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HW8: M Chap 15: Question B, Exercises 2, 6 M Chap 16: Question B, Exercises 1 M Chap 17: Questions C, D Wavelength of 1 eV electron From Last Time… • For an electron, • Essay topic and paragraph due Friday, Mar. 24 "= • Light waves are particles and matter particles are waves! ! • Matter particles are waves with wavelength λ=h/p. Phy107 Lect 23 1240 eV # nm 2 $ 0.511 MeV 1 kinetic energy • 1 eV electron, • 10 eV electron • 100 eV electron Wed. Mar. 22, 2006 λ=1.23 nm λ=0.39 nm λ=0.12 nm Phy107 Lect 23 2 Can this be correct? Question A 10 eV electron has a wavelength of ~ 0.4 nm. What is the wavelength of a 40 eV electron? A. 0.2 nm • If electrons are waves, they should demonstrate wave-like effects – e.g. Interference, diffraction • A 25 eV electron has wavelength 0.25 nm, similar to atomic spacings in crystals B. 0.4 nm C. 0.8 nm Wavelength = Wed. Mar. 22, 2006 1 1.23 eV 1/ 2 # nm = E kinetic E kinetic rest energy • Electromagnetic radiation (e.g. light) made up of photon PARTICLES with energy E=hf and momentum p=E/c=h/ λ Wed. Mar. 22, 2006 constant constant rest energy Kinetic energy Phy107 Lect 23 3 Wed. Mar. 22, 2006 Phy107 Lect 23 4 ! Wave reflection from crystal Reflection from next plane Constructive & Destructive Interference Reflection from top plane • Interference arises when waves change their ‘phase relationship’. • Can vary phase relationship of two waves by changing physical location of speaker. ‘1/2 λ phase diff’ ‘in-phase’ side view • Interference of waves reflecting from different atomic layers in the crystal. • Difference in path length ~ spacing between atoms Wed. Mar. 22, 2006 Phy107 Lect 23 5 Constructive Wed. Mar. 22, 2006 Destructive Phy107 Lect 23 6 1 X-ray diffraction Davisson-Germer experiment Molecular structure • Diffraction spot arrangement indicates atomic arrangement • Used to determine atomic arrangements of complex molecules. • Diffraction of electrons from a nickel single crystal. Bright spot: constructive interference • Established that electrons are waves – e.g. DNA X-ray diffraction pattern Davisson: Nobel Prize 1937 54 eV electrons (λ=0.17nm) Wed. Mar. 22, 2006 Phy107 Lect 23 7 Wed. Mar. 22, 2006 Phy107 Lect 23 8 Analogy with sound Suppose an electron is a wave… • Sound wave also has the same characteristics • But we can often locate sound waves • Here is a wave: "= h p – E.g. echoes bounce from walls. Can make a sound pulse • Example: x – – – – λ ! …where is the electron? Hand clap: duration ~ 0.01 seconds Speed of sound = 340 m/s Spatial extent of sound pulse = 3.4 meters. 3.4 meter long hand clap travels past you at 340 m/s – Wave extends infinitely far in +x and -x direction Wed. Mar. 22, 2006 Phy107 Lect 23 9 Beat frequency: spatial localization Wed. Mar. 22, 2006 Phy107 Lect 23 10 Making a particle out of waves • What does a sound ‘particle’ look like? 440 Hz + 439 Hz – One example is a ‘beat frequency’ between two notes – Two sound waves of almost same wavelength added. 440 Hz + 439 Hz + 438 Hz Constructive interference Large amplitude Wed. Mar. 22, 2006 Destructive interference Small amplitude Phy107 Lect 23 Constructive interference Large amplitude 11 Wed. Mar. 22, 2006 Phy107 Lect 23 440 439 438 437 436 Hz Hz Hz Hz Hz + + + + 12 2 Spatial extent of localized sound wave Same occurs for a matter wave • Construct a localized particle by adding together waves with slightly different wavelengths. 8 4 • Since de Broglie says λ = h /p, each of these components has slightly different momentum. 