Intro to Modern Physics
Intro to Modern Physics PHYS 3141
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Lecture 10 More about the particle nature of light Last class The photoelectric effect Energy transfer from photons to electrons is particlelike Each photon carries energy Ehf Today The Com ton ex eriment Momentum Transfer from photons to electrons is particlelike Provides more evidence of particlelike nature of EM radiation Photons undergo particlelike collisions with electrons The EIecTr oMagne ric Spec rr um E Xr ays EM radia rion of k 102 10nm Discovered from Ca rhode ray Tube 1 mn rn Compfon effecf shows parficle nafur e of xrays and yr ays The Compton Experiment 5 attend P 311am u lundent XRay Graphite Shine xray beam onto graphite block Electrons are knocked out by the xray Wavelength of scattered radiation changes 9mmquot at resl I q 639 X pnown l A 2 AA h 1 cos9 mac lf Pho ron scaTTer39s off elec rr on like a par l icle Apply rules of momentum and energy conservation To find shif l in wavelength 1 5hif r is small Comp l39on wavelength m C That39s why its only no l39iceable in Xrays and ymys 000243 rim Ligh r carries momentum 393 39 Fr39om r ela rivi ry r13 Fr39om Planck and Eins rein Because 1 U Derivafion of Compfon39s Equafion ConservaTion of momen rum F 3 F rearranged Magnifude p 23914 9 17gt E 51139 l le Z PLCaSQ 1 Conserva rion of Energx Be vra Louis 1m Enerm 5 Electra1 mot Emma o PinoJCDM 1 RC 51 HEPr 0 L215 Can p m E a Lectrm mC Jl MFA E 0 Wth 3 FC Cowservn39tion requires mc 2 mc 1ltFzCL FLO Rearrange m 51 R quot91 c MDCLV P M 6 L Square bofh sides malv mm c 1 macme WWCl ALY 2 Solveforpz quot7392 F11FLLZFPL ImpLC FL Se r equa rions 1 and 2 equal To eacho rher39 21 31 PJ39 ZPIPLCDSQ I 1 Jet 1 Fzquot Zpp lmaLC f1 2 ff Z c gt59 a Xf lgfz 1mc Pr VW CJL P2gt PJ LCILose mcL I FL a310CGSS F i AL A152040s9 r x L 1059 me Examgle 3 9 IncidenT wavelengTh is shifTed by 15 when 9120 a WhaT is The incidenT wavelen Th b WhaT is The scaTTered wavelengTh when 990 r x L 1059 VLC Problem 3 37 990 wavelength shift is 1 wha r is The incident wavelength gtr gt 5 L 0 5959 ma onclusion 0 Cha far 3 C Some Things To ponder over until we yet To Chapter 5 EM rodio rion has bo rh por ricle and wovelike proper ries How can bo rh be reconciled Who r is The significance of Planck39s cons ron r h Who r would The world be like if h lJs Do The problem 52139 NOW Nex r class review and problem solving Lecture 18 The Schrodinqer Equation Wave equa rion Thaf governs The propaga rion of ma r rer waves Derived in parT by in rui rion There39s no derivaTion like New ron39s second law Fzma accep rance based on correcT predic rions Non rela rivis ric Dirac derived rela rivis lic version laTer We39ll only do The one dimensional case for now aside no re39 an e uivalen r me rhod ma rrix mechanics was worked ou r by Heisenberg The Schrodinger Equation Analogous to classical wave equation 6221 2 6221 2 C 2 0t 6x where U is the amplitude of the wave and c is its speed Dispersion relation Relation between Frequency 0 and Wavenumber k Relation between Energy E and Momentum p For light in vacuum the dispersion relations are a ck Epc What is the dispersion relationship for a particle Dispersion relationship for a particle Total nonrelativistic energy 2 Ezp V 2m E kinetic potential Using DeBroglie39s relations EM and phk iisz 2m V ha The Schrodinger equation gives us the correct dispersion relationship for a particle The chr6dinqer39 Equation aZTOCJ 8Txt at 2 a 2 VxtTxt 171 m x S rar r wi rh a simple case A free par Ticle A free par ricle means There39s no ne r for39ce Hence Vx r 2V0 2 cons rcm r Equa rion is sa risfied by a harmonic wave LPxt Adm a This is The