁻ ions are in contact along the body diagonal. If the edge length is 4.11 Å, calculate the ionic radii.
📋 Given:
• CsCl structure (8:8 coordination)
• Edge length a = 4.11 Å
• Ions touch along body diagonal
🎯 Find: rCs⁺ + rCl⁻
Step 1: Body diagonal relationship
Body diagonal = a√3
Step 2: Ion contact condition
Body diagonal = 2(rCs⁺ + rCl⁻)
Step 3: Equate expressions
a√3 = 2(rCs⁺ + rCl⁻)
Step 4: Calculate sum of radii
rCs⁺ + rCl⁻ = a√3/2 = 4.11 × 1.732/2 = 3.56 Å
rCs⁺ + rCl⁻ = 3.56 Å
16
⚡ Lattice Energy Hard
Calculate the lattice energy of NaCl using the Born-Landé equation.
📋 Given:
• Nearest neighbor distance r₀ = 2.82 Å
• Madelung constant M = 1.748
• Born exponent n = 7.5
• z⁺ = +1, z⁻ = -1
• e² = 2.31 × 10⁻¹⁹ J·m
🎯 Find: Lattice energy (kJ/mol)
Step 1: Born-Landé equation
U = -NAMz⁺z⁻e²/4πε₀r₀ × (1 - 1/n)
Step 2: Calculate (1 - 1/n)
(1 - 1/7.5) = (1 - 0.133) = 0.867
Step 3: Calculate constant term
NAe²/4πε₀ = 1.389 × 10⁻⁹ J·m/mol
Step 4: Calculate lattice energy
U = -1.389 × 10⁻⁹ × 1.748 × 1 × 1 × 0.867 / (2.82 × 10⁻¹⁰)
Step 5: Final calculation
U = -7.54 × 10⁵ J/mol = -754 kJ/mol
Lattice Energy = -754 kJ/mol
17
🌡️ Thermal Expansion Hard
A cubic crystal has a lattice parameter of 4.0 Å at 25°C. If the linear expansion coefficient is 12 × 10⁻⁶ K⁻¹, calculate the lattice parameter at 500°C.
📋 Given:
• Initial lattice parameter a₀ = 4.0 Å at T₀ = 25°C
• Linear expansion coefficient α = 12 × 10⁻⁶ K⁻¹
• Final temperature T = 500°C
🎯 Find: Lattice parameter at 500°C
Step 1: Calculate temperature difference
ΔT = 500 - 25 = 475 K
Step 2: Apply linear expansion formula
a = a₀(1 + αΔT)
Step 3: Calculate expansion factor
αΔT = 12 × 10⁻⁶ × 475 = 5.7 × 10⁻³
Step 4: Calculate new lattice parameter
a = 4.0 × (1 + 0.0057) = 4.0 × 1.0057 = 4.023 Å
Lattice parameter = 4.023 Å
18
🔍 Vacancy Concentration Hard
Calculate the equilibrium concentration of vacancies in copper at 1000°C if the vacancy formation energy is 0.9 eV.
📋 Given:
• Temperature T = 1000°C = 1273 K
• Vacancy formation energy Ev = 0.9 eV
• Boltzmann constant k = 8.617 × 10⁻⁵ eV/K
🎯 Find: Vacancy concentration (nv/N)
Step 1: Apply Boltzmann distribution
nv/N = exp(-Ev/kT)
Step 2: Calculate kT
kT = 8.617 × 10⁻⁵ × 1273 = 0.110 eV
Step 3: Calculate exponent
-Ev/kT = -0.9/0.110 = -8.18
Step 4: Calculate concentration
nv/N = exp(-8.18) = 3.0 × 10⁻⁴
Vacancy concentration = 3.0 × 10⁻⁴
19
📊 Diffusion Coefficient Hard
Calculate the diffusion coefficient of carbon in iron at 900°C using the Arrhenius equation.
