Physical Chemistry-3 (Solid State, Surface Chemistry, Spectroscopy and Quantum Mechanics) BCH5T10

Syllabus topics (RTMNU)

• Solids and their classification
• Difference between crystalline and amorphous solids
• Classification of crystalline solids
• Laws of crystallography: Law of constancy of interfacial angles, Law of rationality of indices, Law of symmetry
• Elements of a crystal
• Space lattice, Unit cell, Bravais lattices, crystal systems
• Identification of crystal planes, interplanar distance in cubic systems
• X-ray diffraction by crystal, derivation of Bragg's equation
• Determination of crystal structure of NaCl, KCl and CsCl
• Laue's method and powder method
• Numerical problems

Solids and their Classification; Crystalline vs Amorphous Solids

Solids have definite shape and volume because their particles (atoms, ions, molecules) are held close by strong forces and can only vibrate about fixed positions.

They are classified as crystalline (regular, repeating 3-D arrangement — long-range order; e.g. NaCl, diamond, ice) and amorphous (irregular arrangement — only short-range order; e.g. glass, rubber, plastics). Amorphous solids are sometimes called super-cooled liquids.

Classification of crystalline solids (by bonding):

Laws of Crystallography and Elements of a Crystal

1. Law of constancy of interfacial angles (Steno): the angles between corresponding faces of crystals of a substance are always the same, whatever the size or shape of the crystal.
2. Law of rationality of indices (Haüy): the intercepts of any crystal face on the crystallographic axes are small whole-number multiples (or infinity) of the unit intercepts a, b, c. So Weiss indices and Miller indices are small integers.
3. Law of symmetry: all crystals of a substance have the same elements of symmetry.

Elements of symmetry: (a) plane of symmetry — divides the crystal into two mirror-image halves; (b) axis of symmetry — rotation by 360°/n gives the same appearance (n = 2, 3, 4, 6: diad, triad, tetrad, hexad); (c) centre of symmetry — every face has an identical face at equal distance on the opposite side. A cube has 23 elements of symmetry: 9 planes + 13 axes (3 tetrad, 4 triad, 6 diad) + 1 centre.

Space Lattice, Unit Cell, Bravais Lattices and Crystal Systems

Space lattice: a regular 3-D arrangement of points, each point representing a particle, with identical surroundings. Unit cell: the smallest repeating unit of the lattice which, repeated in three dimensions, gives the whole crystal. It is described by edge lengths a, b, c and angles α, β, γ.

There are 7 crystal systems and 14 Bravais lattices (Bravais, 1848):

Identification of Crystal Planes (Miller Indices) and Interplanar Distance in Cubic Systems

Miller indices (hkl): (1) find intercepts of the plane on the axes in terms of a, b, c; (2) take reciprocals; (3) clear fractions by multiplying with the LCM; (4) write in brackets without commas.

Example: intercepts 2a, 3b, ∞c → reciprocals 1/2, 1/3, 0 → × 6 → (320).

X-ray Diffraction by Crystal and Derivation of Bragg's Equation

Crystals act as 3-D diffraction gratings for X-rays because interplanar spacings (1–3 Å) are similar to X-ray wavelengths.

Derivation: Let X-rays of wavelength λ fall at glancing angle θ on parallel planes separated by distance d. Ray 1 reflects from the upper plane at A; ray 2 from the next plane at B. Draw perpendiculars AC and AD from A on the incident and reflected parts of ray 2. Extra path travelled by ray 2 = CB + BD. In triangles ACB and ADB, CB = BD = d sin θ. So path difference = 2d sin θ. For constructive interference (bright spot), path difference = whole number of wavelengths:

Worked example: X-rays of λ = 1.54 Å give first-order reflection from a NaCl crystal at θ = 15.9°. Find d. d = nλ ÷ (2 sin θ) = 1 × 1.54 ÷ (2 × 0.274) = 2.81 Å (matches d₂₀₀ of NaCl).