simplesT soluTion for This case Aside Brief review of complex numbers Now we demonsTr39aTe ThaT The Schr b dinger equaTion gives us The cor39r39ecT dispersion r39elaTion Tx t Adam a 6211 ik2Aeikx cot k2 P 68 1 iwAeKOCWI ia P 2 X Plug These inTo The Schr b dinger equaTion E 82Wxt 6 I xt th th zh 2m 6x2 t 2 kZ P V011 z h z39co P m i2sz 2m V02ha Cor39r39ecT dispersion r39elaTion for par39Ticles eparation of Time and Space Dependencies of LI xt h262 Pxt OLPxt 2m 6x2 6 If potential energy Vxt doesn39t change with time if the potential is only a function of x Vx then the time and space dependence of P can be separated Vx t Px t 2 171 In this case we onlv need to solve for the xde endent I art and we use Time Independent chr6dinqer Equation me W E W This is the equation you will need to solve for most problems Solving for The Schrodinger equaTion allows us To find The wavefuncTion w The wavefuncTion is The full IuanTum mechanical descri Tion of The s sTem For example squaring The wavefuncTion gives us The probabiliTy funcTion We will solve The Schrodinger equaTion for an infiniTe square well a firsT order approximaTion To describe bound parTicles sTep poTenTials scaTTering experimenTs demonsTraTes quanTum mechanical feaTures ref lecTion aT inTerfaces Tunneling barrier poTenTials Tunneling Condition on acceptable solution for x 1 VX and mus r be con rinuous X 2 x and mus r be fini re X 3 WXand mus r be singlevalued X 4 yx gt O as x gt i 00 for normalization A note about probabilities Recall The square of the wavefunction I PI2 gives the probability distribution 1 referred to as probability amplitude not directly measurable The probability of finding a particle between x and xdx at time t is Pxtdx T m Txtdx L1Jxzlzdx Normalization Condition LP x t LPOC 1 1 This condition determines constants and restricts possible solutions For timeindependent case you only need to normalize Vx 1 X31 POCJ W Delah Mick quotM W X W06 Expectation Values Expec ra rion value Averge value of any measureable quan ri ry For past faw z 5 W Cz 2 gt964 91 on In Senirnli lt ODgt g VHCJO 45quot 7 mix 394 The infinite sguare well vac 0 er 0 lt x lt L Viz DO Par 5 lt o AWL xgt L Pwtiele LaAFCKegk m Oltgtclt L Exac r solu rion To Schrodinger equaTion can be obTained wi rhou r difficulT maThemaTics Illus rra res impor ran r feaTures of quanTum mechanical problems and proper ries of The wavefunc rion Reasonable approxima rion To some real si rua rion for example The mo rion of a free elecTron in a me ral Consider The confined particle as a STANDING WAVE r L i ii m 1 We already used DeBroglie39s relations and The standing 7A2 wave condition To find The particle39s energy and To Show That if is quanfized 0 if A34 Now derive same result and more by solving The Schr39b39dinger39 equa rion V9 V00 3 0 Cor OltxltL Vzod gar DCltO Mot vcgtL 4 0 20 Pwtide can FEaeok in Oltxlt L o L K 2 2 h Liv6 V39x x E x 2m dxz v w 2 22 d2 x For0ltxltLIVXO j 2m dx2 E W05 2 Z 2 212 m ZmE d2 x Let A722 gt kZVx 012 H W General Solutions Mac 2 A sin kx or Mac 2 A coskx Verify 7yx A sin kx Similar for 05 solu rion M Ak coskx dx 2 k2A sin kx k2wx wx A sin kx or wx A cos kx Boundary condi rions require Tha r wxO a r xO Therefore QUOC A sin kx because cos 0 1 Boundary condi rions require Tha r VxO a r xL Therefore k2 Wheren123 No re Tha r k is The wavenumber k 277 Where x1 is the DeBroglie wavelength Now find allowed energies Enemy Eigenvalues 2mE hZ Recall k2 Also k 2 from boundary conditions 2mE 11272392 h2 L2 Se r These equal To eacho rher39 Allowed energies for39 a par ricle in a box 2 2 2 n h 7 n 2mL2 Now find The r39esT of The wavefuncTion So far39 we have n x L yx A sin ConsTanT A is deTermined by