P460 - Early Modern1 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 Modern2 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 need::number of states (“nodes”) for any  energy of any state - probability versus energy the number of states = number of standing waves = N( )d  = 8  V/  d  with V = volume Classical (that is wrong) assigned each node the same energy E = kt and same relative probability this gives energy density u( ) = 8  4 *kT wavelength u -> infinity as wavelength ->0 u

P460 - Early Modern3 Blackbody Radiation II Modern, Planck, correct: E = h  = hc/  Energy and frequency are the same From stat. Mech -- higher energy nodes/states should have smaller probability try 1: Prob = exp(-h /kt) -wrong try 2: Prob(E) = 1/(exp(h /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  = 8  /  * hc/ * 1/(exp(hc/ kt) - 1) wavelength u Agrees with experimental observations Higher Temp

P460 - Early Modern4 Photoelectric effect Photon absorbed by electron in a solid (usually metal or semiconductor as “easier” to free the electron). Momentum conserved by lattice if E  >  electron emitted with E e = E    is work function) Example = 4.5 eV. What is largest wavelength (that is smallest energy) which will produce a photoelectron? E  =  or = hc/  = 1240 eV*nm/4.5 eV = 270 nm EeEe EE  E e in solid 0 Conduction band 

P460 - Early Modern5 Compton Effect  + e ->  ’+ e’ electron is quasifree. What are outgoing energies and angles? Conservation of E and p E initial = E final or E + m e = E’ + Ee’ x: p  = p(e’)cos  +  p(  ’)  cos  y: 0 = p(e’)sin  - p (  ’)sin  4 unknowns (2 angles, 2 energies) and 3 eqns. Can relate any 2 quantities 1/  ’ - 1/E  = (1-cos  )/mec 2  e  Feymann diag e   e e

P460 - Early Modern6 Compton Effect ·if.66 MeV gamma rays are Comptoned scattered by 60 degrees, what are the outgoing energies of the photon and electron ? 1/  ’ - 1/E  = (1-cos  )/m e c 2 1/Egamma’ = 1/.66 MeV + (1-0.5)/.511 or Egamma’ = 0.4 MeV and Te = kinetic energy = =.26 MeV  e  Z ’’

P460 - Early Modern7 Brehmstrahlung + X-ray Production · e+Z ->  ’+ e’+Z electron is accelerated in atomic electric field and emits a photon. Conservation of E and p. atom has momentum but E atom =p 2 /2/M atom. And so can ignore E of atom. Einitial = Efinal or Egamma = E(e)- E(e’) Ee’ will depend on angle -> spectrum e Z z e  E+Z->e+Z+  Brem e +  -> e +  Compton e+Z+  -> e+Z photoelectric Z+  -> Z+e+e pair prod energy e brehm   e

P460 - Early Modern8 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 = E e +E pos > 2m e > 1 MeV and atom conserves momentum  + Z -> e + + e - + Z can then annihilate electron-positron pair. Need 2 photons to conserve energy e + + e - ->   Z Particle antiparticle Usually electron positron ALSO: Mu-mu pairs p+cosmic MWB

P460 - Early Modern9 Electron Cross section Brehmstrahlunf becomes more important with higher energy or higher Z from Rev. of Particle Properties

P460 - Early Modern10 Photon Cross Section vs E From Review of Particle properties

P460 - Early Modern11 Rutherford Scattering off Nuclei ·First modern. Gave charge distribution in atoms. Needed << atomic size. For 1 MeV alpha, p=87 MeV, =h/p = cm kinematics: if M target >> M alpha 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? P target = P in +P recoil ~ 2P in (at 180 degrees) K target = (2P in )2/ 2/M target = 4*2*M  *K  /2/Mau = 4*4/197*8MeV=.7MeV A  in recoil target

P460 - Early Modern12 Rutherford Scattering II ·Assume nucleus has infinite mass. Conserve E  = T  +2eZe/(4  r) conserve angular momentum L  = mvr = mv(at infinity)b E+R does arithmetic gives cot(  /2) = 2b/D where D= zZe*e/(4  K  ) 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 d  = 2  bdb (plus some trigonometry) gives d  /d  =D 2 /16/sin 4  Z  b  b=impact parameter

P460 - Early Modern13 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 Modern14 Particles as Waves EM waves (Maxwell Eqs) are composed of individual (massless) particles - photons - with E=hf and p = h/ 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 = h/p What is wavelength of K = 5 MeV proton ? Non-rel p=sqrt(2mK) = sqrt(2*938*5)=97 MeV/c =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 Modern15 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 ·a 0 is the Bohr radius =.053 nm = ~atomic size

P460 - Early Modern16 Bohr Model II E n = K + V = E 0 /n 2 where E 0 = eV for H E 0 = -m/2*(e*e/4  hbar) 2 = -m/2   where  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 Modern17 Bohr Model III E 0 = -m/2*(e*e/4  hbar) 2 = -m/2    easily extend Bohr model. He + atom, Z=2 and E n = 4*(-13.6 eV)/n 2 (have (zZ) 2 for 2 charges) reduced mass. 2 partlces (a and b)  =ma*mb/(ma+mb) if other masses E n =  /(m e )*E 0 (zZ/n) 2 Atom mass E(n=1 ) e p.9995me eV  p 94 MeV 2.6 keV  MeV 1.6 keV bb quarks q=1/3 2.5 GeV.9 keV