Physical Chemistry II CHEM 3340 Structure of Matter Quantum Theory PDF

Physical Chemistry II CHEM 3340 Structure of Matter Quantum Theory: Introduction Structure of Matter Classical View Matter consists of particles: molecules, atoms, subatomic particles (proton, neutron, electron) The particles are Very small (atomic distance: 0.1 nm = 10-10 m) Extremely large amount...

Physical Chemistry II CHEM 3340 Structure of Matter Quantum Theory: Introduction Structure of Matter Classical View Matter consists of particles: molecules, atoms, subatomic particles (proton, neutron, electron) The particles are Very small (atomic distance: 0.1 nm = 10-10 m) Extremely large amount in macroscopic quantities (Avogadro number, 6.022x1023) A particle travels in a trajectory: a path with precise position and momentum at each instant The trajectory can be obtained from the classical physics (Isaac Newton, F=ma): all we need is to characterize the forces between the particles (e.g., gravitational, electrostatic) A particle can be accelerated to any energy level (kinetic energy, Ek = 1/2 mv2, momentum: p=mv, Ek =p2/(2m)) Kinetic vs. Potential Energy Energy: System’s capability to do some work Potential Energy: The energy of a particle as a result of its position Some work was done to put the particle from a reference position to the present position. This work is the potential energy (V) Example 1: Gravitational energy: work to move a particle from a reference position (surface of earth) to a height of h : w = V = mgh Example 2: Potential energy of two charged particle (q1, q2) at distance r : V = q1 q2 / (4πε0r) (this is the work required to move the two particles from in nite distance to r, ε0 is the vacuum permittivity) Total energy: sum of kinetic and potential energy, E = V + Ek fi Electromagnetic Field Waves are distinct from particles Electromagnetic eld: oscillating electric and magnetic disturbance that spreads as a harmonic wave through empty space (vacuum). Such waves are generated by the acceleration of electric charge (motion of electrons in the antenna of a radio transmitter) It has two components: Electric eld: acts on charged particles Magnetic eld: acts only on moving charged particles fi Wave travels at speed of light: c=3x108 m/s fi fi Electromagnetic eld Wavelength, λ: the distance between the neighboring peaks [m] Frequency, ν: number of times per second at which its displacement at a xed point returns to its original value [1/s]=[Hz] Speed, wavelength, and frequency are interrelated: λν=c Short wavelength: -> high frequency Wavenumber: number of waves per unit length [1/cm] ⇥ 1 fi ⇥˜ = = c fi Electromagnetic Spectrum Classi cation of the electromagnetic eld according to its frequency (wavelength) White light: mixture of radiation with wavelengths 380 nm - 700 nm There is an intimate relationship between matter and wave: Objects emit electromagnetic radiation!!! fi fi Classical physics cannot interpret important features of the spectrum! Systems can take up energy only in discrete amounts Black Body Radiation Hot objects emit electromagnetic radiation: the atoms and electrons are ceaselessly being accelerated: the atoms vibrate around their mean positions Glowing iron bar: color indicative of temperature As temperature is increased, more blue is mixed into the red -> white Power emitted by a black body Two observations 1. The maximum shifts to small wavelength (Wien’s displacement law) T λmax = constant=2.9 K mm 2. The overall power divided by the surface area of the black body increases with temperature (StefanBoltzmann Law ) M = a T4 M: emmitance [nW/m2] a= 56.7 nW/(m2K4) At 3000K: 460 W/cm2 Blue Red Classical Physics Approach Rayleigh-Jeans law ⇤= 8⇥kT 4 k=1.381x10-23 J/K: