2The classical view of light as an electromagnetic wave.An electromagnetic wave is a traveling wave with time-varying electric and magneticFields that are perpendicular to each other and to the direction of propagation.
3 Light as a waveTraveling wave description E y ( x, t ) = E o sin( kx − ωt )Intensity of light wave 1 I = cε oE o 2 2
4Schematic illustration of Young’s double-slit experiment.
5Diffraction patterns obtained by passing X-rays through crystals can only beexplained by using ideas based on the interference of waves. (a) Diffraction of X-rays from a single crystal gives a diffraction pattern of bright spots on aphotographic film. (b) Diffraction of X-rays from a powdered crystalline materialor a polycrystalline material gives a diffraction pattern of bright rings on aphotographic film.
6(c) X-ray diffraction involves constructive interference of waves being"reflected" by various atomic planes in the crystal .
7 Bragg’s LawBragg diffraction condition 2d sinθ = nλ n = 1, 2, 3, ...The equation is referred to as Bragg’s law, and arises from theconstructive interference of scattered waves.
9(a) Photoelectric current vs. voltage when (b) The stopping voltage and therefore thethe cathode is illuminated with light of maximum kinetic energy of the emittedidentical wavelength but different electron increases with the frequency ofintensities (I). The saturation current is light υ. (Note: The light intensity is notproportional to the light intensity the same) Results from the photoelectric experiment.
10The effect of varying the frequency of light and the cathode material in the photoelectricExperiment. The lines for the different materials have the same slope h but different intercepts
11 Photoelectric EffectPhotoemitted electron’s maximum KE is KEm KEm = hυ − hυ 0 Work function, Φ0The constant h is called Planck’s constant.
12The PE of an electron inside the metal is lower than outside by an energy called theworkfunction of the metal. Work must be done to remove the electron from the metal.
13Intuitive visualization of light consisting of a stream of photons (not to be takentoo literally).SOURCE: R. Serway, C. J. Moses, and C. A. Moyer, Modern Physics, Saunders CollegePublishing, 1989, p. 56, figure 2.16 (b).
14 Light Intensity (Irradiance)Classical light intensity 1 I = cε oE o 2 2Light Intensity I = Γph hυPhoton flux ∆ ph N Γ = ph A∆t
16X-rays are photonsX-ray image of an American one-cent coin captured using an x-ray a-Se HARP camera.The first image at the top left is obtained under extremely low exposure and thesubsequent images are obtained with increasing exposure of approximately one order ofmagnitude between each image. The slight attenuation of the X-ray photons by Lincolnprovides the image. The image sequence clearly shows the discrete nature of x-rays, andhence their description in terms of photons.SOURCE: Courtesy of Dylan Hunt and John Rowlands, Sunnybrook Hospital, Universityof Toronto.
17Scattering of an X-ray photon by a “free” electron in a conductor.
19Schematic illustration of black body radiation and its characteristics.Spectral irradiance vs. wavelength at two temperatures (3000K is about the temperature ofThe incandescent tungsten filament in a light bulb.)
20 Black Body RadiationPlanck’s radiation law 2π hc 2 Iλ = 5 hc λ exp −1 λkT Stefan’s black body radiation law PS = σ S T 4 Stefan’s constant 2π 5 k 4 σS = 2 3 = 5.670 × 10 − 8 W m − 2 K − 4 15c h
Stefan’s law for real surfaces 21Electromagnetic radiation emitted from a hot surface Pradiation = total radiation power emitted (W = J s -1) Pradiation = Sεσ S [T − T ] 4 0 4 σS = Stefan’s constant, W m-2 K-4 ε = emissivity of the surface ε = 1 for a perfect black body ε < 1 for other surfaces S = surface area of emitter (m2)
22Young’s double-slit experiment with electrons involves an electron gun and two slits in aCathode ray tube (CRT) (hence, in vacuum).Electrons from the filament are accelerated by a 50 kV anode voltage to produce a beam thatIs made to pass through the slits. The electrons then produce a visible pattern when they strikeA fluorescent screen (e.g., a TV screen), and the resulting visual pattern is photographed.SOURCE: Pattern from C. Jonsson, D. Brandt, and S. Hirschi, Am. J. Physics, 42, 1974, p.9,figure 8. Used with permission.
