Particles and Waves

Early Atom Models

Democritus (2500 years ago)—tiny indivisible particles, atmos

Dalton (1803)Each element composed of extremely small particles

Atoms of any element are identical and different from others

Atoms not created or destroyed

Compounds always have same relative numbers of kinds of atoms

The Plum Pudding Model of the Atom

• The atom was once thought to be a solid ball of positive material of pudding-like consistency in which were embedded discrete negatively charged objects called electrons.

• In this (incorrect) model of the atom, the atom is a ball of uniform density.

Ernest Rutherford

Ernest Rutherford won1908 Nobel Prize forstudies in radioactivity;discovered nucleus in 1911.

• Ernest Rutherford was the first to show thatthe atom does nothave uniform density

• Most of the itsmass is located at itscenter.

The Nucleus

Rutherford Model of the Atom

Based on his alpha-particlescattering experiment on gold, Rutherford concludedthat the atom consisted ofa hard central core wheremost of the mass of theatom rested.

Nuclear Structure

Facts about the nucleus:

• Protons and neutrons have about thesame mass, and each is about 2000times as massive as the electron.

• The number of protons is the sameas the number of electrons (not shown)which orbit the nucleus.

• For any given element, all nuclei have the same number of protons,but the number of neutrons willvary.

Nuclei, Nucleon, and Quarks

Structure of Sub-Atomic Particles

Quarks

Murray Gell-Mann tookthe name quark from "Three quarks for musterMark", in James Joyce'sbook Finnegan's Wake.(1963)

Whimsical names--called "flavors"--for the quarks

There Are Two Families of Elementary Particles

QUARKS

up charm top

down strange bottom

LEPTONS

electron muon tau

e-neutrinomuon-neutrino

tau-neutrino

Families of Particles

Hadrons

Sample Fermionic Hadrons

Baryons (            ) and Anti-baryon (            )

Symbol NameQuark

Content

Electric

Charge

Mass(GeV/c2) Spin

        proton         1 0.938 1/2

        anti-proton         -1 0.938 1/2

        neutron         0 0.940 1/2

        lambda         0 1.116 1/2

        omega         -1 1.672 3/2

Neutral Atoms and Ions

For neutral atoms

# electrons = # of protons

For ions

# electrons ≠ # protons

Neutral atomIon

Isotopes

Atoms can have the same number of protons but different numbers of neutrons

These are the isotopes of an element

Isotopes Can Be Common or Rare

Physics of the Early 20th Century

The Radiation Spectrum from Hot Objects

This filament is not yet hot enough  to emit significant amounts of blue  light.

This filament is emitting all of the colors of the rainbow, which makes it "white" hot.

Blackbody Radiation

Prism disperses electromagneticenergy into its component parts.

Hot objects emit red and orange, while very hot objects emit red, orange, and blue light.

Predicting Black-Body Spectra

• 19th century concern--predict the intensity of radiation emitted by a black body at a specific wavelength.

• Wilhelm Wien’s theory predicted the overall form of the curve but at long wavelengths his theory disagreed with experimental data.

• Rayleigh and Jeans theory fitted the experimental data at long wavelengths but it had major problems at shorter wavelengths.

• The problem was a term in the denominator. It meant that as the wavelength tended to zero, the curve would tend to infinity.

• However, experimental data shows there is a peak wavelength for each temperature, and the energy emitted at either side of this peak drops. The Rayleigh-Jeans Law predicted no peak wavelength.

Problems of Classical Physics

Classical physics predicted a continual increase in radiated energy with increasing frequency, the “ultraviolet catastrophe,” which did not occur.

Planck's Quantum

Max Planck (1858-1947)German physicist1918 Nobel Prize

• Planck explained the blackbody radiationspectrum by postulating that the radiationwas emitted by oscillating atoms, and furthermore that the energy was quantized.

• The energy of these "atomic oscillators"had to be an integer multiple of hf, wheref is the frequency of vibration of the atoms and h = 6.63 x 10-34 J-s.

E = nhf       n = 0, 1, 2, 3, .........

• Albert Einstein later applied a similar quantum concept to light.