Boltzmann constant Problem: Ultraviolet catastrophe Black body radiation should consist of mainly small wavelength radiation: ultraviolet radiation, x-rays (’Truly black’ bodies do not exist --There is no darkness in the world!) Quantization of Energy Max Planck proposed (1900) that not all energy levels are accessible to the electromagnetic oscillators Permitted energies E = nh n = 0, 1, 2,... ν: frequency of the oscillator h: Planck constant: 6.626x10-24 Js Based on this hypothesis ⇤= 8⇥hc 5 (ehc/( kT 1) Planck’s hypothesis E = nh n = 0, 1, 2,... Atoms with low energies are not excited (cannot vibrate) until they reach an energy of hν Interprets the black body radiation (Maximum in power density, Wien’s displacement law, Stefan-Boltzmann law) In the limit of large energies (E is large, n is large) recovers the Rayleigh-Jeans law Introduced a fundamental property of matter: quantum levels Quantum levels in atomic and molecular spectra Spectroscopy: detection and analysis of the electromagnetic radiation absorbed, emitted, or scattered by a substance Spectrum: Light intensity transmitted (or scattered) by a molecule as a function frequency (wavelength, wavenumber) Features of atomic and molecular spectra Absorption of SO2 Emission of excited iron atoms Appearance of discrete frequencies! Classical physics failed to account for these spectra! Spectroscopic transitions among quantized energy levels Atoms/molecules are con ned to discrete energy values When the energy decreases by an amount of ∆E, the energy is carried away as radiation frequency ν, and an emission ‘line’ is generated. fi Bohr frequency condition: E=h Wave-particle duality Photons: ‘Light particles’ Planck’s hypothesis: electromagnetic oscillator of frequency ν can possess only energies of nhν Transitions occur in multiples of hν: maybe the electromagnetic radiation consists of particles? Photons: particles of electromagnetic radiation Radiation: a ray of light consists of a stream of photons, each one having an energy of hν and speed c As the light intensity is increased, the number of photons is increased, but each photon has energy of hν Test of photon hypothesis: photoelectric effect Ejection of electrons from metals with ultraviolet radiation No electrons are ejected (regardless of the intensity of the radiation) unless the frequency exceeds a threshold value characteristic of the metal The kinetic energy of the ejected electrons varies linearly with the frequency of incident radiation but it is independent of the intensity Even at low light intensities, electrons are ejected immediately if the frequency is above the threshold value The photoelectric effect Particle like collision: a photon of energy hν collides with the surface The kinetic energy (EK = 1/2 me v2) of the electron is the energy of the photon - the energy required to remove the electron from the metal (ϕ) EK = h ϕ: work function Interpretation When hν < ϕ photoejection cannot occur When hν > ϕ the electron carries all the kinetic energy of the photons We should expect to see electrons appear as soon as collisions begin Intensity of light does not affect the energy of electrons Albert Einstein Particles behaving as waves A photon has an energy of E = hν = hc/λ According to relativity theory: E = mc2 = pc (p=mv is the momentum) De Broglie hypothesis: hc/λ = pc λ = h/p Each particle of momentum p, has a wavelength h/p How to test that particles behave as waves? DIFFRACTION Diffraction of Waves When waves reach a narrow slit, the water in the slit vibrates like a point source. The waves thus sent out from secondary sources along the slit are nearly in-phase when arriving any point in the forward direction. The diffracted wave resembles a circular