The normalizaTion condiTion H 2121 5L HA7 jmeHTT 3 0L1 L a g quotat jv jfvs x a w Salve gt f7 3 L HXJ ElgerrfuncTIons are Yuan L 1 n2 n3 Comparison wifh c assica Newfonianl resu fs Classically Parficle can have any energy including zero Parficle has equal probabilify of being found anywhere in The box P1L Correspondence principle For large n Iuanfum sfarfs To look like classical For large n Energy difference between adjacent states is a smaller fraction of The Total energy Sfarfs To look con finuous like classical case Probabilify dis39fribufion sfarfs To look classical For have m ascll mhs st 0 take awemg z l l l7ltlL 14 L L Example 62 WhaT is The ground sTaTe energy of an elecTr39on in a 1 cm long wir39e WhaT if The wire was 01nm 222 11117 n 2mL2 Example 63 WhaT is The probabiliTy of finding The par39Ticle in a region beTween x0 and xL4 Par39Ticle is in The ground sTaTe Pmdx w x wxdx lwx2dx mm L 2 x Sll 1 M L Useful integral 3 53quotqu 39quot L 75 1 11515 C955 quot We Dilat iw mavfns dock run Slaw At 2 7 whyr t LengtL Contracti am moving 9556565 M5 skur tw alums tne Airactim 9 ww tfaw E proper time L53 7 where LPE grapex length WM lyxvmt mt measLres amiab tj ASquot Atquot AXL39r A51 4 A21 The Doppler Effec r f H gram 2 Wave increases frequency decreases wavelenq rh when approaching Wave decreases frequency increases wavelength when receding hffpwwwyoufubecomwafchvqWquevDussampNR1 txamples39 hffpwwwyoufubecomwafchvimochnZSqoampNR1 Click here if you wanf This explained again Classical descrip rion of The Doppler effect does noT work for light in ciassfcal39 Dagger e ct Frequency detecfezi an obSamcr Aepmds on whether Sou 06 is min3 13 or gt195va is moving maxwa goyrcc moving OBSEVEK Q 2 A Q 4 t t g 4 4 v WorKs gar wan 5 39Craveltinj rm3L a medium but M r Light Jijht has no medium cm39t maKe 0L distinction setV05 moving Source A MDVEMj observer Kekqtivist39ic D er Eerect I 7 i gt 5 A 5 in fast TW 0 source marMa S Fregqucy is f Mi WMCIEAg A is X What frequencnj aLo obsexvors P moL B 5563 QE macaw momx a Ems moc nm n w guru 9mm mf n cmlan be 3 39wuns urim mxampmiltkxho Z 595mm 9 mggmew d 53059an ya u n 01H 2 mmovv Zignu onocxu anxwlL Em vxom nxq Rmo N can amp Swim SmmWOr r3 02ng ha momu q Rxnna Z igm a lnmqp NM 9 HANG zo m x W arm rm mammow Lava I gmm PR gm wmoh 195 rm maxim mam w oj From the pvtVt 0 eru 0 ab crxcr H Source 5 f fv cl fmg emitted in N WINES acctfr a atZS39tamce N wowas we t i39me At 4 375 5 i9 m JQ Ax J39 ADC CA39t VA x XL 1 wave emitteA lms tramdad ta positfm 12 cAt N mm was Wittevi at pas th Xx VAt WM nat1 ACCural Lns to PA A 55 c VAt N N CAtVKx Vc an ns q Annnonlxymw do mg x u Int an T g R A it nmnypknhwmmcms q Pnn 3963 n mmq nm n R 1 v m9 939 influmnif Em Fang bduw x u 1n TEE H5 Kim 545 RH mm 40 b zli b 7 ECG W S 2100 9 2m mama s hu 49 13925 n bu h ltn ln yum L yn ye Tltn I ltv EmmNQSOYWSQ Ser I rung r01 M nhltnsnu mymNmew Kin Twiadr l a cwmrmh ax Doms mx l Mmu v mm dnmumaxd hH 931 y H V Tn rlt ltr N0ltR 3th ms u 33rd gto m rmiqnwr Iv mkmrm w Relativity Part 2 Chapter 2 Par r 1 Rela rivis ric Kinema rics Description of mo rion in r ela rivi ry posi rion Time veloci ry Par r 2 Rela rivis ric Dynamics Description of forces in r ela rivi ry energy momen rum Why do we need To r39edefine momen rum and energy ClaggfcaHgJ E lmj M3 zit Prahlem Lamstamt Force E gt constwt Lislerattan E gt 31 no ii 2i reLaLH 3390 pastulwtc no th fnj taxval ratev UAW c u39n need to mdc39 fne mamin39tum Wefmf v tf auy Ano rher reason we need To redefine momentum and energy To preserve The conservafion laws Conserva rion of energy Embm tan 6 39 1 Q A onserva riono momen rum pmtm Fm To show Tha r The classical form of momenTum is no r conserved