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Chem 4502 & Spectroscopy (Jason Goodpaster) By 1900, many advances in physics Chap. 1: Dawn of Quantum Theory Text: McQuarrie and Simon's "Physical Chemistry: A Molecular Approach", 1997. • 1897 - JJ Thomson discovered (All figures and tables are from our text unless otherwise noted.) , measured its mass/charge ratio Lecture 2: pp. 1-6 Blackbody radiation; Planck's constant, h • Maxwells equations completed in pp. 7-10 ; 1873-4 partial differential equations that Einsteins explanation; Ephoton = h ν Lecture 3: can solve all problems in classical pp. 10-14 atom spectrum electricity and magnetism Rydberg formula But there were a few experimental results pp. 15-18 de Broglie waves that could not be explained... 1 2

Blackbody Radiation Fig. 1.1 p. 3 Measured frequency distribution of emitted blackbody radiation

Ideal Blackbody: As T increases: Absorbs and emits radiation without • maximum shifts to higher favoring particular frequencies frequency (ν) E.g., pinhole in sealed container • total radiated energy (area Radiation emitted is characteristic of the under curve) increases temperature of the body. Figure: Engels Quantum Chem

3 4 Problem.... Blackbody radiation: Plancks theory

Dashed line: theoretical prediction (Rayleigh – Jeans Law) Planck (unwillingly) introduced the concept of energy quantization into physics.

Catastrophe: E = n h ν energy density of radiation -1 predicted to diverge as ν2 ν frequency of light (sec ) absorbed or emitted Predicts infinite total by an electron oscillating in the blackbody energy emitted! n positive integer (0, 1, 2, ...) E energy of an oscillating electron; for a given ν, the energies of the oscillators are discrete h a new constant; units must be energy x time 5 Note: Planck didnt quantize the light 6

Blackbody radiation: Plancks theory

Using statistical thermodynamics, he derived a formula for the energy density of the radiation. The value of h was fit to the data.

Planck was able to fit the data, using the value: h = 6.63 x 10-34 J·sec

This success alone did not convince people (including Planck) that energy is really quantized.

7 8 Photoelectric Effect Early measurements of the photoelectric effect: Schematic diagram of a typical apparatus Ejection of an electron (e-) from a metal by radiation. Experiment is:

Shine light on a metal surface. light Measure the kinetic energies of the ejected (if any). See how their energies vary with the intensity and frequency of light.

9 10 from Karplus & Porter Atoms and Molecules p. 53

Photoelectric Effect: Fig. 1.4 p. 8 Experimental Observations: Predictions of Classical Physics Photoelectric Effect (Sodium)

Electron kinetic energy (eKE): eKE :

• increases linearly • should increase with intensity of light with frequency of 2 (proportional to Amplitude of electric field) light

does not depend • should not depend on frequency of light • on intensity of light

As amplitude of oscillating electric field increases, • no photoelectrons electrons on metal surface should oscillate more detected below a and escape with more energy. certain ν

11 12

Photoelectric Effect: Einsteins Theory Photoelectric Effect: Einsteins Theory

One electron in the metal absorbs one photon, Particle model of light: light exists as small in a sort of collision. packages of energy, now called photons

Conservation of energy: Ephoton = hν 2 Ephoton = hν = Φ + ½ mev Ephoton energy in one photon E energy in one photon absorbed by one ν frequency of the photon photon electron in the metal

Φ work function of the metal This applies to light, not to matter (minimum energy needed to free an electron) (not to electrons, atoms, etc.) 2 ½ mev kinetic energy of the photodetached electron 13 14

Quantum Quote of the Day Chap. 1: Dawn of Quantum Theory (continued) Lecture 3:

pp. 15-18 de Broglie waves The next revolution in physics pp. 10-14 Hydrogen atom spectrum will occur when the properties of Rydberg formula mind will be included in the equations of quantum theory. Lecture 4: pp. 18-23 Bohr theory of the H atom 1963 Physics Nobel Laureate pp. 23-25

15 16 de Broglies Hypothesis (1924) After the first World War, I gave a great deal Einstein showed that the energy of a of thought to the theory relativistic particle moving freely in space is: of quanta. It was then that I had a sudden E = (m2c4 + p2c2)½ where m = mass, p = momentum inspiration. Einsteins For a photon, m = 0 and E = hν, so wave-particle dualism h = ( 0 + p2c2)½ = p c for light was an ν absolutely general p = hν / c photon momentum (in the direction of phenomenon extending propagation), another particle aspect of radiation 1892-1987 to all physical nature. Also, ν = c / λ and 1 / λ = ν / c, where λ is the of light and is its frequency, so have: (Fig. 18.2 p. 38) ν

