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Pre-quantum mechanics Modern Physics

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## Pre-quantum mechanics Modern Physics

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**Pre-quantum mechanics Modern Physics**• Historical “problems” were resolved by modern treatments which lead to the development of quantum mechanics • need special relativity • EM radiation is transmitted by massless photons which have energy and momentum. Mathematically use wave functions (wavelength, frequency, amplitude, phases) to describe • “particles” with non-zero mass have E and P and use wave functions to describe SAME P460 - Early Modern**Blackbody Radiation**• Late 19th Century: try to derive Wien and Stefan-Boltzman Laws and shape of observed light spectra • used Statistical Mechanics (we’ll do later in 461) to determine relative probability for any wavelength l • need::number of states (“nodes”) for any l - energy of any state - probability versus energy • the number of states = number of standing waves = N(l)dl = 8pV/l4 dl with V = volume • Classical (that is wrong) assigned each node the same energy E = kt and same relative probability this gives energy density u(l) = 8p/l4*kT u infinity as wavelength 0 u wavelength P460 - Early Modern**Blackbody Radiation II**• Modern, Planck, correct: E = hn = hc/ l Energy and frequency are the same. Didn’t quite realize photons were a particle • From stat. Mech -- higher energy nodes/states should have smaller probability try 1: Prob = exp(-hn/kt) - wrong try 2: Prob(E) = 1/(exp(hn/kt) - 1) did work • will do this later. Planck’s reasoning was obscure but did get correct answer…..Bose had more complete understanding of statistics • Gives u(l) = 8p/ l4 * hc/ l * 1/(exp(hc/lkt) - 1) Agrees with experimental observations u Higher Temp wavelength P460 - Early Modern**Photoelectric effect**• Photon absorbed by electron in a solid (usually metal or semiconductor as “easier” to free the electron) . Momentum conserved by lattice • if Eg > f electron emitted with Ee = Eg - f (f is work function) • Example = 4.5 eV. What is largest wavelength (that is smallest energy) which will produce a photoelectron? Eg = f or l = hc/f = 1240 eV*nm/4.5 eV = 270 nm 0 f Ee in solid Conduction band Ee f Eg P460 - Early Modern**Compton Effect**• g + e g’+ e’ electron is quasifree. What are outgoing energies and angles? • Conservation of E and p photon is a particle • Einitial = Efinal or E + me = E’ + Ee’ • x: py = p(e’)cosf + p( g’)cosq • y: 0 = p(e’)sinf - p (g’)sinq • 4 unknowns (2 angles, 2 energies) and 3 eqns. Can relate any 2 quantities • 1/Eg’ - 1/Eg = (1-cosq)/me c2 Feymann diag g g g q f e e e e P460 - Early Modern**Compton Effect**• if .66 MeV gamma rays are Comptoned scattered by 60 degrees, what are the outgoing energies of the photon and electron? • 1/Eg’ - 1/Eg = (1-cosq)/mec2 1/Egamma’ = 1/.66 MeV + (1-0.5)/.511 or Egamma’ = 0.4 MeV and Te = kinetic energy = .66-.40 = .26 MeV g’ Z g q f e P460 - Early Modern**Brehmstrahlung + X-ray Production**• e+Z g’+ e’+Z electron is accelerated in atomic electric field and emits a photon. • Conservation of E and p. atom has momentum but Eatom =p2/2/Matom. And so can ignore E of atom. • Einitial = Efinal or Egamma = E(e)- E(e’) • Ee’ will depend on angle spectrum E+Ze+Z+ g Brem e +g e + g Compton e+Z+g e+Z photoelectric Z+g Z+e+e pair prod energy e e e z e g g Z brehm g P460 - Early Modern**Pair Production**• A photon can convert its energy to a particle antiparticle pair. To conserve E and p another particle (atom, electron) has to be involved. • Pair is “usually” electron+positron and Ephoton = Ee+Epos > 2me > 1 MeV and atom conserves momentum g + Z -> e+ + e- + Z can then annihilate electron-positron pair. Need 