Determination of Crystal Structure of NaCl, KCl and CsCl; Laue and Powder Methods

Laue method: a stationary single crystal is exposed to continuous (white) X-rays; a photographic film records a pattern of spots (Laue pattern) showing the symmetry of the crystal.

Powder method (Debye–Scherrer): a finely powdered sample (crystallites in all orientations) is exposed to monochromatic X-rays; reflections form cones recorded as arcs on a strip film (or peaks in a diffractometer). From θ values, d is calculated and the ratio of d values gives the lattice type.

Bragg's study of NaCl and KCl (rotating crystal): for planes (100), (110), (111), ratio d₁₀₀ : d₁₁₀ : d₁₁₁ = 1 : 0.707 : 1.154 for fcc, compared with 1 : 1.414 : 0.577 for bcc and 1 : 0.707 : 0.577 for simple cubic (taking the first-order reflections seen). NaCl and KCl gave the fcc ratio — face-centred cubic, coordination number 6:6, 4 formula units per unit cell. In KCl, K⁺ and Cl⁻ scatter X-rays almost equally (same number of electrons, 18), so it behaves like a simple cubic lattice of edge a/2.

CsCl: body-centred arrangement — Cs⁺ at the body centre and Cl⁻ at the corners (two interpenetrating simple cubic lattices); coordination 8:8; 1 formula unit per cell.

Likely exam questions

1. (Long, 8 marks) Derive Bragg's equation. X-rays of wavelength 1.54 Å give a first-order reflection at 11.2° from a crystal; calculate the interplanar spacing.
2. (Long, 8 marks) Explain the laws of crystallography and describe the seven crystal systems with their Bravais lattices.
3. (Short, 4 marks) Distinguish between crystalline and amorphous solids.
4. (Short, 4 marks) Find the Miller indices of a plane with intercepts a, 2b and ∞c.
5. (Short, 4 marks) Describe the structure of CsCl.

Syllabus topics (RTMNU)

• Adsorption: Introduction
• Factors affecting adsorption of gases by solids
• Difference between physical and chemical adsorption
• Adsorption Isotherms: Freundlich Adsorption Isotherm, Langmuir Adsorption Isotherm
• B.E.T. Equation (no derivation), Application of B.E.T. Equation in Determination of Surface Area of Adsorbent
• Applications of Adsorption
• Catalysis: Introduction, General characteristics of catalyst
• Positive and negative catalysis, Promoters and Inhibitors, Action of catalytic promoters and inhibitors
• Homogeneous & Heterogeneous Catalysis, Industrial applications (Haber process), auto catalysis
• Kinetics of Enzyme-Catalysed Reactions - Michaelis-Menten Equation

Adsorption: Introduction and Factors Affecting Adsorption of Gases by Solids

Adsorption is the accumulation (higher concentration) of molecules of a substance at the surface of another substance, compared with the bulk. The substance adsorbed is the adsorbate (e.g. H₂, dye); the surface is the adsorbent (e.g. charcoal, silica gel). The reverse is desorption. In absorption the substance spreads uniformly through the bulk (e.g. water in a sponge). Adsorption is exothermic (ΔH negative) and decreases entropy.

Factors affecting adsorption of gases by solids:

1. Nature of gas: easily liquefiable gases (high critical temperature, e.g. NH₃, SO₂) are adsorbed more than permanent gases (H₂, N₂).
2. Nature and surface area of adsorbent: porous, finely divided solids (activated charcoal) adsorb more.
3. Pressure: adsorption increases with pressure (until saturation).
4. Temperature: physical adsorption decreases with rise in temperature (Le Chatelier, exothermic); chemisorption first increases then decreases.
5. Activation of adsorbent — heating charcoal in steam increases pores and surface area.

