Giáo trình vật lý lượng tử tóm tắt lý thuyết và bài tập

ĐẠI HỌC QUỐC GIA HÀ NỘI KHOA CỎMG NGHỀ HỆ ĐÀO TẠO CHẤT LƯỢNG CAG NGÀNH VẬT LÝ KỸ THUẬT GIÁO lld ữ m VẬT IvÝ I.ƯỢIVG TỂ Tóm tắt lý thuyết tập D e d ic a ted to P ro fesso r lUarcus M ainardi an extraordinary leacher PART I TH E S P E C IA L T H EO R Y F R E L A n V I T Y C H A P TER G A L IL E A N TR A N SP O R M A TIO N S 1.1 1.2 1.3 1.4 1.5 C H A P TER T H E P O S TU LA T ES F E IN S T E IN 2.1 2.2 2.3 2.4 C H A P TER 3.4 T he Constancy o f the Speeđ o f Light The Invariancc o f M axw eU’s Equations G eneral Considerations in Solvin g Problem s Involving Lorentz Transformations Sim ultaneity RELATIVISnC 4.1 C H A P TER A b solute Space and the Ether The M ich elson -M orley Experiment Length and Tim e M easurem ents— A Q uestion o f Principle T he Postulates o f Einstein TH E LO R E N T Z C O O R D IN A T E TRA N SFO R M A TIO N S 3.1 3.2 3.3 C H A P TER Events and Coordinates G alilean Coordinate Transíormations G alilean V elocity Transformations G alilean Acceleration Transíormations Invariance o f an Equation LEN G TH C O N T R A Q IO N The D ìn ition o f Length R E L A H V I S n C H M E D ILA T IO N 5.1 5.2 Proper Tim e T im e Dilation C H A P TER RELATIVISnC CH A P TER R E L A n V I S n C V E L O C IT Y TR A N SP O R M A n O N S 7.1 7.2 7.3 S P A C E - n M E M EASU R EM EN TS T he Lorentz V elocity Transformations and the Speed o f Light G eneral Considerations in S olvin g V elocity Problems T he R elativistic D oppler E íĩect 10 10 10 10 11 15 15 16 20 20 23 23 23 27 37 37 38 38 CHAPTER M A SS, EN ER G Y , A N D MOMENTUM IN R E L A n V I T Y 8.1 8.2 8.3 8.4 8.5 8.6 8.7 T he N e ed to Redefìne C lassical M om entum The Variation o f M ass with V elocity N e w to n ’s Second Law in Rclativity M ass and Energy Relationship: E = mcP' M om entum and Energy Relationship U nits for Energy and M om entum General Considerations in S olvin g M ass-E n ergy Problem s 45 12,5 45 45 46 46 46 47 47 C H A P TER 13 59 C H A P TER 12.7 ELEC TR O N S P IN 13.1 13.2 13.3 13.4 13.5 PART I I THE QUANTUM T H EO R Y F ELEC T R O M A G N E T IC R A D IA T IO N Q uantization o f the Direction o f ửie Orbital Aiigular M om entum Explanation o f the Z eem an Eữect T he S tem -G erlach Experiment Electron Spin Spin -O rb it C oupling Fm e Structure Total Angular M om entum (The Vector M odel) 120 120 126 126 127 128 128 129 AND M A H ER PART I V M A N Y-ELEC TR O N ATOMS CHAPTER ELEC T R O M A G N E T IC R A D IA H O N — PHOTONS 9.1 9.2 9.3 9-4 9.5 T he Theory o f Photons T he Photoelectric E ffect The Com pton Effect Paứ Production and Aiưiihilation Absorption o f Photons 59 60 61 62 63 14.1 14.2 14.3 14.4 C H A P TER CHAPTER 10 M ATTER W AVES 10.1 10.2 10.3 10.4 D e B roglie W aves Experimental V'erification o f D e B roglìe’s H ypothesis The Probability Interpretation o f D e B roglie Waves T he H eisenberg U ncenain ty Principle 80 CHAPTER 11 TH E B O H R ATOM 11.1 The H ydrogen Spectrum 11.2 11.3 11.4 11.5 The Bohr T heory o f the H ydrogcn Atom E m ission o f Radiation in B o h rs Theory Energy Level Diagram s H ydrogenic Atom s CHAPTER 12 E L E Q R O N O R B IT A L MOTION 12.1 12.2 12.3 12.4 12.5 Orbital Angular M om entum from a C lassical Vievvpoint C lassical M agnetic D ip ole M om ent C lassical Energy o f a M agnetic D ip ole M om ent in an Extem al M agnetic Field T he Zeem an Experiment Q uantization o f the M agnitude o f the Orbital Angular M om entum C H A P TER 103 103 104 107 107 117 117 118 119 119 S pectroscopic N otation for E lecừon Coníìgurations in A tom s T he Periodic Table and an A tom ic Shell M odel S pectroscopic Notation for A tom ic States A tom ic E xcited States and L S C oupling T he A n om alou s Z eem an Efifect X -R A Y S 16.1 16.2 16.3 16.4 16.5 16.6 16.7 103 Q