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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 functioms to describe SAME P460 - Early Modern 1 Blackbody Radiation • Late 19th Century: try to derive Wien and StefanBoltzman 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 u wavelength ->0 wavelength P460 - Early Modern 2 Blackbody Radiation II • Modern, Planck, correct: E = hn = hc/ l Energy and frequency are the same • 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 3 • 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 Ee in Ee f Eg f Conduction band solid P460 - Early Modern 4 Compton Effect g + e -> g’+ e’ electron is quasifree. What are outgoing energies and angles? • Conservation of E and p 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)/mec2 g g Feymann diag q f e e P460 - Early Modern g e e 5 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’ g Z q f e P460 - Early Modern 6 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+Z->e+Z+ g Brem e +g -> e + g Compton e+Z+g -> e+Z photoelectric Z+g -> Z+e+e pair prod energy e e z e e g g Z P460 - Early Modern brehm g 7 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 energy ALSO: Mu-mu pairs e+ + e- -> g + g p+cosmic MWB Particle antiparticle Z Usually electron g positron P460 - Early Modern 8 Electron Cross section • Brehmstrahlunf becomes more important with higher energy or higher Z • from Rev. of Particle Properties P460 - Early Modern 9 Photon Cross Section vs E From Review of Particle properties P460 - Early Modern 10 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 a in A P460 - Early Modern target 11 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 b Z P460 - Early Modern q 12 Rutherford Scattering III · 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). Fall with increasing velocity until a minimum then a relativistic rise P460 - Early Modern 13 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 14 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 2 2 n m e 2 K mv / 2 mr 2 8pe0 r 4pe0 2 2 2 solve for radius rn n a n 0 2 me · radius is quantized · a0 is the Bohr radius = .053 nm = ~atomic size P460 - Early Modern 15 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) e2 1 1 a 4pe0 c 137 0.007297352568 0.000000000024 • Bohr model quantizes energy and radius and 1D angular momentum. Reality has energy, and 2D angular momentum (one component and absolute magnitude) • for transitions 1 1 E E0 ( 2 - 2 ) ni n j P460 - Early Modern levels i, j 16 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 ep m p mass .9995me 94 MeV pm 60 MeV bb quarks q=1/3 2.5 GeV P460 - Early Modern E(n=1) -13.6 eV 2.6 keV 1.6 keV .9 keV 17