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Phys 321: Lecture 3 The Interacon of Light and Maer

Prof. Bin Chen Tiernan Hall 101 [email protected] Fraunhofer Lines • By 1814, German opcian Joseph von Fraunhofer cataloged 475 of dark lines in the solar spectrum, known today as the “Fraunhofer Lines” • A video story Kirchhoff’s Laws

1. A hot, dense gas or hot solid object produces a connuous spectrum 2. A hot, diffuse gas produces bright spectral lines (emission lines) 3. A cool, diffuse gas in front of a source of a connuous spectrum produces dark spectral lines (absorpon lines) in the connuous spectrum Measuring Spectra: Prism

Astronomers use spectrographs to separate the light in different wavelengths and measure the spectra of stars and galaxies Measuring Spectra: Diffracon Grang Spectra from different elements Lab measurements

Hδ Hγ Hβ Hα

Fraunhofer lines in Solar Spectrum Spectroscopy: Applicaons

• Idenfy elements on distant astronomical bodies: Chemical Composion • The element Helium is first discovered from the solar spectrum, not Earth. • Measure physical properes: temperature, density, pressure, magnec fields, high-energy parcles, etc. • Measure velocies of distant objects via Doppler shi Doppler Shi

“It’s the apparent change in the frequency of a wave caused by relave moon between the source of the wave and the observer.” --- Sheldon Cooper

A short animated explanaon (Youtube) Reference spectrum in lab

Observed spectrum from Star A

Observed spectrum from Star B

Queson: Which star is moving towards us?

A. Star A

B. Star B The Interaction of Light and Matter

• A hot, dense gas or hot solid object produces a continuous spectrum with no dark spectral lines.1

• A hot, diffuse gas produces bright spectral lines (emission lines).

• Acool, diffuse gas in front of a source of a continuous spectrum produces dark spectral lines (absorption lines) in the continuous spectrum.

Applications of Stellar Spectra Data An immediate application of these results was the identification of elements found in the Sun and other stars. A new element previously unknown on Earth, helium,2 was discovered spectroscopically on the SunDoppler in 1868; it wasShi not found on Earth until 1895. Figure 1 shows the visible portion of the solar spectrum, and Table 1 lists some of the elements responsible for producing the darkλobs > absorptionλrest : Redshi lines.ed -> Object moving away from observer Another rich line of investigation was pursued by measuring the Doppler shifts of spectral λ < λ : Blueshifted -> Object moving towards observer lines. For individual stars, vr obcs , andrest so the low-speed approximation of the following equation, ≪

λobs λrest "λ vr − , (1) λrest = λrest = c can be utilized to determine their radial velocities. By 1887 the radial velocities of Sirius, Procyon, Rigel, and Arcturus hadvr is the radial component of the objects velocity relave been measured with an accuracy of a few kilometers per to the observer. Posive means the object is moving second. away and vice versa.

Example 1.1. The rest wavelength λrest for an important spectral line of hydrogen (known as Hα) is 656.281 nm when measured in air. However, the wavelength of the Hα absorption line in the spectrum of the star Vega in the constellation Lyra is measured to be 656.251 nm at a ground-based telescope. Equation ( 1) shows that the radial velocity of Vega is

c(λobs λrest) 1 vr − 13.9 km s− = λrest = − ; the minus sign means that Vega is approaching the Sun. However, stars also have a proper motion, µ, perpendicular to the line of sight. Vega’s angular position in the sky changes by µ 0.35077 yr 1. At a distance of r 7.76 pc, this proper motion is related = ′′ − = to the star’s transverse velocity, vθ . Expressing r in meters and µ in radians per second results in

1 vθ rµ 12.9 km s− . = =

1In the first of Kirchhoff’s laws, “hot” actually means any temperature above 0 K. However, according to Wien’s displacement law a temperature of several thousand degrees K is required for λmax to fall in the visible portion of the electromagnetic spectrum. It is the opacity or optical depth of the gas that is responsible for the continuous blackbody spectrum. 2The name helium comes from Helios, a Greek Sun god.

The Interaction of Light and Matter

• A hot, dense gas or hot solid object produces a continuous spectrum with no dark spectral lines.1

• A hot, diffuse gas produces bright spectral lines (emission lines).

• Acool, diffuse gas in front of a source of a continuous spectrum produces dark spectral lines (absorption lines) in the continuous spectrum.

Applications of Stellar Spectra Data An immediate application of these results was the identification of elements found in the Sun and other stars. A new element previously unknown on Earth, helium,2 was discovered spectroscopically on the Sun in 1868; it was not found on Earth until 1895. Figure 1 shows the visible portion of the solar spectrum, and Table 1 lists some of the elements responsible for producing the dark absorption lines. Another rich line of investigation was pursued by measuring the Doppler shifts of spectral lines. For individual stars, vr c, and so the low-speed approximation of the following equation, ≪

λobs λrest "λ vr − , (1) λrest = λrest = c can be utilized to determine their radial velocities. By 1887 the radial velocities of Sirius, Procyon, Rigel, and Arcturus had been measured with an accuracy of a few kilometers per second. Radial Velocity of Vega

Example 1.1. The• The rest wavelength of Hα is 656.281 nm as rest wavelength λrest for an important spectral line of hydrogen (known as Hα) is 656.281measured in the lab. From a ground-based nm when measured in air. However, the wavelength of the Hα absorption line intelescope, the Hα line from star Vega in the spectrum of the star Vega in the constellation Lyra is measured to be 656.251 nm at a ground-basedconstellaon Lyra telescope. is measured at 656.251 nm. Equation ( 1) shows that the radial velocity What is its radial velocity? of Vega is

c(λobs λrest) 1 vr − 13.9 km s− = λrest = − ; the minus sign means that Vega is approaching the Sun. However, stars also have a proper motion, µ, perpendicular to the line of sight. Vega’s angular position in the sky changes by µ 0.35077 yr 1. At a distance of r 7.76 pc, this proper motion is related = ′′ − = to the star’s transverse velocity, vθ . Expressing r in meters and µ in radians per second results in

1 vθ rµ 12.9 km s− . = =

1In the first of Kirchhoff’s laws, “hot” actually means any temperature above 0 K. However, according to Wien’s displacement law a temperature of several thousand degrees K is required for λmax to fall in the visible portion of the electromagnetic spectrum. It is the opacity or optical depth of the gas that is responsible for the continuous blackbody spectrum. 2The name helium comes from Helios, a Greek Sun god.

Back to nature of light discussion… Young’s double-slit experiment Light = Waves Maxwell’s Equaons

Ultraviolet Catastrophe of blackbody radiaon Something is wrong! Discrete emission/absorpon lines (Link to a Youtube video) Photoelectric Effect

• Whether or not an is produced depends only on the frequency of the light, not the flux of the light

• Condion for producing an electron: Ephoton > φ • A single photon has energy: E = hν Planck’s quanzed photon energy of light waves!

Atomic Structure

Most of the space within an is empty!