wave with centre at the slit. Water wave A wave goes straight when the size of the slit is much larger than the wavelength. Diffraction through multiple slits Interference pattern develops! Light Diffraction Interference pattern Electron Diffraction ‘Particle’ ‘Wave’ Wave-Particle Duality λ = h/p Electron Diffraction Electron build-up in a twoslit experiment Electrons behave as waves!!!! Wave-particle Duality Electromagnetic radiation consists of a ray of photons: waves are particles Ray of electrons behave as wave Joint particle and wave character is a fundamental property of matter: wave-particle duality Problem 1 What is the wavelength of a baseball (mass 150 g) thrown at 95 miles/hour (~ 150 km/hour) Problem 2 What is the wavelength of an electron (mass 9.109x10-31 kg) accelerated by a potential difference of V = 40 kV ? Failure of Classical Physics Classical Physics Particles Newton Waves Maxwell Equations F=ma How to describe wave-particles? MODERN PHYSICS Quantum Mechanics A particle is distributed through space like a wave Instead of trajectory: wave function, Ψ(x): it contains all the dynamical information (position, momentum, energy) about the system The wave function can be complex number Goal: calculate Ψ ( and thus position, momentum, energy) of a wave-particle for a certain potential energy Operators ◊ The new physics equations are based on advance in new mathematical formulations about operators. ◊ Operator: Doing something with a function, represented by Ĥ ◊ Examples: d ̂ o Derivation operator :H = dx 2 d o Second derivative and multiply by constant k, Ĥ = k dx 2 ◊ When on operator acts on a function, a new function is obtained: (operator) (function) = (new function) ̂ 1 = Ψ2 HΨ ◊ Sometimes the new function is the same as function multiplied by a constant: Eigenfunction (that corresponds to a given operator) ◊ Eigenvalue equations: (operator) (function) = (eigenvalue) (function) These functions are called eigenfunctions ̂ = λΨ HΨ where λ is the eigenvalue. Operator Physics ◊ New approach to physics: the state of an object is fully characterized by a wavefunction, Ψ(x) ◊ For each physical property, there will be an operator: the eigenvalue of the operator on the eigenfunction Ψ(x) will be the value of the property. ◊ Assume a property o (scalar number). There will be an operator, such that ̂ = oΨ OΨ ◊ Example 1: Is function Ψ(x) = kx is eigenfunction for Ĥ d ̂ HΨ = kx = k dx = d/dx ? The right side is not a multiple of Ψ(x) = kx, therefore, it is not an eigenfunction. ◊ Example 2: Is function Ψ(x) = k exp(x) is eigenfunction for Ĥ = d ̂ = HΨ exp(kx) = k exp(kx) = kΨ(x) dx Yes, it is an eigenfunction, with eigenvalue k. k would be value the property. d/dx ? Complex numbers 40.5 = +- 2, (-1)0.5=i, i: imaginary unit, i2 = -1 Complex number z = a + bi a=Re(z), real part b=Im(z), imaginary part z1 + z2 = [a1 + b1i] + [a2 + b2i] = (a1 + a2) + (b1 + b2)i z1 x z2 = [a1 + b1i] x [a2 + b2i] = (a1a2 -b1b2) + (a1b2 + a2b1)i Complex conjugate: replace i -> -i z1 = a1 + b1i , z1* = a1 - b1i, conjugate of eix = e-ix Euler notation: ebix=cos(bx) + i sin(bx) Absolute value: |z|=(z z*)0.5, |1+2i| = [(1+2i)(1-2i)]0.5=50.5 Ψ = eikx , |Ψ| = [eikx e-ikx ]0.5= 1 Schrödinger Equation ◊ Equation with which we can calculate the wavefunction Ψ and the energy E of a wave-particle given the potential energy V(x) Time independent equation in one spatial dimension d2 + V (x) = E 2m dx2 where: m: mass 2 ħ: h/(2π) = 1.05457x10-34 Js Or: ̂ = EΨ HΨ 2 2 ℏ d Ĥ = − + V(x) 2m dx 2 ◊ The Schrödinger equation is the energy eigenvalue equation for the object. ◊ Ĥ : Hamiltonian operator Interpretation of Schrödinger Equation d2 + V (x) 2m dx2 2 Kinetic Energy + =E Potential Energy =Total Energy Assume V(x) = 