1. What’s coming up??? Oct 25The atmosphere, part 1Ch. 8 Oct 27Midterm … No lecture Oct 29The atmosphere, part 2Ch. 8 Nov 1Light, blackbodies, BohrCh. 9 Nov 3Postulates of QM, p-in-a-boxCh. 9 Nov 5,8Hydrogen atomCh. 9 Nov 10,12Multi-electron atomsCh.10 Nov 15Periodic propertiesCh. 10 Nov 17Periodic propertiesCh. 10 Nov 19Valence-bond; Lewis structuresCh. 11 Nov 22Hybrid orbitals; VSEPRCh. 11, 12 Nov 24VSEPRCh. 12 Nov 26MO theoryCh. 12 Nov 29MO theoryCh. 12 Dec 1Putting it all together Dec 2Review for exam

2. Light: wave-like (diffraction) and particle-like (photoelectron effect; momentum transfer) behaviour Electrons: particle-like (mass; electric charge) and wave-like (diffraction) behaviour

3. Planck Postulated “forced to have only certain discrete values” “Energy can only be transferred in discrete quantities.” is the frequency of the energy h is Planck’s constant, 6.626 x 10 -34 J s. Energy is not continuous Planck……. Energy is quantized

4. Compton collided X rays with electrons X rays electron Now along comes de Broglie! for a photon momentum Path of electrons deflected!!! Photons have momentum…. As predicted!!!!!

5.  electron gun electrometer- detects electrons as current angle (  current constructive interference destructive interference Davisson and Germer verified de Broglie’s ideas by measuring electron reflection off a piece of nickel metal: The diffraction of the electron beam shows that electrons really do have wave properties! Thomson passed an electron beam through a sheet of gold….

6. Atomic spectra: Discrete wavelegths (frequencies) of light emitted (or absorbed) For Period I (H, Li, Na, …) atoms, the emission spectrum is well predicted by an equation determiined by Rydberg = R c { 1/n f 2 - 1/n i 2 }

7. SPECTROSCOPY EMISSION Sample heated. Many excited states populated

8. = R c { 1/n f 2 - 1/n i 2 } where is the frequency of light emitted, R is a constant which depends on the atom, n f and n i are integers > 1; n f is associated with a lower state

9. Bohr suggested an explanation for this observation: only certain orbits of an electron about a nucleus are allowed. These each have a discrete energy, given by E n = R c / n 2 The energy levels (allowed orbits) are numbered using different values of n. Transitions between 2 of these levels follow the observed behaviour

10. Although this works very well for H, Li, Na, … it does not predict the spectrum of any other atom. It also does not give any reason why only certain energies are allowed. A more general approach involves using our description of matter (electrons and nuclei for now) as having some wave-like properties. We will start off slowly …

11. We will describe atoms and molecules using wavefunctions, which we will give symbols … like this:  These wavefunctions contain all the information about the item we are trying to understand Since they are waves, they will have wave properties: amplitude, frequency, wavelength, phase, etc.

12. Waves are described by mathematical expressions … to learn something about their world we need to perform mathematical operations on these expressions Multiplication and division Differentiation Waves cannot be precisely located without losing all information about the wavelength

13. For a particle, its total energy (E) is the sum of its kinetic energy (KE) and potential energy (V) E = KE + V The potential energy depends of position; the kinetic energy on motion. To describe the energetics of particles treating them as waves, we use an energy operator, called the Hamiltonian, H

14. We obtain the energy by performing the “energy operation” on the wavefunction – the result is a constant (the energy) times the wavefunction H  This equation is called the Schrodinger wave equation (SWE) Let’s see how this might work

15. Ignoring V for now (we will get back to it later!) we can write a formula for the kinetic energy: KE = ½ m v 2 = p 2 / 2m = (2m) -1 {p x 2 +p y 2 +p z 2 } It turns out that in the quantum mechanical world the momentum of a particle (when we describe it as a wave) is determined by a mathematical operation: p x = i (h /2  d /dx p y = i (h /2  d /dy p z = i (h /2  d /dz

16. Then KE = (2m) -1 {p x 2 +p y 2 +p z 2 } becomes KE =  (h 2 /8   m) {d 2 /dx 2 + d 2 /dy 2 +d 2 /dz 2 } = 2  (h 2 / 8   m)

17. So H = KE operator + PE operator H = H  2  (h 2 / 8   m) + V 2  (h 2 / 8   m) + V {} 