CHEM 101 1nd Edition Lecture 24Outline of Last Lecture I. Radiation II. Electromagnetic SpectrumIII. Electromagnetic RadiationIV. Quantization of EnergyOutline of Current Lecture I. Einstein and the Photoelectric Effect II. Atomic Line Emission Spectra and Neils BohrIII. The Bohr Model of the AtomIV. Energy AbsorptionV. Energy CalculationsVI. Origin of Line SpectraVII. The Wave-Like Behavior of MatterCurrent LectureI. Einstein and the Photoelectric Effect a. Classical theory said that E (energy) of ejected electron should increase with increase in light intensity - not observedi. No e- (electrons) observed until light of a certain minimum E is usedii. Number of e- ejected depends of light intensityb. We can understand experimental observation if light consists of particles called photos of discrete energyi. Photons can be thought of as little quantized packets of energyii. The energy of each photon is proportion to the frequency of the radiationas defined by E= hvc. Einstein describes light as behaving with both particle and wave character i. In the photoelectric effect, photons striking the metal surface will only cause electrons to be ejected if the photons have enough energyd. Problem:i. Molybdenum metal must absorb radiation with a minim frequency of 1.09 x 1015 s-1 before it can emit an electron from its surface via the photoelectric effect. What is the minim energy required to produce the photoelectric effect? What wavelength will provide a photon of this These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.energy?1. Question 1a. E = hvb. = (6.626 x 1034 Js)(1.09 x 1015 s-1)c. = 7.22 x 10-19 J per photoni. (6.022 x 1023 photon / 1 mol)(1 / J per photon)d. = 435 kJ / moli. Watch for units2. Question 2a. c = λvi. λ = c/vb. = (2.998 x 108 m/s) / (1.09 x 1015 s-1)i. Converted m to nmc. = 275 nm (or smaller)i. Watch for unitsII. Atomic Line Emission Spectra and Neils Bohra. Bohr's greatest contribution to science was in building a simple model of the atomi. It was based on an understanding of the Sharp Line Emission Spectra of excited atoms1. Excited atoms smith light of only certain wavelengths2. The wavelengths of emitted light depend on the elementa. Used as a fingerprint for the elementii. A goal of scientists in the late 19th century was to explain why excited gaseous atoms emitted light of only certain vb. Rydberg and the Balmer Equationi. Johann Balmer and Johannes Rydberg came up with the Balmer equation to describe this pattern 1. 1 / = R [(1 / 22) - (1 / n2)] when n > 2a. For Hydrogenii. n is an integeriii. R is called the Rydberg constant 1. 1.0974 x 107 m-1c. Atomic Spectra and Bohri. Bohr proposed a model for the electronic structure of atoms and with it an explanation or the emission spectra of excited atoms1. Electrons traveled about the nucleus in an orbit2. Problem: this contradicts laws of classical physicsa. Any orbit should be possible and so is any energyb. But a charged particle moving in an electric field should emit energyi. The end result should be destructionIII. The Bohr Model of the Atoma. Electrons in an atom can only occupy certain orbitsi. Corresponding to certain energiesb. Electron in permitted orbits have allowed energies; these energies will not be radiated from the atomc. Energy is only absorbed or emitted in such a way as to move an electron from one "allowed" energy state to anotheri. The energy is defined by E = hvd. Potential energy of electron in the nth leveli. En = - Rhc / n21. R is the Rydberg constant2. h is Planck’s constant3. c is the speed of lightii. The potential energy of an electron is negativeiii. As n increases energy of electron become more positive1. There is no energy level that will ever be positiveiv. A n increases the gap between n and n + 1 deceasese. Ground State: Lowest possible energy configuration. i. For the H atom n = 1f. Excited State: electron occupies higher energy level. i. For the H atom n >1IV. Energy Absorptiona. Absorption of energy can excite electron from ground state to excited stateb. Energy is emitted when the electron returns from the excited state to the ground statec. Origin of Line Spectrai. Δ E = Efinal - E initial = -NARhc {(1 / n2final) - (1 / n2initial)}1. nfianal < ninitial, (2<3), Δ E = – , then energy emitted2. NA is Avogadro’s NumberV. Energy Calculationsa. Use this form if you want to calculate the nervy in joules/atomi. En = - Rhc / n2b. If you want to calculate change in energy (in kJ/mol) then multiply this by NA and divide by 1000 (to convert from J to kJ)i. Δ E= Efinal - E initial = -NARhc {(1 / n2final) - (1 / n2initial)}VI. Origin of Line Spectraa. Problem: Calculate the energy emitted when going form the n = 3 to n = 2 level for the Hydrogen atomi. First Way1. For n = 3, E = -Rhc / 32a. E = -(1.097 X 107 m-1)(6.626 X 10-34 Js)(2.998 X 108 m/s) / 9b. E = -(2.42 X 10-19 J/atom)(6.022 X 1023 atoms/mol)i. E = -145800 J/mol = -145.8 kJ/mol2. For n = 2, E = -Rhc / 22a. E = -(1.097 X 107 m-1)(6.626 X 10-34 Js)(2.998 X 108 m/s) / 4b. E = -(5.448 X 10-19 J/atom)(6.022 X 1023 atoms/mol)i. E = -328000 J/mol = -328.0 kJ/mol3. Energy difference is -328 - (-145.8) kJ/mol = -182.2 kJ/molii. Second Way1. DE = -NARhc(1/4 - 1/9)2. DE = -(6.022 X 1023 atoms/mol) (1.097 X 107 m-1) (6.626 X 10-34 J s) (2.998 X 108 m/s) (1/4 - 1/9)3. DE = -(1.312 X 106 J/atom)(1/4 - 1/9)a. DE = -182263 J /molb. DE = -182.3 kJ /moliii. Emitted is a negative valueiv. For this transition what wavelength of energy would be emitted?1. DE = -182.2 kJ/mol a. (-182.2 kJ/mol)(1000 J/kJ)(1 mol/6.022 X 1023 atoms)i. or 3.026 X 10-19 J/atom2. E = hν, so E/h = v and h/E = 1/v ; c = lv, so c/v = l ; a. l = ch/Eb. l = (2.998 X 108 m/s)(6.626 X 10-34 J s) / (3.026 X 10-19 J/atom)i. l = 656 nmVII. The Wave-Like Behavior of Mattera. Photoelectric effect demonstrated that light, usually considered a wave, can also have the properties of particlesb. Can matter exhibit the properties of a wave?i. For light: 1. E = mc22. E = hn = hc / lii. Therefore, mc = h /l and for particles (mass)(velocity) = h / lc. de Broglie Wavelength i. l = h /mvii. The revolutionary idea inked the particle properties of electron (mass andvelocity) with a wave property (wavelength)d. The Heisenberg Uncertainty Principlei. If an electron is wavelike, how does one describe its position?1. Cannot simultaneously define the position