2 photons to conserve momentum e+ + e- g + g ALSO: Mu-mu pairs p+cosmic MWB Z Particle antiparticle Usually electron positron g P460 - Early Modern**Electron Cross section**• Brehmstrahlung becomes more important with higher energy or higher Z • from Rev. of Particle Properties P460 - Early Modern**Photon Cross Section vs E**P460 - Early Modern**Rutherford Scattering off Nuclei**• First modern. Gave charge distribution in atoms. Needed l << atomic size. For 1 MeV alpha, p=87 MeV, l=h/p = 10-12cm • kinematics: if Mtarget >> Malpha then little energy transfer but large possible angle change. Ex: what is the maximum kinetic energy of Au A=Z+N=197 after collision with T=8Mev alpha? • Ptarget = Pin+Precoil ~ 2Pin (at 180 degrees) • Ktarget = (2Pin)2/2/Mtarget = 4*2*Ma*Ka/2/Mau = 4*4/197*8MeV=.7MeV recoil target A a in P460 - Early Modern**Rutherford Scattering II**• Assume nucleus has infinite mass. Conserve Ea = Ta +2eZe/(4per) • conserve angular momentum La = mvr = mv(at infinity)b • E+R does arithmetic gives cot(q/2) = 2b/D where D= zZe*e/(4pe Ka) is the classical distance of closest approach for b=0 • don’t “pick” b but have all ranges 0<b<atom size all alphas need to go somewhere and the cross section is related to the area ds = 2pbdb (plus some trigonometry) gives ds/dW =D2/16/sin4q/2 b=impact parameter a q b Z P460 - Early Modern**Rutherford Scattering III**• already went over kinematics • Rutherford scattering can either be off a heavier object (nuclei) change in angle but little energy loss “multiple scattering” • or off light target (electrons) where can transfer energy but little angular change (energy loss due to ionization, also produces “delta rays” which are just more energetic electrons). P460 - Early Modern**Particles as Waves**• EM waves (Maxwell Eqs) are composed of individual (massless) particles - photons - with E=hf and p = h/l and E = pc • observed that electrons scattered off of crystals had a diffraction pattern. Readily understood if “matter” particles (with mass) have the same relation between wavelength and momentum as photons • Bragg condition gives constructive interference • 1924 DeBroglie hypothesis: “particles” (those with mass as photon also a particle…) have wavelength l = h/p • What is wavelength of K = 5 MeV proton ? Non-rel p=sqrt(2mK) = sqrt(2*938*5)=97 MeV/c l=hc/pc = 1240 ev*nm/97 MeV = 13 Fermi • p=50 GeV/c (electron or proton) gives .025 fm (size of proton: 1 F) P460 - Early Modern**Bohr Model I**• From discrete atomic spectrum, realized something was quantized. And the bound electron was not continuously radiating (as classical physics) • Bohr model is wrong but gives about right energy levels and approximate atomic radii. easier than trying to solve Schrodinger Equation…. • Quantized angular momentum (sort or right, sort of wrong) L= mvr = n*hbar n=1,2,3... (no n=0) • kinetic and potential Energy related by K = |V|/2 (virial theorem) gives • radius is quantized • a0 is the Bohr radius = .053 nm = ~atomic size P460 - Early Modern**Bohr Model II**• En = K + V = E0/n2 where E0 = -13.6 eV for H • E0 = -m/2*(e*e/4pehbar)2 = -m/2 a2 where a is the fine structure constant (measure of the strength of the EM force relative to hbar*c=197 ev nm) • Bohr model quantizes energy and radius and 1D angular momentum. Reality quantizes energy, and 2D angular momentum (one component and absolute magnitude) • for transitions P460 - Early Modern**Bohr Model III**• E0 = -m/2*(e*e/4pehbar)2 = -m/2 a2 (H) • easily extend Bohr model. He+ atom, Z=2 and En = 4*(-13.6 eV)/n2 (have (zZ)2 for 2 charges) • reduced mass. 2 partlces (a and b) m =ma*mb/(ma+mb) if other masses En = m/(me)*E0(zZ/n)2 Atom mass E(n=1) e p .9995me -13.6 eV m p 94 MeV 2.6 keV p m 60 MeV 1.6 keV bb quarks q=1/3 2.5 GeV .9 keV P460 - Early Modern