Midterm results will be posted downstairs (by the labs) this afternoon No office hours today

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

displacement The wavelength, frequency and speed of electromagnetic radiation are all related by: direction of propagation FREQUENCY AND WAVELENGTH RELATED  = c

Excited atoms emit light of different frequencies..  = c Wavelength  (nanometers) Energy VISIBLE SPECTRUM

The wavelength of the yellow light from a sodium lamp is 589 nm. EXAMPLES Calculate the frequency of electromagnetic radiation from its wavelength and velocity. What is the frequency of the radiation?

At the beginning of the 20th century: Matter: Discrete particles Electromagnetic radiation: Continuous waves The two were thought to be quite separate….. CLASSICAL PHYSICS

ELECTRONS in ATOMS Classical physics predicts: An electron will crash into the nucleus Rotating mass is accelerating Accelerating charge emits radiation, lowering its energy Lower energy  shorter radial distance Therefore, electron will collapse into nucleus

Planck studied blackbody radiation profiles HEAT ANALYSE Heat hollow object Light emitted by surface and absorbed Pin-hole lets out some light for analysis

Planck studied blackbody radiation profiles The only way to explain this was…….. Classical theory does not fit!!!!!!!!! Ultraviolet catastrophe !! The Sun is close to this

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, x J s. Energy is not continuous Planck……. Energy is quantized

THE PHOTOELECTRIC EFFECT light electron metal Shine light on a piece of metal and electrons are emitted What was observed: Electrons were emitted only if the frequency of the light is greater than a minimum value depending on the metal. Minimum value

THE PHOTOELECTRIC EFFECT light electron metal KE of electron Frequency of light ( ) 0 When < 0, no electrons are ejected at any light intensity. KE of the ejected electrons depends only on the light’s frequency When > 0, the number of electrons is proportional to the light intensity. This lead Einstein to use Planck’s idea of quanta

EINSTEIN POSTULATED “Electromagnetic radiation can be viewed as a stream of particle-like units called photons.” Energy of a Photon: The energy of the photon depends upon the frequency

EINSTEIN’S THEORY OF RELATIVITY The photon has zero rest mass (m 0 = 0) Using... So that Thus….. E(pc) (m c2)c2)  E(pc) 2 2 

PHOTONS HAVE MOMENTUM MOMENTUM It is this momentum that gives the photon its energy The momentum depends upon the wavelength of the radiation.

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!!!!!

de Broglie posed the question: for a photon for matter with mass m kg moving at v m/s “If light energy has particle-like properties, does matter have wave-like properties?” WAVE-PARTICLE DUALITY! So de Broglie postulated So that

A wave property which matter might exhibit is interference Constructive Destructive

 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….

 electrometer Thomson passed an electron beam through a sheet of gold foil rather than reflecting it off a metal surface: gold foil electron gun angle (  current He observed………. INTERFERENCE………...

Diffraction of an electron beam…. We can relate these spacings to the electron wavelength

All matter and energy shows both particle-like and wave- like properties. WAVE-PARTICLE DUALITY….. MASS INCREASES Example.... WAVELENGTH GETS SHORTER. MASS DECREASES WAVELENGTH GETS LONGER.

Example: What are the wavelengths of a 0.10 kg ball moving at 35 m/s and an electron moving at 1.0 x 10 7 m/s? 1J = kg m 2 s -2 = 1.9 x m Ball: h = x J s Solution: Now do the electron…...

= 7.3 x m Electron: In summary…... = 1.9 x m = 7.3 x m Electron: Ball:

All matter and energy shows both particle-like and wave-like properties. WAVE-PARTICLE DUALITY….. Large pieces of matter are mainly particle-like, with very short wavelengths. Small pieces of matter are mainly wave-like with longer wavelengths.

Electrons have both wavelike and particle like properties: The first attempt was by Niels Bohr………... ELECTRONS in ATOMS their wavelike properties must be taken into account when describing the electronic structure of atoms.