uantum -M echanical System s with M ore Than One E lecưon T he Pauli E xclusion Principle A S in gle Particle in a O ne-D im ensional B ox M any Particles in a O ne-D im ensional B ox M A N Y -ELEC TR O N ATOMS AN D TH E P E R IO D IC T A B L E 15.1 15.2 15.3 15.4 15.5 80 81 82 83 PART I I I H Y D R O G EN LIK E ATOMS T H E P A U L I E X C L U S IO N P R IN C IP L E X -R ay Apparatus Production o f Brem sstrahlung Production o f Characteristic X-R ay Spectra T he M o se le y Relation X -R ay A bsorption E đges A u ger EflFect X -R ay Pluorescence 135 135 135 135 136 140 140 141 142 142 143 157 157 157 158 160 160 161 161 PART V N U C L E A R P H Y S IC S CH A P TER P R O P E R T IE S F N U C L E I 17.1 17.2 17.3 17.4 17.5 17.6 T h e N u cleon s N u cleo n Porces T h e Deuteron N u clei T he N u cleu s as a Sphere N u clear B inding Energy 173 173 174 174 174 175 175 CHAPTER 18 N U C L E A R M O DELS 18.1 18.2 CHAPTER 19 Liquid Drop M odel Shell M odel TH E D ECA Y F U N S T A B LE N U C L E I 19.1 19.2 19.3 19.4 19.5 N uclear D ecay The Slatistical Radioactive D ecay Law G amma Decav A lp D ecay Beta D ecay and the Neutrino 181 CH A P TER 181 182 D IS T R IB U T IO N P U N C n O N S 24.1 24.2 24.3 D iscrete Distribution Punctions C ontinuous Distribution punctions Pundamental Distribution Punctìons and Density o f States N U C L E A R R EA CTIO N S 20.1 20 2 0.3 0.5 20 20 CHAPTER 21 N otaiion C lassitìcation o f Nuclear R eactions Laboratory and Center-of-M ass S ystem s Energetics o f Nuclear R eactions N u clear Cross Sections N u clear Pission N u clear Pusion P A R T IC L E P H Y S IC S 21.1 2 1.3 21.5 21.7 21.8 21 1 Particle G enealogy P an icle Interactions Conscrvation Laws Conservation o f Leptons C onservation o f Baryons Conservation o f Strangeness Conservation o f Isotopic Spin and Parity Short-Lived Particles and the R esonances The E ightíold Way Quarks 193 193 194 194 195 C H A P TER C L A S S IC A L S T A n S n C S : T H E M A XVVELLBỌLTZM ANN D IS T R IB U n O N C H A P TER 207 207 208 209 210 210 211 224 224 224 226 227 227 227 227 228 229 230 CH A P TER M O LECU LES 22.1 2.2 CHAPTER 23 M olecular Bonding Excitations o f Diatom ic M olecu les K IN E T IC TH EO R Y 23.1 23.2 Average Values in a Gas T he Ideal Gas Law 245 245 246 259 259 259 F erm i-D ac S taóstics B o se-E in stein Statistics High-Temperature Lim it Two U se íu l Integrals B lackbody Radiation Free Electron Theory o f M etals S peciíic H eats o f Crystalline Solids T he Q uantum -M echanical Ideal Gas Derivation o f the Q uantum Distribution Punctions S O L ID S 27.1 27.2 T he Band T heory o f Solids Superconductivity 287 287 288 288 2S9 289 292 296 301 305 309 309 318 325 A p p e n đ ix S om e Pundamental Constants in Convenient Units S om e U seíiil C onversions M asses o f S om e Particles M asses o f N euư al A tom s In d e x 276 QUANTUM S T A H S n C S : P E R M I- D IR A C AND 26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8 26.9 207 PART V I ATOMIC SYSTEM S CHAPTER 22 268 269 270 193 B O S E - E IN S T E IN D IS T R IB U n O N S CHAPTER 20 268 325 325 325 326 333 Galllean Transformations I ì Ể1 E V E ÌST S A N D C O O R D IN A T E S \Ve b e ^ n by considering the cept o f a physical cvent The cvcnt niisiht be the striking o f a tree by a ligluiiing bó)t or the collision o f tw o particles, and happens at a point in space and at an instant in time The particular ẹvent is speciíìed by an obser\’cr by assicn in g to it four coordinales: the three position coorđinates >r, y , z that measure the distance from the oricin o f a coordinatc sỵstcm \vhere the observer is located, id the tim e coordinate í that the obser\'cr rccords wiih his clock Consiđer now tu’0 observers, o and \ where o ' travels with a constant velocit\' V \vith respect to o along th eừ cồm m on ĩ — x ' axis (Fig 1-1), Both o b s e n ers are equipped \vlih melcrsticks and clocks so that they can m ẹ ^ u r e coordinates o f events Purther, suppose that both observers adjiist their clock s so that wh en they'*p^s each other at J = y = 0, the clocks rcad' I = i' = Aiiy gÌN cn eveni p will have eight num bers assợciated with it, the four coordinates assiuncd by o and tlic four coordinates ự , / / , i') aisign ed (to the sam e evem ) by O ' O Evenl (X /.