The Photoelectric Effect—The Question

The Photoelectric Effect—The Answer

Einstein found an answer to the question by assuming that the energy of light was different for light of different frequenciesUnless the light was of high enough frequency to have the energy to release an electron, nothing happenedEinstein won the Nobel Prize for this discovery, NOT for the theory of relativity

To cause the release of an electron light has to have energy equal at least to the Work Function W

Light Comes in PacketsA “packet of light is called a photon

A photon is a “quantum” of light—the smallest unit

Photons have no “rest energy”

• They move at the speed of light but have no mass

The energy of a photon is given by:

E = hf

Lower energy

Higher energy

Example

KEmax = hf - W0

-----------------------------------------------Problem:  

A metal has a workfunction of 8 x 10-19 J.

What is the maximum kinetic energy of electrons emitted from the metal when light of frequency f = 2 x 1015 Hz is shone on the surface?

Solution:

KEmax = 5.26 x 10-19 J

Photon Energy Example

How many photons stream forth in onehour from a light bulb radiating 100 watts

of light energy?  Assume = 500 nm.--------------------------------------------------------

c = ff = c /

  = (3 x 108 m/s) / (500 x10-9 m)  = 6 x 1014 Hz

E = h f    = (6.63 x 10-34 J-s) (6 x 1014 s-1)

    = 3.98 x 10-19 J    

(Solution continued)

100 watts = 100 J/s One hour = 3600 s

Total energy radiated = 3.6 x 105 J

 N = 3.6 x 105 J / 3.98 x 10-19 J

       = 9 x 1023

The ElectronVolt (eV) and the Rule of 1240

The kinetic energy of an electronaccelerated across a potentialdifference of one volt is one

electronvolt (eV).

The eV is not a unit of charge, or a unit of voltage; it is a unit

of energy.

The energy E in electronvolts (eV) of a photon is related to its

wavelength in nanometers (nm) through the following

relationship:

E = (1240 eV-nm) /

Photoelectric Effect

Problem Light of wavelength   = 400 nm shineson a metal surface whose work function

W0 is 2.0 ev.

What is the maximum kinetic energyof the photoemitted electrons?

Solution

E = 1240 / 400    = 3.1 eV

KE = 3.1 - 2.0      = 1.1 eV

Photocells in Garage Door Openers

Photocells in Movie Projection

Compton Scattering

A.H. Compton discovered that x-ray photons when collided with electrons in a graphite crystal displayed particle properties of momentum equal to h / .

• When the x-rays emerged from the crystal they had a wavelength longer than their original value.

• The lengthening of their wavelength showed that they had lost energy. The lost energy was accounted for in the KE of an ejected electron from the crystal.

Compton Scattering--explanation

Eo = Ef + KEelectron

KEelectron = Eo - Ef

KEelectron = hfo - hff

KEelectron = h ( fo - ff )

KEelectron = hc ( 1/o - 1/f ) where f = c /

' = (o) + [(h / mc)(1-cos)]

where is the angular deviation of the x-rays from their original path.

from special relativityE = mc2

momentum = p = (E / c2)(v)but v = c, therefore p = E / c

since E = hf = h (c /),p = (hc /) / c

p = h /

In 1927, Arthur H. Compton was awarded the Nobel Prize in Physics for his discovery of the particle properties of x-rays.

Deriving the Momentum of a Photon

Photon MomentumWhat power (in Watts) must the beam of a 0.5 kg flashlight deliver if the recoil momentum of the outgoing photons is to suspend the flashlight against the earth's gravity? (Assume that all photons from the bulk are visible photons with wavelength =5000 Å.)

A certain flashlight bulb consumes 10 Watts of electrical power. If the conversion efficiency from electrical power to light energy is 10%, what mass could the outgoing beam of light be able to suspend against the earth's gravity? By what percentage is this force, when averaged over a period of one second, expected to fluctuate?

Fundamental Particle Relationships

Why Do Excited Atoms Emit Light In Narrow Bands (lines)

Atomic Spectra

• Atoms in heated gases emit and absorb light of certain wavelengths.

• Shown above are three emission spectra and one absorption spectrum.

Line Spectrum of Atomic Hydrogen

In 1885 Johann Balmer discovered an equation which describes the emission-absorption spectrum of atomic hydrogen:

1 / = 1.097 x 107 (1 / 4 - 1 / n2)

        where n = 3, 4, 5, 6, ...