in rela rivis ric siTua rions consider 5 5144 7 7 Vi u 5w 0 39 VL v in M m39 w G Md SC ELK togathzr upon cofifsim midi mew O In gramTia 5 me MM muz mevj mL v O Pam ZMDV O 39 VJ 0 BC re 39 ingrazme S mm5nd tspeed v I V390 1 k v AQEer e use Dram trWS V mmtt m my 11 v 1 V VCL ampv ZV t NJL 2 v14 VH Pine H MV mlvl 1mv Firm 2M V I VVLL dac Ht Wark 2M Neeak mtaivzs c wpressfaws 9w mnmrm vm mat maxg g SEnLC ER PL 1 Relativistic Momen rum mus r be conserved in all reference frames 539 m L7 70 9mi YOU W K V where Rivelae39f g 039 o j ec t rekadr ive to observw r edvce5 to rm1 pr ultltc Liqzjf CGJ vaLie NewtoyI5 1nd Law gt 51 air 4 X037quot a f At 66 J42 Yin 00 Few reqv5res more ar ce ta acaeiexo fe aw o ject as if gets aster a In fnite Maratj Pequi39feal to reqc h 39 Relativistic Energx Kinet fc EnergLTsf worK aim1 accelera h mg am a j ect rm UL0 ta speed u EKzgk id EEJ dp LLOLP A9 diYm Ym m v wwde a y 11 an 41 amp Wow m G dp7WWULL1X3J ow L a Simplf j de 0H0 ImrL Xz oLM Xm1y21M a noamp1 C 1 741 wal quot78 30W 1 7LL 71m W obi l VLL YXMM quot mu EK 8 up uLf ddu SK lwWZyL M 7 M EK m WW1quot mc 7544 mc VL a dad n e gt Relativist c Kinetic Ema 3 3 Note gar ULltlt c X I 21 I binamtal WPMSZOV ZCL I ny ts 1 H Z YVLLquotVVLL 2339 Lm ud olafsicql 150 Lil also ND CC39 as w gtc2 990 EEK gt00 secad rst mumquot szwdcnt D695 2 total anergg gt Ev Ex rest energy E L i E EKmC1 ch mo wkcn algkzpl fi gt Em1 quotrest Cnerg 1 Significance Energy and mass are equivalent Energy can be conver red in ro mass and vice versa LasT class NewTonian reaTiviTy breaks down wiTh Maxwell39s equaTions EampM LighT musT have a 39preferred39 frame of reference LighT propagaTes Through 39eTher39 Frame of eTher is preferred frame How can we deTecT This 39eTher39 Recall MosT experimenTs noT Sufficienle sensiTive SoluTion InTerferomeTry MichelsonMorley experimenT Now MichelsonMorley Experimem Circa 1887 used an inTerferomeTer To deTecT eTher inTerferomeTer39s sensiTiviTy is on The order of wavelengTh of lighT 9 reIaTive phase shifT of M2 Takes you from consTrucTive To desTrucTive Review Interference Wikipedia definition quotThe addi rion of Two or more waves Tha r resul rs in a new wave pa ernquot Constructive Destructive A 39 A 39 Result of addition v v v v i 9 3 3 r f f x x J v f V n Wme5 Mg in phase waves we wt u Finest Quick digression Concepfual layou r of The experiment See example 11 Boa r race Each oarsmcm rows of speed c in a river wi rh curren r of speed v Boa rs will no r arrive back of poin r A of The same rime Boa r 1 arrives firs r by a Time difference of AT k L pi Ground The MichelsonMorley Interferometer a M2 v b 1 Fringe width F 6 Rotation L Beam splitter Compensator M2 l quot H gt02 A 0 Sodium light source diffuse quot 39 O hTTpwwwyouTubecomwaTchv8QUthaxWaoampfeaTure relaTed The MichelsonMorley Experiment l l in l B M1 8 U 5 O lt h 9 U D Rotation E l E q 8 Beam e splitter Compensator r 5 M2 EEtg l D w Ovegt l A 0 Sodium 39 l light source lt gtl diffuse L g 3 A 34 5 3 1 390 quot O A 0 335 W i M U E Expected motion of the earth through the ether to introduce delay At just like Boat example httpwwwyoutubecomwatchvZ8 K3quQiqkampfeature related When you rotate interferometer 90 At switches sign Expect to see shift in fringes Resulf NO SHIFT WAS FOUND ImplicaTions No eTher39 Speed of lighT is invar39ianT RelaTiviT does noT a I I To Einstein39s Pos rula res 1 The laws of Physics are The same in all i ner rial frames all laws 2 The speed of Iigh l in vacuum is ALWAYS c independen r of source39s mo rion 3 No rhing rr39avels fas rer39 rhan c speed of ligh r Na rur e39s speed limi r No 39speciul39 inertial frame gt no ether All observers measure 6 as speed of light in vacuum