The radius the of nucleus was 10,000 mes smaller than the radius of the atom The Interaction ofThe Light Interaction and Matter of Light and Matter to the nucleus. Rutherfordto the nucleus. coined Rutherford the term coinedproton theto termreferproton to the nucleusto refer of to the the hydrogen nucleus of the (Z 1), 1836atom times (Z more1), 1836 massive times than more the massive electron. than But the how electron. were Butthese how charges were these charges = arranged? arranged?=

The WavelengthsThe of Wavelengths Hydrogen of Hydrogen The experimentalHydrogenThe data experimental were abundant. Spectral data wereThe wavelengths abundant. Lines The of wavelengths 14 spectral lines of 14 of spectral hydrogen lines of hydrogen had been preciselyhad determined. been precisely Those determined. in the visible Those region in the of visible the electromagnetic region of the electromagnetic spectrum spectrum are 656.3 nm• (red, In 1885, a Swiss teacher, Johann are H 656.3α), 486.1 nm (red, nm (turquoise,Hα), 486.1 H nmβ), (turquoise, 434.0Balmer nm H (blue,β found an ), 434.0 Hγ ), nm and (blue, 410.2 H nmγ ), and 410.2 nm (violet, Hδ). In 1885empirical formula to reproduce the wavelengths of (violet, a Swiss Hδ). In school 1885 teacher, a Swiss Johann school teacher, Balmer Johann (1825–1898), Balmer had (1825–1898), found, by had found, by trial and error, ahydrogen lines (in visible wavelengths), today formulatrial and to error, reproduce a formula the to wavelengths reproduce the of wavelengthsthese spectral of lines these of spectral hydrogen, lines of hydrogen, today called the Balmer series or Balmer lines: today called theknown as the Balmer series orBalmerBalmer series lines: 1 1 11 1 1 RH ,RH , (7) (7) λ = 4 − nλ2= 4 − n2 ! " ! " 7 1 where n 3, 4, 5where,...,n and R3, 4, 5,...1.09677583, and RH 101.7096775831.3m 110is the experimentally1.3m− is the experimentally de- de- = H = ×− ± = termined Rydberg= constant7 for× hydrogen.± 7 Balmer’s formula was very accurate, to within termined Rydbergis the experimentally determined Rydberg constant for hydrogen constant for hydrogen. Balmer’s formula was very accurate, to within a fraction of a percent. Inserting n 3 gives the wavelength of the Hα Balmer line, n 4 a fraction of a percent. Inserting n 3 gives the wavelength= of the Hα Balmer line, n 4 = gives Hβ, and so on.= Furthermore, Balmer realized2 that since 22 4, his= formula could be gives Hβ, and so on. Furthermore, Balmer realized that since 2 4, his formula= could be generalized to generalized to =

1 1 1 1 1 1 RH , (8) RH , 2 2 (8) λ = m2 −λn=2 m − n ! " ! " with m

7 RH is named in honor of Johannes Rydberg (1854–1919), a Swedish spectroscopist. 7 RH is named in honor of Johannes Rydberg (1854–1919), a Swedish spectroscopist.

The Interaction of Light and Matter to the nucleus. Rutherford coined the term proton to refer to the nucleus of the hydrogen atom (Z 1), 1836 times more massive than the electron. But how were these charges arranged?=

The Wavelengths of Hydrogen The experimental data were abundant. The wavelengths of 14 spectral lines of hydrogen had been precisely determined. Those in the visible region of the electromagnetic spectrum are 656.3 nm (red, Hα), 486.1 nm (turquoise, Hβ), 434.0 nm (blue, Hγ ), and 410.2 nm (violet, Hδ). In 1885 a Swiss school teacher, Johann Balmer (1825–1898), had found, by trial and error, a formula to reproduce the wavelengths of these spectral lines of hydrogen, today called the Balmer series or Balmer lines:

1 1 1 RH , (7) λ = 4 − n2 ! "

7 1 where n 3, 4, 5,..., and RH 1.09677583 10 1.3m− is the experimentally de- termined= Rydberg constant for hydrogen.= 7 Balmer’s× formula± was very accurate, to within a fraction of a percent. Inserting n 3 gives the wavelength of the Hα Balmer line, n 4 gives Hβ, and so on. Furthermore,= Balmer realized that since 22 4, his formula could= be generalized to Hydrogen Spectral Lines = 1 The Interaction1 of Light1 and Matter More generally, RH , (8) TABLEλ 2 = The wavelengthsm2 − ofn selected2 hydrogen spectral lines in air. (Based on Cox, (ed.), Allen’s Astrophysical! Quantities, Fourth" Edition, Springer, New York, 2000.) with m

1 q q F 1 2 r, (9) 2 = 4πϵ0 r ˆ

8The quantity ! h/2π 1.054571596 10 34 J s and is pronounced “h-bar.” ≡ = × −

Bohr’s Semiclassical Atomic Model

Model: • Nucleus is a point mass sing at the center of the atom • move in circular orbits Bohr’s Postulaons: • Only a discrete number of orbits of the electrons (those with angular momentum that is an integral mulple of ħ = h/2π) is allowed • Radiaon of a discrete of energy is emied/absorbed only when the electrons jumps from one orbit to another • The radiated energy equals to the energy difference between the orbits SolutionsforAnIntroductiontoModernAstrophysics 25

can be used to replace Ee in the preceding equation, 2 4 2 2 2 2 m c E 2EiEf cos ! E .Ei Ef me c / : e C i ! C f D ! C This simplifies to 1 1 1 2 .1 cos !/: Ef ! Ei D mec ! Using Eq. (5.5) for the photon energy, E hc=",resultsin D h #" "f "i .1 cos !/: D ! D mec ! 5.6 The change of wavelength is given by the Compton scattering formula, Eq. (5.6), but with the electron mass replaced by the proton mass, The unit of Planck’s constant h #" "f "i .1 cos !/: D ! D mpc ! 6 The characteristic change in wavelength is h=mpc 1:32 10! nm. This is smaller than the value of the D " 4 Compton wavelength, "C h=mec 0:00243 nm, by a factor of me=mp 5:4 10 . D D D " ! 5.7 Planck’s constant has units of Js [kg m2 s 2]s kg [m s 1]m; D ! D ! which are the units of angular momentum (mvr). 5.8 (a) For an atom with Z protons in the nucleus (charge Ze)andoneelectron(charge e), Coulomb’s law (Eq. 5.9) becomes ! 1 Ze2 ; F 2 r D !4$%0 r O and the electrical becomes 1 Ze2 U : D !4$%0 r This means that the quantity “e2”shouldbereplaced by “Ze2”whereveritappearsinanexpressionfor the hydrogen atom. Similarly, “e4”shouldbereplaced by “Z2e4.” The result will be the corresponding expression for a one-electron atom with Z protons. From Eq. (5.13), the orbital radii are therefore

2 4$%0! 2 a0 2 rn n n ; D &Ze2 D Z

where a0 0:0529 nm. The energies are D &Z2e4 1 Z2 En 13:6 eV 2 2 2 2 2 D !32$ %0 ! n D ! n from Eq. (5.14). (b) Z 2 for singly ionized helium (He II), and so for the ground state (n 1), D D a0 2 r1 .1 / 0:0265 nm D 2 D and 22 E1 13:6 eV 54:4 eV: D ! 12 D ! The ionization energy is 54:4 eV. The Interaction of Light and Matter

FIGURE 4 Niels Bohr (1885–1962). (Courtesy of The Niels Bohr Archive, Copenhagen.)

12 1 9 where ϵ0 8.854187817 ... 10− Fm− is the permittivity of free space and r is a unit vector directed= from charge 1× toward charge 2. ˆ Consider an electron of mass me and charge e and a proton of mass mp and charge e in circular orbits around their common center of− mass, under the influence of their mutual+ 19 electrical attraction, e being the fundamental charge, e 1.602176462 10− C. This two-body problem may be= treated as an equivalent× one-body problem by usingThe the Interaction reduced massof Light and Matter