Difference between Physical and Chemical Adsorption

Adsorption Isotherms: Freundlich and Langmuir

An adsorption isotherm is a graph of the amount adsorbed per gram of adsorbent (x/m) against pressure (or concentration) at constant temperature.

Freundlich isotherm (empirical): x/m = k P^1/n (n > 1). Taking log: log (x/m) = log k + (1/n) log P. A plot of log (x/m) vs log P is a straight line with slope 1/n and intercept log k. For solutions, P is replaced by concentration C. Limitation: fails at high pressure (does not show saturation).

Langmuir isotherm (1916) — assumptions: (1) adsorption forms a monolayer; (2) the surface has a fixed number of identical sites; (3) no interaction between adsorbed molecules; (4) dynamic equilibrium between adsorption and desorption.

Derivation: let θ = fraction of surface covered. Rate of adsorption = k_a P (1 − θ); rate of desorption = k_d θ. At equilibrium k_aP(1 − θ) = k_dθ, so θ = k_aP ÷ (k_d + k_aP) = bP ÷ (1 + bP), where b = k_a/k_d. Since x/m ∝ θ, x/m = aP ÷ (1 + bP).

B.E.T. Equation and Determination of Surface Area

Brunauer, Emmett and Teller (1938) extended Langmuir's idea to multilayer physical adsorption. The B.E.T. equation (no derivation):

Surface area determination: N₂ adsorption is measured at 77 K for several P/P₀ (about 0.05–0.3). A plot of P/[V(P₀ − P)] vs P/P₀ is a straight line; from slope and intercept, V_m = 1/(slope + intercept). Then surface area = (V_m/22414 cm³) × N_A × σ, where σ = area of one N₂ molecule (about 0.162 nm²). Used for catalysts, charcoal and powders.

Applications of Adsorption

• Gas masks (charcoal adsorbs poisonous gases); removal of moisture by silica gel.
• Decolourising sugar solution with animal charcoal; water purification by alum and activated carbon.
• Heterogeneous catalysis (reactants adsorb on catalyst surface).
• Chromatography (adsorption chromatography, TLC).
• Creating high vacuum (charcoal in Dewar flasks); froth flotation of sulphide ores; adsorption indicators in titration; dyeing.

Catalysis: Introduction and General Characteristics

Catalyst: a substance that changes the rate of a chemical reaction without being consumed, by providing an alternative path of lower activation energy. The phenomenon is catalysis (term by Berzelius, 1835).

Characteristics: (1) remains unchanged in mass and chemical composition at the end (may change physically); (2) small amount is enough; (3) does not start a reaction (mostly) but speeds it; (4) does not change equilibrium position or ΔG, only helps reach equilibrium faster; (5) is specific; (6) has optimum temperature; (7) activity is increased by promoters and decreased by poisons.

Positive and Negative Catalysis; Promoters and Inhibitors

• Positive catalysis: rate increases — MnO₂ in 2KClO₃ → 2KCl + 3O₂.
• Negative catalysis: rate decreases — glycerol slows the decomposition of H₂O₂; tetraethyl lead acts as anti-knock in petrol.
• Promoter: substance that increases the activity of a catalyst though not a catalyst itself — Mo (or K₂O, Al₂O₃) promotes Fe in the Haber process. Action: creates more active sites, changes lattice spacing, increases adsorption.
• Inhibitor/poison: substance that decreases catalyst activity — CO or H₂S poisons Fe in the Haber process; As₂O₃ poisons Pt in the contact process. Action: preferential adsorption blocking active sites.

Homogeneous and Heterogeneous Catalysis; Haber Process; Autocatalysis

Homogeneous: catalyst and reactants in the same phase — 2SO₂(g) + O₂(g) → 2SO₃(g) with NO(g) (lead chamber process); acid-catalysed hydrolysis of esters. Heterogeneous: different phases — usually a solid catalyst with gases; works by adsorption on the surface (adsorption theory: diffusion → adsorption → reaction → desorption → diffusion away).

Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ. Catalyst: finely divided iron with molybdenum/K₂O–Al₂O₃ promoters; about 450–500 °C and 200 atm. Other industrial examples: V₂O₅ in the contact process for H₂SO₄; Pt–Rh in Ostwald process for HNO₃; Ni in hydrogenation of oils (vanaspati).

Autocatalysis: a product of the reaction itself acts as catalyst. Example: in the titration of oxalic acid with KMnO₄, Mn²⁺ formed catalyses the reaction — the first drops decolourise slowly, then fast. Rate–time curve is S-shaped (sigmoid).

Kinetics of Enzyme-Catalysed Reactions: Michaelis–Menten Equation

Mechanism: E + S ⇌ ES (k₁, k₋₁); ES → E + P (k₂). Applying the steady-state approximation to [ES]: k₁[E][S] = (k₋₁ + k₂)[ES]. With [E]₀ = [E] + [ES] and rate v = k₂[ES], we get

Likely exam questions

1. (Long, 8 marks) State the assumptions of Langmuir adsorption isotherm and derive it. How does it differ from the Freundlich isotherm?
2. (Long, 8 marks) Derive the Michaelis–Menten equation for enzyme-catalysed reactions and give the significance of Km.
3. (Short, 4 marks) Distinguish between physisorption and chemisorption.
4. (Short, 4 marks) What are promoters and catalytic poisons? Give examples.
5. (Short, 4 marks) Explain autocatalysis with an example.

Syllabus topics (RTMNU)

• Rotational Spectroscopy: Rotational spectra of diatomic molecules, Energy levels of rigid rotor
• Selection rule for transition between energy levels
• Expression for wave number of spectral lines in terms of rotational constant (B) and rotational quantum number (J)
• Intensity of spectral lines
• Types of molecules showing rotational spectra
• Application of rotational spectra for determination of moment of inertia and bond length
• Introduction to non-rigid rotor, Numerical problems
• Vibrational Spectroscopy: Vibrational energy levels of simple harmonic oscillator, selection rules
• Types of molecules showing vibrational spectra
• Vibrational energy level of anharmonic oscillator, selection rule, idea of overtones
• Vibrational-Rotational spectra: P, Q and R branches
• Structural information from infrared spectra

Rotational Spectra of Diatomic Molecules: Rigid Rotor Energy Levels

Rotational (microwave) spectroscopy studies absorption of microwave radiation (about 1–100 cm⁻¹) that changes the rotational energy of molecules.

Rigid rotor model: a diatomic molecule is treated as two masses m₁ and m₂ joined by a rigid bond of length r. Moment of inertia I = μr², where reduced mass μ = m₁m₂/(m₁ + m₂).

Solving the Schrödinger equation gives energy levels:

Levels are at 0, 2B, 6B, 12B, 20B … so the gaps increase: 2B, 4B, 6B, 8B …

Selection Rule and Wavenumber of Spectral Lines

Selection rules: (1) the molecule must have a permanent dipole moment; (2) ΔJ = ±1.

For absorption J → J + 1: ν̄ = ε_J+1 − ε_J = B[(J + 1)(J + 2) − J(J + 1)] = 2B(J + 1) cm⁻¹.

Lines appear at 2B, 4B, 6B, 8B …; spacing between successive lines is constant = 2B. So the spectrum is a series of equally spaced lines.

Intensity of Spectral Lines

Intensity depends on the population of the lower level. Population N_J/N₀ = (2J + 1) e^−E_J/kT. The (2J + 1) degeneracy factor increases with J while the Boltzmann factor decreases, so intensity first rises, reaches a maximum and then falls. The most populated level: J_max = √(kT/2hcB) − ½.