0, freely moving particle dsin(kx) = k cos(kx) dx dk cos(kx) = dx k 2 sin(kx) = k2 d2 = 2 dx 2mE 2 = d2 =E 2 2m dx Solution: = sin(kx) 2 d2 = 2 dx k2 k = 2 2mE 2 Freely moving particle d2 + V (x) 2 2m dx 2 Solution: =E = sin(kx) V(x) = 0 k = 2 2mE 2 Each wavelength (λ = 2π/k) has different energy, E=p2/2m 2 2 4⇥ 2mE 2m p k2 = 2 = = 2 2 2m 4⇥ 2 2 = p2 h=p 2 ħ = h/(2π) h The solutions satisfy the = p experimentally obtained De Broglie wavelengths Physical Meaning of Wavefunction Once we solved the Schrödinger equation, we get energy and Ψ. What is the meaning of Ψ? Where is the particle and how fast it is? Born interpretation for one dimension: If the wavefunction of a particle has the value Ψ at some point x, then the probability of nding the particle between x and x+dx is proportional |Ψ|2dx=ΨΨ* dx Value of Ψ Value of |Ψ|2 fi Probability Probability density amplitude Value of |Ψ|2 dx Probability Born interpretation in three dimensions Ψ is a function of x, y, z fi fi If the wavefunction of a particle has value of Ψ at some point r, then the probability of nding the particle in an in nitesimal volume dτ=dx dy dz at that point is proportional to |Ψ|2 dτ Normalization Direct consequence of Born-interpretation: the probability of nding the particle in the entire (available) space is 1 If function Ψ is a solution to the Schrödinger equation, then NΨ is also a solution (N: normalization constant) Probability of nding the particle at position x: (NΨ*) (NΨ)dx =N2Ψ*Ψ dx Integration over the entire space should equal to 1 N 2 ⇤ ⇥ ⇤ dx = 1 N= 1 ( ⇤ ⇤ ⇥ dx)0.5 fi fi This is the way to determine N Quantization d2 + V (x) 2 2m dx 2 There =E are many solutions to the Schrödinger equation. Each solution Ψ comes with an energy E However, the Born interpretation puts severe restrictions on the acceptability of the wavefunctions Restrictions: Ψ cannot be in nite anywhere (Ψ must be nite everywhere) Ψ is is single value (it cannot have two values at some point) Ψ is continuous (no jump) Ψ has continuous slope (no kinks) Ψ cannot be zero everywhere (the particle is somewhere) |Ψ|2 can be integrated: square-integrable fi of these restrictions, only some solutions are allowed -> only some energies are allowed: the energy of a particle is quantized fi Because Not acceptable wavefunctions not continuous fi not single valued derivative not continuous in nite Average position of a particle ◊Obtaining position of on object: x̂ = x × i.e., xΨ = xΨ (trivial) ◊ Quantum mechanical rule about the average position of a particle < x >= ⇤ ⇤ ⇥ x dx How fast is the particle? What is the momentum, p? E = p2/(2m) d2 + V (x) 2 2m dx 2 =E Kinetic energy -> curvature of Ψ is proportional to kinetic energy and momentum Equation for momentum Quantum mechanical way of obtaining momentum ℏ d d p̂ = =p i dx i dx Procedure: 1. Calculate Ψ from Schrödinger equation 2. Substitute Ψ in the equation, and obtain p (p should a number) Problem: The wavefunction often does not give momentum For example, Ψ = cos (kx) dcos(kx) k = sin(kx) i dx i i k sin(kx) = pcos(kx) fi p is not a number, p does not have a de nite value Superposition of states However, Ψ = cos (kx) can be thought of a superposition of two wavefunctions, exp(ikx) and exp(-ikx) cos (kx) = 0.5 exp(ikx) + 0.5 exp(-ikx) = 0.5 [cos (kx) + i (sin kx)] + 0.5 [cos (kx) - i (sin kx)]= cos (kx) Both exp(ikx) and exp(-ikx) has de nite momentum: For exp(ikx) d exp(ikx) ik = exp(ikx) i dx i k exp(ikx) = pexp(ikx) p=kħ For exp(-ikx): p=-kħ fi The wavefunction Ψ = cos (kx) exists as a superposition of two functions [exp(ikx) and exp(-ikx)] with p=kħ and p=-kħ, respectively Physical interpretation of superposition When a measurement is made half of the time we will get p = kħ (the