25The diffraction of electrons by crystals gives typical diffraction patterns that would beExpected if waves being diffracted as in x-ray diffraction with crystals [(c) and (d) fromA. P. French and F. Taylor, An Introduction to Quantum Mechanics (Norton, New York,1978), p. 75; (e) from R. B. Leighton, Principles of Modern Physics, McGraw-Hill, 1959),p. 84.
26 De Broglie RelationshipWavelength λ of the electron depends on its momentum p h λ= pDe Broglie relations h h λ= OR p= p λ
28 Time-Independent Schrodinger EquationSteady-state total wave function jEt Ψ ( x,t ) = ψ ( x)exp − Schrodinger’s equation for one dimension d 2ψ 2m 2 + 2 ( E −V )ψ = 0 dx Schrondinger’s equation for three dimensions ∂ ψ ∂ ψ ∂ ψ 2m 2 2 2 + 2 + 2 + 2 ( E − V )ψ = 0 ∂x 2 ∂y ∂z
29Electron in a one-dimensional infinite PE well.The energy of the electron is quantized. Possible wavefunctions and the probabilitydistributions for the electron are shown.
30 Infinite Potential WellWavefunction in an infinite PE well nπx ψ n ( x) = 2 Aj sin a Electron energy in an infinite PE well (πn) 2 h n 2 2 2 En = 2 = 2 2ma 8maEnergy separation in an infinite PE well h (2n + 1) 2 ∆E = En +1 − En = 2 8ma
31 Heisenberg’s Uncertainty PrincipleHeisenberg uncertainty principle for position and momentum ∆x∆p x ≥ Heisenberg uncertainty principle for energy and time ∆E∆t ≥
32(a) The roller coaster released from A can at most make it to C, but not to E. Its PE at A is less thanthe PE at D. When the car is at the bottom, its energy is totally KE. CD is theenergy barrier that prevents the care from making it to E. In quantum theory, on the otherhand, there is a chance that the care could tunnel (leak) through the potential energy barrierbetween C and E and emerge on the other side of hill at E.(b) The wavefunction for the electron incident on a potential energy barrier (V0). The incidentAnd reflected waves interfere to give ψ1(x). There is no reflected wave in region III. In regionII, the wavefunction decays with x because E < V0.
33 Tunneling Phenomenon: Quantum LeakProbability of tunneling 2 ψ III ( x) C12 1 T= = 2 = ψ I ( x) 2 A1 1 + D sinh 2 (αa )Probability of tunneling through 16 E (Vo − E ) T = To exp(−2αa ) To = where 2 VoReflection coefficient R 2 A2 R = 2 =1 −T A1
35Scanning Tunneling Microscopy (STM) image of a graphite surface wherecontours represent electron concentrations within the surface, and carbon rings areclearly visible. Two Angstrom scan. |SOURCE: Courtesy of Veeco Instruments,Metrology Division, Santa Barbara, CA.
37STM image of Ni (100) surface STM image of Pt (111) surfaceSOURCE: Courtesy of IBM SOURCE: Courtesy of IBM
38Electron confined in three dimensions by a three-dimensional infinite PE box.Everywhere inside the box, V = 0, but outside, V = ∞. The electron cannot escapefrom the box.
39 Potential Box: Three Quantum NumbersElectron wavefunction in infinite PE well n1πx n2πy n3πz ψ n1n2 n3 ( x, y, z ) = A sin sin sin a b c Electro energy in infinite PE box En1n2 n3 = ( h 2 n12 + n2 + n3 2 2 = ) h2 N 2 2 8ma 8ma 2 N =n +n +n 2 2 1 2 2 2 3
40The electron in the hydrogenic atom isatom is attracted by a central force thatis always directed toward the positiveNucleus.Spherical coordinates centered at thenucleus are used to describe the positionof the electron. The PE of the electrondepends only on r.
41Electron wavefunctions and the electron energy are obtained by solving the Schrödinger equationElectron’s PE V(r) in hydrogenic atom is used in the Schrödinger equation − Ze 2 V (r ) = 4πε o r
42(a) Radial wavefunctions of the electron in a hydrogenic atom for various n and values.(b) R2 |Rn,2| gives the radial probability density. Vertical axis scales are linear in arbitraryunits.