17 p = h / λ λ = h / p for light 18

de Broglies Hypothesis de Broglies Hypothesis How to test this strange hypothesis? de Broglie generalized this equation to matter as well: Recall 2-slit experiment with light: λ = h / p But: to detect

where p = classical momentum = m v (mass x velocity) interference effects, must detect λ = de Broglie wavelength of the material particle differences on the Matter waves ?!? order of λ between the lengths of He [de Broglie] has lifted one corner of the great veil. alternate paths. Einstein Maxima on detector occur where Is this possible?? 19 20 r2 – r1 = nλ (n = 1, 2, 3, ...) de Broglies Hypothesis de Broglies Hypothesis Can we observe a de Broglie wavelength? • Problem: What is the de Broglie wavelength of Problem: What is the de Broglie wavelength of a golf an electron (m = 9.11 x 10-31 kg) accelerated ball (50 grams) traveling at 30 meters/second? e through a potential of 54.0 Volts? λ = h / p h = 6.63 x 10-34 J∙s λ = h / p h = 6.6 x 10-34 J∙s e = 1.60 x 10-19 C λ = 4.4 x 10-34 meters = 4.4 x 10-22 picometers (pm) λ = 1.67 x 10-10 m = 1.67 Å = 167 pm This wavelength is on the scale of atomic diameters, -15 The size of a nucleus of an atom is about 10 meters. and it is in the x-ray region. Conclude: golf ball displays only particle-like properties. (What if its at rest? For this, need uncertainty Slits: spacings between atoms in a crystal. 21 22 principle.)

Davisson-Germer experiment, 1927: For single-plane : n λ = a sin θ Scattering of an electron beam from a nickel a = distance between Ni atoms = 2.15 x 10-10 m crystal shows constructive and destructive (known from x-ray diffraction experiments) interference (Atkins Physical Chemistry 6th Ed. p. 293) θ = angle at which intensity maximum is observed Constructive n = 1 for the strong 1st order peak interference occurs at angles at which For electrons accelerated to 54.0 V, D&G measured the electron waves the maximum scattering angle (n=1) to be 50.1o. scattered by different Ni atoms What is the wavelength, λ, of the electrons ? have path lengths λ = a sin θ / n = 2.15 x 10-10 m (sin 50.1o) / 1 differing by nλ. λ = 1.65 x 10-10 m Destructive: nλ/2, n odd → nodes. 23 Same as we calculated from the de Broglie formula ! 24 George Thomson - used ≈ 10 keV electrons to These electron diffraction experiments confirmed penetrate thin metal films, recorded diffraction patterns the de Broglie wavelength: Fig. 1.8 p. 17 Similarity between X-ray and λ = h / p Electron Diffraction Patterns (Aluminum Foil)

Nobel Prizes in Physics: 1929 de Broglie 1937 G. P. Thomson and C. J. Davisson

1906 JJ Thomson (Georges Dad) discovery of the electron as a subatomic particle (1987) X-rays Electrons 25 26

What would we see if we passed http://www.charfac.umn.edu/instruments/ a single electron through Davisson and Germers or Thomsons experiments ? IT Characterization Facility 12 Shepherd Labs Transmission electron The electron would land at a localized spot on the microscope (TEM) with an detector (no interference pattern). operating voltage range of 20 to 120 kV. Allows magnifications of thin samples (<500 nm) up to 700,000 times. Point-point resolution: 0.34 nm This is like the single photon 2-slit experiments – Line resolution: 0.2 nm 27 28 wave-particle duality applies to both light and matter. SEE DOWN TO 0.6 Å Chemical and Engineering News Electron microscope Sept. 2004 achieves direct sub-angstrom imaging of a crystal

A milestone in electron microscopy--the first direct sub- angstrom imaging of a crystal lattice--has been reported by researchers at Oak Ridge National Laboratory and Nion, a company in Kirkland, DUMBBELLS STEM image Wash., that specializes in of a silicon crystal in the advanced electron-microscope [112] orientation reveals optics [Science, 305, 1741 pairs of atom columns in (2004)]. The researchers fitted a which the intrapair 300-kV scanning transmission electron microscope (STEM) at separation is 0.78 Å. ORNL with a Nion aberration 29 30 corrector...