Í I) Fig 1-1 GALILEAN TRANSPORMATIONS G a LILEAN c o o r d in a t e t r \n sfo [CHAP Ans.' r m a t i o n s The relationship betwecn the measurements (,v.y /) o f o and the measurements (.r ',y , a palicular event is obtaincd by exaniining Pic 1- 1: y' V! t ’ = -V - CHAP 1] = V r' = GALILEAN TRANSPORMATIONS The coordinates assigned to the bird by the man on the station platform are {x y ,z , /) = (8 0 m ,0 , s) t') o f O ' The passenger measures the distance y to the bird as z x' = x - V I = 0 m - (3 m /s)(20s) = 200m Therefore the birds coordinales as determined by the passenger are additon, in classical phvsics it is im plicitly assumed that (,í',y ,z './') = {2 0 m 0, 0,2 s) í' = / :se fcur equations are called the G alilean coordinate transform atĩons 1.2 G ALILEAN V E L O C IT Y T R \N S F R M A T N S In aidition lo the coorđinates o f an event, the velocity o f a particle is o f interest Observers o and O' describe the particles velocity by assigning three com ponents to it, with (u ,, u.) being the velocity nponents as measured by o and ịú^, U y, uỊ) being the velocit>' com ponents as meãsured by O ' The relationship benveen (Uj, u ) and (u^, iiỊ) is obtained from the lim e diữerentiation o f the lilear, coordinate transformations Thus, from v' = X — 17, , dx d ^ di (cix Reíer to Problem 1.1 Five second s afker making the íìrst coordinate measurement, the man on th platfom i determines that the bird is 850 m away From these data find the velocity o f the bir (assum ed constant) as delcrmined by the man on the platform and by the passenger on the traứi Ans The coordinates assigned to the bird at the second position by ứie man on the platíorm are f c y - ' 2) = (8 m, ,0 , 25 s) Hence, the velocity u, o f the bird as measured by the man on the platform is ' \ Í2 - / | ' s-2 s The positive sign ìndicates the bird is Aying in the positive jr-direction The passeriger finds that at tl second position the dỉsĩance y*, to the bird ỉs ogether, the Gaiiìean velocit}- lransfo n n a tio n s are —V úy •rí = ,12 - i-ụ = 850 m - (30m /s)(25s) = 100 m = ú = u Thus (JCÍ VỊ.r'^, ) = (lOOm, 0, 0, 25 s), and the velocity u', o f the bird as measured by the passeng on the train is -^ 100i n - 20 m G AL ILE AN A C C E L E R A T IO N T R A iN SPO R M A T IO N S The acceleration o f a particle is the time derivative o f its vclocity, i.e., a = du /d i , etc To find the lilean accelem tion :ransformarions we diíTcrentiate the velocity transformations and use the facts that = l and L' = constant to obtain = u, - 1) = lO m /s - 30m /s = -2 m /s a', = a JS the measurcd acceleration com poncm s are thc same for all observers moving with uniform rclalive Dcity INVARL-VNCE O F AN E Q U A T IO N A sam ple o f radioactive material, at rest in the laboratory, ejects two electrons in opposi direclions, One o f the electrons has a speed o f c and the other has a spced o f 0.7c, as m easun by a laboratory observer According to classical velocỉty transfonĩiations, what w ill be the speed ( one electron as measured from the othcr? Ans B y invariance o f an equation it is meant that the equation vvill have the same form vvhen determined by I observers In classical theory It is assum ed ihai space and time measurements o f two observers a r é Ited by the Galilean transformations Thus, when a particular fom i