43 Electron energy is quantized Electron energy in the hydrogenic atom is quantized. n is a quantum number, 1,2,3,… 4 2 me Z En = − 2 2 2 8ε o h nIonization energy of hydrogen: energy required to remove the electron from the ground state in the H-atom 4 me −18 E I = 2 2 = 2.18 ×10 J = 13.6 eV 8ε o h
46(a) The polar plots of Yn,(θ, φ) for 1s and 2p states.(b) The angular dependence of the probability distribution, which is proportional to| Yn,(θ, φ)|2.
47The energy of the electron in the hydrogenatom (Z = 1).
48 The physical origin of spectra.(a) Emission(b) Absorption
49An atom can become excited by a collision with another atom.When it returns to its ground energy state, the atom emits a photon.
50 Electron probability distribution in the hydrogen atomMaximum probability for = n − 1 2 n ao rmax = Z
51The Li atom has a nucleus with charge +3e, 2 electrons in the K shell , whichis closed, and one electron in the 2s orbital. (b) A simple view of (a) wouldbe one electron in the 2s orbital that sees a single positive charge, Z = 1The simple view Z = 1 is not a satisfactory description for the outer electronbecause it has a probability distribution that penetrates the inner shell. Wecan instead use an effective Z, Zeffective = 1.26, to calculate the energy of theouter electron in the Li atom.
Ionization energy from the n-level for an outer electron 52 2 Z (13.6 eV) EI ,n = effective 2 n
53a) The electron has an orbital angular momentum, which has a quantized component L along an externalMagnetic field Bexternal.b) The orbital angular momentum vector L rotates about the z axis. Its component Lz is quantized; herefore, the L orientation, which is the angle θ, is also quantized. L traces out a cone.c) According to quantum mechanics, only certain orientations (θ ) for L are allowed, as determined by nd m
54An illustration of the allowedPhoton emission processes.Photon emission involves∆ = ± 1,
55Orbital Angular Momentum and Space QuantizationOrbital angular momentum L = [( +1)] 1/ 2where = 0, 1, 2, ….n−1Orbital angular momentum along Bz Lz = m Selection rules for EM radiation absorption and emission ∆ = ±1 and ∆m = 0, ± 1
56Spin angular momentum exhibits spacequantization. Its magnitude along z isquantized, so the angle of S to the z axisis also quantized.
57 Electron Spin and Intrinsic Angular Momentum SElectron spin 1 S = [ s( s + 1) ] s= 1/ 2 2Spin along magnetic field 1 S z = ms ms = ± 2 the quantum numbers s and ms, are called the spin and spin magnetic quantum numbers.
59(a) The orbiting electron is equivalent to a current loop that behaves like a bar magnet.(b) The spinning electron can be imagined to be equivalent to a current loop as shown.This current loop behaves like a bar magnet, just as in the orbital case.
60 Magnetic Dipole Moment of the ElectronOrbital magnetic moment e μ orbital =− L 2meSpin magnetic moment e μ spin =− S ms
61Energy of the electron due to its magnetic moment interacting with a magnetic fieldPotential energy of a magnetic moment E BL = −µorbital B cos θwhere θ is the angle between µorbital and B. A magnetic moment in a magnetic field experiences a torque that tries to rotate the magnetic moment to align the moment with the field. A magnetic moment in a nonuniform magnetic field experiences force that depends on the orientation of the dipole.
62(a) Schematic illustration of the Stern-Gerlach experiment.A stream of Ag atoms passing through a nonuniform magnetic field splits into two.
63(b) Explanation of the Stern-Gerlach experiment. (c) Actual experimental result recorded on aphotographic plate by Stern and Gerlach (O. Stern and W. Gerlach, Zeitschr. fur. Physik, 9, 349,1922.) When the field is turned off, there is only a single line on the photographic plate. Theirexperiment is somewhat different than the simple sketches in (a) and (b) as shown in (d).
