Chem 14A

1. Chem 14A Friday, June 29, 2007

2. Lecture Outline Brief review of quantum theory experiments Photoelectric effect Wave-particle Duality Heisenberg Uncertainty Principle Schrödinger Equation Atomic Spectra and Energy

3. Early Quantum Mechanics Max Planck postulated that energy is quantized based on blackbody radiation experiments where continuous energy absorption was not seen Einstein used this information to develop a theory that light is also quantized, or delivered in photons, particle-like energy packets with energy value E=hn

4. Early Quantum Mechanics Millikan performed a series of experiments to prove Einstein’s theory that light is quantized by using the photoelectric effect

5. Photoelectric Effect A Metal surface is illuminated with incident radiation, exciting electrons which are ejected from the surface with kinetic energy

6. Photoelectric Effect

7. Photoelectric Effect Electrons were only ejected under certain conditions: Radiation frequency must be above a certain level (threshold value) Electrons ejected immediately without regard to radiation intensity (number of photons) Kinetic energy of ejected electrons increases linearly with frequency of incident radiation

8. Photoelectric Effect What do the results mean? If light is wave-like, the energy contained in one of those waves should depend only on its amplitude--that is, on the intensity of the light. Lower intensity light would give fewer electrons ejected more slowly. If light is particulate, changing the frequency will result in different energy values for the electrons, but a different intensity will have no effect, since each particle has the same energy value

9. Photoelectric Effect As a result of the photoelectric effect, Einstein’s theory that electromagnetic radiation is quantized was validated, and he was awarded the Nobel prize. Light now has a particulate nature!

10. Work Function Photons of light collide with the metal surface electrons, ejecting them Energy required to eject en electron from the surface= work function (F) Energy of the colliding photon E=hn

11. Work Function The work function describes the difference in energy between the photon itself and the energy needed to eject the electron E< F If the photon’s energy is smaller than the energy needed to eject the electron, the electron will never be ejected, regardless of the light intensity (number of photons)

12. Work Function E>F If the photon’s energy is greater than the energy needed to eject an electron, the electron is ejected with positive kinetic energy [KE=1/2mv2] KE= E- F

13. Work Function KE= E- F Substituting into the equation gives: 1/2mv2=hn- F

14. Quick Exercise I. It takes 7.21x10-19J of energy to remove an electron from an iron atom. What is the maximum wavelength of light that can accomplish this? 1/2mv2=hn- F ?=c/n c= 3.00x108 m/s

15. Answers to Quick Exercise I. 1/2mv2=hn- F We want to find the smallest possible frequency in order to get the largest possible wavelength, since ?=c/n

16. Answers to Quick Exercise I. Re-writing the equation in terms of frequency, and solving for the lowest possible frequency value by setting the velocity to zero:

17. Answers to Quick Exercise I. Solve for the lowest frequency value Then plug into equation for wavelength

18. Answers to Quick Exercise I. Wavelength= 2.76x10-7 m or 276 nm

19. Wave-Particle Duality Electromagnetic Radiation also has evidence of wave-like properties Diffraction: Light will show a diffraction pattern when passed through a diffraction grating

20. Wave-Particle Duality What is diffraction? Diffraction is a property unique to waves, based on the addition and subtraction of several wave forms. Constructive Interference= when waves are added together the amplitude increases Destructive Interference= when waves are subtracted from each other the amplitude decreases

21. Wave-Particle Duality

22. Wave-Particle Duality The collection of high and low intensities formed from wave constructive and destructive interference produces a pattern of lines behind the diffraction grating

23. Wave-Particle Duality Duality applies to electromagnetic radiation as seen through experiments, what about matter? DeBroglie proposed that all particles have wave-like properties too The DeBroglie Equation ?=h/mv mv=p linear momentum

24. Wave-Particle Duality Wave-like properties of matter are strongly mass dependent Using the DeBroglie Equation: Baseball wavelength: ?=h/mv= 6.626x10-34/(.1kg)(35m/s)=1.9X10-34 m

25. Wave-Particle Duality Electron wavelength: ?=h/mv= 6.626x10-34/(9.1x10-31kg)(1x107m/s) =7.3X10-11 m It is hard to observe wave properties of macroscopic objects, but they do exist

26. Quick Exercise II. Calculate the de Broglie wavelength for: A proton with a velocity 5% of the speed of light An electron with a velocity 15% of the speed of light c=3x108 m/s ?=h/mv me= 9.1x10-31kg h= 6.626x10-34 Js mp= 1.67x10-27kg

27. Answers to Quick Exercise II. 1. 2.6x10-5 nm 2. 1.6x10-2 nm

28. Heisenberg Uncertainty Principle Matter has both wave and particulate properties Waves are delocalized- Their exact location cannot be described Particles are localized- Their exact location is easily quantifiable

29. Heisenberg Uncertainty Principle Since matter has both wave and particulate properties, how do we define its location and momentum?