📋 Given:
• Pre-exponential factor D₀ = 2.0 × 10⁻⁶ m²/s
• Activation energy Q = 80 kJ/mol
• Temperature T = 900°C = 1173 K
• Gas constant R = 8.314 J/(mol·K)
🎯 Find: Diffusion coefficient D (m²/s)
Step 1: Apply Arrhenius equation
D = D₀ exp(-Q/RT)
Step 2: Calculate RT
RT = 8.314 × 1173 = 9.75 × 10³ J/mol
Step 3: Calculate exponent
-Q/RT = -80 × 10³/9.75 × 10³ = -8.21
Step 4: Calculate diffusion coefficient
D = 2.0 × 10⁻⁶ × exp(-8.21) = 2.0 × 10⁻⁶ × 2.7 × 10⁻⁴
Step 5: Final answer
D = 5.4 × 10⁻¹⁰ m²/s
D = 5.4 × 10⁻¹⁰ m²/s
20
🎯 Scherrer Equation Hard
Calculate the crystallite size using the Scherrer equation for a powder with peak broadening.
📋 Given:
• X-ray wavelength λ = 1.54 Å
• Peak broadening β = 0.5° (in 2θ)
• Bragg angle θ = 22.5°
• Scherrer constant K = 0.9
🎯 Find: Crystallite size D (nm)
Step 1: Convert β to radians
β = 0.5° × π/180 = 8.73 × 10⁻³ rad
Step 2: Apply Scherrer equation
D = Kλ/(β cos θ)
Step 3: Calculate cos θ
cos(22.5°) = 0.924
Step 4: Calculate crystallite size
D = (0.9 × 1.54 × 10⁻¹⁰)/(8.73 × 10⁻³ × 0.924)
Step 5: Final answer
D = 1.72 × 10⁻⁸ m = 17.2 nm
Crystallite size = 17.2 nm
21
⚡ Band Gap Energy Hard
A semiconductor absorbs light with a threshold wavelength of 1100 nm. Calculate the band gap energy.
📋 Given:
• Threshold wavelength λ = 1100 nm = 1.1 × 10⁻⁶ m
• Planck's constant h = 6.626 × 10⁻³⁴ J·s
• Speed of light c = 3.0 × 10⁸ m/s
• 1 eV = 1.602 × 10⁻¹⁹ J
🎯 Find: Band gap energy (eV)
Step 1: Apply photon energy formula
E = hc/λ
Step 2: Calculate energy in Joules
E = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/(1.1 × 10⁻⁶)
Step 3: Calculate
E = 1.81 × 10⁻¹⁹ J
Step 4: Convert to eV
E = 1.81 × 10⁻¹⁹ J ÷ 1.602 × 10⁻¹⁹ J/eV
Step 5: Final answer
Eg = 1.13 eV
Band gap energy = 1.13 eV
22
🌡️ Phase Transition Hard
Calculate the transition temperature for the α-Fe → γ-Fe phase transformation.
📋 Given:
• Enthalpy change ΔH = 837 J/mol
• Entropy change ΔS = 7.6 J/(mol·K)
• At equilibrium ΔG = 0
🎯 Find: Transition temperature (K and °C)
Step 1: At equilibrium condition
ΔG = ΔH - TΔS = 0
Step 2: Rearrange for temperature
T = ΔH/ΔS
Step 3: Substitute values
T = 837 J/mol ÷ 7.6 J/(mol·K)
Step 4: Calculate
T = 110 K
Step 5: Convert to Celsius
T = 110 - 273 = -163°C
Transition Temperature = 110 K (-163°C)
23
🔬 Complex Structure Expert
Calculate the density of spinel (MgAl₂O₄) which has a cubic unit cell with 8 formula units and edge length 8.08 Å.
📋 Given:
• Spinel structure with Z = 8 formula units
• Edge length a = 8.08 Å = 8.08 × 10⁻⁸ cm
• Molecular mass of MgAl₂O₄ = 142.3 g/mol
• NA = 6.022 ×