WHY THE ELECTRON DOES NOT CRASH INTO THE NUCLEUS! Bohr postulated that the wavelength of the electron just fits the radius of the orbit. This why the electron does not crash into the nucleus!!!

WHY THE ELECTRON DOES NOT CRASH INTO THE NUCLEUS! IF the wavelength of the electron does not fits the radius of the orbit. The electron waves interfere destructively The number of wavelengths leads to….. NOT STABLE!

n= 1,2,3,4……... n = 4 n = 3 n = 2 n = 1 Each orbit has a quantum number associated with it. THE BOHR ATOM “Electrons move around the nucleus in only certain allowed circular orbits” QUANTUM NUMBERS and the ENERGY the energy of an orbit……..

BOHR ATOM ENERGY LEVEL DIAGRAM n=1 -A n=2 -A/4 EnEn 0 n=3 -A/9 n=4 Energy -A/16 e-e- Now provide energy to the atom (for instance, by absorbing a photon) and excite electron to a higher energy level … we say the atom is in an excited state

BOHR ATOM ENERGY LEVEL DIAGRAM n=1 -A n=2 -A/4 EnEn 0 n=3 -A/9 n=4 Energy -A/16 e-e- ELECTRON DE-EXCITATION Emission of energy as a photon e-e-

The energy of the photon emitted or absorbed is given by the energy difference between the energy levels and Planck’s relationship! ATOMIC SPECTRA:INTERPRETATION by BOHR’S MODEL  E = energy of final state - energy of initial state

nini nfnf ABSORPTION OF A PHOTON

SPECTROSCOPY EMISSION Sample heated. Many excited states populated n = 1 Ground state n = 2 n = 3 n = 4 n = Ion 8 Excited states... Energy The spectrum…..

SPECTROSCOPY EMISSION Sample heated. Many excited states populated

LiNaK EXCITED GROUP 1 ELEMENTS

The hydrogen emission spectrum can be broken into series: n = 1 Ground state n = 2 n = 3 n = 4 n = Ion 8 Excited states... Energy For the Lyman series, n f = 1 and n i = 2,3,4… For the Balmer series, n f = 2 and n i = 3,4,5… For the Paschen series, n f = 3 and n i = 4,5,6…

For the Balmer series, n f = 2 and n i = 3,4,5… n = 1 Ground state n = 2 n = 3 n = 4 n = Ion 8 Excited states... Energy THE BALMER SERIES EMISSION

For the Balmer series, n f = 2 and n i = 3,4,5… n = 1 Ground state n = 2 n = 3 n = 4 n = Ion 8 Excited states... Energy THE BALMER SERIES EMISSION

is the energy required to remove an electron from a gaseous atom or ion. First ionization energy of X: Higher ionization energies indicate greater difficulty in removing electron. X X + + e – Second ionization energy of X: X+X+ X 2+ + e – IONIZATION ENERGY UNITS: kJ mol -1 Back to the energy level diagram…….

BOHR ATOM ENERGY LEVEL DIAGRAM n=1 -A n = 2 -A/4 EnEn n=3 -A/9 n=4 ENERGY -A/16 e-e- 0 the electron is JUST free And the energy of the electron is ZERO……. IF we choose a photon so that Then…... E final = 0

We can estimate the IONIZATION ENERGY for a hydrogen atom. Final state has n =  The initial state has n=1  E= energy of final state - energy of initial state The positive sign tells you that you need energy to remove the electron! IONIZATION ENERGY We need to calculate the IE for one mole….. = 0 - (-A) = A = x J for one atom! E= -A / (  2 )=0 THIS DEFINES IONIZATION. E=-A/(1 2 )= -A THIS IS THE GROUND STATE.

We can estimate the IONIZATION ENERGY for a hydrogen atom.  E= energy of final state - energy of initial state The positive sign tells you that you need energy to remove the electron! The ionization energy for one mole is IONIZATION ENERGY = 2.178x J atom -1 x 6.022x10 23 atoms mol -1 =13.12 x 10 5 J mol -1 = 1312 kJ mol -1 = 0 - (-A)= A= x J for one atom