Types of Molecules Showing Rotational Spectra

Only molecules with a permanent dipole moment are microwave active: HCl, CO, NO, H₂O, NH₃. Homonuclear diatomics (H₂, N₂, O₂) and symmetric molecules (CO₂, CH₄, C₆H₆) show no pure rotational spectrum. Molecules are also classed by moments of inertia: linear, symmetric top (NH₃, CH₃Cl), spherical top (CH₄), asymmetric top (H₂O).

Determination of Moment of Inertia and Bond Length; Non-rigid Rotor; Numerical Problems

From the line spacing (2B) we get B → I = h/(8π²Bc) → r = √(I/μ).

Problem 1: The rotational lines of ¹²C¹⁶O are 3.86 cm⁻¹ apart. Find I and r. (h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10¹⁰ cm s⁻¹, 1 amu = 1.6605 × 10⁻²⁷ kg)

• 2B = 3.86 → B = 1.93 cm⁻¹.
• I = h/(8π²Bc) = 6.626 × 10⁻³⁴ ÷ (8 × 9.870 × 1.93 × 2.998 × 10¹⁰) = 1.45 × 10⁻⁴⁶ kg m².
• μ = (12 × 15.995)/(27.995) amu = 6.856 amu = 1.138 × 10⁻²⁶ kg.
• r = √(I/μ) = √(1.45 × 10⁻⁴⁶ ÷ 1.138 × 10⁻²⁶) = 1.13 × 10⁻¹⁰ m = 1.13 Å.

Problem 2: For H³⁵Cl, B = 10.59 cm⁻¹. μ = (1.008 × 34.97)/(35.978) amu = 0.980 amu = 1.627 × 10⁻²⁷ kg. I = 6.626 × 10⁻³⁴ ÷ (8π² × 10.59 × 2.998 × 10¹⁰) = 2.64 × 10⁻⁴⁷ kg m²; r = √(2.64 × 10⁻⁴⁷ / 1.627 × 10⁻²⁷) = 1.27 Å.

Non-rigid rotor: in a real molecule the bond stretches as rotation becomes faster (centrifugal distortion), so I increases and B effectively decreases at high J. ε_J = BJ(J + 1) − DJ²(J + 1)², where D is the centrifugal distortion constant (very small). Lines come slightly closer at high J: ν̄ = 2B(J + 1) − 4D(J + 1)³.

Vibrational Spectroscopy: Simple Harmonic Oscillator and Selection Rules

Vibrational spectra lie in the infrared region (about 400–4000 cm⁻¹). For a simple harmonic oscillator, restoring force F = −kx (k = force constant). Classical frequency ω̄ = (1/2πc)√(k/μ) cm⁻¹.

Force constant example: HCl absorbs at 2886 cm⁻¹. k = 4π²c²ω̄²μ = 4π² × (2.998 × 10⁸ m s⁻¹)² × (2.886 × 10⁵ m⁻¹)² × 1.627 × 10⁻²⁷ kg ≈ 481 N m⁻¹.

Types of Molecules Showing Vibrational Spectra; Anharmonic Oscillator and Overtones

IR active: vibrations that change the dipole moment — HCl, CO, H₂O, asymmetric stretch and bending of CO₂. IR inactive: H₂, N₂, O₂, symmetric stretch of CO₂. A non-linear molecule of N atoms has 3N − 6 vibrational modes; a linear one 3N − 5.

Anharmonic oscillator: real bonds do not obey Hooke's law at large stretching and can break (dissociate). The Morse curve describes it. Energy: ε_v = (v + ½)ω̄_e − (v + ½)²ω̄_ex_e, where x_e is the anharmonicity constant. Levels come closer as v increases. Selection rule: Δv = ±1, ±2, ±3 … (higher ones weaker). Lines: fundamental (0 → 1) ≈ ω̄_e(1 − 2x_e); first overtone (0 → 2) ≈ 2ω̄_e(1 − 3x_e); second overtone (0 → 3) ≈ 3ω̄_e(1 − 4x_e) — overtones lie at slightly less than 2 and 3 times the fundamental.