Emission spectra of atoms (e.g., hydrogen): another experiment that baffled classical physics. Fig. 1.7 p. 13 Hydrogen Atom

31 32 Emission spectrum of H atom Rydberg formula for H atom ∼ -1 2 2 ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 )

What are n1 and n2 for the shortest wavelength The observed emitted (lines) were emission line of the H atom? found to fit the Rydberg formula: What is this wavelength, in nm? In what region of the electromagnetic spectrum is it? ∼ -1 2 2 ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 ) n1=1, n2 = ∞ ν∼ = 109,680 cm-1

where n1 and n2 are integers (n2 > n1) λ = 1 / ν∼ = 1 / 109, 680 cm-1 = 91.2 x 10-7 cm

λ = 91.2 x 10-9 m = 91.2 nm UV region Why would H atom care about integers?33 34

Note on use of wavenumbers (cm-1) Chap. 1: Dawn of Quantum Theory (continued) Spectroscopists often refer to cm-1 as an "energy" unit. Lecture 4: Really cm-1 is Energy / hc pp. 18-23 Bohr theory of the H atom

What cm-1 of light corresponds to an energy of 1. eV? pp. 23-25 Uncertainty Principle

For light, E = hc / λ . For E = 1. eV : Lecture 5: ν∼ = 1 / λ = E / hc = 1.6022 x 10-19 J Chap. 2 Classical Wave Equation: (6.6261 x 10-34 Js)(2.9979 x 1010 cm/s) pp. 39-49 Classical wave equation (1-dim.)

ν∼ = 8066 cm-1 corresponds to 1. eV of energy MathChapter A pp. 31-34 Complex numbers 35 36 p. 190 1885-1962 Bohr s Model of the H Atom • The electron can orbit around the nucleus only at discrete values of r, the orbital radius. r • These stationary orbits must satisfy the condition: quantum mev r = n ℏ where n = 1, 2, 3,.. number Bohr designed his coat of Here, ℏ ( h-bar ) is shorthand for h/2π.

arms with a yin-yang symbol. ( me v r ) is the angular momentum (L) of the electron - Inscription: CONTRARIA (with me = e mass, v = velocity, r = orbital radius). SUNT COMPLEMENTA Energy levels, Principle 37 Thus, Bohr assumed that the angular momentum is quantized.38 of Complementarity (Opposites are Complements)

Can obtain Bohrs m vr = nℏ by considering e of the H Atom Fig. 1.9 p. 19 de Broglie Waves in Bohr Orbits Total energy (E) of the electron in the H atom: 2 2 E = ½ mev + (-e / (4πεo r)) kinetic potential

For a stable orbit, must have a balance of forces: Stable Unstable: wave progressively disappears → 2 2 2 e /(4πεo r ) = mev /r Coulomb attraction between centrifugal (pseudo) Stable orbit - circumference must be: 2πr = nλ electron and outward force

Combine with de Broglie relation: λ = h / mev Multiply both sides by ½ r: 2πr = n (h / m v) m vr = n (h/2π) = n Same as e2/(8πε r) = ½ m v2 e e ℏ above o e 39 Substitute into equation for E above. 40 Bohr Model of the H Atom Bohr Model of the H Atom

2 After substituting, the total energy of the electron is: E = -e / (8πεo r) 2 2 Next: Obtain r in terms of fundamental constants. E = e /(8πεor) + (-e / (4πεo r)) kinetic potential We had: e2/(8πε r) = ½ m v2 o e 2 Simplifying: E = - e / (8πε r) Also: mev r = n h / 2π o Negative energy → bound states Solve for v, then substitute here The electron is bound to the nucleus. 2 2 2 2 Result: r = εoh n / π me e r increases as n Recall: only discrete values of r are possible Substituting values (p. 20 eqn. 1-18), obtain for n=1 the radius of the first (lowest energy) Bohr orbit:

r1 = 0.529 Å ao, Bohr radius 41 42 Now used as basic distance unit in atomic units.

Bohr Model of the H Atom Bohr Model of the H Atom 4 2 2 2 Bohr prediction: E = [ - m e / (8 εo h ) ] / n We had: E = -e2 / (8πε r) o Evaluate the constants in [brackets] 2 2 2 -31 Substitute: r = εoh n / π m e (n = 1, 2, ...) (use m = 9.1044 x 10 kg, reduced mass of electron and proton)

