Plan for Wed, 29 Oct 08

• Lecture– Electromagnetic radiation (7.1)– The nature of matter, and evidence for the

quantization of matter and energy (7.2)

• Friday Dr. Villarba subs…you still have a quiz!!• Get Exp 6 from her website:

– http://seattlecentral.edu/faculty/mvillarba/CHEM161

Electromagnetic Radiation

Just like we use visible light to see things, we can use any region of the EM spectrum to “see” objects.

Radar: Radio WavesWeather systems

Radio telescopes. This is the Very Large Array (VLA) in NM.

A galaxy imaged in the visible spectrum.

The same galaxy imaged in the radio spectrum at the VLA.

Thermal Imaging: Detecting IR radiationMuch of a person’s energy is radiated away from the body in the form of infrared (IR) energy.

You can produce more IR energy to warm yourself by moving around…this is why you shiver when you go outside in the cold with no coat on. This process is called thermogenesis.

Why does your mother insist you wear a hat in the winter?

“IR” Photography

Image obtained with “IR” film, which is really film that is activated by light at 700-900 nm.

Electromagnetic Radiation

• Electromagnetic radiation or “light” is a form of energy.

• Characterized by:–Wavelength ()–Amplitude (A)

• Has both electric (E) and magnetic (H) components.

Electromagnetic Radiation (cont.)• Wavelength (): The distance between two

consecutive peaks in the wave.

Increasing Wavelength

1 > 2 > 3

Unit: length (m)

Electromagnetic Radiation (cont.)• Frequency (): The number of waves (or cycles)

that pass a given point in space per second.

Decreasing Frequency

1 < 2 < 3

Units: 1/time (1/sec) or, Hertz (Hz)

Electromagnetic Radiation (cont.)• The product of wavelength () and frequency

() is a constant.

() () = c

Speed of light

c = 3 x 108 m/s

c is a constant, independent of

What statement is true when comparing red light to blue light?

A. Red light travels at a greater speed than blue light.

B. Blue light travels at greater speed than red light.

C. The wavelength of blue light is longer.

D. The wavelength of red light is longer.

Back in the old days…• It was generally agreed that matter and light

were distinct.• Matter was particulate in nature, light could be

described using waves.• Physicists circa 1900 had it all figured out…• One famous physicist asserted that within ten

years or so all the major problems in physics would be solved.

• The only thing left, really, was this niggling little problem with black-body radiation…

Blackbody Radiation• Planck performed experiments on the light

emitted from a solid heated to “incandescence”.

As an object is heated, intensity of emission increases, and peak wavelength shifts to smaller wavelengths.

Can “classical” physics reproduce this observation?

Relative Intensity of Radiation from the SunThis kind of intensity response is called “black-body” radiation. This is the general radiation response for a super-heated object, which will emit radiation across the whole of the EM spectrum.

Note that over half the emitted radiation is in the IR...the sun is hot!!

The most intense radiation is in the visible...the sun is bright!!

Only 8% in the UV, but even this small amount can wreak havoc because radiation in this region is so energetic.

Why does the intensity of radiation drop off as we go to higher energy??

Explaining Black-body Radiation

1900 – Rayleigh-Jeans, Ultraviolet catastrophe

1901 – Planck, quantized black-body radiation

Relative Intensity of radiation from the sun.

R-J law agreed with experiment only at very long wavelengths

To fit exp. data, Planck introduced a parameter into a modified R-J law…

E = nh

Light as Energy• Planck found that in order to model this behavior, one has to

envision that energy (in the form of light) is lost in integer values according to:

Energy Change

n = 1, 2, 3 (integers)

frequency

h = Planck’s constant = 6.626 x 10-34 J.s

Light as Energy (cont.)• In general the relationship

between frequency and “photon” energy is

Ephoton = h

• Example: What is the energy of a 500 nm photon?

= c/ = (3x108 m/s)/(5.0 x 10-7 m)

= 6 x 1014 1/s

E = h =(6.626 x 10-34 J.s)(6 x 1014 1/s) = 4 x 10-19 J

h = 6.636 x 10-34 J s

c = 2.9979 x 108 m/s

1 Hz = 1 s-1

Which type of photon will have the largest energy?

A. Ultraviolet

B. X-Ray

C. Microwave

D. Visible

Energy Quantization

The student can stop at any point on the ramp. Her distance from the ground changes continuously.

The student can stop only at certain points on a flight of stairs. Her distance from the ground is quantized.

