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Quantum physics (quantum theory, quantum mechanics)

Quantum physics (quantum theory, quantum mechanics). Part 1:. Outline. Introduction Problems of classical physics emission and absorption spectra Black-body Radiation experimental observations Wien’s displacement law Stefan – Boltzmann law Rayleigh - Jeans Wien’s radiation law

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Quantum physics (quantum theory, quantum mechanics)

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  1. Quantum physics(quantum theory, quantum mechanics) Part 1:

  2. Outline • Introduction • Problems of classical physics • emission and absorption spectra • Black-body Radiation • experimental observations • Wien’s displacement law • Stefan – Boltzmann law • Rayleigh - Jeans • Wien’s radiation law • Planck’s radiation law • photoelectric effect • observation • studies • Einstein’s explanation • Summary

  3. Question: What do these have in common? • lasers • solar cells • transistors • computer chips • CCDs in digital cameras • Ipods • superconductors • ......... • Answer: • They are all based on the quantum physics discovered in the 20th century.

  4. Why Quantum Physics? • “Classical Physics”: • developed in 15th to 20th century; • provides very successful description of “every day, ordinary objects” • motion of trains, cars, bullets,…. • orbit of moon, planets • how an engine works,.. • subfields: mechanics, thermodynamics, electrodynamics, • Quantum Physics: • developed early 20th century, in response to shortcomings of classical physics in describing certain phenomena (blackbody radiation, photoelectric effect, emission and absorption spectra…) • describes “small” objects (e.g. atoms and their constituents)

  5. “Classical” vs “modern” physics

  6. Quantum Physics • QP is “weird and counterintuitive” • “Those who are not shocked when they first come across quantum theory cannot possibly have understood it” (Niels Bohr) • “Nobody feels perfectly comfortable with it “ (Murray Gell-Mann) • “I can safely say that nobody understands quantum mechanics” (Richard Feynman) • But: • QM is the most successful theory ever developed by humanity • underlies our understanding of atoms, molecules, condensed matter, nuclei, elementary particles • Crucial ingredient in understanding of stars, …

  7. Features of QP • Quantum physics is basically the recognition that there is less difference between waves and particles than was thought before • key insights: • light can behave like a particle • particles (e.g. electrons) are indistinguishable • particles can behave like waves (or wave packets) • waves gain or lose energy only in "quantized amounts“ • detection (measurement) of a particle  wave will change suddenly into a new wave • quantum mechanical interference – amplitudes add • QP is intrinsically probabilistic • what you can measure is what you can know

  8. emission spectra • continuous spectrum • solid, liquid, or dense gas emits continuous spectrum of electromagnetic radiation (“thermal radiation”); • total intensity and frequency dependence of intensity change with temperature (Kirchhoff, Bunsen, Wien, Stefan, Boltzmann, Planck) • line spectrum • rarefied gas which is “excited” by heating, or by passing discharge through it, emits radiation consisting of discrete wavelengths (“line spectrum”) • wavelengths of spectral lines are characteristic of atoms

  9. Emission spectra:

  10. Absorption spectra • first seen by Fraunhofer in light from Sun; • spectra of light from stars are absorption spectra (light emitted by hotter parts of star further inside passes through colder “atmosphere” of star) • dark lines in absorption spectra match bright lines in discrete emission spectra • Helium discovered by studying Sun's spectrum • light from continuous-spectrum source passes through colder rarefied gas before reaching observer;

  11. Fraunhofer spectra

  12. Spectroscopic studies

  13. Thermal radiation • thermal radiation = e.m. radiation emitted by a body by virtue of its temperature • spectrum is continuous, comprising all wavelengths • thermal radiation formed inside body by random thermal motions of its atoms and molecules, repeatedly absorbed and re-emitted on its way to surface  original character of radiation obliterated  spectrum of radiation depends only on temperature, not on identity of object • amount of radiation actually emitted or absorbed depends on nature of surface • good absorbers are also good emitters (Kirchhoff) • http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/absrad.html • http://casswww.ucsd.edu/archive/public/tutorial/Planck.html • http://panda.unm.edu/Courses/Finley/P262/ThermalRad/ThermalRad.html • http://csep10.phys.utk.edu/astr162/lect/light/radiation.html • http://www.youtube.com/watch?v=CDncSyDvpdQ • http://www.enotes.com/topic/Kirchhoff's_law_of_thermal_radiation • http://ip.anndannenberg.com/IPHandouts/Heattransfernotes.pdf

  14. warm bodies emit radiation

  15. Black-body radiation • “Black body” • perfect absorber • ideal body which absorbs all e.m. radiation that strikes it, any wavelength, any intensity • such a body would appear black  “black body” • must also be perfect emitter • able to emit radiation of any wavelength at any intensity -- “black-body radiation” • “Hollow cavity” (“Hohlraum”) kept at constant T • hollow cavity with small hole in wall is good approximation to black body (Kirchhoff 1859) • thermal equilibrium inside, radiation can escape through hole, looks like black-body radiation • Kirchhoff’s challenge to theorists: calculate spectrum of hollow cavity radiation (black body radiation)

  16. Studies of radiation from hollow cavity • hollow cavity behaves like black body • In 2nd half of 19th century, behavior of radiation within a heated cavity studied by many physicists, both theoretically and experimentally • Experimental findings: • spectral density ρ(f,T) (= energy per unit volume per unit frequency) of the heated cavity depends on the frequency f of the emitted light and the temperature T of the cavity and nothing else.