The Interaction of Light and Matter memp (me)(1836.15266v me) µ Electron 0.999455679 me = me mp = me 1836.15266e–me =v The Interaction+ of Light and+ Matter Electron r e– and the total mass v r + Electron e – Proton e M me mp me 1836.15266 me 1837.+15266 me 1.0005446 mp = + = + r = e = Proton of the system. Since M mp and µ me,e the+ hydrogen atom may be thought of as being composed of a proton of≃ mass M that≃ isProton at rest and an electron of mass µ that follows a circular orbit of radius r around the proton; see Fig. 5. The electrical attraction between the electron and the protonFIGURE produces 5 theThe electron’s centripetal of the hydrogen acceleration atom. v2/r, as described FIGURE 5 The Bohr model of the hydrogen atom. by Newton’s second law: FIGURE 5 The Bohr model of the hydrogen atom. or Bohr’sor Model ofF Hydrogenµa, Atom = or 1 e2 v2 r 1 µe2 r. v2 2 implying 4πϵ 2r 2 rr µ r. − 1 e0 −ˆ =4πϵ−v r2 ˆ =ˆ − r ˆ Electric force provides centrifugal acceleraon r µ 0 r. 2 −4πϵ0 r ˆ = − r ˆ 2 Canceling theCanceling minus sign the andminus the sign unit1 and vectorq1q the2 unitr, this vector expressionvr, this expression can be solved can be for solved the kinetic for the kinetic 1 2 rˆ µ r, energy,Cancelingµv the: minus1 sign2 and the unit vector2 r, this expressionTheˆ Interaction can be solvedof Light forand theMatter kinetic 2 energy, 2 µv : 4πϵ0 r ˆ = − r ˆ energy, 1 µv2: ˆ 2 v 2 2 9 2 7 2 Electron Formally, ϵ0 is defined as ϵ0 1/µ0c , where 1µ0 24π 1101 − 2eNA−1 ise the permeability of free space and Kinec energy 8 1 ≡ K 1 µv≡K 1× eµv2 . . e– (10) (10) c 2.99792458 10 ms− is the definedK speedµv of2 light. . πϵ r (10) ≡ × = 2 ==82πϵ0 r= 8 0 r = 2 = 8πϵ0 r

10 10 + Now the electricalNow the potential electrical energy potentialU of energy the BohrU of atom the10 Bohr is atom is e Now theElectrical potenal energy electrical potential energy U of the Bohr atom is Proton 2 1 1e2 e2 1 e UU U 2K.2K. 2K. = −4πϵ r == −−4πϵ0 r = − = −4πϵ0 0 r = − Thus the totalThus energy the totalE energyK UEof theK atomU isof the atom isFIGURE 5 The Bohr model of the hydrogen atom. Thus the total energy E = K + U of= the+ atom is Total energy = + 1 e2 1 e2 E K UE KK 2orKU KK 2K 1K. e2 .(11) (11) = + == −+ ==− −= −8πϵ=0−r = −8πϵ r E K U K 2K K . 1 e02 v2 (11) = + = − = − = −8πϵ r r µ r. 0 2 Note that theNote relation that the between relation the between kinetic, potential, the kinetic, and potential,total energies and− is4 totalπϵ in0 accordancer energiesˆ = − r isˆ in accordance with the virial theorem for an inverse-square force, Note thatwith the relation the virial between theorem the for kinetic, an inverse-squareCanceling potential, the minus force, and sign total and the energies unit vector isr in, this accordance expression can be solved for the kinetic E 1 U K. Because the kinetic energy must be positive, the total energy E is negative. with the2 virialE theorem1 U forK an. Because inverse-square the kineticenergy, force, energy1 µv2: must be positive, the totalˆ energy E is negative. This= merely= − indicates2 that the electron and the proton2 are bound.Toionize the atom (that E 1 U ThisK=. merely Because= − indicates the kinetic that energy the electron must be and positive, the proton the are totalbound energy.ToEionizeis negative.the atom (that is,2 to remove the proton and electron to an infinite separation), an amount of1 energy1 of e2 This= merely= is,− indicates to remove that the the proton electron and and electron the proton to an are infinitebound separation),.ToK ionizeµv2 anthe amount atom. (that of energy of (10) magnitude E (or more) must be added to the atom. = 2 = 8πϵ0 r is, toThus remove farmagnitude| the the| derivation protonE and has(or more)been electron completely must to be an added classicalinfinite to theseparation),in nature. atom. At this an point, amount however, of energy of | | 10 magnitudewe can useE Bohr’sThus(or more) quantization far the must derivation be of angular added has to momentum, beenNow the the completelyatom. electrical potential classical energy inU nature.of the Bohr At this atom point, is however, |we| can use Bohr’s quantization of angular momentum, Thus far the derivation has been completely classical in nature. At this1 point,e2 however, L µvr n!, U (12) 2K. = = = −4πϵ0 r = − we can use Bohr’s quantization of angular momentum,L µvr n!, (12) 10 This is found from a derivation analogous to the oneThusleadi theng t totalo=the energygravit=atEional KresultU. Theof the zero atom of po istential energy is taken to be zero at r . ! = + 10 This is found =fro∞m a derivatioLn analµvrogous to tnhe o,ne leading to the gravitational result. The(12) zero of potential = = 1 e2 energy is taken to be zero at r . E K U K 2K K . (11) = ∞ = + = − = − = −8πϵ r 10 This is found from a derivation analogous to the one leading to the gravitational result. The zero of potential0 energy is taken to be zero at r . Note that the relation between the kinetic, potential, and total energies is in accordance = ∞ with the virial theorem for an inverse-square force, 1 E 2 U K. Because the kinetic energy must be positive, the total energy E is negative. This= merely= − indicates that the electron and the proton are bound.Toionize the atom (that is, to remove the proton and electron to an infinite separation), an amount of energy of magnitude E (or more) must be added to the atom. Thus far| the| derivation has been completely classical in nature. At this point, however, we can use Bohr’s quantization of angular momentum, L µvr n!, (12) = = 10 This is found from a derivation analogous to the one leading to the gravitational result. The zero of potential energy is taken to be zero at r . = ∞

The Interaction of Light and Matter

v Electron e– r

e+ Proton

FIGURE 5 The Bohr model of the hydrogen atom. or 1 e2 v2 r µ r. 2 −4πϵ0 r ˆ = − r ˆ Canceling the minus sign and the unit vector r, this expression can be solved for the kinetic 1 2 ˆ energy, 2 µv :

