Light: The Wave Particle Duality

CHEM 107T. Hughbanks

Waves

■ Characterized by Wavelength, Frequency, Amplitude, Speed of Propagation

zDirectionof propagation

Amplitude

Wavelength (λ)

Interference Light as a Wave■ Light displays many wave-like properties:

Interference & DiffractionWavelength, frequency, speed:

c = λν(m/s = m × s-1)

■  c fixed; λ (or ν) specifies color of light■  Until about 1900 the wave model of light

was thought to be complete.

Light bulbs are “polychromatic”

Electromagnetic Waves (Light)

The Electromagnetic Spectrum

Cell phones,MRI’s

Airport scanners

Diagnostic X-rays

X-raycrystallography

Microwaveovens

IRspectra

Cell Phone wavelengths

■ The US and Canada use cell phones that operate in the 850 MHz frequency range and typically transmit ~3W power.

■ What wavelength does this correspond to?

What a molecule “feels”!

= an atom or molecule

What an atom or molecule experiences:

zE

t = 0.50ν

Electric field "down"

Light as a Particle

■ During this century, several discoveries have lead to a particle model of light.

■ Photons = “particles” of light

■ Energy of a photon: E = hν■  h = “Planck’s Constant” = 6.626 × 10–34 J s

■  h is very small, but NOT zero

Photon Energy

■ A He:Ne laser in a laser pointer emits red light with a wavelength of 633 nm. (1 nm = 10–9 m).

■ What is the energy of a photon at this wavelength? One mole of such photons?

■ A particular He:Ne laser has an output of 1 mW (1 mJ/s). How many photons are emitted per second?

Wave-Particle picture of Light

Photons (and electrons) come in "packets" - they have wave and particle-like characteristics

λ

Wave-Particle picture of Light

The center of the wave packet moves through spaceat the speed of the photon (or electron!)

Photoelectric Effect■  Shine light onto a metal surface.■ Under some conditions, electrons emitted.■ Detect # of electrons, kinetic energies.■ Energy is conserved!

Light in Electrons out

Metal Surface

e-hν

Photoelectric Effect

ν

#e–

ν0

Photoelectric Effect

ν > ν0

#e–

I

Photoelectric Effect

ν0

K.E.

ν

Photoelectric Effect

K.E.

I

Photoelectric Effect

ν

# e-

ν

KEΙ

# e-

Ι

KEν0

ν0

ν > ν0

ν > ν0

Photoelectric Effect

■ Experimental results NOT consistent with a simple “wave model” of light.

■  Postulate of photons (“particle model” of light) allows explanation. (Einstein, 1905)

Photoelectric Effect

■ Light with ν = 1.3 × 1015 s–1 ejects electrons from cesium metal. If the kinetic energy of the electrons is 5.2 × 10–19 J, what is the binding energy of electrons in cesium metal?

■ HINT: Conservation of Energy!

Atomic Spectra and Energy Levels

Atomic Spectra

■ Excited atoms emit light (neon signs, etc.)■ Emission from different elements is

different colors.■ Emission of only certain wavelengths■  “Spectral lines”■ Existence of spectral lines implies

“quantized energy levels.”

Atomic Spectra: Emission

Atomic Spectra: Emission

∆E(atom) = E(photon) = hν

+ hν

Excited Atom → De-excited Atom + Photon

Atomic Spectra: Absorption

∆E(atom) = E(photon) = hν

+ hν

Atom + Photon → Excited Atom

Atomic Spectra

■  Spectra reveal quantized energies.

■ Conservation of Energy– Relates atom’s energy levels to photon’s

wavelength

■ Absorption & Emission– Same lines (from same element)– Usually see more lines in emission

Problem - Energy Levels

■ A hypothetical atom has only 4 allowed energy levels.

■ The emission spectrum of this atom shows 6 lines at wavelengths of 100 nm, 114 nm, 150 nm, 300 nm, 480 nm, and 800 nm.

■ Atomic energy levels usually get closer together as energy increases.

Problem - Energy Levels

■ Draw an energy level diagram for this hypothetical atom.

■ Label the 4 states as E1, E2, E3, and E4, with E1 < E2 < E3 < E4.

■ Use arrows to show the 6 observed transitions, and the correct wavelengths.

■ Try to draw your diagram “to scale,” showing the spacings between levels.