V Oi yav A These Lures Obieleis aLU v 5 measure a expect 39nLIII39 result for MichelsonMorley experiment Now eT39s sTarT geTTing inTo The Theory Definitions EvenT SomeThing ThaT happens aT a specific Time amp place example a flash of lighT Does noT 39belong39 To a reference frame Observer 39WiTness39 To an evenT Does belong To a reference frame EvenTs are described wiTh spaceTime coordinaTes x y z TinSframe x39y39z39T39inS39 frame Consequence of Einstein39s Pos rula res cocks inside a given frame are 5 nchronized buf nof befween frames Two spafially separafed evenfs fhaf are simulfaneous in one frame are NOT simulfaneous in any o rher inerfial frame moving rela rive To The firs r See Figure 115 5ee Video hffpwwwyoufubecomwafchvwfeiuxyqfoMampfeafurereafed The Lorentz Transformation Frame 539 is moving of speed v along x r39ela rive To 5 Transform even r x y z T in 5 info x39 y39 z39 T39 in 539 S CM C with ass umtng xKLLt d 23 mver c39 n 1 22 ac Ml v t39 gt Salve DV Time trmj39e wmmtlovx Substitute eqm CD gr x into 7V7 ar X x D B Lgt vtvn t 3 t 2th T if 3 umorch YBquot 41 12 YL v Mom9 Vim is X Set up thought experimeth 5 Fqsl a fight Wk Le OV quV d ay15 s knot 5 WC t O a Kaiat mpaxnds as a Spherical wowC in botk rw s In S xb5 cltl 9 IRS 35 3 7E L I CL Substitute cans amok 39Ln tp sum 9 y Krvt1 5 21 L Wt 1 71 1 2 CSRDVLOL reduce 39to 57V 5 Salvf or r New stmplf j 1qu C3 Q Br 3 MA t 22 X t 3C lt L JG 13v7 Summary of Lorentz Transformations avors x Zfo v15 55 2124 vt 5 g 3 3 2 72 t aLv V Ca me Wyn where vexeluci bd along x yl Mat tlwn39t Dr vltltC J 2ft P GalELEM trms rmathns Relativistic Velocity Transformations Observer in frame 5 sees an ob39ect movin at speed If What does observer in 539 see 74quot Cuba Ltd Jute WagJ L3 LL Si Si Galilean trans orms cannot be used because 5 eed u39 can never exceed c Dan in I mil 3 0H fecal x ll Vt dd it t1 XHZ Vza oLxJ Yltdxvdt 5er 3 oLtZ X dtvdgtcCL Ae 0L my 0Lx Hat voct ow 2mm mixL MAt v i r 17 quotbfax my u1 v L VinaCa Invwse use H x l waxL1 Weortm t Note oLt th Lai A LLJ af m3 m4 3 3 v dt 3 My J L natc atqamdewce n on Ma 41 ML ow u HMa Aga n For vltlt C gt GaLileam tramswcorms Mx xx v N ULLKV Exam Ie 1 4 from Tex r Rela rive Speed of Cosmic Rays S gt S 6 Sm 06 6 z 0 5 quotm 5 Wha r is The speed of 2 rela rive To 1 SI 0 5 495 5 ZC ins 71 2 0EC in 5 Wha r is The speed of 1 relaTive To 2 Lecture 13 Bohr s model of the atom 2 Orbital radius rn 217 an 2 00529 m Z i rtmi n m r n is state of the atom 2 Energy En E0 Z2 E0 1366V n can only be an mteger n A potential Energy Transitions between energy levels Light is emitted or absorbed when the electron makes a transition between the allowed energy levels 232 abswbs C 11 a quot1 WM WW 51 5 T MM A 04 era a v nl o la nl Energy of the emitted or absorbed light 1s equal to energy dmerence 1n levels Transitions between energv levels Wavelenvth of radiated A hoton is determined bi which transition it made Depends on nal and initial states of electron hc Ephoton hfzszi Ef EfEi 2 2 EfEiEOZ ZEOZ 2EOZ2 i2 i2 nf nl nf n l E rv n leoZz 1 1 M l he nf2 2 H M09 Example 46 Compute the wavelength of the light emitted in the transition from ni4 to nf2 1n Hydrogen i E022 1 1 J xi hc n2 n2 f i Ionization Energy Binding Energy Energy required to remove electron from atom For Hydrogen in ground state nl E1l36eV This means it takes l36eV to remove electron from atom recall electron is free when E O and r1 gtoo For Hydrogen in 114 state E2J rc v It takes 34eV to remove electron from atom Rydberg Atoms Giant Atoms E Atoms with large n Very unstable Largest made so far n600 Rydberg Atoms Giant Atoms Why are they unstable At room temperature energy of thermal motion kBT z 0025eV