Obtain: E = [ - m e4 / (8 ε 2 h2) ] / n2 o E = - 2.18 x 10-18 J / n2 = - 13.6 eV / n2 2 Quantized energy levels; energies go as 1/n I.e., 13.6/n2 eV is the energy (with respect to that of the where n () is an integer (1, 2, 3, ...) separated e- and nucleus) by which an electron in a state with quantum number n is bound to the nucleus. Recall: Rydberg Formula for H atom Convert to wavenumbers (and use more sig figs): ∼ ν = 1 / λ (cm) = 109,680 cm-1 ( 1 / n 2 – 1 / n 2) 1 2 E / hc = ν∼ = - 109,680 cm-1 / n2 empirical formula based on measurements Recall: Rydberg Formula SAME # !! 43 44 -1 2 2 ν = 1 / λ (cm) = 109,680 cm ( 1 / n1 – 1 / n2 ) Bohr Model of the H Atom Fig. 1.10 p. 21 Diagram for the Hydrogen Atom Bohr prediction: E / hc = ν∼ = -109,680 cm-1 / n2 n=∞, 0 cm-1 To emit a photon, an H atom must make a transition

from a higher energy (less negative), higher n (n2) state n=2, -1 (E2) -109,680/4 cm r = 0.529*4=2.12 to a lower energy (more negative), lower n (n1) state 2 Å (E1). Ionization Then, the energy of the emitted photon is: potential of H atom: 13.6 eV Ephoton = hν = ΔEsystem = E2 – E1 Bohr Frequency Condition ∼ n=1, Or, in wavenumbers: -109,680 cm-1 -1 2 2 (-13.6 eV) ν = 109,680 cm (1 / n1 – 1 / n2 ) 45 r1 = 0.529 Å 46 in agreement with the Rydberg formula.

Bohr Model of the H Atom Despite its success in predicting the H atom emission spectrum, the Bohr model had major problems: ☹ did not predict the effects of external magnetic fields on the H atom

☹ could not be successfully extended to atoms with more than 1 electron

☹ incompatible with uncertainty principle

In Chap. 6, we will revisit the H atom spectrum 47 48 using the Schrödinger equation. Heisenbergs Uncertainty Principle: Δx Δp ≥ h Heisenbergs Uncertainty Principle (mid 1920s) One form of the uncertainty principle:

Δx Δpx ≥ h (1-dimension) p. 114 Δx is the uncertainty in the particles location Werner Heisenberg Δpx momentum 1901-1976 Δ px = m Δ vx Therefore, the more narrowly the position is measured, the more uncertainty there is in the particles momentum (velocity), and vice versa. If the position were known exactly, the momentum 49 50 would be completely unknown.

Understanding the Heisenberg s Uncertainty Principle uncertainty principle in terms of superpositions Δx Δp ≥ h of waves This can be understood in terms of fundamental Atkins Physical Chemistry th limits on measurements. 5 Ed., p. 384 It is rooted in the wave nature of matter. Δx Δ(1/λ) ≥ 1 (classical) Complementary smaller Δx requires larger range of λs position and momentum Add de Broglie: energy and time 1/ λ = p / h some aspects of angular gives momentum (Ch. 6) 51 Δx Δp ≥ h 52

Uncertainty Principle: Apply to electrons What is the minimum uncertainty in the speed of Uncertainty Principle: Apply to Macroscopic Objects an electron known to be located within 0.529 Å What is the minimum uncertainty in the position of (Bohr radius for n=1) of the nucleus of a H atom? a golf ball (assume m=50. g exactly) known to be -34 -31 traveling at a speed of 30.00000±0.00001 m/s? [ h = 6.626x10 J·s; me = 9.11 x 10 kg ] [ h = 6.626x10-34 J·s ] Δx (mΔv) = h (minimum) so Δv = h / (m Δx) Δx (mΔv) = h (minimum) so Δx = h / m Δv Δv = 6.626x10-34 J·s / ((9.11x10-31 kg)(0.529 X 10-10 m)) Δx = 6.626x10-34 J·s/ (0.050 kg)(0.00001 m) Δv = 1.4 x 107 m/s Δx = 1 x 10-27 m This about 1012 times smaller than the size of a nucleus (∼10-15 m) ! This is much larger than the e- speed in the Bohr model: The uncertainty principle applies to all matter, but its mvr = nℏ so v = n ℏ / mr = 2 x 106 m/s consequences are (usually) not for 53 54 Bohr model isnt consistent with the uncertainty principle. macroscopic objects.

Uncertainty Principle: Questions to Ponder Uncertainty Principle: Questions to Ponder

Does the electron really have a well-defined position, What is the uncertainty in the position of a golf ball speed and momentum at any given moment, but we thrown up in the air, just at the point when it is can only measure it to within these uncertainties? changing direction and not moving? Or, are its position and momentum inherently fuzzy to these extents? Why does the transporter on Star Trek require If you answered the former: what does it mean for a Heisenberg compensators? property to exist, if it cannot possibly be observed? What would the world be like if h were a lot bigger? How does the uncertainty principle relate to the effects of the observer on the properties of that which is observed? 55 56