  17. A small hole in the wall of the cavity approximating an ideal blackbody. Electromagnetic radiation (for example, light) entering the hole has little chance of leaving before it is completely adsorbed within the cavity.

  18. Blackbody radiation – Stefan-Boltzmann • “Global description”, i.e without frequency dependence: • Descriptions successful, i.e. in agreement with observations • Joseph Stefan (1879) first measured temperature dependence of the total amount of radiation emitted by blackbody at all wavelengths and found it varied with absolute temperature • Ludwig Boltzmann: theoretical explanation • total power output of black body: Stefan-Boltzmann law: • For an “ideal” radiator (“black body”), total emitted power (per unit emitting area), P/A P/A = σ·T4σ = 5.672 · 10-8W m-2 K-4 (Josef Stefan, Ludwig Boltzmann 1879, 1884) • http://csep10.phys.utk.edu/astr162/lect/light/radiation.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.htmlhttp://scienceworld.wolfram.com/physics/Stefan-BoltzmannLaw.html • Note that the power per unit area radiated by blackbody depends only on the temperature, and not of other characteristic of the object, such as its color or the material, of which it is composed. • An object at room temperature (300 K) will double the rate at which it radiates energy as a result of temperature increase of only 570.

  19. Black body radiation – Wien’s law • Wien’s displacement law (1893) peak vs temperature: max ·T = C, C= 2.898 · 10-3mK inverse relationship between the wavelength maxof the peak of the emission of a black body and its temperature when expressed as a function of wavelength • http://en.wikipedia.org/wiki/Wien's_displacement_law hyperphysics.phy-astr.gsu.edu/hbase/quantum/wien2.htmlhttp://scienceworld.wolfram.com/physics/WiensDisplacementLaw.htmlhttp://webphysics.davidson.edu/faculty/dmb/blackbody/Wiendemo.html • Example: The wavelength at the peak of the spectral distribution for a blackbody at 4300 K is 674 nm (red). At what temperature would the peak be 420 nm (violet)? Solution: From Wien’s law, we have λ1T1 = λ2T2 (674 x 10-9m)(4300 K) = (420 x 10-9m)(T2) T2=6900 K

  20. Wien’s displacement law • is used to • determine the surface temperatures of stars by analyzing their radiation. • also used to map out the variation in temperature over different regions of the surface of an object. Map = “thermograph” • Example: thermograph can be used to detect cancer because cancerous tissue results in increased circulation which produce a slight increase in skin temperature.

  21. Radiation from the Sun The radiation emitted by the surface of the sun has maximum power at wavelength of about 500 nm. Assuming the sun to be a blackbody emitter, • what is it surface temperature? • Calculate λmax for a blackbody at room temperature, T=300 K.

  22. Attempts to get radiation spectrum • Wilhelm Wien (1896) r(f, T) = af3exp(-bf/T), (a and b constants). • OK for high frequency but fails for low frequencies • http://en.wikipedia.org/wiki/Wien_approximation http://theochem.kuchem.kyoto-u.ac.jp/Ando/planck1901.pdf http://bado-shanai.net/map%20of%20physics/mopWienslaws.htm http://physics.info/planck/ • Rayleigh-Jeans Law (1900) r(f,T) = af2T (a = constant) • (constant found to be = 8pk/c3 by James Jeans, in 1906) • OK for low frequencies, but “ultra – violet catastrophe” at high frequencies, i.e. intensity grows  f2   for f  (corresponding to limit of wavelength  0) • http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/mod6.htmlhttp://scienceworld.wolfram.com/physics/Rayleigh-JeansLaw.html

  23. Ultraviolet catastrophe (Rayleigh-Jeans)

  24. Raleigh – Jeans formula • Power radiated from cavity (out of the hole) is proportional to the total energy density U (energy per unit volume) of the radiation in the cavity. • Spectral distribution of power emitted, ρ()d is proportional to the spectrum of the energy density u()d in the cavity,, proportionality factor = c/4 (c=speed of light) • ρ() = radiation power per unit wavelength interval • u() is calculated from classical electrodynamics and thermodynamics

  25. Energy density in hollow cavity • Energy density of radiation in the cavity modelled by electromagnetic oscillators in the cavity; • Need to find the number of modes of oscillation of the electromagnetic field in the cavity with wavelength λ in the interval dλ and multiply it by average energy per mode. • number of modes of oscillation per unit volume, n(λ), is independent of the shape of cavity and is given by:

  26. Raleigh-Jeans equation • Equipartition principle: energy per degree of freedom =1/2 kT • oscillator has 2 degrees of freedom • Mechanical: kinetic and potential energy • Electromagnetic: electric and magnetic energy • Energy per oscillation mode = kT • Therefore

  27. expt data for T=1600 Comparison of the Rayleigh-Jeans Law with experimental data at T=1600 K. The u(λ) axis is linear.