4 Energy Levels

E4E3E2

E1

E

6 Transitions

E4E3E2

E1

E

Problem - Energy Levels

■ Convert to frequencies using: ν = c/λ. λ = 100 nm ⇒ ν = c/λ = 3 × 1015 m s-1

λ = 114 nm ⇒ ν = c/λ = 2.63 × 1015 m s-1

λ = 150 nm ⇒ ν = c/λ = 2 × 1015 m s-1

λ = 300 nm ⇒ ν = c/λ = 1 × 1015 m s-1

λ = 480 nm ⇒ ν = c/λ = 6.25 × 1014 m s-1

λ = 800 nm ⇒ ν = c/λ = 3.75 × 1014 m s-1

■ Remember, E ∝ν (E = hν)

Emission Wavelengths

E4E3E2

E1

800

300

480

150

114

100 E

Spectra and Analysis

■ Absorption and emission spectra exist for both atoms and molecules.

■ Observed wavelengths are characteristic of a particular substance.

■  Spectra are often used to identify unknown substances, in areas ranging from forensics to astrophysics.

Electrons: More Wave-Particle Duality, Orbitals

CHEM 107T. Hughbanks

Electron Diffraction

■  Electron diffraction from aluminum foil.■  Pattern is similar to pattern obtained in X-ray

(light) diffraction.☛ Electrons have some wave-like properties.

Uncertainty Principle■ A curious result of wave mechanics.

■  “Position and energy (or momentum) can not be specified simultaneously.”

■ Mathematically (one dimension):(∆x)(∆p) > h/4π

☛ Electrons as delocalized waves rather than particles at a specific position.

Properties of Electrons

■ Like light, electrons can show properties of both waves and particles.

■ Electrons bound in atoms can only have certain quantized energies.

■ Electrons in atoms can best be described as “delocalized waves.”

Properties of Electrons■ The position of an electron can not be

specified. The more precisely its position is specified, the less we can know about its momentum.

■ Electrons have magnetic properties.

■  “Spin”2 possible spin values; the image of a

spinning particle is not literally true.

DeBroglie’s inspired guessFor light:Momentum, p = (Energy/velocity) = E/c

using Planck’s formula, E = hν = (hc/λ)or, combining equations: p = (h/λ)DeBroglie assumed the same relationship for

free electrons and used the particle-like definition of momentum, p = mu:

p = mu = h/λ or λ = (h/mu)

Photons vs. Electrons

■  Energy (E) E = hν = h(c/λ)

■  Wavelength (λ) λ = h(c/E) = (h/p)

■  Velocityc = 3 × 108 m s-1

Photons Electrons

E = (1/2)mu2

λ = (h/p) = (h/mu)

u = √(2E/m) = (h/mλ)

h = 6.626 × 10–34 J s

Problems■ A common wavelength used in X-ray

diffraction is λCu = 1.54 Å = .154 nm (This is comparable to the lengths of chemical bonds.) Q: What is the energy of one mole of such photons?

Answer: 7.77 × 105 kJ/mol■ What is the kinetic energy of one mole of

electrons with the same wavelength?Answer: 6.12 × 103 kJ/mol

Electron Diffraction

Electron diffraction from aluminum foil.Patterns similar to those in X-ray (light) diffraction.☛ Electrons have some wave-like properties.

Aluminum Graphite (single crystal)

Electrons

■ Electrons show properties of waves AND particles. (Like light does!)

■ Electrons in atoms best described as “delocalized waves”

Problems■ A common wavelength used in X-ray

diffraction is λCu = 1.54 Å = .154 nm (This is comparable to the lengths of chemical bonds.) Q: What is the energy of one mole of such photons?

■ What is the kinetic energy of one mole of electrons with the same wavelength?

Ψn = sin nπxL

Where n = 1,2,3…n is called a quantum number

Functions for Electron in a Box

■ We can write a general equation for the allowed standing waves:

Standing Waves -Wavefunctions

n = 3

n = 2

n = 1

Here n is a quantum number, identifying the allowed states.

Ψ

amplitudes

(Electrons Confined to small spaces!)

There are some animated standing waves at:http://www.sengpielaudio.com/StandingWaves.htm

Bohr “Orbits” and deBroglie waves (Discarded )

Atomic Energy Levels and Orbitals

CHEM 107T. Hughbanks

Wavefunctions & Intensity

n = 3

n = 2

n = 1

Intensity at any point is proportional to the square of the amplitude of the wave.

Ψn 2

Electrons & ���Quantum Mechanics

■ Experiments show energy levels exist.���How can we explain this?

