Pre-quantum mechanics Modern Physics - PowerPoint PPT Presentation

pre quantum mechanics modern physics n.
Skip this Video
Loading SlideShow in 5 Seconds..
Pre-quantum mechanics Modern Physics PowerPoint Presentation
Download Presentation
Pre-quantum mechanics Modern Physics

play fullscreen
1 / 23
Download Presentation
Pre-quantum mechanics Modern Physics
Download Presentation

Pre-quantum mechanics Modern Physics

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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 Feymann diag g g g q f e e e e P460 - Early Modern

  6. 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

  7. 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 e z e g g Z brehm g P460 - Early Modern

  8. 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 e+ + e- -> g + g ALSO: Mu-mu pairs p+cosmic MWB Z Particle antiparticle Usually electron positron g P460 - Early Modern

  9. Electron Cross section • Brehmstrahlunf becomes more important with higher energy or higher Z • from Rev. of Particle Properties P460 - Early Modern

  10. Photon Cross Section vs E From Review of Particle properties P460 - Early Modern

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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 has energy, and 2D angular momentum (one component and absolute magnitude) • for transitions P460 - Early Modern

  17. 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

  18. Developing Wave Equations • Need wave equation and wave function for particles. Schrodinger, Klein-Gordon, Dirac • not derived. Instead forms were guessed at, then solved, and found where applicable • So Dirac equation applicable for spin 1/2 relativistic particles • Start from 1924 DeBroglie hypothesis: “particles” (those with mass as photon also a particle…)have wavelength l = h/p P460 - Early Modern

  19. Wave Functions • Particle wave functions are similar to amplitudes for EM waves…gives interference (which was used to discover wave properties of electrons) • probability to observe =|wave amplitude|2=|y(x,t)|2 • particles are now described by wave packets • if y = A+B then |y|2 = |A|2 + |B|2 + AB* + A*B giving interference. Also leads to indistinguishibility of identical particles t1 t2 merge vel=<x(t2)>-<x(t1)> (t2-t1) Can’t tell apart P460 - Early Modern

  20. Wave Functions • Describe particles with wave functions • y(x) = S ansin(knx) Fourier series (for example) • Fourier transforms go from x-space to k-space where k=wave number= 2p/l. Or p=hbar*k and Fourier transforms go from x-space to p-space • position space and momentum space are conjugate • the spatial function implies “something” about the function in momentum space P460 - Early Modern

  21. Wave Functions (time) • If a wave is moving in the x-direction (or -x) with wave number k have • kx-wt = constant gives motion of wave packet • the sin/cos often used for a bound state while the exponential for a right or left traveling wave P460 - Early Modern

  22. Wave Functions (time) • Can redo Transform from wave number space (momentum space) to position space • normalization factors 2p float around in Fourier transforms • the A(k) are the amplitudes and their squares give the relative probability to have wavenumber k • could be A(k,t) though mostly not in our book • as different k have different velocities, such a wave packet will disperse in time. See sect. 2-2. Not really 460 concern….. P460 - Early Modern

  23. Heisenberg Uncertainty Relationships • Momentum and position are conjugate. The uncertainty on one (a “measurement”) is related to the uncertainty on the other. Can’t determine both at once with 0 errors • p = hbar k • electrons confined to nucleus. What is maximum kinetic energy? Dx = 10 fm • Dpx = hbarc/(2c Dx) = 197 MeV*fm/(2c*10 fm) = 10 MeV/c • while <px> = 0 • Ee=sqrt(p*p+m*m) =sqrt(10*10+.5*.5) = 10 MeV electron can’t be confined (levels~1 MeV) proton Kp = .05 MeV….can be confined P460 - Early Modern