1 1 e2 K µv2 . (10) = 2 = 8πϵ0 r

Now the electrical potential energy U of the Bohr atom is10

1 e2 U 2K. = −4πϵ0 r = − Thus the total energy E K U of the atom is = + 1 e2 E K U K 2K K . (11) = + = − = − = −8πϵ0 r Note that the relation between the kinetic, potential, and total energies is in accordance with the virial theorem for an inverse-square force, 1 E 2 U K. Because the kinetic energy must be positive, the total energy E is negative. This= merely= − indicates that the electron and the proton are bound.Toionize the atom (that is, to remove the proton and electron to an infinite separation), an amount of energy of magnitude E (orThe more) Interaction must of Light be added and Matter to the atom. | | Thusto rewrite far the the kineticBohr’s derivation energy, Eq. Model (10). has been completely of Hydrogen classical in nature. Atom At this point, however, we can use Bohr’s quantization1 e2 1 of1 angular(µvr)2 momentum,1 (n!)2 µv2 . Now use Bohr’s quanzaon of angular momentum 2 2 8πϵ0 r = 2The Interaction=The2 µrInteraction of Light= 2 and ofµr Light Matter and Matter L µvr n!, (12) Solving this equationto rewrite for the the kinetic radius energy,r shows Eq. that (10). the= only values= allowed by Bohr’s quan- tization condition areto rewrite the kinetic energy, Eq. (10). 10 To rewrite kinec energy This is found from a derivation anal1ogoeu2 s to1the one1l(µvr)eadin2g to1th(ne!g)2ravitational result. The zero of potential 1 µve22 1 1 (µvr)2 . 1 (n!)2 energy is taken to be zero at r . 4πϵ !2 µv2 2 2 . 8πϵ0 r =2 2 2 = 2 µr = 2 2µr 2 =r∞n n8πϵ ar0n =, 2 = 2 µr = 2 (13)µr = µe2 =0 Solving this equation for the radius r shows that the only values allowed by Bohr’s quan- tizationSolving condition this are equation for the radius r shows that the only values allowed by Bohr’s quan- 11 where a0 5.291772083We have the radius of allowed the electron orbits tization10 condition− m 0 are.0529 nm is known as the . Thus the electron can= orbit at a distance× of a ,4=a ,9a ,...from the proton, but no other separations 0 0 0 !2 are allowed. According to Bohr’s hypothesis, when the4πϵ electron0 2 is in2 one of these orbits, rn n4πϵa!02n , (13) the atom is stable and emits no radiation. = µer 2 =0 n2 a n2, (13) n 2 0 Inserting this expression for r into Eq. ( 11) reveals that= theµe allowed= energies of the Bohr atom are 11 whereInserng this to the expression of total energy a0 5.291772083 10− m 0.0529 nm is known as the Bohr radius. Thus the = × = 11 electronwhere can orbita0 at5 a. distance291772083 of a0,410a0−,9am0,...0from.0529 the nm proton, is known but no as other the Bohr separations radius. Thus the electron can= orbitµe4 at a distance1 × of a ,4=a1 ,9a ,...from the proton, but no other separations are allowed.E According to Bohr’s hypothesis,13.60 eV when0. the0 electron is in(14) one of these orbits, n 2 2!2 2 2 n is the principal the atomare is allowed. stable= − and32 According emitsπ ϵ0 non radiation. to= Bohr’s− hypothesis,n when the electron is in one of these orbits, Insertingthe atom this is expression stable and for emitsr into no Eq. radiation. ( 11) reveals that the allowed energies of the Bohr atom are The integer n, known as theInsertingprincipal this quantum expression number for r ,into completely Eq. ( 11) determines reveals the that char- the allowed energies of the acteristics of each orbitBohr of the atom Bohr are atom. Thus, when the electron is in the lowest orbit (the 4 ground state), with n 1 and r1 a0, its energy isµeE1 113.6 eV. With the1 electron in = = En = − 13.6 eV . (14) the ground state, it would take at least 13.6= eV− to32 ionizeπ 2ϵ2! the2 n atom.µe2 =4 − When1 then electron2 is1 in E 0 13.6 eV . (14) the first , with n 2 and r2 4a0, itsn energy is greater2 2!2 than2 it is in the ground2 = = = −32π ϵ0 n = − n state: E2 13.6/4 eV 3.40 eV. If the electron= −The does integer not=n radiate,− known in as any the ofprincipal its allowed quantum orbits, then number what, is completely the origin of determines the the char- spectral linesacteristics observedThe for of integer hydrogen?each orbitn, known of Bohr the Bohrproposed as the atom.principal that Thus, a photon quantum when is the emitted electron number or is absorbed, completely in the lowest determines orbit (the the char- when an electronground makes state a), transition with n from1 and oner1 orbita0 to, its another. energy Consider is E1 an13 electron.6 eV. With as it the electron in acteristics of each= orbit of= the Bohr atom. Thus,= when− the electron is in the lowest orbit (the “falls” from athe higher ground orbit, state,n , it to would a lower take orbit, at leastn , 13.6 without eV stoppingto ionize at the any atom. intermediate When the electron is in groundhigh state), with n 1 andlow r1 a0, its energy is E1 13.6 eV. With the electron in orbit. (This isthenotfirsta fallthe excited in ground the state classical state,, with it sense;n would=2 the and take electronr2 at least=4a is0,never 13.6its energy eVobserved to is ionize greater between= the than− atom. the it is When in the ground the electron is in state: E 13.6/4 eV =3.40 eV. = two orbits.) The electronthe2 first loses excited energy state#E , withEhighn E2low and, andr this4 energya , its energy is carried is greater away than it is in the ground If the electron= − does not= radiate−= in any− of its allowed2 0 orbits, then what is the origin of the from the atom by a singlestate: photon.E Equation13.6/4 ( eV 14) leads=3.40 to an eV. expression= for the wavelength of the emittedspectral photon, lines observed2 for hydrogen? Bohr proposed that a photon is emitted or absorbed If the electron= − does not= radiate− in any of its allowed orbits, then what is the origin of the when an electron makes a transition from one orbit to another. Consider an electron as it spectral lines observed for hydrogen? Bohr proposed that a photon is emitted or absorbed “falls” from a higher orbit, nhigh, to a lower orbit, nlow, without stopping at any intermediate Ephoton Ehigh Elow orbit. (Thiswhen is annot electrona fall in makes= the classical a− transition sense; from the electron one orbit is never to another.observed Consider between an theelectron as it “falls” from a higher orbit, nhigh, to a lower orbit, nlow, without stopping at any intermediate two orbits.) The electron loses energy #E Ehigh Elow, and this energy is carried away or from theorbit. atom (This by a single is not photon.a fall in Equation the classical= ( 14) sense; leads− to the an electron expression is never for theobserved wavelength between the of the emittedtwo orbits.) photon, The electron loses energy #E Ehigh Elow, and this energy is carried away hcfrom the atomµe4 by a single1 photon.µe Equation4 1= ( 14) leads− to an expression for the wavelength , 2 2!2 2 2 2!2 2 λof= the!− emitted32π ϵ0 photon,nhigh " − −32π ϵ0 nlow E#photon Ehigh E$low = −

Ephoton Ehigh Elow or = −

or hc µe4 1 µe4 1 , 2 2!2 2 2 2!2 2 λ = !−32π ϵ0 nhigh " − −32π ϵ0 nlow hc µe4 #1 µe$4 1 , λ = −32π 2ϵ2!2 n2 − −32π 2ϵ2!2 n2 ! 0 high " # 0 low $

