Unit II - Lecture 6ChemistryThe Molecular Nature of Matter and ChangeFifth EditionMartin S. SilberbergCopyright ! The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Quantum Theory and Atomic Structure7.1 The Nature of Light7.2 Atomic Spectra7.3 The Wave-Particle Duality of Matter and EnergyFigure 7.1Frequency and Wavelengthc = " #The Wave Nature of LightFigure 7.2Amplitude (intensity) of a wave.Figure 7.3Regions of the electromagnetic spectrum.Sample Problem 7.1SOLUTION:PLAN:Interconverting Wavelength and Frequencywavelength in units givenwavelength in mfrequency (s-1 or Hz)# = c/"Use c = "#10-2 m1 cm10-9 m1 nm= 1.00x10-10 m= 325x10-2 m= 473x10-9 m# =3x108 m/s1.00x10-10 m= 3x1018 s-1# =# =3x108 m/s325x10-2 m= 9.23x107 s-13x108 m/s473x10-9 m= 6.34x1014 s-1PROBLEM:A dental hygienist uses x-rays ("= 1.00A) to take a series of dental radiographs while the patient listens to a radio station (" = 325 cm) and looks out the window at the blue sky ("= 473 nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108 m/s.)o1 A = 10-10 m1 cm = 10-2 m1 nm = 10-9 mo325 cm473 nm1.00 Ao10-10 m1 AoFigure 7.4Different behaviors of waves and particles.Figure 7.5The diffraction pattern caused by light passing through two adjacent slits.Figure 7.6Blackbody radiationE = n h #$E = $n h #$E = h #when n = 1Smoldering coalElectric heating element Lightbulb filamentFigure 7.7Demonstration of the photoelectric effect.Sample Problem 7.2SOLUTION:PLAN:Calculating the Energy of Radiation from Its WavelengthPROBLEM:A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation?After converting cm to m, we can use the energy equation, E = h# combined with # = c/" to find the energy.E = hc/"E =6.626x10-34 J*s 3x108 m/s1.20 cm10-2 m1 cmx= 1.66x10-23 JFigure 7.8The line spectra of several elements.= RRydberg equation -1"1n221n12R is the Rydberg constant = 1.096776x107 m-1Figure 7.9Three series of spectral lines of atomic hydrogen.for the visible series, n1 = 2 and n2 = 3, 4, 5, ...Figure 7.10Quantum staircase.Figure 7.11The Bohr explanation of three series of spectral lines.Figure 7.12A tabletop analogy for the H atom’s energy.$E = Efinal – Einitial = -2.18 x 10-18 J-1n2initial1n2finalSample Problem 7.3SOLUTION: (a)PLAN:Determining $E and " of an Electron TransitionPROBLEM:A hydrogen atom absorbs a photon of visible light (see Figure 7.11) and its electron enters the n = 4 energy level. Calculate (a) the change in energy of the atom and (b) the wavelength (in nm) of the photon.The H atom absorbs energy, so Efinal > Einitial. Visible light is absorbed when ninitial = 2. Calculate $E using equation 7.4. (b) Use equations 7.2 and 7.1 to calculate wavelength and convert to nm.$E = -2.18 x 10-18 J-1n2initial1n2final = -2.18 x 10-18 J-122142 = -2.18 x 10-18 J-14116= 4.09 x 10-19 J" = hc/$E =6.626x10-34J*s 3x108m/sx= 4.86 x 10-7 m x4.09 x 10-19 J1 nm10-9 m = 486 nm (b)Figure B7.2Figure B7.1 AEmission and absorption spectra of sodium atoms.Flame tests.strontium 38Sr copper 29CuFigure B7.1 BFigure B7.3 The main components of a typical spectrometer.Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected.Sample in compartment absorbs characteristic amount of each incoming wavelength.Computer converts signal into displayed data.Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths.Lenses/slits/collimaters narrow and align beam.Detector converts transmitted radiation into amplified electrical signal.Figure B7.4 Measuring chlorophyll a concentration in leaf extract.Figure 7.13Wave motion in restricted systems.Table 7.1 The de Broglie Wavelengths of Several ObjectsSubstance Mass (g) Speed (m/s)" (m)slow electronfast electronalpha particleone-gram massbaseballEarth9x10-289x10-286.6x10-241.01426.0x10271.05.9x1061.5x1070.0125.03.0x1047x10-41x10-107x10-157x10-292x10-344x10-63" =h /muSample Problem 7.4SOLUTION:PLAN:Calculating the de Broglie Wavelength of an ElectronPROBLEM:Find the deBroglie wavelength of an electron with a speed of 1.00x106m/s (electron mass = 9.11x10-31 kg; h = 6.626x10-34 kg*m2/s).Knowing the mass and the speed of the electron allows to use the equation " = h/mu to find the wavelength." = 6.626x10-34 kg*m2/s9.11x10-31 kg x1.00x106 m/s= 7.27x10-10 mFigure 7.14Comparing the diffraction patterns of x-rays and electrons.x-ray diffraction of aluminum foil electron diffraction of aluminum foilCLASSICAL THEORYMatter particulate, massiveEnergy continuous, wavelikeSince matter is discontinuous and particulate perhaps energy is discontinuous and particulate.Observation TheoryPlanck: Energy is quantized; only certain values allowedblackbody radiationEinstein: Light has particulate behavior (photons)photoelectric effectBohr: Energy of atoms is quantized; photon emitted when electron changes orbit.atomic line spectraFigure 7.15Summary of the major observations and theories leading from classical theory to quantum theory.Since energy is wavelike perhaps matter is wavelikeObservation TheorydeBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electronsDavisson/Germer: electron diffraction by metal crystalSince matter has mass perhaps energy has massObservation TheoryEinstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum.Compton: photon wavelength increases (momentum decreases)after colliding withelectronFigure 7.15 continuedQUANTUM THEORYEnergy same as Matterparticulate, massive, wavelikeThe Heisenberg Uncertainty Principle$ x * m $ u ! h4%Sample Problem 7.5SOLUTION:PLAN:Applying the Uncertainty PrinciplePROBLEM:An electron moving near an atomic nucleus has a speed 6x106 ± 1%. What is the uncertainty in its position ($x)?The uncertainty ($x) is given as ±1% (0.01) of 6x106 m/s. Once we calculate this, plug it into the uncertainty equation.$u = (0.01)(6x106 m/s) = 6x104 m/s$ x * m $ u ! h4%$x !4% (9.11x10-31 kg)(6x104 m/s)6.626x10-34 kg*m2/s! 1 x 10-9