o f an equation is determined by One er\'er, the Galilean transformations can be applled to this fonn to determine the form for the other erver l f both fom is are thc same, the equation is invariant under thc Galilean transformations See blems 1.11 and 12 S o lv ed P ro b lem s A passenger in a train moving at 30 m /s passes a man standing on a station platform at í = í" = Twenty seconds afler the train passes him, the man on the platform determincs that a bird Aying along the tracks in Ihe same direction as the train is 800 m away What are Ihe coordinates o f ứie birđ as determined by Ihe passeimer? = - m /s so that as measured by the passenger, the bird is moving in the negative y-direction Note that th rcsuil is consistent with that obtained from the Galỉỉean velocity transformation: Let obsen'er o be at rest with respect to the laboratory and let observer O' be al rest wiửi respect the panicle moving with specd 0.6c (taken in the positive direetion) Then, from the Galilean veloci tiansformation, u', = u , - v = - c - 0.6c = - 3c This problem demonstrates ihai vclocities greater than Ihe specd of lighl are possible with t Galilcan transformations, a result that is inconsistent with Special Relali\ ity A train m oving with a velocity o f m i/h r passes through a railroad station at 12:00, Twcn seconds later a bolt o f lightning strikes the railroad tracks one m ile from the station in the san direction that the train is m oving Find the coordinatcs o f the lightning flash as measured by i observer at the station and by Ihe engineer o f the train Ans Both observers measure the limc coordinatc as / = /' = (2 s) V3600s^ GALILEAN TRANSPORMATIONS The-observer at the station measures the spatial coordinate to be determined by the engineer of Ihe Irain is [CHAP CHAP 1] The spatial coorđinatc as GALILEAN TRANSPORMATIONS (a) For the boy in the car the ball ơavels straight up and dovra, so y = Vot' + ịa í^ = (2 f t /s ỵ + ị ( - = 20?" - 16f'2 y=V = o y = ,x - VI = mi - (60 mi/hr) (Ề) For the stationaiy observer, One obtains (rom the Galilean transformations 1.5 A hunter on the ground íìres a bullet in the nonheast đirection which sưikes a deer 0.25 miles from Ihe hunter The bullet travels with a speed o f 1800m i/hr At the instam when the bullet is tìred, an airplane is directiy over the hunter at an allitude h o f one mile and is traveling due east with a velocity o f 600m i/hr ViTien the bullet strikes the deer what are the coordinates as determined by an observer in the airplane? Ans t = í 1.9 Using the Galilean transformations, 1800/hr : I0-' hr = r* = Consider a mass anached to a spring and m ovm g on a horizontal, frictionless suríace Show, frorr the classical transíormation laws, that the equations o f motion o f the mass are ứie same detennined by an observer at rest with respect to the suríace and by a second observer moving wiứ constant velocity along the direction o f the spring Ans x' = x - v t = (O.:5m ĩ)cos45“ - (600 mi/hr)( 1.39 X 10”* hr) = 0.094 mi >■' = y = (0.25 mi) sin 45° = 177 mi The equation o f motion of the mass, as determined by an observer at rest with respect to the surface, i: F = ma, or z' = —/ i = — lm i = —Imi 1.6 y = ý = t— l r I = x ' + p /= + lOOí - k ịx - X o ) = m ^ u To determine the equation o f motion as íound by the second observer we use the Galileai transformations to obtain An observer, at restwith respect to the ground, observes the following colHsion A particle o f mass m, = kg moving with velocity Uị = m / s along the x-axis approaches a second particle o f mass m, = kg moving with velocity u, = - m /s along the v-axis After a head-on collision the ground observer finds that has velocity = m /s along the A:-axis Find the velocity o f m, alter the collision d -x X= / + v i' , so the total tímc for ihc