The Interaction of Light and Matter

to rewrite the kineticThe Interaction energy, Eq. of Light (10). and Matter

to rewrite the kinetic energy, Eq.1 (10).e2 1 1 (µvr)2 1 (n!)2 µv2 . 8πϵ r = 2 = 2 µr2 = 2 µr2 1 e2 01 1 (µvr)2 1 (n!)2 µv2 . 2 2 Solving this equation8πϵ0 forr the= 2 radius=r 2showsµr that= the2 µr only values allowed by Bohr’s quan- tization condition are Solving this equation for the radius r shows that the only values allowed by Bohr’s quan- tization condition are 4πϵ !2 r 0 n2 a n2, (13) n 2 0 =!2 µe = 4πϵ0 2 2 rn n a0n , (13) = µe2 = 11 where a0 5.291772083 10 m 0.0529 nm is known as the Bohr radius. Thus the = × − = electron can orbit at a distance11 of a0,4a0,9a0,...from the proton, but no other separations where a0 5.291772083 10 m 0.0529 nm is known as the Bohr radius. Thus the are allowed.= According× to− Bohr’s= hypothesis, when the electron is in one of these orbits, electron can orbit at a distance of a0,4a0,9a0,...from the proton, but no other separations are allowed.the atom According is stable andto Bohr’s emits hypothesis, no radiation. when the electron is in one of these orbits, the atomInserting is stable and this emits expression no radiation. for r into Eq. ( 11) reveals that the allowed energies of the InsertingBohr atom this areexpression for r into Eq. ( 11) reveals that the allowed energies of the Bohr atom are µe4 1 1 En 13.6 eV . (14) =µe−4 32π12ϵ2!2 n2 = − 1 n2 E 0 13.6 eV . (14) n 2 2!2 2 2 = −32π ϵ0 n = − n The integer n, known as the principal quantum number, completely determines the char- The integeracteristicsn, known of each as theorbitprincipal of the Bohr quantum atom. number Thus,The Interaction when, completely the electron of determinesLight is and in the Matter the lowest char- orbit (the acteristicsground of each state orbit), with of then Bohr1 and atom.r1 Thus,a0, itswhen energy the electron is E1 is in13 the.6 lowest eV. With orbit the (the electron in = = = − groundthe state ground), with state,n itto1 would andther nucleus.1 takea0 at, its least Rutherford energy 13.6 is eVE1 coined to ionize13. the6 the eV. term atom. Withproton the When electron theto refer electron in to theis in nucleus of the hydrogen = = = − the groundthe first state, excited it would stateatom take, with ( atZn least1),2 13.6 and 1836 eVr2 to times ionize4a0, its more the energy atom. massive is When greater thanthe than electron the it electron. is is in in the ground But how were these charges = = the firststate: excitedE2 state13, with.6/4n eV 2 and=3r.240 eV.4a0, its energy is greater than it is in the ground = − arranged?= = − = state: E2If the13 electron.6/4 eV does not3.40 radiate eV. in any of its allowed orbits, then what is the origin of the Ifspectral the electron= − lines does observed not= radiate− for inhydrogen? any of its Bohr allowed proposed orbits, then that what a photon is the is origin emitted of the or absorbed spectralwhen lines an observed electronThe for makes hydrogen? Wavelengths a transition Bohr proposed from of Hydrogen one that orbit a photon to another. is emitted Consider or absorbed an electron as it when an electron makes a transition from one orbit to another. Consider an electron as it “falls” from a higherThe orbit, experimentalnhigh, to a lower data orbit, weren abundant.low, without The stopping wavelengths at any intermediate of 14 spectral lines of hydrogen “falls”orbit. from (This a higher is not orbit,a fallnhigh in, to the a lower classical orbit, sense;nlow, without the electron stopping is never at anyobserved intermediate between the had been precisely determined. Those in the visible region of the electromagnetic spectrum orbit.two (This orbits.) is not Thea fall electron in the classical loses energy sense;# theE electronEhigh is Eneverlow, andobserved this energy between is thecarried away Hydrogen spectralThe Interaction lines=of Light −and derived Matter two orbits.)from the The atom electron by aare losessingle 656.3 energy photon. nm#E (red, EquationE Hhighα), ( 486.1E 14)low leads, and nm thisto (turquoise, an energy expression is carried H forβ), the away 434.0 wavelength nm (blue, Hγ ), and 410.2 nm = − fromof the the atom emitted by a single photon,(violet, photon. H Equationδ). In 1885 ( 14) a leads Swiss to an school expression teacher, for the Johann wavelength Balmer (1825–1898), had found, by which gives of theWavelength of the emied photon emitted photon, trial and error, a formula to reproduce the wavelengths of these spectral lines of hydrogen, today called the Balmer series or Balmer lines: Ephoton 1 Ehigh µeElow4 1 1 Ephoton Ehigh= Elow − . (15) = − 3 2!3 2 2 λ = 64π ϵ0 c !nlow − nhigh " or 1 1 1 RH , (7) or λ = 4 − n2 Or Comparing this with Eqs.4 ( 7) and ( 8) reveals4 ! that Eq." ( 15) is just the generalized hc 4 µe 1 4 µe 1 The Interaction of Light andBalmerhc Matter formulaµe for the1 spectral linesµe of hydrogen,1 with, nlow 2 for the Balmer series. whereλ =n −323,π42,ϵ52!,...2 n2, and−R −321π.096775832ϵ2!,2 n2 107 =1.3m 1 is the experimentally de- Insertingλ = − values32π!2ϵ2 into!2 n2 the0 combination−high−"32π#2Hϵ2! of2 n constants2 0 low in$ front of the parentheses− shows that ! =0 high " # 0= low $ 7 × ± which gives Which gives this termtermined is exactly Rydberg the Rydberg constant constant for hydrogen. for hydrogen:Balmer’s formula was very accurate, to within a fraction of a percent. Inserting n 3 gives the wavelength of the Hα Balmer line, n 4 4 4 = 2 = 1 µe gives1 Hβ,1 and so on. Furthermore,µe Balmer realized that since1 2 4, his formula could be . RH (15) 10967758.3m− . = λ = 64π 3ϵ2!3cgeneralizedn2 − n2 to 3 2!3 0 ! low high " = 64π ϵ0 c = Comparing this with Eqs. ( 7) and (This 8) reveals value is that in Eq.perfect ( 15) agreement is just the with generalized the1 experimental1 value1 quoted following Eq. ( 7) Balmer formula for the spectralCompare with linesfor of thehydrogen,Balmer’s hydrogen with empirical expression linesnlow determined2 for the Balmer by Johann series. Balmer,RH and this agreement, illustrates the great (8) = λ = m2 − n2 Inserting values into the combinationsuccess of constants of Bohr’s in front model of the of parentheses the hydrogen shows atom. that 11 ! " this term is exactly the Rydberg constant for hydrogen: with m

11The slightly different Rydberg constant, R , found in many texts assumes an infinitely heavy nucleus. The ∞ reduced mass, µ, in the expression for RH is replaced by the electron mass, me, in R . ∞

The Interaction of Light and Matter

which gives

1 µe4 1 1 . (15) 3 2!3 2 2 λ = 64π ϵ0 c !nlow − nhigh "

Comparing this with Eqs. ( 7) and ( 8) reveals that Eq. ( 15) is just the generalized Balmer formula for the spectral lines of hydrogen, with nlow 2 for the Balmer series. Inserting values into the combination of constants in front of the= parentheses shows that this term is exactly the Rydberg constant for hydrogen:

µe4 The Interaction of Light and Matter 1 R 10967758.3m− . H 3 2!3 = 64π ϵ0 c = which gives