Similarly, atomic energy levels are like steps…the energies available to an atom do not form a continuum, they are quantized.

Evidence of Quantization• Black-body Radiation (Planck)

– A system can transfer energy in “packets” of size h– These packets are called quanta (singular: quantum)– Prior to this discovery it was thought that systems could absorb or emit

any amount of energy.

Other observations supported this “quantum” view: • Atomic Emission spectra (Balmer, Rydberg, Bohr)

– Light emitted from excited atoms occurs in discrete lines rather than a continuum.

• Photoelectric Effect (Einstein)– Energy itself is actually quantized into packets called photons.– This means energy has a particle-like nature as well as a wave-like

nature.• Electron Diffraction Patterns (Davisson & Germer)

– Matter also has a wave-like nature!!

Atomic EmissionWhen we heat a sample of an element, the atoms become excited. When the atom

relaxes it emits visible light. The color of the light depends on the element.

Li Na K Ca Sr

H

Li

Ba

When the light emitted from excited atoms is passed through a prism, we see discrete bands of color at specific wavelengths.

Continuous vs. Discrete SpectraH

Li

Ba

When the light emitted from excited atoms is passed through a prism, we see discrete bands of color at specific wavelengths.

White light passed through a prism gives a continuous spectrum ...all visible wavelengths are present.

Photon Emission

Em

issi

on

of p

hoto

n

An excited atom relaxes from high E to low E by emitting a photon.

We can determine the energy difference (E) between levels by measuring the wavelength of the emitted photon.

E = hc/= hc/ E

If = 440 nm, = 4.5 x 10-19 J

Quantized vs. Continuous

“Continuous” spectrum “Quantized” spectrum

Energy levels are so close together that any E ispossible

Energy levels are discretized, so only certain E are allowed

EE

The following visible emission spectrum was obtained from Rubidium:

A. Relative to hydrogen, the spectrum is less complex.

B. Rubidium is characterized by a continuum of energy levels.

C. No energy levels exist for which the difference in energy is equal to a 650 nm photon (red).

D. The energy levels of Rubidium are equivalent to those of hydrogen.

Which statement is true?

The Photoelectric Effect

Shine light on a metal and observe electrons that are released.

You need some minimum amount of photon energy to see electrons (“o”).

Further, for ≥ o, number of electrons increases linearly with light intensity.

The Photoelectric Effect (cont)Frequency: determines whether e- are ejected, and their KE (velocity).

Intensity: determines the number of e- that are ejected…but they will all have the same velocity!!

The Photoelectric Effect (cont.)• As frequency of incident light is increased, kinetic energy of emitted e-

increases linearly.

1

2me

2 h photon

= h0 Workfunction: energy

needed to release e-

• Light apparently behaves as a particle.

00

Frequency ()

The Photoelectric Effect (cont.)• For Na with = 4.4 x 10-19 J, what wavelength corresponds to o?

1

2me

2 h photon 0

h = = 4.4 x 10-19 J

hc = 4.4 x 10-19 J

hc

4.4x10 19J

6.626x10 34 J.s 3x108m /s 4.4x10 19J

= 4.52 x 10-7 m = 452 nm

00

Frequency ()

In a workfunction experiment using 300 nm light, the electrons ejected from Potassium (K) have a greater velocity that those ejected from Sodium (Na). Therefore:

A. Na > K

B. K > Na

C. K = Na

D. K = 0

metal

Wave Interference Patterns

Diffraction of LightLight is shined through a crystal, and its waveforms are “scattered.” When they come out the other side, they create interference patterns on a detector plate.

• Diffraction can only be explained by treating light as a wave instead of a particle.

Diffraction of Particles?

• Turns out we can get similar interference patterns by bombarding crystals with beams of high energy electrons also…

• This can only be explained by treating matter as a wave.

de Broglie Wavelength• If matter exhibits wave-like properties, we should be able

to determine the wavelength of a particle.• Recall the energy of a photon, and the definition of the

speed of light:

• Substituting,

• Employing Einstein’s relationship, E = mc2,

photonE h c c

hcE

2 2

/E hc hm

c c c

h

mc

de Broglie Wavelength• We can generalize this relationship to any velocity:

• What is the de Broglie wavelength of an electron traveling at the speed of light? (melectron = 9.31 x 10-31 kg)

• What is the de Broglie wavelength of a 80 kg student walking across campus at 3 m/s?

hm

c h

mv