0 – We say that there is some ‘uncertainty’ in the momentum -4 -8 Δx -15 -10 -5 0 5 10 • And still don’t know exact location of the particle! 15 J – Wave still is spread over Δx (‘uncertainty’ in position) – Can reduce Δx, but at the cost of increasing the spread in wavelength (giving a spread in momentum). • Δx = spatial spread of ‘wave packet’ • Spatial extent decreases as the spread in included wavelengths increases. Wed. Mar. 22, 2006 Phy107 Lect 23 Interpreting Wed. Mar. 22, 2006 13 Phy107 Lect 23 14 8 Heisenberg Uncertainty Principle 4 0 • Using -4 – Δx = position uncertainty – Δp = momentum uncertainty -8 -15 -10 -5 0 5 10 15 J • For sound, we would just say that the sound pulse is centered at some position, but has a spread. • Heisenberg showed that the product • Can’t do that for a quantum-mechanical particle. ( Δx ) • ( Δp ) is always greater than ( h / 4π ) • Many measurements indicate that the electron is indeed a point particle. Often write this as • Interpretation is that the magnitude of electron ‘wavepulse’ at some point in space determines the probability of finding the electron at that point. Wed. Mar. 22, 2006 Phy107 Lect 23 Planck’s constant 15 where h = ! Wed. Mar. 22, 2006 ("x )("p) ~ h /2 h is pronounced ‘h-bar’ 2" Phy107 Lect 23 16 ! Thinking about uncertainty Uncertainty principle question ("x )("p) ~ h /2 For a classical particle, p=mv, so an uncertainty in momentum corresponds to an uncertainty in velocity. ! ("x )("v ) ~ h /2m This says that the uncertainty is small for massive objects, but becomes important for very light objects, such as electrons. ! Suppose an electron is inside a box 1 nm in width. There is some uncertainty in the momentum of the electron. We then squeeze the box to make it 0.5 nm. What happens to the momentum? A. Momentum becomes more uncertain B. Momentum becomes less uncertain C. Momentum uncertainty unchanged Large, massive objects don’t show effects of quantum mechanics. Wed. Mar. 22, 2006 Phy107 Lect 23 17 Wed. Mar. 22, 2006 Phy107 Lect 23 18 3 Planetary model of atom Using quantum mechanics • Positive charge is concentrated in the center of the atom ( nucleus ) • Quantum mechanics makes astonishingly accurate predictions of the physical world • Atom has zero net charge: • Can apply to atoms, molecules, solids. – Positive charge in nucleus cancels negative electron charges. • An early success was in understanding – Structure of atoms – Interaction of electromagnetic radiation with atoms Wed. Mar. 22, 2006 Phy107 Lect 23 electrons 19 nucleus • Electrons orbit the nucleus like planets orbit the sun • (Attractive) Coulomb force plays role of gravity Wed. Mar. 22, 2006 Phy107 Lect 23 20 Difference between atoms • No net charge to atom – number of orbiting negative electrons same as number of positive protons in nucleus – Different elements have different number of orbiting electrons • • • • • Hydrogen: 1 electron Helium: 2 electrons Copper: 29 electrons Uranium: 92 electrons! Organized into periodic table of elements Wed. Mar. 22, 2006 Phy107 Lect 23 Elements in same column have similar chemical properties 21 Wed. Mar. 22, 2006 Phy107 Lect 23 22 Atoms and photons Planetary model and radiation • Circular motion of orbiting electrons causes them to emit electromagnetic radiation with frequency equal to orbital frequency. • Experimentally, atoms do emit electromagnetic radiation, but not just any radiation! • Same mechanism by which radio waves are emitted by electrons in a radio transmitting antenna. • In fact, each atom has its own ‘fingerprint’ of different light frequencies that it emits. • In an atom, the emitted electromagnetic wave carries away energy from the electron. 