  28. Frequencies in cavity radiation • cavity radiation = system of standing waves produced by interference of e.m. waves reflected between cavity walls • many more “modes” per wavelength band  at high frequencies (short wavelengths) than at low frequencies • for cavity of volume V, n = (8πV/4)  or n = (8πV/c3) f2  f • if energy continuous, get equipartition, <E> = kT  all modes have same energy  spectral density grows beyond bounds as f • If energy related to frequency and not continuous (E = nhf), the “Boltzmann factor” e-E/kT leads to a suppression of high frequencies

  29. Planck’s quantum hypothesis • Max Planck (Oct 1900) found formula that reproduced the experimental results • derivation from classical thermodynamics, but required assumption that oscillator energies can only take specific values E = 0, hf, 2hf, 3hf, … • for a multi-state system of particles in thermal equilibrium, probability for particle to be in state with energy E, P(E) = (1/Z)e-E/kT(“Boltzmann factor” ) <Eosc> is the average energy of a cavity “oscillator”

  30. Planck’s calculation • Classically, the electromagnetic waves in the cavity are produced by accelerated electric charges in the walls vibrating like simple harmonic oscillators • Classical thermodynamics: average energy for simple harmonic oscillator from Maxwell-Boltzmanndistribution function: f(E) = Ae-E/kT where A is a constant (normalization)and f(E) is the fraction of oscillators with energy E.

  31. Average energy of oscillator • Maxwell-Boltzmann distribution: A is normalization factor so that • Average energy • Planck: to derive function which agrees with data, need to assume that energy of oscillators is a discrete variable • M.B. factor • Average energy

  32. Planck’s calculation • Normalization condition • Average energy • Using geometric series formula and

  33. Get average energy • and finally • Multiplying with number of oscillators, get Planck formula

  34. Planck’s radiation formula • Multiplying average oscillator energy by the number of oscillators per unit volume in the interval dλ given by n(λ)=8πcλ-4 (the number of modes of oscillation per unit volume), finally obtain the energy distribution function for the radiation in cavity: • And for blackbody radiation:

  35. Measurements of Lummer and Pringsheim (1900) calculation Black-body radiation spectrum

  36. Consequences of Planck’s hypothesis • oscillator energies E = nhf, n = 0,1,…; • h = 6.626 10-34Js = 4.13 10-15eV·s now called Planck’s constant •  oscillator’s energy can onlychange by discrete amounts, absorb or emit energy in small packets – “quanta”; Equantum = hf • average energy of oscillator <Eosc> = hf/(ex – 1) with x = hf/kT; for low frequencies get classical result <Eosc> = kT, k = 1.38 · 10-23 J·K-1

  37. Problems • estimate Sun’s surface temperature assuming: • Earth and Sun are black bodies • Stefan-Boltzmann law • Earth in energetic equilibrium (i.e. rad. power absorbed = rad. power emitted) , mean temperature T = 290K • Sun’s angular size Sun = 32’ • show that for small frequencies, Planck’s average oscillator energy yields classical equipartition result <Eosc> = kT • show that for standing waves on a string, number of waves in band between  and +is n = (2L/2) 

  38. Homework problems, cont’d • Thermal Radiation from the Human Body. The temperature of the skin is approximately 35°C. What is the wavelength at which the peak occurs in the radiation emitted from the skin? • The Quantized Oscillator. A 2-kg mass is attached to a massless spring of force constant k=25N/m. The spring is stretched 0.4m from its equilibrium position and released. (a) Find the total energy and frequency of oscillation according to classical calculations. (b) Assume that the energy is quantized and find the quantum number, n, for the system. (c) How much energy would be carried away in one-quantum change? • The Energy of a “Yellow” Photon. What is the energy carried by a quantum of light whose frequency equals 6 x 1014 Hz yellow light? What is the wavelength of this light? • Bonus problem: fill in the gaps in the derivation of Planck’s formula

  39. Summary • classical physics explanation of black-body radiation failed • Planck’s ad-hoc assumption of “energy quanta” of energy Equantum = h, modifying Wien’s radiation law, leads to a radiation spectrum which agrees with experiment. • old generally accepted principle of “natura non facitsaltus” violated • Opens path to further developments

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