Wave Nature of Light

• Rutherford’s model of the atom could not explain chemical behavior

• Bohr and others described the arrangement of electrons around the nucleus

• These arrangements could account for differences between the elements

• Bohr’s model was based on spectroscopic evidence

Electromagnetic Radiation

• Light can be described as though it is a wave

Parts of a wave• Amplitude and wavelength

Crest

Trough

Wave Stuff

• Amplitude: Vertical distance from crest to midline (brightness/intensity)

• Wavelength: Horizontal distance from crest to crest (meters) (color)

• Velocity: Rate at which wave travels (3.00x108 m/s in a vacuum) (constant)

• Frequency: # waves that pass a given point per unit time (waves/sec, cps, hertz, s-1)

Electromagnetic Spectrum

Low frequency

High frequency

Wave Math

• Velocity = wavelength x frequency

• Velocity = c

Wavelength = Frequency = • c = • Since c is constant, and are inversely

related

Example problems

• Find the frequency of purple light (wavelength 455 nm).

• Solution: c = , or = c/Wavelength must be in meters, since c is in m/s: 455 nm = 4.55x10-7 m

= (3.00x108 m/s)/(4.55x10-7 m) = 6.59x1014s-1

Example #2

• Find the wavelength of red light with a frequency of 4.56x1014s-1.

• c = or = c/• = (3.00x108 m/s)/(4.56x1014s-1) =

6.58x10-7m (658 nm)

Particle Nature of Light

• Light also acts as a particle: proposed by Newton

• Max Planck and glowing blackbodies: energy is quantized

• Smallest energy unit available is a quantum

• Energy of quanta depends on the frequency of the energy:

E = h

Particle Nature of Light

• h is Planck’s constant: 6.626x10-34Js

• Energy is directly proportional to frequency

• Energy is inversely proportional to wavelength:

E = hand = c/so E = hc/

Light Particle Sample Problem

• Find the energy of a microwave photon having a wavelength of 3.42x10-2m.

• Solution: E = hc/= (6.626x10-34Js)(3.00x108m/s)/3.42x10-2m

= 5.81x10-24J

Photoelectric Effect

• Certain metals will eject electrons when exposed to light

• Number of electrons ejected depends only on the intensity of light

• No electrons ejected by light below certain frequency

Photoelectric Effect

• Could not be explained by wave model of light

• Einstein explained effect using quantum theory of light

• He won the Nobel Prize for his work (not for relativity)

Photoelectric Effect Simulator

Atomic Emission Spectra

• When elements are zapped with energy, they give off light

• Light is first shone through a slit

• When light is shone through a prism, colors are separated

• Only some of the colors appear as fine lines against a dark background

Emission spectra of common fluorescent bulbs

Emission Spectrum Setup

Absorption Spectra

• In absorption spectra, light is shone through cold gas or liquid

• Light then goes through a slit and prism or grating

• Resulting spectrum is continuous except for dark lines

Absorption Spectrum

Absorption Spectrum of Chlorophyll

Liquids tend to have less distinct absorption spectra

Gas Absorption Spectra

Absorption Spectrum Setup

Spectra Types

Bohr Model of the Atom

• Bohr wanted to explain the presence of sharp lines in the hydrogen spectrum

• He proposed that hydrogen’s electron could only have certain distinct energies

• These energies were integral multiples of some minimum energy

• The energy levels correspond to differently sized orbits

Bohr’s Atom

• Spectral lines were due to electrons jumping from one level to another.

• Incoming energy promotes an electron to a higher energy level

• When the electron returns to the lower level it releases energy

Quantum Numbers

• Bohr assigned the energy levels numbers

• The Principle Quantum Number (n) represents the main energy level

• n can only have non-zero integral values

• n = 1, 2, 3, ...

Why quantum numbers?

• Louis de Broglie: Wave-particle duality

• As a wave: E = hc/• As a particle: E = mc

2

• Combined: hc/ = mc 2

= hc/mc 2 = h/mc

• For objects moving slower than light, replace c with v (velocity): = h/mv

Particle wavelength problems

• Every particle has a wavelength• The larger the particle, the shorter the

wavelength• Example: Calculate the wavelength of an

electron moving at 0.80c (mass = 9.109×10-31 kilograms).