This means that a collision with a room temperature atom imparts 0025eV of energy Example Suppose we have a Hydrogen atom with n600 E600 2 136eV 1 2 38gtlt10 5 600 kBTgtgtE600 So atom is easily ionized by random collisions with molecules in air Problem 425 a What is the diameter of Hydrogen atom in n600 b What is the speed of the electron in that orbit c What is the speed in the nl orbit b Angular Momentum L nh for an electron in an atom L 2 mW for circular motion in general Reduced Mass Correction So far we have assumed that nucleus is xed and that only the electron orbits This is equivalent to assuming that the mass of the nucleus M is in nite note not entirely a bad assumption since m z 1000 x m proton electron To account for the motion of the nucleus we need to consider the total kinetic energy For a nucleus of mass M and an electron of mass m total kinetic energy 2 2 2 mM k p p mMp2 p Where u 2M 21 2mM 21 m M nucleus electron reduced mass To make the reduced mass correction replace m by u mee4 16264 E0 2 gt E0 2 271 271 Reduced mass correction is small For Hydrogen u909X103931kg While m911x103931kg correction is 02 difference Correspondence Principle In the classical limit classical and quantum calculations should agree note classical limit usually means high energies At high 11 energy levels are very closely spaced gt classical continuous In classical physics we expect that the frequency of radiation should be equivalent to the frequency of revolution of the electron We can verify this for large n in Bohr s model At large n frequency of emitted radiation szkze4 l l szkze4 Zn l hf 2 2 2 2 2 2 Ln l n J 2 in 11 1 J f szkze4 Zn l W szkze4 szkze4 472713 n2n l2j 472713 n3 27Ih3n3 Classical frequency of revolution h hZ 2 frevzi gt use vzn and r 2 2727 ke Z f nqr nh nh szkze Critique of Bohr TheorV Old Quantum Mechanics 17 1 ant mm hani quota amixt r f laial lhyi ith a h quantum assumptions Assumptions lacked foundations Replaced by a more complete theory DeBroglie Schrodinger Heisenberg Pauli Dirac etc Old quantum mechanics was incomplete Example Bohr predicted Hydrogen atom transitions but not rate of transitions Bohr picture of electrons orbiting nucleus is incorrect Problem set is posted online De nitions you need to do textbook problems Lyman series all transitions from n to nl Lyman 0c is n2 to nl Lyman B is n3 to nl Balmer series all transitions from n to n2 quotquotMI I h39lnnl39il a K r39 j K f rquot Equot I LJELL If quota Emii 39 u 39 g 1 Lyman Series I A I f r39 39N e x N ff all 3 f r 39 x P 139 I I39 If 5 39 l f I A Hi i I i V w x r 9 5 1 i 1 l l I a l 5 l x a W I E l l uuFd state B lhwr Series I 1 55 x x ln IR quot3 a In awhile 39 3quot 1 N i K 3 5 a I 3 4 I 3 xx 2nd excited slam xxquot 3 N quot f quotiquot39quotirT 7fT x xx on I u s n c If x ifV quot Ljuni1allnn requot M Las r Class Early Models of The A rom Ru rherford39s model of The A rom Small dense posi rively charged nucleus Mos r of The a rom is emp ry space r a l39om gtgt r nucleus 103910m 1014m Al39omic ine spec rra A roms emi r ligh r a r very specific frequencies Frequency is differen r for each Type of a rom Issues we s rill haven39T addressed Don39T know wha r elec rrons are doing No explana rion for al39omic ine spec rra Bohr395 Model of The Atom Assumed Tha r elec rr ons or39bi r nucleus in circular or39bi rs UWIEUS Elec rr39ons are bound by elec rr39os ra ric a r rr ac rion Felec l39r39osfa l39ic Fcen l r ipe ral Iii urblts 2 2 26 mv 2 r r NOT SO FAST THERE39S A PROBLEM HERE Problem conflict with classical EampM Electron in circular motion is by definition constantly accelerating Accelerating charges emit radiation Electron loses energy as it radiates