Vibrational–Rotational Spectra: P, Q and R Branches

A vibrational transition is accompanied by rotational changes, so in the gas phase each IR band shows fine structure. Selection rules: Δv = ±1, ΔJ = ±1 (ΔJ = 0 allowed only for some molecules, e.g. NO).

• R branch (ΔJ = +1): lines on the high-wavenumber side: ν̄ = ν̄₀ + 2B(J + 1).
• P branch (ΔJ = −1): lines on the low-wavenumber side: ν̄ = ν̄₀ − 2BJ.
• Q branch (ΔJ = 0): at ν̄₀; absent in HCl, so a gap of 4B appears at the band centre.

Structural Information from Infrared Spectra

IR spectra identify functional groups by characteristic absorption bands; the 1500–400 cm⁻¹ region is the fingerprint region, unique to each compound.

IR also shows hydrogen bonding (broadening, shift to lower wavenumber), purity of samples and progress of reactions (e.g. disappearance of C=O on reduction).

Likely exam questions

1. (Long, 8 marks) Derive the expression for the energy levels of a rigid diatomic rotor and show that rotational lines are equally spaced. The line spacing in the rotational spectrum of a molecule is 20.68 cm⁻¹ and its reduced mass is 1.63 × 10⁻²⁷ kg; calculate its bond length.
2. (Long, 8 marks) Discuss the vibrational energy levels of harmonic and anharmonic oscillators. Explain overtones and the P, Q, R branches of vibrational–rotational spectra.
3. (Short, 4 marks) Which molecules show pure rotational spectra? Why?
4. (Short, 4 marks) What is zero-point energy?
5. (Short, 4 marks) Give the IR absorption regions of any four functional groups.

Syllabus topics (RTMNU)

• Failure of classical mechanics in explaining Black body radiation, Photoelectric effect, and heat capacity of solids
• Bohr's model of Hydrogen atom and hydrogen spectrum (Derivation Excluded)
• Planck's quantum theory
• De Broglie's hypothesis (Derivation and experimental proof)
• Heisenberg's uncertainty principle (Explanation and experimental proof), Numerical problems
• Introduction to wave functions (ψ), Schrödinger wave equation
• Eigen values and Eigen functions, well behaved wave functions
• Interpretation of wave function (ψ) and its square (ψ²), Normalized and orthogonal wave functions
• Linear Operators, Commutation of operators
• Postulates of quantum mechanics, Derivation of Schrödinger wave equation from postulates of quantum mechanics
• Application of Schrödinger wave equation for a particle in a one-dimensional box, Numerical problems

Failure of Classical Mechanics: Black Body Radiation, Photoelectric Effect, Heat Capacity of Solids

1. Black body radiation: a perfect black body absorbs all radiation falling on it and emits radiation of all wavelengths when heated. Experimentally, the energy distribution curve rises to a maximum at a particular wavelength (λ_max), which shifts to shorter wavelength as temperature rises (Wien's law, λ_maxT = constant). The classical Rayleigh–Jeans law predicted that energy keeps increasing at short wavelengths without limit — the "ultraviolet catastrophe". Classical theory could not explain the fall at short wavelengths.

2. Photoelectric effect: when light of suitable frequency falls on a metal surface, electrons are emitted. Observations: (a) no emission below a threshold frequency ν₀, however intense the light; (b) kinetic energy of electrons depends on frequency, not intensity; (c) emission is instantaneous; (d) number of electrons depends on intensity. Classical wave theory predicted that energy depends on intensity and that any frequency would work with enough time — it failed.

3. Heat capacity of solids: classical theory (Dulong–Petit law) gives molar heat capacity C_v = 3R ≈ 25 J K⁻¹ mol⁻¹ for all solids at all temperatures. Experiment shows C_v falls towards zero at low temperature (and diamond has low C_v even at room temperature). Einstein and Debye explained this using quantised vibrational energy.

