Plan for Wed, 29 Oct 08

1. 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

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

3. Weather systems Radar: Radio Waves A galaxy imaged in the visible spectrum. Radio telescopes. This is the Very Large Array (VLA) in NM. The same galaxy imaged in the radio spectrum at the VLA.

4. Thermal Imaging: Detecting IR radiation Much 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?

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

7. Electromagnetic Radiation • Electromagnetic radiation or “light” is a form of energy. • Has both electric (E) and magnetic (H) components. • Characterized by: • Wavelength (l) • Amplitude (A)

8. Electromagnetic Radiation (cont.) • Wavelength (l): The distance between two consecutive peaks in the wave. Increasing Wavelength l1 > l2 > l3 Unit: length (m)

9. Electromagnetic Radiation (cont.) • Frequency (n): The number of waves (or cycles) that pass a given point in space per second. Decreasing Frequency n1 < n2 < n3 Units: 1/time (1/sec) or, Hertz (Hz)

10. Electromagnetic Radiation (cont.) • The product of wavelength (l) and frequency (n) is a constant. (n) (l) = c Speed of light c = 3 x 108 m/s c is a constant, independent of l

11. 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.

12. 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…

13. 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?

14. Relative Intensity of Radiation from the Sun This 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. The most intense radiation is in the visible...the sun is bright!! Note that over half the emitted radiation is in the IR...the sun is hot!! 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??

15. Explaining Black-body Radiation Relative Intensity of radiation from the sun. 1900 – Rayleigh-Jeans, Ultraviolet catastrophe R-J law agreed with experiment only at very long wavelengths 1901 – Planck, quantized black-body radiation To fit exp. data, Planck introduced a parameter into a modified R-J law…

16. 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: DE = nhn frequency Energy Change n = 1, 2, 3 (integers) h = Planck’s constant = 6.626 x 10-34 J.s

17. Light as Energy (cont.) h = 6.636 x 10-34 J s c = 2.9979 x 108 m/s 1 Hz = 1 s-1 • In general the relationship between frequency and “photon” energy is Ephoton = hn • Example: What is the energy of a 500 nm photon? n = c/l = (3x108 m/s)/(5.0 x 10-7 m) n = 6 x 1014 1/s E = h n =(6.626 x 10-34 J.s)(6 x 1014 1/s) = 4 x 10-19 J

18. Which type of photon will have the largest energy? A. Ultraviolet C. Microwave B. X-Ray D. Visible

19. Energy Quantization The student can stop only at certain points on a flight of stairs. Her distance from the ground is quantized. The student can stop at any point on the ramp. Her distance from the ground changes continuously. Similarly, atomic energy levels are like steps…the energies available to an atom do not form a continuum, they are quantized.

20. Evidence of Quantization • Black-body Radiation (Planck) • A system can transfer energy in “packets” of size hn. • 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 itselfis 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!!

21. K Ca Sr Li Na H Li Ba Atomic Emission When 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. When the light emitted from excited atoms is passed through a prism, we see discrete bands of color at specific wavelengths.

22. H Li Ba Continuous vs. Discrete Spectra 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.

23. Photon Emission An excited atom relaxes from high E to low E by emitting a photon. We can determine the energy difference (DE) between levels by measuring the wavelength of the emitted photon. Emission of photon DE = hc/l  l = hc/ DE If l = 440 nm, DE = 4.5 x 10-19 J E = hn = hc/l Planck’s constant: h = 6.636 x 10-34 J s

24. “Continuous” spectrum “Quantized” spectrum Quantized vs. Continuous Energy levels are so close together that any DE is possible Energy levels are discretized, so only certain DE are allowed DE DE

25. The following visible emission spectrum was obtained from Rubidium: Which statement is true? A. Relative to hydrogen, the spectrum is less complex. B. Rubidium is characterized by a continuum of energy levels. 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.

26. 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 (“no”). Further, for n ≥ no, number of electrons increases linearly with light intensity.

27. 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!!

28. The Photoelectric Effect (cont.) • As frequency of incident light is increased, kinetic energy of emitted e- increases linearly. • = hn0 Workfunction: energy needed to release e- • Light apparently behaves as a particle.

29. The Photoelectric Effect (cont.) • For Na with F = 4.4 x 10-19 J, what wavelength corresponds to no? 0 hn = F = 4.4 x 10-19 J hc/l = 4.4 x 10-19 J l = 4.52 x 10-7 m = 452 nm

30. 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 C. K = Na B. K > Na D. K = 0

31. Wave Interference Patterns

32. Diffraction of Light Light 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.

33. 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.

34. 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,

35. 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?