If electron were radiating and losing energy it would spiral into the nucleus within 10us Bohr39 made 3 assumptions on which To build model 1 Elec rr39ons can move wi rhin cer rain or39bi rs wi rhou r r adia ring s ral39ionar39y sl39a resquot 2 A roms r39adia re only when an electron makes a Transition from one s ra rionar39y s ra re ro ano rher Radia rion ligh r is emi r red only when an electron moves from a higher or39bi r ro lower39 or39bi r A rom radia res when Ei gt E where E Energy of ini rial s ra re and E Energy of final s ra re Energy of pho ron released pho l39on Ei Ef Recall from Planck39s rheory Emuquot hf where h is Planck39s cons ran r and f is The frequency ll z L H Wm hf El Ef W nil This explains a romic line spec rra Similarly an a rom absorbs radiation when Ei lt E E a 1 T camm 139 1 B absor39p rion occurs only when hf Ef E 3 Correspondence principle In The limi r of large orbi rs and large enel39yles quunlurn cuwum l39IOn ShOLlld agree Wl l39h classical calcula rion Now le r39s s rar r cons rruc ring The model Implication of Bohr39s assumptions Angular momentum of electron can only have certain values quantized Angular momentum L 391 1 2 3 integers 27Z h Planck39s constant Define n 39hbar39 h 2 L h 1055 x1o34 Js 2g 6582xurmevg LGh Use this to derive the orbital radii and energy of electron Learn this derivation it WILL be on the test STep 1 Consider Anqular MomenTum L 7172 for an elecTron in an aTom L mvr for circular moTion in general where m and v are The mass and velociTy of The elecTron and r is The orbiTal radius PuT The Two TogeTher and solve for v nh mvr nh V 1 WW S rep 2 NOW bGCk 1390 Felec rr osfa l39ic Fcen l r ipe ral NoTe This model is only good for 1 elecTr39on 2 2 aToms kZe mv 2 I I 2 Again solve for39 v v lkze 2 mr S rep Se r equa rions 1 and 2 equal ro eacho rher39 kZe2 WW mr Now square bo rh sides and solve for39 r39 72th kZe2 mzr2 mr 2 2 nh 2 n kae Radial position of electron 39rquot is quantized 712 Define The Bohr39 radius 610 2 00529nm mke Enemy of electron Derivation E Kine ric Po ren rial 1 2 E mv2 kZe 2 r 2 Use 6 lua rion 2 or39 v v 211626 mr 1 2 1 kZe2 v 2 2 r Subs ri ru re back i n ro energy equation l ch62 kZe2 2 r r E l kZe2 E 2 r nzhz Plug in radius r 2 kae E mk264 Z2 E Z2 n 2h2 n2 0 n2 2 4 where E mkze 2h 0 2 218gtlt1018J 136eV Detailed illus rr39a rion Summary Allowed orbi ral radii of elec rron 2 n r a0 a0 00529nm Z Allowed Energy of elec rron 22 En E0 2 E0 21366V n Bo rh r and E are quan rized Higher n larger r higher energy lim En 0 I l OO Example Wha r is The energy and orbi ral radius of an elec rron in a Hydrogen a rom in The 1 n1 s ra re b n2 s ra re rn a0 a0 00529nm 22 En E0n 2 E0 136eV Problem 413 1 Compu re The radius of Hydrogen39s n6 or39bi r b Compu re The radius of n6 orbit of He r z a a0 00529nm n Z 0 22 En E0n 2 E0 136eV LecTur39e 16 Wavelike ProperTies of ParTicles OuTline of Topics in ecTur39es 14 17 Review of waves par39T 1 DiffracTion and InTer39fer39ence InTr39oducTion To Young39s double siT experimenT Wave naTur39e of MaTTer39 The DeBr39oglie HypoThesis ExperimenTs demonsTr39aTing The wave naTur39e of maTTer39 Pr39obabilisTic inTer39pr39eTaTion of The wavefuncTion Young39s double siT exper39imenT r39evisiTed Review of waves par39T 2 Wave packeTs Par39Ticle wave packeTs Review 0 waves I ar39T 3 Classical uncer39TainTy r39eIaTions Heisenberg uncer39TainTy r39eIaTions amp consequences Young39s double siT exper39imenT r39evisiTed in lighT of uncer39TainTy InTer39pr39eTaTions of quanTum mechanics Review so far 0 DeBroglie postulates that matter has a wave nature39 derives wavelength 7 for matter waves 0 Experiments confirm that electrons and atoms have wave