1. 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

2. 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 460) 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

3. P460 - Early Modern3 Blackbody Radiation II Modern, Planck, correct: E = h  = hc/  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 wrong but did get correct answer….. Gives u  = 8  /  * hc/ * 1/(exp(hc/ kt) - 1) wavelength u Agrees with experimental observations

4. 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 

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

6. 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  )/mec*c 1/Egamma’ = 1/.66 MeV + (1-0.5)/.511 or Egamma’ = 0.4 MeV and Te = kinetic energy =.66-.40 =.26 MeV  e  Z ’’

7. 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 Eatom =p*p/2/Matom. And so can ignore E of atom. Einitial = Efinal or Egamma = Ee- Ee’ 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

8. 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 = Ee+Epos > 2me > 1 MeV  + 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

9. P460 - Early Modern9 Rutherford Scattering off Nuclei ·First modern. Gave charge distribution in atoms. Needed << atomic size. For 1 MeV alpha, p=87 MeV, =h/p = 10 -12 cm 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 180 degree Ktarget = (2Pin)*(2Pin)/2/Mtarg = 4*2*M  *K  /2/Mau = 4*4/197*8MeV=.7MeV A  in recoil target

10. P460 - Early Modern10 Rutherford Scattering II ·Assume nucleus has infinite mass. Conserve Ealpha = Talpha +2eZe/(4  r) conserve angular momentum Lalpha = mvr = mv(infinity)b E+R does arithmetic gives cot(  /2) = 2b/D where D= zZe*e/(4  Kalpha) 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 trig) gives d  /d  =D*D/16/sin 4  Z  b  b=impact parameter