64Stern-Gerlach memorial plaque at the University of Frankfurt. The drawing shows the original Stern-Gerlachexperiment in which the Ag atom beam is passed along the long- length of the external magnet to increase thetime spent in the nonuniform field, and hence increase the splitting. The photo on the lower right is OttoStern (1888-1969), standing and enjoying a cigar while carrying out an experiment. Otto Stern won the Nobelprize in 1943 for development of the molecular beam technique. Plaque photo courtesy of Horst Schmidt-Böcking from B.Friedrich and D. Herschbach, "Stern and Gerlach: How a Bad Cigar Helped Reorient Atomic Physics", Physics Today, December 2003, p.53-59.
65Orbital angular momentum vector L and spin angular momentum vector S can add eitherIn parallel as in (a) or antiparallel, as in (b).The total angular momentum vector J = L + S, has a magnitude J = √[j(j+1)], where in(a) j = + ½ and in (b) j = - ½
66(a) The angular momentum vectors L and S precess around their resultant total angularMomentum vector J.(b) The total angular momentum vector is space quantized. Vector J precesses about the zaxis, along which its component must be mj
67A helium-like atomThe nucleus has a charge +Ze, where Z = 2 for He. If one electron is removed, we havethe He+ ion, which is equivalent to the hydrogenic atom with Z = 2.
68Energy of various one-electron states.The energy depends on both n and
70Electronic configurations for the first five elements. Each box represents an orbitalψ (n, , m)
71Electronic configuration for C, N, O, F and Ne atoms.Notice that in C, N, and O, Hund’s rule forces electrons to align their spins. For the Neatom, all the K and L orbitals are full.
72 The Helium AtomPE of one electron in the He atom 2 2 2e e V (r1 , r12 ) = − + 4πε o r1 4πε o r12
73Absorption, spontaneous emission and stimulated emission Absorption, spontaneous emission, and stimulated emission.
74The principle of the LASER. (a) Atoms in the ground state are pumped up to the energy level E 3by incoming photons of energy hυ13 = E3-E1. (b) Atoms at E3 rapidly decay to the metastablestate at energy level E2 by emitting photons or emitting lettice vibrations. hυ32 = E3-E2.
75(c) As the states at E2 are metastable, they quickly become populated and there is a populationinversion between E2 and E1. (d) A random photon of energy hυ21 = E2-E1 can initiate stimulatedemission. Photons from this stimulated emission can themselves further stimulate emissionsleading to an avalanche of stimulated emissions and coherent photons being emtitted.
79The principle of operation of the HeNe laser. Important HeNe laser energy levels (for 632.8 nmemission).
80(a) Doppler-broadened emission versus wavelength characteristics of the lasing medium.(b) Allowed oscillations and their wavelengths within the optical cavity.(c) The output spectrum is determined by satisfying (a) and (b) simultaneously.
Laser Output Spectrum 81Doppler effect: The observed photon frequency depends on whether the Ne atom is moving towards (+vx) or away (− vx) from the observer vx vx v2 = v0 1 + v1 = v0 1 − c c Frequency width of the output spectrum is approximately υ2 – υ1 2v0v x ∆v = cLaser cavity modes: Only certain wavelengths are allowed to exist within the optical cavity L. If n is an integer, the allowed wavelength λ is λ n = L 2
82Energy diagram for the Er3+ ion in the glass fiber medium and light amplification byStimulated emission from E2 to E1.Dashed arrows indicate radiationless transitions (energy emission by lattice vibrations).
83A simplified schematic illustration of an EDFA (optical amplifier). The erbium-ion doped fiber is pumped by feeding the light from a laser pump diode,through a coupler, into the erbium ion doped fiber.
84(a) The retina in the eye has photoreceptors that can sense the incident photons on them and henceprovide necessary visual perception signals. It has been estimated that for minimum visualperception there must be roughly 90 photons falling on the cornea of the eye. (b) The wavelengthdependence of the relative efficiency ηeye(λ) of the eye is different for daylight vision, or photopicvision (involves mainly cones), and for vision under dimmed light, (or scotopic vision represents thedark-adapted eye, and involves rods). (c) SEM photo of rods and cones in the retina.SOURCE: Dr. Frank Werblin, University of California, Berkeley.
85Some possible states of the carbon atom, not in any particular order.