properties Wave aspect of matter describes probability of measurement result 0 Particle wavepackets describe free particles 5 Review of waves part 3 Classical Uncertaintx Relations A wavepacket or a pulse IS a sum or harmonic waves M111 g A colltx Ath Conjugate variables time t and frequency to position x and wavenumber k The spatial extent or duration of a pulse depends on the range of wavelengths or frequencies combined To see how This works consider our simple example mid 2 WonCS w itk 4166 th gt MA 0 K Qlt ncsL a7 wardz n b lg cas4llt24 A3513 LOS EDL Jt 7 L W vwe39w i I eAEA J em wttathewaupc Le r s look of TO Le consider only The spa rial par r mix 6 lg COSEQSJEJCDi i ig mph4L3 9m mm mm binc Lap K lwiattwormckgt mag 7 AXAKZZW If The exTenT Ax is big The range of wavenumber39s Ak is small 9 bf A U gt jmali AK Small ADL 73953 AK HUMP hum VMCJ If The dur39aTion AT is long The range of frequencies Am is small r 5 5 A t 3 Small At v 5Vnall A07 bf Awe lgtspl l LWmauoL vac t For a general wave packe r one can show ADOZA K gt Classtcm met fm39tj AM At gt 1 rcIa Cmf In Quan rum Mechanics particles are described by wavepacke rs Use DeBr oine r elafions To find uncer rain ry in momen rum and energy Ap hAk and AE hAa Facel 39 i Heisenberg Uncertainy Principle Elm Await mheie pa em AXA0 AE At Me the Uncetafm tfcs En Fasftion momentum ma a mat time Implica rion Hr Canno r measure posi rion and momen rum arbi rr ar ily well of The same Time Example The Gamma Ray Microscope How well can we measure both momentum and position of an electron To measure its position need to scatter light off the electron One photon is the minimum amount of light possible recall photon carries momentum ph I 1 504M f 5mm Q All opfical sys rems have a minimum uncer rain ry due 1390 diffracfion of ligh r LWJ u 59 39 quot X finchall 3 can Sonia I E mag r ekctran erme Fruit Minimum uncer rain ry in posifion minimum separa rion between me objecfs which can be disfinguished wid rh of central maximum of diffrac rion pa r rern xi sin 6 Measurement of momentum x com onent of hoton39s recoil momentum p 7 h Sin 939 X x component of electron39s recoil momentum 1 Lv s N I I i I Uncertainty from lens 9 I Muzak i 9 hsing phntan AP x eketrm A 406 i39 A some a Combined uncertainties AxApxhgtg we could improve Ax by decreasing A bu r pho ron would kick39 elec rron harder bigger Ap we could improve Ap by increasing A bu r image of elec rron would ge r fuzzier bigger Ax we could improve Ax by increasing 9 lens size bu r we would be accep ring pho rons which kick elec rrons harder bigger Ap Tradeoff be rween Ax and Apt measuring one well means The other is measured poorly Back To Young39s Double Sli r Experimen i Fringe on screen is r r r i To 7L gives us a measure of momentum of The par i39icle as if wen i39 Through sli rs posi rion of par i39icle is completely uncer rain we donquott know which inT if wen i39 139hrough cmffrv new uxFuenth Now if we make a posi rion measuremen r ro de rer mine which sli r rhe elec rr39on wen r rhr ouh usin a I ho ron as in amma r a example momen rum of elec rr39on is al rer ed by The measurement adds uncer TamTy To momenTum owashes ou r in rer fer39ence pa r rer39n Al v 4x Threw r 77 fl 7 Conseguences of the Uncertainty Princigle AxAp 2 and AEAt 2 Uncertaint rinci Ie States that you cannot know both position and momentum or both energy and time arbitrarily well Important consequence Cannot have zero kinetic energy Example Minimum Enemy of a particle in a 39box39 UncerToinT39 in I osiTion is The lenjh of The box L Ax L UncerToinTy in momenTum Ap is defermined as follows DefiniTion of variance APY39 p a Par Ticle moves equally back and forTh so fj O 5 Al P