400 nm 600 nm 500 nm 700 nm Hydrogen – Electron predicted to continually lose energy. – The electron would eventually spiral into the nucleus – However most atoms are stable! Mercury Wavelength (nm) Wed. Mar. 22, 2006 Phy107 Lect 23 23 Wed. Mar. 22, 2006 Phy107 Lect 23 24 4 Hydrogen emission spectrum • Hydrogen is simplest atom n=4 – One electron orbiting around one proton. Hydrogen emission n=3 • The Balmer Series of emission lines empirically given by n = 4, λ = 486.1 nm • This says hydrogen emits only photons of a particular wavelength, frequency • Photon energy = hf, so this means a particular energy. • Conservation of energy: n = 3, λ = 656.3 nm – Energy carried away by photon is lost by the orbiting electron. Hydrogen Phy107 Lect 23 25 Wed. Mar. 22, 2006 The Bohr hydrogen atom • Radiation emitted only when electron jumps from one stable orbit to another. 26 Energy levels • Retained ‘planetary’ picture: one electron orbits around one proton • Only certain orbits are stable Phy107 Lect 23 • Instead of drawing orbits, we can just indicate the energy an electron would have if it were in that orbit. Zero energy Einitial n=4 n=3 Photon E3 = " 13.6 eV 32 E2 = " 13.6 eV 22 E1 = " 13.6 eV 12 Efinal n=2 • Here, the emitted photon has an energy of Einitial-Efinal ! ! Stable orbit #2 n=1 Stable orbit #1 Wed. Mar. 22, 2006 Energy axis Wed. Mar. 22, 2006 Phy107 Lect 23 27 Wed. Mar. 22, 2006 Phy107 Lect 23 28 ! Emitting and absorbing light Photon emission question Zero energy n=4 n=3 n=2 Photon emitted hf=E2-E1 n=1 ! E3 = " 13.6 eV 32 E2 = " 13.6 eV 22 ! n=2 Photon absorbed hf=E2-E1 E1 = " 13.6 eV 12 n=1 ! E3 = " 13.6 eV 32 E2 = " 13.6 eV 22 ! E1 = " 13.6 eV 12 A. 10.2 eV B. 3.4 eV C. 1.7 eV Absorbing a photon of correct energy makes!electron jump to higher quantum state. Photon is emitted when ! electron drops from one quantum state to another Wed. Mar. 22, 2006 n=4 n=3 An electron can jump between the allowed quantum states (energy levels) in a hydrogen atom. The lowest three energy levels of an electron in a hydrogen atom are -13.6 eV, -3.4 eV, -1.5 eV. These are part of the sequence En = -13.6/n2 eV. Which of the following photons could be emitted by the hydrogen atom? Phy107 Lect 23 29 Wed. Mar. 22, 2006 The energy carried away by the photon must be given up by the electron. The electron can give up energy by dropping to a lower energy state. So possible photon energies correspond to differences between electron orbital energies. The 10.2 eV photon is emitted when the electron jumps from the -3.4 eV state to the -13.6 eV state, losing 10.2 eV of energy. Phy107 Lect 23 30 5 Example: the Balmer series Energy conservation for Bohr atom • Each orbit has a specific energy En=-13.6/n2 • Photon emitted when electron jumps from high energy to low energy orbit. Ei – Ef = h f • Photon absorption induces electron jump from low to high energy orbit. Ef – Ei = h f • Agrees with experiment! Wed. Mar. 22, 2006 • All transitions terminate at the n=2 level • Each energy level has energy En=-13.6 / n2 eV • E.g. n=3 to n=2 transition – Emitted photon has energy ## 13.6 & # 13.6 && E photon = %%" 2 ( " %" 2 (( = 1.89 eV $$ 3 ' $ 2 '' – Emitted wavelength ! Phy107 Lect 23 E photon = hf = Wed. Mar. 22, 2006 31 hc hc 1240 eV # nm , "= = = 656 nm " E photon 1.89 eV Phy107 Lect 23 32 ! Spectral Question Compare the wavelength of a photon produced from a transition from n=3 to n=1 with that of a photon produced from a transition n=2 to n=1. A. λ31 < λ 21 n=3 n=2 B. λ31 = λ 21 C. λ31 > λ 21 E31 > E21 Wed. Mar. 22, 2006 so λ31 < λ 21 n=1 Phy107 Lect 23 33 6