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UW-Madison PHYSICS 208 - LECTURE NOTES

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The Description of the microscopic worldPrevious Lecture: Quantization of light, photonsPhotoelectric effectParticle-Wave dualismThis Lecture: Matter wavesUncertainty Principle Wave functionsStart the atomMTE3 on Wed 23, 5:30-7:00 pm, Ch 2103 and SH 180, same sessions in each roomAlternate Exams on Wed2:30-4:00 pm6:00-7:30 pmin the Lab roomSend your requests for Alternate Examsby email specifying the class conflictMTE 3 Contents Ampere’s Law and force between wires with current (32.6, 32.8)Faraday’s Law and Induction (ch 33, no inductance 33.8-10)Maxwell equations, EM waves and Polarization (ch 34, no 34.2)Photoelectric effect (38.1-2-3)Matter waves and De Broglie wavelength (38.4)Atom (37.6, 37.8-9, 38.5-7)Wave function and Uncertainty (39)3Quantization of lightLight is made of ‘quanta’ of energy called photonsquantum’ of energy: E=hf f = frequency of lightPhoton is a particle, but moves at speed of light!This is possible because it has zero mass.Zero mass, but it does have momentum:Photon momentum p=E/c (remember U/c for radiation)Photoelectric effect (1905) No matter how intense is the light: until the light wavelength passes a certain threshold, no electrons are ejected.Kmax = hƒ – ϕHow much is a quantum of green light?One quantum of energy for 500 nm light (green)€ E = hf =hcλ=6.634 ×10−34Js( )× 3 ×108m / s( )500 ×10−9m= 4 ×10−19JWe need a convenient unit for such a small energy!1 electron-volt = 1 eV = |charge on electron|x (1 volt) = 1.602x10-19 JEnergy of an electron accelerated in a potential difference of 1 VIn these units, E(1 green photon) = (4x10-19 J)x(1 eV / 1.602x10-19 J) = 2.5 eVhc =1240 eV nmso you can use 1240 eV nm/500nm = 2.5 eVQuiz on photoelectric effectWhich of the following is not true of photoelectric emission?A. increasing the light intensity causes no change in the kinetic energy of photoelectronsB. the max energy of photoelectrons depends on the frequency of light illuminating the metalC. increasing the intensity of the light will increase the KE of photoelectronsD. Doubling the light intensity doubles the number of photoelectrons emitted6A. is true because the intensity is connected to the number of electrons not to the energy of each onesB. is true because Kmax 󲰮fC. is false because Kmax depends on fD. Intensity = Power/Area = energy/(time x A) = (Nhf/time x Area)Is Light a Wave or a Particle?WaveElectric and Magnetic fields act like wavesSuperposition, Interference and DiffractionParticlePhotonsPhoto-electric effectCompton EffectSometimes Particle Sometimes Wavedepending on the experiment!Wave properties of particles1924 graduate studentNobel Prize in 1929 de Broglie wavelengthDe Broglie postulated that it holds for any object with momentum- an electron, a nucleus, an atom, a baseball,…...Should be able to see interference and diffraction for any material particle!!Wavelength of an electron of 1 eVde Broglie wavelengthResult: λ = 1.23 nm Solve forIf m of particle is large λ small and wave properties not noticeableme = 9.1 x 10-31 = 0.511 MeV/c2De Broglie question10Compare the wavelength of a bowling ball with the wavelength of a golf ball, if each have 10 Joules of kinetic energy.A) λbowling > λgolfB) λbowling = λgolfC) λbowling < λgolfThe largest the mass of the object the less noticeable arethe quantistic effects!Football launched by Brett Favre can go at 30m/s and m = 0.4kg€ λ=hp=6.6 ×10−34Js0.4kg × 30m / s= 5.5 ×10−35mLight properties as a waveLight has wavelength, frequency, speedfλ = speedLight shows interference and diffraction phenomenaIs an Electron a Particle or a Wave?Particle: you can “bounce” things off them.How would know if electron was a wave?Look for interference!This is what we observe like if 1 electron passes through both slits!!We destroy interference illuminatingto determine the slit they go throughElectron Diffraction: the experimentParallel beams of mono-energetic electrons on a double slitslit widths << λelectronelectron detector far from slitsComputersimulationphotographDavisson-Germer experimentDiffraction of electrons from a nickel single crystal.Found pattern by heating just by chance. Nichel formed a crystalline structure.Established that electrons are waves. 54 eV electrons (λ=0.17nm) Bright spot: constructive interferenceDavisson: Nobel Prize 193715What’s your view of atoms?16Hystory of Atoms• Thompson’s classical model - raisin-cake (1897): cloud of + charge with embedded e-• Problem: charges cannot be in equilibrium Planetary modelPositive charge concentrated in the nucleus (∼ 10-15 m)Electrons orbit the nucleus (r~10-10 m)Problem1: emission and absorption at specific frequenciesProblem2: electrons on circular orbits radiate17Rutherford’s experiment (1911)Most of the alpha-particles (ionized He nuclei made of 2p+2n) not deflected much but a few bounced back18Bohr’s Model of Hydrogen Atom (1913)•Postulate 1: Electron moves in circular orbits where it does not radiate (stationary states) •Postulate 2: radiation emitted in transitions between stationary statesOrbit radius: rn = n2 a0•orbital angular momentum quantized L = mvr = n h/2π€ Ei− Ef= hfNotice: Sometimes fis indicated by € νEfEiEmitted photon has energy hfa0 = 0.053 nmEn = -13.6 eV/n219Emitting and absorbing lightZero energyn=1n=2n=3n=4€ E1= −13.612 eV€ E2= −13.622 eV€ E3= −13.632 eVn=1n=2n=3n=4€ E1= −13.612 eV€ E2= −13.622 eV€ E3= −13.632 eVPhoton absorbed hf=E2-E1Photon emittedhf=E2-E1Energy axis20QuestionThis quantum system has equally-spaced energy levels as shown. Which photon could possibly be absorbed by this system?E1=1 eVE2=3 eVE3=5 eVE3=7 eV€ Ephoton=hcλ=1240 eV ⋅ nmλA. 1240 nmB. 413 nmC. 310 nmD. 248 nm€ 2eV ≠ 1240 / 4132eV ≠ 1240 / 3102eV ≠ 1240 /2484eV ≠ 1240 /12404eV ≠ 1240 / 4134eV = 1240 / 31021Why “quantized” orbits?Electron is a wave.Its ‘propagation direction’ is around circumference of orbit.Wavelength = h / pWaves on a circular orbit?Incorporating wave nature of electron gives an intuitive understanding of ‘quantized orbits’22Standing waves on a stringFundamental, wavelength 2L/1=2L, frequency f 1st harmonic, wavelength 2L/2=L, frequency 2f 2nd harmonic, wavelength 2L/3,frequency 3f λ/2λ/2λ/2€


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UW-Madison PHYSICS 208 - LECTURE NOTES

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