particle travels from left to right), the other half of time p = -kħ (the particle travels from right to left) Subatomic particles can exist in superposition of states: combination of possible states Observation settles the superposition of states into a de nite state fi Schrödinger’s cat Heisenberg uncertainty principle Assume we have a wavefunction with very well de ned position (position is known) This is a superposition of a very large number of states with different momentum: the momentum is uncertain (measurements give very different values) fi It is impossible to specify simultaneously with arbitrary precision, both the momentum and position of a particle Heisenberg uncertainty principle Quantitative version Position-momentum uncertainty relation ∆p ∆ x ≥ ħ/2 ∆ p: uncertainty in the linear momentum ∆ x: uncertainty in the position Position and momentum are complementary; not simultaneously speci able fi In 3D, ∆ p and ∆ x shall be considered in the same direction Problem Estimate the uncertainty of speed of electron (mass 9.109x10-31 kg) in a region of 100 pm! Problem Estimate the uncertainty of speed of electron (mass 9.109x10-31 kg) in a region of 100 pm! ∆p ∆ x ≥ ħ/2 ∆p = ħ/(2 ∆ x) = 1.05457x10-34 Js /2 / 100x10-12 m =5.2728x10-25kgm/s ∆v = ∆p /m=5.2728x10-25kgm/s / 9.109x10-31 kg = 5.79 x 105 m/s = 579 km/s The ‘speed’ of an electron in atomic distances is very inaccurate!!! Translational Motion We have shown earlier that: Freely moving particle: d2 + V (x) 2 2m dx 2 Solution: = sin(kx) k2 2 E= 2m =E V(x) = 0 k = 2 2mE 2 Particle in a Box fi fi ◊Translational motion of a particle in a box: a particle of mass m that can travel in a straight line in one dimension but is con ned between to walls separated by distance L ◊Potential energy: zero inside the box, in nity outside the box ◊ Many real world examples: ˜ gas molecule in small container ˜ electrons in metal ˜ electron in a conjugated large organic substance Particle in a Box: Schrödinger Equation d2 + V (x) 2 2m dx 2 =E V (x) = 0 for 0 < x < L V (x) = ⇤ for x 0, x ⇥ L General form of solution in between the walls (x) = C sin(kx) + D cos(kx) Because of V, not all solutions acceptable. Boundary conditions: (0) = 0 (L) = 0 The particle cannot exist where the potential is in nity. Let’s take Ψ(0) = 0 fi Ψ(0) = C sin(k0) + D cos(k0) = D cos(1) = D Thus, Ψ(0) = D = 0 Ψ(x) = C sin(kx) Particle in a Box: Schrödinger Equation d2 + V (x) 2 2m dx 2 (0) = 0 V (x) = 0 for 0 < x < L =E V (x) = ⇤ for x Ψ(x) = C sin(kx) (L) = 0 0, x ⇥ L Now let’s take Ψ(L) = 0 Ψ(L) = C sin(kL) = 0 C cannot be zero because then Ψ(x) = 0 sin(kL) = 0 kL =nπ, n=1,2,... n cannot be zero, because then Ψ(x) = C sin(0x)=0 Solution to the equation Ψ(x) = C sin(kx) k=nπ/L, Ψ(x) = C sin(nπx/L) n=1,2,... Particle in a Box: Normalization n x (x) = C sin( ) L The probability of nding the particle in the box is 1 L L 2 n x 2 2 dx = C sin dx = 1 L 0 0 sin2 ax = 0.5x 0 n x sin dx = [0.5x L 2 n 2n x L 0.25 sin( )]0 = [0.5x]L 0 = 0.5L L L C 0.5L = 1 2 (x) = fi L 0.25asin(2ax) + const 2 n x sin L L C= 2 L n=1,2,... Particle in a Box: Energies d2 + V (x) 2 2m dx 2 =E Inside the box the particle is free moving -> we solved it earlier = sin(kx) Now we have: (x) = 2 n x sin L L n=1,2,... Energy of particle in a box k2 = 2mE 2 n ⇥2 2mE = 2 L n2 2 2 n 2 2 h2 n 2 h2 E= 2 = 2 = 2 L 2m L 2m 4 8mL2 n 2 h2 E= 8mL2 n=1,2,... Particle in Box: Interpretation of solutions probably find that particle at the middle of that waves n 2 h2 E= 8mL2 (x) = 2 n x sin n=1,2,... L L Only certain solutions with integer quantum numbers, n, are possible Larger curvature of Ψ, -> larger energy Particle in Box: Interpretation of solutions n 2 h2 E= 8mL2 n=1,2,... Only certain energies are allowed