1 Thisµe4 value1 is in1 perfect agreement with the experimental value quoted following Eq. ( 7) The Interaction. of Light and(15) Matter for3 the2!3 hydrogen2 2 lines determined by Johann Balmer, and this agreement illustrates the great λ = 64π ϵ0 c !nlow − nhigh " 11 successwhich gives of Bohr’s model of the hydrogen atom. Comparing this with Eqs. ( 7) and ( 8) reveals that Eq. ( 15) is just the generalized Balmer formula for the spectral lines of hydrogen, with n 2 for the Balmer series. low 4 Inserting values into theExample: combination of constants Hα in frontline of the= parentheses1 showsµe that 1 1 Example 3.1. What is the wavelength of the photon emitted. when an electron(15) makes a this term is exactly the Rydberg constant for hydrogen: λ = 64π 3ϵ2!3c n2 − n2 transition from the n 3tothen 0 2 orbit! low of thehigh Bohr" hydrogen atom? The energy lost 4 • µe 1= = Hα line is produced when the electron in R by the electron10967758 is.3m carried− . away by the photon, so H Comparing3 2!3 this with Eqs. ( 7) and ( 8) reveals that Eq. ( 15) is just the generalized the hydrogen atom makes the transion = 64π ϵ0 c = Balmer formula for the spectral lines of hydrogen, with nlow 2 for the Balmer series. The Interactionfrom the n = 3 to the n = 2 orbit. What is of Light and Matter = This value is in perfect agreement withInserting the experimental values into value theE quotedphoton combination followingEhigh Eq. of ( constantsE 7)low in front of the parentheses shows that for the hydrogen lines determined by Johann Balmer, and this agreement illustrates= the− great which gives the wavelength? this term is exactly the Rydberg constant for hydrogen: success of Bohr’s model of the hydrogen atom.11 hc 1 1 4 13.46 eV 2 13.6 eV 2 1 µe 1 1 λ = −µe n − − 1 n Example 3.1. What is the wavelength of the photon emitted. whenR an electron makes(15) a 10967758high .3m− . low λ = 64π 3ϵ2!3c n2 − n2 H 3 2!3 # $ transition from the n 3tothen 20 orbit! oflow the Bohrhigh hydrogen" atom?= The64π energyϵ0 lostc = by the electron is carried= away by the= photon, so 1 1 Comparing this with Eqs. ( 7) and (This 8) reveals value is that in Eq.perfect ( 15) agreement is just the with13 generalized.6 the eV experimental value. quoted following Eq. ( 7) 2 2 Balmer formula for theE spectralphoton linesEhigh ofE hydrogen,low with nlow 2 for the= Balmer− series. 3 − 2 = for− the hydrogen lines determined= by Johann Balmer,# and this$ agreement illustrates the great Inserting values into the combination of constants in front of the parentheses shows that 11 hc success1 of Bohr’s model1 of the hydrogen atom. this term is exactly the Rydberg constant13Solving.6 eV for hydrogen: for the13 wavelength.6 eV gives λ 656.469 nm in a vacuum. This result is within 0.03% λ = − n2 − − n2 high # low $ = of4 the measured value of the Hα spectral line, as quoted in Example 1.1 and Table 2. µeExample 3.1. What1 is the wavelength of the photon emitted when an electron makes a R 1109677581 .3m− . H 133.62! eVThe3 discrepancy. between the calculated and the observed values is due to the measure- = 64π ϵtransition0 c = 2 from2 the n Comparing to the measured 3tothen 2 orbit of the Bohr hydrogen atom? The energy lost = − 3 − 2 λair = 656.275 nm mentsby the# being electron made$ is carried in= airwavelength in the air: away rather by than the= photon, in vacuum. so Near sea level, the speed of light is slower This value is in perfect agreement with the experimental value quoted following Eq. ( 7) Solving for the wavelength gives λ 656.469 nm in a vacuum. This result is within 0.03% Why? for the hydrogen lines determined by Johannthan= in Balmer, vacuum and this by agreement a factor illustrates of approximately the great 1.000297. Defining the index of refraction of the measured value of the Hα spectral line, as quoted in ExampleE 1.1E and TableE 2. success of Bohr’s model of the hydrogento be atom.n 11 c/v, wherephotonv is thehigh measuredlow speed of light in the medium, nair 1.000297. The discrepancy between the calculated and= the observed values is= due to the− measure- = ments being made in air rather thanGiven in vacuum. that Nearλν sea level,v for the wavehc speed propagation, of light is slower1 and since ν cannot1 be altered in moving from one than in vacuum by a factor of approximately 1.000297.= Defining the index13 of. refraction6 eV 13.6 eV Example 3.1. What is the wavelengthmedium of the to photon another emitted without whenλ = an resulting− electron makes inn2 unphysical a − − discontinuitiesn2 in the electromagnetic transitionto be n fromc/v the, wheren 3tothev is then measured2 orbit speed of the of Bohr light hydrogen in the medium, atom?nair The energy1.000297. losthigh # low $ Given that= λν v for wave propagation,field and of since the lightν cannot wave, be altered the in measured moving= from wavelength one must be proportional to the wave speed. by the electron is carried= away by the= photon, so medium to another= without resulting in unphysical discontinuities in the electromagnetic1 1 Thus λair/λvacuum vair/c 113/n.6air eV. Solving for the. measured wavelength of the Hα line field of the light wave, the measured wavelength must be proportional= = to the wave speed. 2 2 Ephoton Ehigh inE airlow yields = − 3 − 2 Thus λair/λvacuum vair/c= 1/nair−. Solving for the measured wavelength of the Hα line# $ = = in air yields hc 1 1 13.6Solving eV for the13 wavelength.6 eV gives λ 656.469 nm in a vacuum. This result is within 0.03% λ = − of then2 measured− − λ valueair n of2λvacuum the H/nα spectralair= 656 line,.469 as quoted nm/1. in000297 Example656 1.1.275 and nmTable. 2. λair λvacuum/nair 656.469high nm/#1.000297 =656low.$275 nm. = = = = The discrepancy= between the calculated and the observed values is due to the measure-continued 1 1 continued 13.6ments eV being made. in air rather than in vacuum. Near sea level, the speed of light is slower 11 2 2 The slightly different Rydberg= constant,− R11 , found3 in− many2 texts assumes an infinitely heavy nucleus. The ∞The# slightly different$ Rydberg constant, R , found in many texts assumes an infinitely heavy nucleus. The reduced mass, µ, in the expression for RH isthan replaced in by vacuum the electron by mass, am factore, in R . of approximately 1.000297. Defining the index of refraction ∞ ∞ Solving for the wavelength gives λ reduced656to be.469n mass, nmc/v inµ a,, vacuum. in where the expression Thisv is result the for measured isR withinH is replaced 0.03% speed by of the light electron in the mass, medium,me, in Rnair. 1.000297. ∞ of the measured value of the Hα spectral= Given line, that as= quotedλν inv Examplefor wave propagation, 1.1 and Table and 2. since ν cannot be altered in moving= from one The discrepancy between the calculatedmedium and the to another observed= without values is resulting due to the in measure- unphysical discontinuities in the electromagnetic ments being made in air rather than infield vacuum. of the Near light sea level,wave, the the speed measured of light wavelength is slower must be proportional to the wave speed. than in vacuum by a factor of approximately 1.000297. Defining the index of refraction Thus λair/λvacuum vair/c 1/nair. Solving for the measured wavelength of the Hα line to be n c/v, where v is the measured speed of light in= the medium,= nair 1.000297. Given that= λν v for wave propagation,in air and yields since ν cannot be altered in moving= from one medium to another= without resulting in unphysical discontinuities in the electromagnetic λair λvacuum/nair 656.469 nm/1.000297 656.275 nm. field of the light wave, the measured wavelength must be proportional= to the= wave speed. = Thus λair/λvacuum vair/c 1/nair. Solving for the measured wavelength of the Hα line continued = = in air yields 11The slightly different Rydberg constant, R , found in many texts assumes an infinitely heavy nucleus. The ∞ λair λvacuum/nair 656reduced.469 nm mass,/1.µ000297, in the expression656.275 for nmRH. is replaced by the electron mass, me, in R . = = = ∞ continued

11The slightly different Rydberg constant, R , found in many texts assumes an infinitely heavy nucleus. The ∞ reduced mass, µ, in the expression for RH is replaced by the electron mass, me, in R . ∞

Which of the following transions of the hydrogen atom has the shortest wavelength?

A. n = 3 –> n = 2 (Hα) B. n = 4 -> n = 2 (Hβ) C. n = 2 -> n = 1 (Lyman α) D. n = 5 -> n = 3 (Paschen γ) E. n = 4 -> n = 3 (Paschen α) The Interaction of Light and Matter

This result differs from the quoted value by only 0.0009%. The remainder of the discrepancy is due to the fact that the index of refraction is wavelength dependent. The index of refraction also depends on environmental conditions such as temperature, pressure, and humidity.12 Unless otherwise noted, throughout the remainder of this text, wavelengths will be as- sumed to be measured in air (from the ground).

The reverse process may also occur. If a photon has an energy equal to the difference in energy between two orbits (with the electron in the lower orbit), the photon may be absorbed by the atom. The electron uses the photon’s energy to make an upward transition from the lower orbit to the higher orbit. The relation between the photon’s wavelength and the quantum numbers of the two orbits is again given by Eq. (15). After the quantum revolution, the physical processes responsible for Kirchhoff’s laws (discussed in Section 1) finally became clear. • A hot, dense gas or hot solid object produces a continuous spectrum with no dark spectral lines.Hydrogen This is the continuous spectrum spectral of blackbody radiationlines emitted and at any Kirchhoff’s Law temperature above absolute zero and described by the Planck functions Bλ(T ) and Bν (T ). The wavelength λmax at which the Planck function Bλ(T ) obtains its maximum value is given by Wien’s displacement law. 1. A hot, dense gas or hot solid object The Interaction of Light and Matter • A hot, diffuse gas produces bright emission lines. Emission lines are produced when produces a connuous spectrum an electron makes a downward transition from a higher orbit to a lower orbit. The n = • 0 ... energy lost2. byA thehot, diffuse electron is gas produces bright carried away by a single photon. For example, the n = 4 hydrogen Balmerspectral lines ( emission linesemission lines are produced by) electrons “falling” from higher n = 3 orbits down to the n 2 orbit; see Fig. 6(a). –2 ␣␤␥␦ 3. A cool, diffuse= gas in front of a source Paschen (Pa) • A cool, diffuse gas in front of a source of a continuous spectrum produces dark n = 2 of a connuous spectrum produces –4 ␣␤␥␦ absorption linesdark spectral lines ( in the continuous spectrum.absorpon lines Absorption lines) are produced when Balmer (H) an electron makes a transition from a lower orbit to a higher orbit. If an incident photon in thein the connuous spectrum continuous spectrum has exactly the right amount of energy, equal–6 to

