Key experiments in deducing the structure of atomsI. Discovery of the electron−cathode+anodefluorescent screenglass tubeCathode ray experimentsto vacuum pumpObservations:● Produced a glow in the tube with a small amount of air present.● Produced an intense spot of light on screen when all air was removed.● Solid objects placed between the cathode and screen cast shadows.● A magnet deflected the light spot.● Above effects were independent of the metal used to make the cathode.−cathode+anodefluorescent screenglass tubeto vacuum pumpelectron beammagnet(field ⊥ to page)+electric field3. Turn off electric field.2. Turn on magnetic fieldand compensate beamdeflection.1. Turn on electric field.magnet(field ⊥ to page)II. Electron charge to mass ratiomagnet(field ⊥ to page)electron beam+electric fieldmagnet(field ⊥ to page)Thomson’s experiment−Result: q/m = -1.76x108C/gNStop viewIII. Charge of the electronmist of oil dropsmist of oil dropsatomizer+−electric fieldradioactive sourceMillikan oil dropexperimentResult: qe= -1.60x10−19Cme= 9.09x10−31kgDiscovery of radioactivity1. Alpha particles – high energy helium nuclei (5 to 15 MeV) emitted from minerals containing radium, thorium, and uranium.charge = +2e, mass = 4 amu2. Beta particles – energetic electrons and positrons (0.1 to 8 MeV)charge = ±e, mass = 5.486x10-4amu3. Gamma rays – high energy electromagnetic radiation (0.1 to 20 MeV)charge = 0, mass = 0232Th 228Ra + 4He137Cs 137Ba*+ e−+ ν137Ba* 137Ba + γN Sfront viewphotographic plateside view (from N-pole)β+β−αγThe structure of the atom1. The “plumb pudding“ modelelectronuniform distributionof positive chargealpha particleThe Rutherford scattering experimentθdetectorscattered alpha particlesthin metal foilcollimatorcollimated radioactivesourcescattering angle θimaginary target discimpact parameter xalpha particle trajectorytarget nucleusdxθ (degrees)0 20 40 60 80 100 120 140 160 180dF/dθ0246810Angular distribution for Rutherford scattering of 5.00 MeValpha particles from a 0.520 μm thick Au foil)2/(sinsin8WtNdθdF42221Aθθπρ⋅⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=EeZZθ (degrees)0 20 40 60 80 100 120 140 160 180dF/dθ10-610-510-410-310-210-1100101constant = 2.49x10-5The Coulomb forceElectrostatic force between two point charges.Magnitude:q2rq12021rε4πqqF =Direction: + + or −− F is positive (repulsive)− + or + − F is negative (attractive)e2= qe2/(4πε0) = 14.4 eV-Ǻ = 2.31x10−28J-m.Exercise:Calculate the distance of closest approach for 180oscattering of 5.00 MeV alpha particles from Au nuclei.REK= 5.00 MeVV = 5.00 MeVExercise:Calculate the potential energy of a system composed of an alpha particle and a gold nucleus separated by a distance of 10 fm.Amplitude-40-2002040Amplitude-40-20020X0.0 0.4 0.8 1.2 1.6 2.0Amplitude-80-40040Interference of two waves having slightly different values of kX/λ0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Amplitude-6-4-20246A standing wave shown for three different values of tClassical behavior of lightAn electromagnetic waveLight sourceShutter with two slitsScreenLight bandDark bandLight bandDark bandLight bandDark bandLight bandDark bandLight bandDiffraction of light passing through two slitsEmission spectrumAbsorption spectrumROY G BIVThe visible atomic spectra of some elements⎟⎟⎠⎞⎜⎜⎝⎛−=2221n1n1CνC = 3.29x1015s−1Significant events in the development of quantum theoryI. Discovery of quantization – black body radiationII. Discovery of particle properties of light –Photoelectric effectCompton scatteringIII. Discovery of wave properties of particles –de Broglie wavesIV. Discovery of atomic energy levels –Bohr modelovenspectrographU(ν)νBlack body radiation experimentFrequency distribution of black body radiationFrequency (1013 s-1)0 1020304050U(ν) (10-17J-s/m3)02468T = 2000 KTheory (Rayleigh-Jeans)ExperimentVisible lightThe photoelectric effectIncident lightEjected electronMetal plateEeνν0BEe= hν−BEphoton= hνEeBExercise:Gamma rays from the radioactive decay of 241Am, having anenergy of 60.0 keV, are used to eject electrons from Ag atoms. If the binding energy of the ejected electrons is 25.5 keV, what is their kinetic energy?Calculate the wavelength of the gamma rays.Incident waveScattered waveScattered electronCompton Scattering)cos1(0θλλ−=−′cmhThe de Broglie hypothesisphλ =Exercise: Calculate the wavelength of an electron with a kineticenergy of 13.6 eV.Diffraction of electrons passing through a gold foilBasic formulas you should knowkinetic energy of a particle EK= ½ mv2momentum of a particle p = mvalso Ek= p2/2m p = (2mEk)½force due to an electric field Fel= qEforce due to a magnetic field Fmag= q(v x B)= qvB when v ⊥ BCoulomb force between two nuclei FCoul= Z1Z2e2/r2Coulomb potential energy ot two nuclei VCoul= Z1Z2e2/rExercise: Calculate the wavelength of a baseball weighing 200 gand traveling at 90 mi/hr.The Bohr model for one-electron atomsrL−qe+ZqepE1E2L1L2The Bohr equations and atomic transitionsZnamZe4πnhrnZEnhmZe2πE20222222022242n==−=−=E0= 13.606 eV a0= 0.529 ÅE1E2ΔE = E2− E1= hν = hc/λΔE = −13.6Z2/n22− [−13.6Z2/n12]= 13.6Z2(1/n12− 1/n22)Pfund seriesBrackett seriesPaschen seriesBalmer seriesLyman seriesSpectral series of the hydrogen atomn = 1n = 2n = 3n = 4n = 5n = ∞−13.6 eV−3.39 eV−1.51 eV−0.85 eV−0.54 eV(−13.6/4) (−13.6/9)(−13.6/16)(−13.6/25)(−13.6/1)Exercise: Calculate the radius of an Ar17+ion when its electron is in the 3rdenergy level.Exercise: Calculate the energy, frequency, and wavelength of a transitionbetween the n = 3 and n = 2 energy levels in hydrogen.Orbital electron wave interferencerequirement fora standing wave:nλ = 2πrExercise: Suppose we wish to define the trajectory of an electron in the first Bohr orbit of a hydrogen atom (r = 0.529 Ǻ). We must establish both its position and velocity at a particular instant in time. If we make a measurement to determine the location of theelectron with an uncertainty of ±0.05 Ǻ (i.e., ±10%), what is the minimum simultaneous uncertainty in the velocity of the electron?The Heisenburg uncertainty principleΔxΔλΔpqΔq ≥ħ/2The CMS atthe LHCCMS = CompactMuon SolenoidLHC = LargeHadron Collider“There was a time when the newspapers said that only twelve men understood the theory of
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