Larger box size and heavier particle gives lower energy The lowest energy (n=1), called zero-point energy, E0 = h2/(8mL2) is not zero Separation of Energy Levels n=1,2,... n 2 h2 E= 8mL2 The energy difference between adjacent Cv=3/2R energy levels: E = En+1 h2 En = (n + 1) 8mL2 2 h2 h2 (n) = (2n + 1) 8mL2 8mL2 2 The energy difference is small for ◊ large boxes ◊ heavy particles fi Macroscopic objects (earth, ball) do not exhibit signi cant energy quantum levels, because m and L are large Delocalized π electrons in a box 1 β carotene 11 π bonds 294 nm 11 energy levels On each level: 2 electrons Photon of energy (hν) can be absorbed when one electron is kicked from n = 11 -> 12 h2 E = (2n + 1) 8mL2 n=11, m=me, L = 294 nm -> ∆E = 1.6 x 10-19 J This corresponds to an electromagnetic Absorbs blue light -> wave of ν = ∆E/h = 2.41 x 1014 Hz. eyes will detect the lack The experimental value: 6.03x1014 1/s (λ = 497 nm) of blue light as orange Visible light, it has color. color! Tunneling What happens if the potential energy of a particle does not rise to in nity? Classical view: ◊ if the energy of the particle E > V, then it can overcome the barrier. ◊ if V> E, the particle re ects V Text x Like free particle Like free particle Ψ = Asin(kx)+Bcos(kx) Ψ = Csin(kx)+Dcos(kx) ? constant d2 + V (x) = E 2m dx2 d2 2m = (E V ) 2 2 dx Solutions to Schrödinger equation if E0 The solution of this is exponential! d eax /dx = a eax d2 eax /dx2 = a2 eax Tunneling Solution ◊The particle can overcome potential energy barrier that is larger than its energy: the wavefunction does not drop to zero at the right side ◊ Leakage by penetration through a classically forbidden region: tunneling ◊ Tunneling probability decreases with the width wall width and the with mass of the particle-> it is important with light particles (electron, proton), not so important with heavy particles ◊ Example: acid base reactions (proton transfer) are very fast because of tunneling Scanning Tunneling Microscopy Pt/Rh/W tip Xe atoms on Ni surface Tunneling current is very sensitive to distance -> it can map atomic heights of a surface Atomic resolution! Rotation: particle on a ring Classical Physics Rotating particle has angular momentum Moment of inertia: J=pxr Jz =±pr I =mr2 Energy: E = p2/(2m)=Jz2/(2mr2)=Jz2/(2I) Translational motion p = momentum J = Iω Rotational motion J Angular momentum m x mass = I moment of inertia v velocity x ω angular velocity Rotation: particle on a ring Quantum mechanics Angular momentum Jz =±pr De Broglie wavelength λ = h/p Jz =±hr/λ Not all wavelengths are possible! ◊ We can solve the Schrödinger equation: freely moving particle on a ring! ◊ It is convenient to write the wavefunction as a function of angle on the circumference of the ring Ψ(ϕ) = C cos(k ϕ) Boundary condition: Ψ(ϕ+2π) = Ψ(ϕ) fi The wavefunction should perfectly ‘ t’ on the circumference: k= ml= 0,±1, ±2,... p = h/λ Quantized angular momentum What is the wavenumber? Circumference: 2rπ ‘Number’ of waves: ml = 0, ±1, ±2,... Ψ(ϕ) = C cos(k ϕ) λ = 2rπ/ml Angular momentum is quantized Jz =±hr/λ = hrml/(2rπ) = mlh/(2π) = ml ħ Jz = ml ħ ml = 0, ±1, ±2,... ml>0: clockwise rotation ml 0 states are double degenerate, because there are two states with the same energy The ml = 0 state is non-degenerate Rotational motion has no zero point energy (E = 0 for ml =0) Particle on a Sphere Solution to the Schrödinger equation is more complicated: two ‘phase’ variable Φ: azimuth, θ: colatitude Solution results in boundary conditions with two quantum numbers Orbital angular momentum quantum number Magnetic quantum number l = 0, 1, 2,... ml = l, l-1, l-2,..., -l For each [l,ml] pair there is a solution Ψ l,ml(ϕ, θ) Spherical Harmonics Ψ Blue: Ψ>0 White: Ψ