–8 H␤ H␤ (eV) Energy –10 Emission line Absorption line

H␣ H␣ n = 1 n = 1 –12 n = 2 n = 2 n = 1 n = 3 n = 3 –14 ␣␤␥␦ Lyman (Ly) n = 4 n = 4 (a) FIGURE(b) 7 diagram for the hydrogen atom showing Lyman, Balmer, and Paschen FIGURE 6 Balmer linesEmission Lines produced by the Bohr hydrogenAbsorpon Lines lines atom. (downward (a) Emission lines. arrows (b) Absorptionindicate emission lines; upward arrow indicates absorption lines). lines.

the difference in energy between a higher orbit and the electron’s initial orbit, the 12See, for example, Lang, Astrophysical Formulae, 1999, page 185 for a fitting formula for n(λ). photon is absorbed by the atom and the electron makes an upward transition to that higher orbit. For example, the hydrogen Balmer absorption lines are produced by atoms absorbing photons that cause electrons to make transitions from the n 2 orbit to higher orbits; see Figs. 6(b) and 7. =

Despite the spectacular successes of Bohr’s model of the hydrogen atom, it is not quite correct. Although angular momentum is quantized, it does not have the values assigned by Bohr.13 Bohr painted a semiclassical picture of the hydrogen atom, a miniature Solar System with an electron circling the proton in a classical circular orbit. In fact, the electron orbits are not circular. They are not even orbits at all, in the classical sense of an electron at a precise location moving with a precise velocity. Instead, on an atomic level, nature is “fuzzy,” with an attendant uncertainty that cannot be avoided. It was fortunate that Bohr’s model, with all of its faults, led to the correct values for the energies of the orbits and to a correct interpretation of the formation of spectral lines. This intuitive, easily imagined model of the atom is what most physicists and astronomers have in mind when they visualize atomic processes.

4 AND WAVE–PARTICLE DUALITY

The last act of the quantum revolution began with the musings of a French prince, Louis de Broglie (1892–1987; see Fig. 8). Wondering about the recently discovered wave–particle duality for light, he posed a profound question: If light (classically thought to be a wave)

13As we will see in the next section, instead of L n!, the actual values of the orbital angular momentum are = L √ℓ(ℓ 1) !, where ℓ, an integer, is a new quantum number. = +

More quantum physics…

Useful for later lectures The Interaction of Light and Matter

FIGURE 8 Louis de Broglie (1892–1987). (Courtesy of AIP Niels Bohr Library.) could exhibit the characteristics of particles, might not particles sometimes manifest the properties of waves? de Broglie’s Wavelength and Frequency In his 1927 Ph.D. thesis, de BroglieWave-parcle extended the duality wave–particle of all dualityparcles to all of nature. Photons carry both energy E and momentum p, and these quantities are related to the frequency ν and wavelength λ of the light wave by Eq. (5): If a light-wave could also act • In his 1927 Ph.D. thesis, de like a parcle, why shouldn’t Broglie’s extended the wave- maer-parcles also act like parcle duality to E all parcles waves? ν The Interaction of Light and Matter (16) • de Broglie wavelength = h h λ . (17) = p

• Everything exhibits wave de Broglie proposed that these equationsproperes! be used to define a frequency and a wavelength for all particles. The de Broglie wavelength and frequency describe not only massless pho- tons but massive electrons, protons, neutrons, atoms, molecules,FIGURE 8 Louis people, de Broglie (1892–1987). planets, (Courtesy of AIP stars, Niels Bohr and Library.) galaxies as well. This seemingly outrageous proposalcould of exhibitmatter the characteristics waves of particles, has might been not particles confirmed sometimes manifest the in countless experiments. Figure 9 shows the interferenceproperties of waves? pattern produced by electrons de Broglie’s Wavelength and Frequency in a double-slit experiment. Just as Thomas Young’s double-slitIn his 1927 Ph.D. thesis, experiment de Broglie extended the established wave–particle duality to the all of nature. Photons carry both energy E and momentum p, and these quantities are related to the wave properties of light, the electron double-slit experimentfrequency ν and canwavelength beλ of explained the light wave by Eq. (5): only by the E 14 wave-like behavior of electrons, with each electron propagating throughν both slits. The (16) = h h wave–particle duality applies to everything in the physical world; everythingλ . exhibits its (17) = p wave properties in its propagation and manifests its particle nature in its interactions. de Broglie proposed that these equations be used to define a frequency and a wavelength for all particles. The de Broglie wavelength and frequency describe not only massless pho- 14 tons but massive electrons, protons, neutrons, atoms, molecules, people, planets, stars, and See Chapter 6 of Feynman (1965) for a fascinating description ofgalaxies the as details well. This seemingly and profound outrageous proposal implications of matter waves has of been the confirmed in countless experiments. Figure 9 shows the interference pattern produced by electrons electron double-slit experiment. in a double-slit experiment. Just as Thomas Young’s double-slit experiment established the wave properties of light, the electron double-slit experiment can be explained only by the wave-like behavior of electrons, with each electron propagating through both slits.14 The wave–particle duality applies to everything in the physical world; everything exhibits its wave properties in its propagation and manifests its particle nature in its interactions.

14See Chapter 6 of Feynman (1965) for a fascinating description of the details and profound implications of the electron double-slit experiment.

Electron double-slit experiment

Electrons exhibit wave-parcle duality

Each electron passes both slit

The Interaction of Light and Matter

FIGURE 9 Interference pattern from an electron double-slit experiment. (Figure from Jönsson, Zeitschrift für Physik, 161, 454, 1961.)

6 1 Example 4.1. Compare the wavelengths of a free electron moving at 3 10 ms− and 1 × a 70-kg man jogging at 3 m s− . For the electron, h h λ 0.242 nm, = p = mev = which is about theExample: size of an atom Wavelength and much shorter of thana jogger the wavelength of visible light. Electron microscopes utilize electrons with wavelengths one million times shorter than visible wavelengths• What is the wavelength of a 70-kg man jogging at 3 to obtain a much higher resolution than is possible with optical micro- scopes. m/s? The wavelength of the jogging man is

h h 36 λ 3.16 10− m, = p = mmanv = × which is completely negligible on the scale of the everyday world, and even on atomic or nuclear scales. Thus the jogging gentleman need not worry about diffracting when returning home through his doorway!

Just what are the waves that are involved in the wave–particle duality of nature? In a double-slit experiment, each photon or electron must pass through both slits, since the interference pattern is produced by the constructive and destructive interference of the two waves. Thus the wave cannot convey information about where the photon or electron is, but only about where it may be. The wave is one of probability, and its amplitude is denoted by the Greek letter " (psi). The square of the wave amplitude, " 2, at a certain location describes the probability of finding the photon or electron at that location.| | In the double-slit experiment, photons or electrons are never found where the waves from slits 1 and 2 have 2 destructively interfered—that is, where "1 "2 0. | + | =