• Solution: = h/mv = 6.626x10-34Js/[(9.109×10-31kg)(0.80)(3.00x108m/s)]= 3.0x10-12m (smaller than an atom, bigger than a nucleus)

Particle wavelength problems

• Find the wavelength of a baseball (145g) thrown toward home plate at 95.0 mph (42.5 m/s)

• Solution: = h/mv = 6.626x10-34Js/[(0.145kg)(42.5m/s)]= 1.08x10-34m (much smaller than a nucleus)

Back to quantum numbers

• Only certain orbits are allowed because they are the only ones in which an integral number of wavelengths can “fit”.

• “In-between” orbitals would require a fractional number of wavelengths.

“I think it is safe to say that no one understands quantum mechanics.” Physicist Richard P. Feynman

Heisenberg Uncertainty Principle

• It is impossible to know both the position and momentum of an electron simultaneously

• Electrons are both particles and waves• It’s in their nature to be indeterminate• Can be thought of as being “smeared out”

over a region of space• Indeterminacy is related to Planck’s

constant

More Energy Levels

• The fine lines in emission spectra are actually made up of several even finer lines

• Each energy level has sublevels• Each sublevel has a shape• Each sublevel has one or more orbitals• Each orbital holds two electrons• How do we sort all this out?

Using Quantum Numbers!

• Four quantum numbers are needed in Schrödinger’s equation to describe the probability function of an electron

• n = principle quantum number = 1, 2, 3, ...Main energy level – determines size of

orbital• l = azimuthal quantum number = 0, 1, ...

n-1Sublevel – determines orbital shape

Well-used quantum numbers

• s: l = 0 (first two columns of PT)p: l = 1 (last six columns of PT)d: l = 2 (middle ten columns of PT)f: l = 3 (bottom two rows of PT)

• m = magnetic quantum number = - l to +lSpecifies orbital – determines

orientation• s = spin quantum number = ±½

Specifies spin

Orbital shapes

• s orbital: spherical• Every energy level has

an s orbital: 1s, 2s, etc.• Higher level s orbitals

are lobed• Nodes are areas of

minimum electron density

• One node is added for each level

• s sublevel: one orbital, two electrons

p orbital

• p orbitals are dumbbell shaped

• p sublevel (l = 1) consists of three orbitals: px py pz (six electrons)

• Three p orbitals are orthogonal to each other

• Only present after first main energy level (n>1)

d orbital

• d orbital is cloverleaf-shaped

• Five orbitals, ten electrons make up d sublevel

• Only available when n>2

f orbitals

• Complicated shape

• Seven orbitals, fourteen electrons

• Only available when n>3

Allowed quantum number combinations

• Pauli exclusion principle: no two electrons can have the same set of four quantum numbers

• Aufbau principle: electrons fill the lowest energy state available first

• Lower numbers mean lower energy (n and l)

• Various m and s states are degenerate (of equal energy)

Allowed Quantum Number Combinations

n l m s1 0 0 +½1 0 0 -½1s sublevel – 2e-

2 0 0 +½2 0 0 -½2s sublevel – 2e-

n l m s2 1 -1 +½2 1 0 +½2 1 1 +½2 1 -1 -½2 1 0 -½2 1 1 -½2p sublevel – 6e-

Allowed Quantum Number Combinations

3s and 3p are similar to 2s and 2p

3d sublevel:n l m s3 2 -2 +½3 2 -2 -½3 2 -1 +½3 2 -1 -½

n l m s3 2 0 +½3 2 0 -½3 2 1 +½3 2 1 -½3 2 2 +½3 2 2 -½3d sublevel - 10e-

Electron configurations

• Electron configurations show the location of every electron in the atom

• Electrons follow three rules: Pauli exclusion principle, Aufbau principle, Hund’s rule

• Each orbital is represented by a box and a symbol, and each electron by an arrow.

Electron configurations

More Electron Configurations

Hund’s rule: When putting electrons into degenerateorbitals, do not pair them until necessary.

More Electron Configurations

More Electron Configurations

Add non-standard configurations

• And reason for 4s-3d order

Orbital filling diagrams

Orbital filling diagrams

Noble Gas Shorthand Structures

• Noble gas symbols can be used to represent the core electron structure

• Manganese (25 e-): • Equivalent to [Ar] 4s2 3d5

1s22s2 2p63s23p64s2 3d5

Electron Dot Structures

• Electron dot structures represent only the valence electrons

• Valence electrons are the electrons in the outermost energy level (highest value of n)

• The maximum number of electrons allowed in the valence shell is 8

Electron dot structures

• Consists of the element’s symbol and dots representing the valence electrons

Hydrogen: Helium:Lithium:

Beryllium: Nitrogen: Neon:

Sodium: Iron: Lead:

H He Li

Be N Ne

Na Fe Pb