Planck's Quantum Theory

Max Planck (1900) explained black-body radiation by assuming:

1. Energy is emitted or absorbed not continuously but in small discrete packets called quanta (for light, photons).
2. Energy of one quantum is proportional to frequency: E = hν = hc/λ, where h = 6.626 × 10⁻³⁴ J s (Planck's constant).
3. A body can emit or absorb energy only in whole-number multiples of hν: E = nhν.

Einstein (1905) used it to explain the photoelectric effect: hν = hν₀ + ½mv², i.e. photon energy = work function + kinetic energy of electron.

Bohr's Model of the Hydrogen Atom and the Hydrogen Spectrum

Postulates (Bohr, 1913): (1) electrons revolve in fixed circular orbits (stationary states) without radiating energy; (2) angular momentum is quantised: mvr = nh/2π; (3) energy is emitted or absorbed only when an electron jumps between orbits: ΔE = E₂ − E₁ = hν.

Results (derivation excluded): radius r_n = 0.529 n² Å; energy E_n = −13.6/n² eV (−2.18 × 10⁻¹⁸/n² J) for H.

Limitations: fails for multi-electron atoms, fine structure, Zeeman effect, and contradicts the uncertainty principle.

de Broglie's Hypothesis: Derivation and Experimental Proof

de Broglie (1924) proposed that moving material particles also have wave nature (dual nature of matter).

Derivation: from Einstein, E = mc²; from Planck, E = hν = hc/λ. Equating: mc² = hc/λ → λ = h/mc. For a particle moving with velocity v, replace c by v:

Experimental proof: Davisson and Germer (1927) directed a beam of electrons on a nickel crystal and observed a diffraction pattern, with wavelength matching h/mv; G.P. Thomson obtained diffraction rings by passing electrons through thin metal foil. Application: electron microscope.

Heisenberg's Uncertainty Principle: Explanation, Proof and Numerical Problems

Statement (1927): it is impossible to determine simultaneously and exactly both the position and momentum of a microscopic particle.

Explanation (γ-ray microscope thought experiment): to locate an electron precisely we must use light of very short wavelength (γ-rays). Such a photon has high momentum and, on striking the electron, changes its momentum by an uncertain amount. Using longer wavelength disturbs momentum less but gives poor position. So accuracy in one is lost in the other. Consequences: the idea of fixed orbits (Bohr) is not valid; we speak of probability of finding the electron.

Problem: Uncertainty in the velocity of an electron is 1 × 10⁵ m s⁻¹. Find the minimum uncertainty in its position. Δx = h/(4π m Δv) = 6.626 × 10⁻³⁴ ÷ (4 × 3.1416 × 9.109 × 10⁻³¹ × 10⁵) = 5.79 × 10⁻¹⁰ m (about 5.8 Å — larger than the atom itself).

Wave Functions, Schrödinger Wave Equation, Eigenvalues and Eigenfunctions

The state of a particle is described by a wave function ψ (psi), a function of coordinates (and time). The time-independent Schrödinger equation in one dimension:

An equation of the form Âψ = aψ is an eigenvalue equation: ψ is an eigenfunction of operator  and the constant a is its eigenvalue. Example: d/dx (e^3x) = 3e^3x, so e^3x is an eigenfunction of d/dx with eigenvalue 3.

Well-behaved wave function: must be (1) single-valued, (2) finite everywhere, (3) continuous, with a continuous first derivative, and (4) square-integrable (normalisable).

Interpretation of ψ and ψ²; Normalised and Orthogonal Wave Functions

ψ itself has no physical meaning (it can be negative or complex). Born interpretation: ψψ* (or ψ² for real ψ) gives the probability density — ψ²dτ is the probability of finding the particle in volume element dτ.