The Interaction of Light and Matter

Heisenberg’s Uncertainty Principle The wave attributes of matter lead to some unexpected conclusions of paramount importance for the science of astronomy. For example, consider Fig. 10(a). The probability wave, !, is a sine wave, with a precise wavelength λ. Thus the momentum p h/λ of the particle described by this wave is known exactly. However, because ! 2 =consists of a numberThe of Interaction equally high of Light peaks and extending Matter out to x , the particle’s location| | is perfectly uncertain. The particle’s position can be narrowed=± down∞ if several sine waves with different Heisenberg’s Uncertaintywavelengths Principle are added together, so they destructively interfere with one another nearly The wave attributeseverywhere. of matter Figure lead to some 10(b) unexpected shows the conclusions resulting of combination paramount importance of waves, !, is approximately for the sciencezero of astronomy. everywhere For except example, at one consider location. Fig. Now 10(a). the The particle’s probability position wave, may be determined !, is a sine wave,with with a greater a precise certainty wavelength becauseλ. Thus! 2 is the large momentum only forp ah/ narrowλ of the range of values of x. particle describedHowever, by this the wave value is known of the exactly.particle’s| However, momentum| because has become! 2 =consists more of uncertain a because ! is | | number of equallynow high a combination peaks extending of waves out to x of various, the wavelengths. particle’s location This is is perfectly nature’s intrinsic trade-off: =±∞ uncertain. The particle’sThe uncertainty position in can a beparticle’s narrowed position, down if# severalx, and sine the waves uncertainty with different in its momentum, #p, are wavelengthsHeisenberg’s areinversely added together, related. As so theyone Uncertainty decreases, destructively the interfere other must with Principle increase. one another This nearly fundamental inability of a everywhere. Figureparticle 10(b) to simultaneously shows the resultinghave combination a well-defined of waves, position!, is and approximately a well-defined momentum is a zero everywhere except at one location. Now the particle’s position may be determined direct result of the wave–particle duality of nature. A German physicist, Werner Heisenberg with a greater certainty because ! 2 is large only for a narrow range of values of x. (1901–1976), placed this inherent “fuzziness” of the physical world in a firm theoretical However, the value of the particle’s| momentum| has become more uncertain because ! is now a combinationframework. of waves He of demonstrated various wavelengths. that the This uncertainty is nature’s in a intrinsic particle’s trade-off: position multiplied by the ! The uncertaintyuncertainty in a particle’s in itsposition, momentum#x, and must the uncertaintybe at least as in its large momentum, as /2: #p, are inversely related. As one decreases, the other must increase. This fundamental inability of a particle to simultaneously have a well-defined position and a well-defined1 momentum is a #x #p !. (18) direct result of the wave–particle duality of nature. A German physicist,≥ 2 Werner Heisenberg (1901–1976), placed this inherent “fuzziness” of the physical world in a firm theoretical framework. HeToday demonstrated this is known that the as uncertaintyHeisenberg’s in a particle’s uncertainty position principle multiplied. The by equality the is rarely realized uncertainty in its momentum must be at least as large as !/2: in nature, and the form often employed for making estimates is 1 #x #p !. #x #p !. (18) (19) ≥ 2 ≈

Today this is knownA similar as Heisenberg’s statement relates uncertainty the uncertainty principle. The of an equality energy is rarely measurement, realized #E, and the time in nature, and theinterval, form often#t, over employed which for the making energy estimates measurement is is taken:

Forms oen used: #x #p !. #E #t !. (19) (20) ≈ ≈ A similar statement relates the uncertainty of an energy measurement, #E, and the time interval, #t, over which the energy measurement is taken:

#E #t !. (20) ≈

(a) (b) FIGURE 10 Two examples of a probability wave, !: (a) a single sine wave and (b) a pulse composed of many sine waves.

(a) (b) FIGURE 10 Two examples of a probability wave, !: (a) a single sine wave and (b) a pulse composed of many sine waves.

More quantum numbers • Heisenberg’s uncertainty principle implies that electrons cannot have well-defined circular orbits and angular momentum at the same me Bohr’s model must be updated! • Electron orbits are “fuzzy” • More than one quantum number • Principle quantum number n: n = 1, 2, 3 … Ø Energy and Size of the orbital • Angular momentum quantum number l: l = 0, …, n-1 Ø Shape of the orbital • Magnec quantum number ml: ml= -l, …, 0, … +l Ø Orientaon of the orbital

• ms: ms = + ½ or - ½ Ø Orientaon of the spin axis of the electron Atomic Orbitals: Probability Map l = 0 l = 1 l = 2 l = 3 l = 4

n = 1 Comparing with experimental results

n = 2

n = 3

n = 4 The Interaction of Light and Matter

Field off Field on m, = +1

2p DE m, = 0

Zeeman Effect DE m, = –1 • Energy states correspond to the magnec quantum number ml (which describes the orientaon of the electron’s moon) are separated under the presence of external magnec field 1s FIGURE 13 Splitting of absorption lines by the Zeeman effect.

However, the atom’s surroundings may single out one spatial direction as being different from another. For example, an electron in an atom will feel the effect of an external magnetic field. The magnitude of this effect will depend on the 2ℓ 1 possible orientations of the + electron’s motion, as given by mℓ, and the magnetic field strength, B, where the units of B are teslas (T).15 As the electron moves through the magnetic field, the normally degenerate orbitals acquire slightly different energies. Electrons making a transition between these formerly degenerate orbitals will thus produce spectral lines with slightly different frequencies. The splitting of spectral lines in a weak magnetic field is called the Zeeman effect and is shown in Fig. 13. The three frequencies of the split lines in the simplest case (called the normal Zeeman effect) are

eB ν ν0 and ν0 , (22) = ± 4πµ

where ν0 is the frequency of the spectral line in the absence of a magnetic field and µ is the reduced mass. Although the energy levels are split into 2ℓ 1 components, electron transi- tions involving these levels produce just three spectral lines+ with different polarizations.16 Viewed from different directions, it may happen that not all three lines will be visible. For example, when looking parallel to the magnetic field (as when looking down on a sunspot), the unshifted line of frequency ν0 is absent. Thus the Zeeman effect gives astronomers a probe of the magnetic fields observed around sunspots and on other stars. Even if the splitting of the spectral line is too small to be directly detected, the different polarizations across the closely spaced components can still be measured and the magnetic field strength deduced.

Example 4.3. Interstellar clouds may contain very weak magnetic fields, as small as 10 B 2 10− T. Nevertheless, astronomers have been able to measure this magnetic field.≈ Using× radio telescopes, they detect the variation in polarization that occurs across the

15 4 Another commonly used unit of magnetic field strength is gauss, where 1 G 10− T. Earth’s magnetic field is

5 =

roughly 0.5 G, or 5 10− T. , ×, 16See the S ection, The Complex Spectra of Atoms,, , concerning selection rules.

Zeeman Effect and the Sun’s magnec field TheThe ComplexInteraction of Light and Spectra Matter of Atoms

1S 1P 1D 1F 3S 3P 3D 3F 24 n = 4 4 n = 3 33 22 n = 2 n = 2 n = 2 20 n = 2 18 16 14 12

Energy (eV) Energy 10 8 6 4 2 0 n = 1 FIGURE 14 Some of the electronic energy levels of the helium atom. A small number of possible Some of the electronic energy levels of the helium atom and possible allowed transions allowed transitions are also indicated. (Data courtesy of the National Institute of Standards and Technology.) of stars. (It is beyond the scope of this text to discuss the detailed physics that underlies the existence of selection rules.) The revolution in physics started by Max Planck culminated in the quantum atom and gave astronomers their most powerful tool: a theory that would enable them to analyze the spectral lines observed for stars, galaxies, and nebulae.21 Different atoms, and combina- tions of atoms in molecules, have orbitals of distinctly different energies; thus they can be identified by their spectral line “fingerprints.” The specific spectral lines produced by an atom or molecule depend on which orbitals are occupied by electrons. This, in turn, depends on its surroundings: the temperature, density, and pressure of its environment. These and other factors, such as the strength of a surrounding magnetic field, may be determined by a careful examination of spectral lines.

SUGGESTED READING

General Feynman, Richard, The Character of Physical Law, The M.I.T. Press, Cambridge, MA, 1965. French,A. P.,and Kennedy, P.J. (eds.), Niels Bohr: A Centenary Volume, Harvard University Press, Cambridge, MA, 1985. Hey, Tony, and Walters, Patrick, The New Quantum Universe, Cambridge University Press, Cambridge, 2003.

21Nearly all of the physicists mentioned in this chapter won the Nobel Prize for physics or in recognition of their work.