30. Heisenberg Uncertainty Principle We cannot know both the position and momentum of a particle to a high degree of precision at the same time The more precisely we know a particles position, the less precisely we know its momentum and vice versa ?x*?p=h/4p

31. Heisenberg Uncertainty Principle For large particles, such as a baseball the level of uncertainty is small. For small particles, such as an electron, the level of uncertainty is large- we don’t know the exact path an electron travels in around the nucleus (orbitals are 90% probability representations)

32. Quick Exercise III. The hydrogen atom has a radius on the order of 0.05 nm. Assuming we know the position of an electron to an accuracy of 1% of the radius, calculate the uncertainty in the velocity of the electron. me=9.11x10-31 kg h= 6.626x10-34 Js

33. Answer to Quick Exercise III. ?x*?p=h/4p ?x= 0.01x0.05nm= 0.0005nm= 5x10-13m ?p=h/4p?x= 1.05x10-22 kg m/s ?p=m?v ?v= ?p/m=(1.05x10-22 kg m/s)/9.11x10-31kg ?v= 1.15X108 m/s

34. The Schrödinger Equation Austrian Physicist Erwin Schrödinger adopted the use of wave functions to describe the path of a particle, since it has wave-like character A wave function ? is a mathematical function of the position of a particle in three-dimensional space (x,y,z)

35. The Schrödinger Equation A wave function squared describes the probablity density: the probability of finding a particle within a certain space Schrodinger developed an equation for calculating wave functions H?= E? where H=Hamiltonian operator

36. The Schrödinger Equation Solutions to the Schrodinger Equation: There are many different solutions to the Schrödinger equation Each solution to the Schrödinger equation represents an orbital For example, There are separate wave functions for the H 1s, 2s, and 2p orbitals

37. Atomic Spectra and Energy Levels Atomic spectra give further evidence that energy is quantized: Emission spectra are produced when light emitted by atoms is passed through a prism White light passed through a prism gives a continuous spectrum (rainbow) Light emitted by excited atoms (ie hydrogen) gives discrete spectral lines

38. Atomic Spectra and Energy Levels Absorption spectra are created by shining light through an atomic vapor Absorption spectra exhibit discrete quantities of energy being absorbed by an atom and give the same spectra lines as emission spectra, only they are black against a rainbow background

39. Atomic Spectra and Energy Levels

40. Atomic Spectra and Energy Levels Because an atom can emit only certain wavelengths of light as indicated by the discrete spectral lines, energy is quantized Atoms lose energy in specific amounts, implying that electrons are present only in certain energy states During an electronic transition, an electron moves from a higher energy level to a lower energy level and releases energy in terms of a photon

41. Atomic Spectra and Energy Levels Bohr Frequency Condition: ?E=hn transition energy released as a photon Since each line on the spectrum corresponds to a particular electronic transition, the spectrum can be used to develop an energy-level diagram

42. Atomic Spectra and Energy Levels Rydberg constant: all spectral lines ?=R(1/n12 – 1/n22) n1= 1,2… n2= n1+1, n1+2… R=Rydberg constant=3.29x1015 Hz Balmer series: visible spectral lines n1=2 and n2=3,4,… Lyman series: ultraviolet spectral lines n1=1 and n2=2,3,…

43. Atomic Spectra and Energy Levels Using the Rydberg equation to identify spectral lines Calculate the wavelength of radiation emitted by a hydrogen atom during an electronic transition with n2=3 and n1=2 What color is the light?

44. Atomic Spectra and Energy Levels n=R(1/n12 – 1/n22) R=3.29x1015 Hz n=R(1/22 – 1/32)= 4.57x1014 Hz ?=c/n= (3x108m/s)/(4.57x1014 Hz) ?=6.57x10-7m or 657nm red light

45. Quick Exercise IV. Calculate the wavelength for a transition between n2=5 and n1=1 What region of the electromagnetic spectrum is the light found in ? n=R(1/n12 – 1/n22) R=3.29x1015 Hz ?=c/n

46. Answers to Quick Exercise IV. Wavelength= 94.9nm Lyman Series Ultraviolet region n=R(1/12 – 1/52)= 3.158x1015 Hz ?=c/n= (3x108m/s)/(3.158x1015 Hz)