■  “Quantum Mechanics”���Application of wave concepts

■  Schrödinger’s Wave Equation:

H Ψ = EΨ

Schrödinger EquationH Ψ = EΨ

■ This is a second order differential equation in disguise.

■ H is a “differential operator,” Ψ is a “wave function,” and E is an energy.

■ H includes terms for kinetic and potential energies.

Schrödinger Equation

∂2Ψ∂x2 +

∂2Ψ∂y2 +

∂2Ψ∂z2 −

e2

4πε0rΨ = EΨ

For a hydrogen atom:

Schrödinger Equation■  Solving the equation gives Ψ and E for allowed

states. For atoms, this is usually done in spherical coordinates.

Ψ = Ψ (r, θ, φ)

■  Solution contains 3 different quantum numbers. Each valid solution is called an “orbital.”

θ

φ

r

x

y

z e- position

nucleus

Rydberg Formula for H-atom

Energy levels for the Hydrogen Atom:

E =−2.18 × 10-18 J

n2

n - principal quantum no.

E = −2.18 × 10-18 Jn2

n - principal quantum no.

Hydrogen atom Orbital Energies

× (2.18 × 10-18 J)

(3/4)

(8/9)

1s

2s 2p

3s 3p 4s 4p

3d 4d 4f

(15/16)

(3/4) = 12 - (1/2)2

(8/9) = 12 - (1/3)2

(15/16) = 12 - (1/4)2

Orbitals & Quantum Numbers■ Quantum numbers are called n, l, and ml .

■ Also called “principal,” “azimuthal,” and “magnetic” quantum numbers.

■ A set of these 3 defines an orbital.

■ An orbital is the wave representation of an electron in an atom.

Quantum Numbersn - principal quantum number

influences energy and size of the orbitaln = 1, 2, 3, ...

l - azimuthal quantum numbershape of orbital (mainly)l = 0, 1, 2, ..., (n-1)

ml - magnetic quantum numberorientation of orbital (mainly)ml = -l, ..., 0, ... +l

Allowed combinations – LEARN!

n l ml # oforbitals

type oforbitals

1 0 0 1 1s2 0

10

-1,0,+113

2s2p

3 012

0-1,0,+1

-2,-1, 0,+1,+2

135

3s3p3d

The meaning of Ψ■ Orbitals are wavefunctions, defined in

mathematical terms.

■  Physical interpretation?

☛ Ψ 2 tells us the probability of finding the electron at some point in space.

■  “Pictures” of orbital shapes are actually graphs of Ψ 2 .

Some Orbital Wavefunctions ���(don’t try to memorize these!)

Ψ2s = (2a)−3/2 (2 − ra) exp(−r /2a)

14π

Ψ2pz =13

(2a)−3/2 ra

exp(−r /2a) 3

4π cosθ

Ψ2px =13

(2a)−3/2 ra

exp(−r /2a) 3

4π sinθ cosφ

Ψ2py =13

(2a)−3/2 ra

exp(−r /2a) 3

4π sinθ sinφ

radial extent shape, orientation

1s orbital

Distance from nucleus

Elec

tron

prob

abilit

y

Most probable distance: Bohr radius = 53 pmr2Ψ2

Representing Orbitals, Ψ & Ψ2

2s orbital

r2Ψ2Ψ

Ψ2s2 - what the pictures mean Shapes of p-orbitals

px py pz■  3 p-orbitals for each n-value (2p, 3p, ...)■  all same shape■  lobes point in perpendicular directions

Ψ

Ψ & Ψ2 fora 2p-orbital

Electron Density in a 2p Orbital

2py

2px and 2pz Orbitals

2px 2pz

3p Orbitals

3py not shown

3px 3pz

Shapes of 3d Orbitals Ψ

3dxy 3dyz

3dx2–y2 3dz2

3dxz

Hydrogen orbitals’ radial extent

1s most probable distance: a0 = Bohr radius = 53 pm

Elec

tron

prob

abilit

y

r2Ψ2

Distance from nucleus

From orbitals to atoms■ Each orbital can “hold” 2 electrons,

provided they have opposite spins.

■ Build up atoms by filling orbitals with appropriate # of electrons.

■  Start at low energy, work toward high energy.

■  “Electron configurations”

Orbital Filling

■ Low energy orbitals fill first.

■ Orbital energy increases– rapidly as n increases– more slowly as l increases

Hydrogen���Orbital ���Energies ���⇓ ���

Many-e–���Orbital ���Energies