• Midterm results will be posted downstairs (by the labs) this afternoon

• No office hours today

What’s coming up???• Oct 25 The atmosphere, part 1 Ch. 8• Oct 27 Midterm … No lecture• Oct 29 The atmosphere, part 2 Ch. 8• Nov 1 Light, blackbodies, Bohr Ch. 9• Nov 3 Postulates of QM, p-in-a-box Ch. 9• Nov 5,8 Hydrogen atom Ch. 9• Nov 10,12 Multi-electron atoms Ch.10• Nov 15 Periodic properties Ch. 10• Nov 17 Periodic properties Ch. 10• Nov 19 Valence-bond; Lewis structures Ch. 11• Nov 22 Hybrid orbitals; VSEPR Ch. 11, 12• Nov 24 VSEPR Ch. 12• Nov 26 MO theory Ch. 12• Nov 29 MO theory Ch. 12• Dec 1 Putting it all together• Dec 2 Review 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.

EXAMPLESCalculate the frequency of electromagnetic radiation from its wavelength and velocity.

What is the frequency of the radiation?

mnm

mnm 7

9

1089.51101

589

1147

18

1009.51089.510998.2

s

mmsc

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 acceleratingAccelerating charge emits radiation, lowering its energyLower energy shorter radial distanceTherefore, 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.”

hE 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

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.

0vv Minimum value

THE PHOTOELECTRIC EFFECT

light electron

metal

KE

of

elec

tro

n

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: hE

hc

hE

The energy of the photon depends upon the frequency

c

EINSTEIN’S THEORY OF RELATIVITY

The photon has zero rest mass (m0 = 0)

E pcE hvUsing ...

pc hv

phvc

h

So that

Thus…..

E (pc) (m c2)2 2

0

2

E (pc)2 2

PHOTONS HAVE MOMENTUM

ph

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 rayselectron

Now along comes de Broglie!

h

pmc

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

h

mc

h

m vph

mh v

“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 mightexhibit is interference

Constructive

Destructive

electron gun

electrometer- detects electrons as current

angle (

curr

ent

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 (

curr

ent

He observed……….

INTERFERENCE………...

Diffraction of an electron beam….

ph

mh v

We can relate these spacings to the electronwavelength

All matter and energy shows both particle-like and wave-

like properties.

WAVE-PARTICLE DUALITY…..

MASS INCREASES

Example....

h

m vph

mh v

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 107 m/s?

vmh

)/35)(10.0(10626.6 34

smkgJs

1J = kg m2 s-2

= 1.9 x 10-34 m

Ball:

h = 6.626 x 10-34 J sSolution:

Now do the electron…...

vmh

)/101)(1011.9(10626.6

731

34

smkgJs

= 7.3 x 10-11 m

Electron: kgm 311011.9

In summary…...

= 1.9 x 10-34 m

= 7.3 x 10-11 mElectron:

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

2

2

n

AZEn

the energy of an orbit……..

BOHR ATOM ENERGY LEVEL DIAGRAM

n=1-A

n=2-A/4

En

0n=3-A/9n=4

En

erg

y

-A/16

e- 2nA

En 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

2nA

En

n=1-A

n=2-A/4

En

0n=3-A/9n=4

En

erg

y

-A/16

e-

ELECTRON DE-EXCITATION

Emission of energy as a photon

e-

The energy of the photon emitted or absorbed

is given by the energy difference between the

energy levels and Planck’s relationship!

hEEphoton

ATOMIC SPECTRA:INTERPRETATION

by BOHR’S MODEL

E = energy of final state - energy of initial state

hE photon

2

2

2

2

if n

AZ

n

AZE

222 11

fi nnAZE

ni

nf

hEEE if ABSORPTION OF A PHOTON

SPECTROSCOPYEMISSION

Sample heated.

Many excited states populated

n = 1 Ground state

n = 2

n = 3n = 4n = Ion8

Excited states

...

En

erg

y

The spectrum…..

SPECTROSCOPYEMISSION

Sample heated.

Many excited states populated

Li Na K

EXCITED GROUP 1 ELEMENTS

The hydrogen emission spectrum can be broken

into series:

n = 1 Ground state

n = 2

n = 3n = 4n = Ion8

Excited states

...

En

erg

yFor the Lyman series, nf= 1 and ni = 2,3,4…

For the Balmer series, nf = 2 and ni = 3,4,5…

For the Paschen series, nf = 3 and ni = 4,5,6…

For the Balmer series, nf = 2 and ni = 3,4,5…

n = 1 Ground state

n = 2

n = 3n = 4n = Ion8

Excited states

...

En

erg

y

THE BALMER SERIES

22

11

if nnhA

hE

v

EMISSION

For the Balmer series, nf = 2 and ni = 3,4,5…

n = 1 Ground state

n = 2

n = 3n = 4n = Ion8

Excited states

...

En

erg

y

THE BALMER SERIES

2234

18 121

10626.610179.2

inJsJ

v

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+ X2+ + e–

IONIZATION ENERGY

UNITS: kJ mol-1

Back to the energy level diagram…….

BOHR ATOM ENERGY LEVEL DIAGRAM

2nA

En

n=1-A

n = 2-A/4

En

n=3-A/9n=4

EN

ER

GY

-A/16e-0

the electron is JUST free

And the energy of the electron is ZERO…….

IF we choose a photon so that

Then…...

Efinal = 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 = 2.178 x 10-18 J for one atom!

E= -A / (2)=0THIS DEFINES IONIZATION.

E=-A/(12)= -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 10-18 J atom-1 x 6.022x1023 atoms mol-1

=13.12 x 105 J mol-1

= 1312 kJ mol-1

= 0 - (-A) = A = 2.178 x 10-18 J for one atom