• Normalised: ∫ψψ* dτ = 1 over all space (the particle must be somewhere).
• Orthogonal: ∫ψ_iψ_j* dτ = 0 for i ≠ j. Functions both normalised and mutually orthogonal are orthonormal.

Linear Operators, Commutation of Operators and Postulates of Quantum Mechanics

Operator: a symbol for a mathematical operation (d/dx, ×x, √). Linear operator: Â(f + g) = Âf + Âg and Â(cf) = cÂf — e.g. d/dx is linear; √ and squaring are not. Commutation: Â and B̂ commute if ÂB̂f = B̂Âf, i.e. commutator [Â, B̂] = ÂB̂ − B̂Â = 0. Example: x and d/dx do not commute: [x, d/dx]f = −f. Non-commuting operators (position and momentum) cannot be measured exactly together — the basis of the uncertainty principle.

Postulates:

1. The state of a system is completely described by a well-behaved wave function ψ; ψψ*dτ gives probability.
2. Every observable (measurable quantity) has a linear (Hermitian) operator: position x → x; momentum p_x → (h/2πi) d/dx; kinetic energy → −(h²/8π²m) d²/dx².
3. The only possible measured values of an observable are the eigenvalues of its operator: Âψ = aψ.
4. The average (expectation) value: ⟨a⟩ = ∫ψ*Âψ dτ ÷ ∫ψ*ψ dτ.
5. The wave function evolves with time according to the time-dependent Schrödinger equation.

Derivation of the Schrödinger equation from postulates: total energy = kinetic + potential, so Ĥ = T̂ + V. Using T̂ = p̂_x²/2m with p̂_x = (h/2πi) d/dx, p̂_x² = −(h²/4π²) d²/dx², so T̂ = −(h²/8π²m) d²/dx². By postulate 3, Ĥψ = Eψ: −(h²/8π²m) d²ψ/dx² + Vψ = Eψ, which rearranges to d²ψ/dx² + (8π²m/h²)(E − V)ψ = 0.

Particle in a One-Dimensional Box; Numerical Problems

A particle of mass m is confined in a box of length L with V = 0 inside (0 < x < L) and V = ∞ at the walls and outside, so ψ = 0 at x = 0 and x = L.

Inside: d²ψ/dx² + k²ψ = 0, with k² = 8π²mE/h². General solution ψ = A sin kx + B cos kx. Boundary ψ(0) = 0 gives B = 0; ψ(L) = 0 gives sin kL = 0, so kL = nπ, k = nπ/L (n = 1, 2, 3 …; n = 0 is not allowed as ψ would vanish).

Problem: An electron is in a 1-D box of length 1 nm. Find E₁ and the wavelength absorbed for n = 1 → 2. E₁ = h²/8mL² = (6.626 × 10⁻³⁴)² ÷ (8 × 9.109 × 10⁻³¹ × 10⁻¹⁸) = 6.02 × 10⁻²⁰ J (0.376 eV). ΔE = E₂ − E₁ = 3E₁ = 1.807 × 10⁻¹⁹ J. λ = hc/ΔE = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 1.807 × 10⁻¹⁹ = 1.10 × 10⁻⁶ m (≈ 1100 nm).

Applications: explains colour of conjugated dyes (π electrons in a "box"), quantisation and zero-point energy.

Likely exam questions

1. (Long, 8 marks) Apply the Schrödinger wave equation to a particle in a one-dimensional box and obtain expressions for energy and normalised wave function. Calculate the energy of an electron in the first level of a 0.5 nm box.
2. (Long, 8 marks) Derive de Broglie's equation and give its experimental proof. State Heisenberg's uncertainty principle and calculate the uncertainty in position of an electron whose velocity is uncertain by 2 × 10⁴ m s⁻¹.
3. (Short, 4 marks) How did classical mechanics fail to explain black-body radiation?
4. (Short, 4 marks) What are the conditions for a well-behaved wave function?
5. (Short, 4 marks) Explain linear operators and commutation of operators with examples.