Balmer found this by trial and error, andhad no understanding of the physicsunderlying his equation.

Why are there lines in the first place?To answer this question we have to understand a few things

Neils Bohr Explains the Hydrogen Atom and Balmer’s Results

Neils Bohr, a Danish physicist, treated thehydrogen atom as if it were an electron ofcharge -e orbiting in a circular path abouta proton of charge +e.

Bohr Model

Energy Levels in Hydrogen

Hydrogen atom explained

Energy Levels in Hydrogen

En = -13.6 eV /n2 -----------------------  E1 = - 13.6   eV E2 = -   3.40 eV E3 = -   1.51 eV E4 = -   0.85 eV E5 = -   0.54 eV E6 = -   0.38 eV

Energy Transitions in Atoms 

Energy of photon = Energy lost by electron                            hf  = Ei - Ef

Calculating Wavelengths of Emitted Light

hf  = Ei - Ef E3  --->  E2:Ei  =  - 1.51 eVEf  =  - 3.40 eV------------------------------hf   = - 1.51 - (-3.40)      = 1.89 eV------------------------------ = (1240 eV-nm) / E

   = 1240 / 1.89   = 656 nm

Balmer Series

Wavelength(nm)

Relative Intensity

Transition Color

383.5384 5 9 -> 2 Violet

388.9049 6 8 -> 2 Violet

397.0072 8 7 -> 2 Violet

410.174 15 6 -> 2 Violet

434.047 30 5 -> 2 Violet

486.133 80 4 -> 2 Bluegreen (cyan)

656.272 120 3 -> 2 Red

656.2852 180 3 -> 2 Red

Other Energy Transitions

The final state in the energytransitions is n = 3 for thePaschen series, n = 2 for the Balmer series, and n = 1 for the Lyman series.

Recalling that the range ofvisible wavelengths isapproximately 300-700 nm,one can see that only transitions ending at n = 2 emit light in the visible range

The De Broglie Wavelength

Louis deBroglie     (1892-1987)

 1929 Nobel Prize

p = momentum  

= h/p

Electron beam produces a pattern similar to the one produced by light 

Electron's de Broglie Wavelength

What is the wavelength of an electronmoving at a speed v = 2 x 104 m/s?-----------------------------------------------------m = 9.1 x 10-31 kgp = 18.2 x 10-27 kg-m/sh = 6.63 x 10-34 J-s= h / p   = 3.6 x 10-8 m   = 360 x 10-10 m   = 36 x 10-9 m   = 36 nm

Baseball de Broglie Wavelength

What is the wavelength of baseballmoving at a speed v = 2 x 104 m/s?-----------------------------------------------------m = 0.5 kgmomentum = 1 x 104 kg-m/sh = 6.63 x 10-34 J-s= h / momentum   = 6.63 x 10-38 m   = very small

Louis deBroglie: Hydrogen Electrons Behave Like Waves

An integral multiple of wavelengths must fit in the length r, otherwise destructive interference would occur.

DeBroglie Waves in Bohr's Model

Schrödinger's Wave Mechanics and Probability Clouds

s orbital p orbital

The Heisenberg Uncertainty Principle

Werner Heisenberg        (1901-1976)        1932 Nobel Prize Observing an object

changes its momentum and energy.

electron

The Heisenberg Uncertainty Relations

px  > h/2 

If one attempts to locate particle to within an uncertainty x, then the least uncertainty  p there will be in the particle's momentum is

p = (h/2) / x

Et  > h/2

If a particle exists for a time period, then the least uncertainty  E there will be in the particle's energy is

E = (h/) / t

Understanding Uncertainty

Understanding Uncertainty

Particle Confinement

Confinement Calculation

Heisenberg Uncertainty Example

Example:  An electron is observed for a period of time long enough to ascertain its position to within one hundred angstroms.  What is the least uncertainty in the electron's velocity?

x = 1 x 10-8 m

p = (6.63 x 10-34) / (2) / (1 x 10-8) = 1.055 x 10-26 kg-m/sm = 9.1 x 10-31 kgmv = pv = 1.2 x 104 m/s