1Lecture 20Reminders:Reading for this week TZD Chapter 5, and then 6 for next week.Homework #7 due next WednesdayNext week schedule – I am gone Tuesday-Friday Professor Stenson will lecture on W and FHelp room only on Monday!Today’s topic: Bohr Model of the AtomNiels Bohr 1885 – 1965Neon_Lights_and_Other_Discharge_LampsWhat are these pairs?“In optics, stimulated emission is the process by which an electron, perturbed by a photonhaving the correct energy, may drop to a lower energy level resulting in the creation of another photon. The perturbing photon is seemingly unchanged in the process (cf. absorption), and the second photon is created with the same phase, frequency, polarization, and direction of travel as the original.”http://en.wikipedia.org/wiki/Stimulated_emission−2 eV−3 eV−5 eV0 eV 0 eV−7 eV−5 eV−8 eVWhat energy levels for electrons are consistent with this spectrum?Electron Energy levels: 5ev3ev2evPhoton energy0 100 200 300 400 500 600 700 800 nm−5 eV−7 eV−10 eV0 eV5 eV7 eV10 eV0 eV−10 eVABCD5 eV3 eV2 eV0 eVEClicker questionA. n=1B. n=1, n=2, or n=3C. n=3D. n=2 or n=3E. Any of the statesClicker questionAn atom with the energy levels shown is initially in the ground state. A free electron with an energy of 16.0 eV hits the atom. What possible states could the atom be in some time after the interaction?−2 eV−20 eV−5 eV−10 eV−1 eVn=4n=1n=3n=2n=5Electron jumps Electron jumps back to low energyback to low energyAtomic electron excited to Atomic electron excited to higher energy and free electron higher energy and free electron loses same amount of energy.loses same amount of energy.Free electron hits atomeeLess KEeIf the atom goes to n=1, 2, or 3, the free electron will lose 0 eV, 10 eV, or 15 eV of kinetic energy.A. n=1B. n=1, n=2, or n=3C. n=3D. n=2 or n=3E. Any of the statesClicker questionAn atom with the energy levels shown is initially in the ground state. A photon with an energy of 16.0 eV hits the atom. What possible states could the atom be in after the interaction?−2 eV−20 eV−5 eV−10 eV−1 eVn=4n=1n=3n=2n=5For the free electron, whatever energy is absorbed by the atom is deducted from the free electron’s kinetic energy.For photons, the photon is absorbed and so it must transfer allof its energy to the atom. Otherwise energy would not be conserved. Therefore, atoms can only absorb photons with an energy that will exactly move the atom to another energy level.A. 1 eVB. 1 eV, 2 eV, 5 eV, 10 eV C. 10 eVD. 10 eV, 15 eV, 18 eVE. 10 eV, 15 eV, 18 eV, 25 eVClicker questionAn atom with the energy levels shown is initially in the ground state. Which of the following is the most complete list of photon energies that can be absorbed by the atom?−2 eV−20 eV−5 eV−10 eV−1 eVn=4n=1n=3n=2n=5Any photon with an energy ≥ 20 eV will ionize the atom. The electron will escape. Since a free electron can have any energy, any photon with energy ≥ 20 eV can be absorbed by the atom.Photons with energy of 10 eV, 15 eV, 18 eV, 19 eV will cause the electron to jump to the n=2, n=3, n=4, n=5 energy level.20 eV will go into ejecting the electron and the rest will go into the free electron’s kinetic energy. Similar to the photoelectric effect.21. 1/λ= R (1/n′2-1/n2) – from Balmer2. Gravity –Gm1m2/r2force between planets and sun gives orbits. Coulomb −ke2/r2force between electron and proton could be expected to give orbits as well. Keep Rutherford Planetary picture.3. Classical EM says electron going in circle should radiate energy, and spiral in (accelerating charge radiates).+-protonBohr’s additional postulate:Electrons orbit at only particular radii which have particular energies.Bohr Model of the AtomBut WHY?!+-rvA. mvB. mv2/rC. v2/r2D. mvrE. ½mv2If the electron orbits the proton at a constant speed, the magnitude of the net force on the electron is…Clicker question This force comes from the Coulomb force:221rqkqFC=Setting the net force (from Coulomb) equal to the mass times acceleration (mv2/r) for circular motion gives us:222rkermv=which we can also write as:rkemv22=What does this say about total energy?v+-rFdistance from proton0potentialenergyWe now put together three pieces:1. mv2= ke2/r (just derived)2. electrostatic potential energy is U = -ke2/r3. non-relativistic kinetic energy is K = ½mv2This means K = -½U and the total energy is rkeUUUKUE2212121−==−=+=The total energy, radius, and velocity are all related. Knowing just one of the three determines the other two!++++-NucleusElectron-HigherEnergyEnergylevelsA force is applied to the electron bringing it to a larger radius orbit.What can we say about the energy?A. Total, potential, and kinetic energy increaseB. Total, potential, and kinetic energy decreaseC. Total and potential energy decrease, kinetic energy increasesD. Total and potential energy increase, kinetic energy decreasesE. Some other combinationrkemv22=rkeU2−=221mvK =KUE+=Clicker questionrkeE221−=Increasing r increases potential and total (less negative), but lowers kinetic (smaller v).Electron energy levelsground level1stexcited level2ndexcited level3rdexcited leveldistance from proton0potentialenergyIn the Bohr model, each energy level corresponds to a certain radius and velocity.rkeE221−=rkemv22=Bohr Model Energy Levelsv+-rFdistance from proton0potentialenergy-Only certain energy levels exist.Æ just an assumption so farElectron can hop down energy levels, releasing a photon on each hop.But what determines these “special” energies?Bohr Model Energy Levels32212 nakeEBn−=is the Bohr radiusBohr supposed that electrons could only be in certain energy levels but then he needed to justify this in some way.Bohr postulated that angular momentum was quantizedFor electron at radius r the angular momentum isvrmLe=whereπ2/h=hQuantizing angular momentum leads to a quantization of radius:Bnanr2=nm 053.022==kemaeBhWhy are only certain energy levels allowed?Quantizing radius leads to a quantization of energy:Remember angular momentum isprLrrr×=Quantizing, Bohr found:hnvrmLe==()()()211219229221103.52106.1/1099.812 nmCCNmnakeEBn−−×××−=−=()218221102.212 nJnakeEBn−×−=−=()22216.1312 neVnakeEBn−=−=Should be a familiar number!Hydrogen energy levelsUsing the formula for energyBohr could calculate the various transitions and they agreed with the generalized Balmer formula.()()2222'1'1eV 6.131'1nnnnEEEERnn−−=−−=−=γBohr’s Calculation of Hydrogen