CHEM 1211 Lecture 15Outline of Last Lecture I. Hess’s law continuedII. Chapter 6: Atoms Electronic StructureIII. Color in Neon Lights, Fireworks, etc…A. Properties of Light (photon)IV. Constant Frequency & Variable WavelengthA. Photoelectric EffectB. Light & EnergyOutline of Current Lecture I. Electromagnetic RadiationA. Emission SpectrumB. Absorption SpectrumC. Bohr Model of AtomII. Predicting Emission Lines of HydrogenA. Rydberg EquationIII. Electronic TransitionsA. Wave nature of electronsCurrent Lecture: I. Electromagnetic RadiationA. Emission Spectrum: formed by electric current passing through a gas in a vacuumtube (at a very low pressure) which causes gas to emit lightEx: PrismLight Prism Detector- If we add lots of energy to a sample of elemental gas, we get an emission spectrum (elemental specific)1. In the hydrogen emission spectrum a band is observed at 4860A. What is the frequency and energy of this emission line?C = ƛv V = C / ƛ4860A 1 x 10 -10 m = 4.860 x 10-7m 1AV (visible: green light): 3.00 x 10 8 m/s = 6.17 x 1014 s-1 nHz4.860 x 10-7mE= hvE= (6.626 x 10-34 J*s)(6.17 x 108 m/s) = 4.088 x 10-19 J / photon4.088 x 10 -19 J 6.02 x 10 23= 2.46 x 105 J / mol photon photon Mol photonE = 246 KJ / mol photonB. Absorption Spectrum: shine a bean of white light through gas- Indicates wavelengths of light that have been ABSORBEDC. Bohr Model of Atom: only works for hydrogen, explains atomic spectra- Energy of quantized state: - RhC / n2- N: integral numbers only ( 1 to infinity )- Radius of orbitals: n2 (0.0529nm)II. Predicting Emission Lines of HydrogenA. Rydberg Equation1 = RH 1 - 1ƛ n12 n22- RH : 1.097 x 107 m-1- RH: 1.097 x 10-2 nm-1- n refers to one of the infinite possible energy state of the electron in hydrogen- requirement: n1 < n2III. Electronic Transitions1. Move an electron from a lower energy level to a higher energy level, E= -RhC 1 - 1 n12 n222. Start a rotational mode of motion3. Begin a vibrations mode of motion4. Gain kinetic energy (translational motion) ex: waterA. Wave nature of electrons2πr = nƛDE Broglie said the wavelength of an object = h / mv2πr = nh / mv- Angular momentum = mvr – nh /
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