1Lecture 21 (Monday)Reminders:Reading for this week TZD Chapter 6.Homework #7 due next WednesdayThis 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 Atomand ExperimentsNiels Bohr 1885 – 1965Hydrogen 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 Energy LevelsPhoton energy is given by:2222eV 6.1312 nnEnakeERBn−=−=−=A. Opposite charges attract with a force inversely proportional to the square of the distance between them.B. The force on an object is equal to its mass times its acceleration.C. Accelerating charges radiate energy.D. Particles have a well-defined position and momentum.E. All of the above.Clicker questionWhich of the following principles of classical physics is violated in deriving the Bohr model of the atom?Note that A & B were used in the derivation of the Bohr model.Hydrogen-Like IonsAtoms which have only one electron can be analyzed much like the hydrogen atom.An atom with atomic number Z with Z-1 electrons removed is a hydrogen like ionThe (Coulomb) force on the electron is22221rkZerqkqFC==The increase in the force results in tighter orbits and a deeper potential well, reducing the energy (more negative).ZanrBn2=2222eV 6.13nZnEZERn−=−=A. 3.4 eVB. 13.6 eV C. 27.2 eVD. 54.4 eVE. None of the aboveClicker questionA single electron is in the n=2 energy level around a helium nucleus (He+). What is the minimum energy photon that can remove this electron?The atomic number of helium is Z=2and the electron is in the n=2 energy level. So the energy of the state is2222eV 6.13nZnEZERn−=−=ZanrBn2=eV 6.132eV 6.132eV 6.132222−=−=−=nZEnThe value |En| is also referred to as the binding energy since it is a measure of how bound the electron is. It takes that amountof energy to free the electron (break its bond to the proton).Franck-Hertz ExperimentA+−+1.5 VCathode/FilamentGrid Anode+−VIElectrons boil off the cathode and accelerate toward the gridIf it has enough kinetic energy after passing the grid it may make it past the 1.5 V retarding voltage and hit the anode causing a current.Mercury atoms inside need 4.9 eV to excite from ground level to the next energy levelIf electron never has 4.9 eV of kinetic energy it will only elastically scatter off the mercury atoms, losing very little energy.HgHgHg2Franck-Hertz ExperimentA+−+1.5 VCathode/FilamentGrid Anode+−VIIf the electron kinetic energy exceeds 4.9 eV, it can inelastically collide with a mercury atom and transfer 4.9 eV to it, losing 4.9 eV of kinetic energy in the process.After the collision, the electron may not gain enough kinetic energy to reach the anode so the current will drop.As the accelerating Voltage increases, the electron can excite multiple mercury atoms.HgHgHgThis experiment gives more proof for the existence of atomic energy levels. Performed in 1914; Nobel prize in 1925.Holst, James Frank and Gustav Hertz around 1924 Current drops when electrons have enough energy (4.9 eV) to lose energy by exciting Mercury from the ground state to the first excited state.Now the electrons have enough energy to hit two different Mercury atoms (2 x 4.9 eV).Summary and Implications of Bohr ModelElectrons orbit the nucleus at particular radii corresponding toparticular energies. These energies are called energy levels orstates.There always exists one lowest energy state called the ground state to which the electron will always return.The only allowed electron energy transitions are between these energy levels.Free electrons with enough kinetic energy can excite atomic electrons. From conservation of energy, the free electron losesthe same amount of kinetic energy as the atomic electron gains.Photons are emitted and absorbed only with energies corresponding to transitions between energy levels.Successes of the Bohr ModelExplains the Balmer formula and predicts the empirical constant R using fundamental constants:⎟⎠⎞−⎜⎝⎛=212'11nnRλ3222)(2chkemRπ=Explains the spectrum for other single electron atoms like singly ionized helium or double ionized lithium.Predicts the approximate size of the hydrogen atom (orbit radius)Sort of explains why atoms emit discrete spectral lines (?)Sort of explains why electrons don’t spiral into the nucleus (?)Remaining Issues with the Bohr ModelWhy is angular momentum quantized?Why don’t electrons radiate when they are in fixed orbitals?How does electron know which level to jump to? (i.e. how to predict intensities of spectral lines)Can’t be generalized to more complex (multi-electron) atomsShapes of molecular orbits and how bonds workCan’t explain doublet spectral linesLouis de Broglie: 1892 – 1987Ideas for How to Resolve these Problems?3Waves• Physicists at this time may have been confused about atoms, but they understood waves.•• They understood that for standing waves, boundary conditions mean that waves only have discrete modes.•• e.g. guitar stringsLλ1=2L f1=c/2Lλ2=L f2=c/Lλ4=L/2 f4=2c/Lλ3=2L/3 f3=3c/2Lλ5=2L/5 f5=5c/2Lλn=2L/n fn=nc/2L…= node = fixed point that doesn’t move.So going around the circle must take an integral number of wavelengths.A. r = λB. r = nλC. πr = nλD. 2πr = nλE. 2πr = λ/nn = 1, 2, 3, …Clicker questionWhat about standing waves in a ring?Just like a standing wave on a string but now the two ends of the string are joined together.What are the restrictions on the wavelength?Circumference is 2πr so the condition is 2πr = nλ. If you start at a peak and go around the circle, you must end up at a peak. Otherwise you will not have a standing wave; it will change.• In 1923, French grad student Louis de Broglie suggested that maybe electrons are actually little waves going around the nucleus.• This seems plausible because…– Standing waves have quantized frequencies, might be related to quantized energies.– Einstein had shown that light, typically thought of as waves, have particle properties. – Might not electrons, typically thought of as particles, have wave properties?de Broglie WavesA. 1B. 5C. 10D. 20E. Cannot determine from picture12345678910What is n in this picture? How many wavelengths does it take to complete a loop?Clicker questionn=1n=2n=3n=10= node = fixed