rounđ trip is c -v^c + v To travel along the other arm a light ray musl be aimeđ such that its resultanl velocity vecior (velocitỊ with respect to the ether plus velociry o f Ihe ether with respccl to the intcrierometer) is perpendicula) to arm A This gives a speed o f for both directions along path /j, so the time for the rount trip ỉs 2lẹ , ‘ - u(a) SÌ^naLv A and B start (b» Slpnol B rcaches 0\ (c) SỉgnaLs.4 and B rcach simulUineously 2/ , / c ự l - (v^/v^) If wc assume v/c stal!ine material has a set o f Bragg planes separatcd by 1.1 Ả For e V neutrons, what is the hichcsl-order Bragg reAection? A n s T(ìe wavelength of the ncuưons is A _ ^ A _ Fig 10-S hc " j2 m a K ~ _ 12.40X lO^eV Ả " v'2(940 X 10»eV)(2eV) The mxximum angle that can be attained is 90° Then, from Bragg’s law, (l.lẢ )s m “ =n(0.202Ả ) or n = 10.89 Numberof ions Volume ~ [2dÝ ~ Equating this to the above result, we have Sĩnce n must be an integer, the highest order is n = 10 ^ = 4 X 10-* m " ’ 10.14 A larae crystal is used to extract single-energy neutrons ữom a beam o f neutrons emerging from a reactor The spacing o f the Bragg planes is 1.1 Ả l f the Bragg angle is set to be ’ , vvhat is the energy o f neuơons seen at this angle for a first-order reAection? Ans d = 2.%2 ; = 2iísinỡ = 2(1.1 Ả )sin30’ = l l Ả X 10‘ '“ m = 2.82À The vvavelenglh of the neutrons is related to their kinetic energy by hc or v /2 (m „ c^ )A : (hcÝ 2(,moC^)X- 2(940 X 10‘ e V )(l.l/ 10.15 Deteưnine the interatomic spacing o f a NaCl crystal if the density o fN aC l is 2.16 the atomic weights o f sodium and chlorine are 23.00 and 35.46, respectivcly Ans 10.16 In One o f theứ experiments, Davisson and Germer used electrons incident normally on a nickel crystal suríace cut parallel to the principal Bragg planes They observed constructive mteríerence at an angle o f 50“ to the normal to the surface Fmd the wavelength associated with the electton beam (The interatomic spacmg o f nickel is 2.15 Ả.) (12.40 X 10^ eV-Â)^ X lO ^kg/m ’ and The molecular weight o f NaCl is 23.00 + 35.46 = 58.46 The number o f molecules per 58.46kg of NaCI is 1km ol 58.46 kg g 6.025 m o le c u l e s X m o le c u le s 58.46 kg kmol Since thcre are two atoms per molecule, we have Numberof atoms _ numberofatoms Volume mass = 4.45 X mass volume X 6.025 X 10^'^atoms 58.46 kg lo ’ — m’ ^ * -^ To relate this to the interatomic spacing d, consider ứie NaCI unit cell shown in Fig 10-5 (diRerences between the Na"*" and Cl“ ions are ignored) The volume o f the cube is (2d) As to the number o f ions to be assigned to the cube, Ihere are: comer ions, each shared by o f the cubes; 12 edge ions, each shared by o f the cubes; face ions, each shared by o f the cubes; and unshared center ion Thus Numberofions = + 1=8 Ans We first finđ the relation between the scattering angle ậ to ứie normal and the interatomic spacing D From Fig 10-6 it is seen that d + ộ / ĩ = 90° so that *;0 MATTRRWAVES [CỈỈAP 10 CHAP 10] •■líií A o Iro m th c íìg u r c w e se e ih c s p a c in g d b e rw c e n th e BraỊ»g p la n e s is MATTER WAVES Thi; dc Broglie vvaveicngth o f the balls is d = D sm ị mv Subitimtmg tlìcse rcsuỉis in the Bragg relation 2J sin = n/., and using the half-ang!c tbrmula ^ ■ ■ 2- ,sin —c o s — = sin 4> From inicríercncc Iheory, the angies patlem are given by (6.625 X 10"'kg)(5 m/s) ' to the lines of zero intensiry in a double-slit intcríerencc '.se obtain The coưesponding v-distance is given from Fig 10-7 as O sin = nk Taking /I = w e then have = i tan 9, Sí i sin 0, = í, - (2.15Ả)sm50.0^ = {I)à or d = 1.65Á 10.17 In the experiment described in Problem 10.16, Davisson and Geưner useđ eV electrons Dctermine the etĩective acceleraiing poteniial o f the nickeỉ crysial ] The dc Brociie W3velcneth o f'5 e V electrons is ^ ■ ~ m„L- ^ ự ĩĩỢ Ì 12.4 X 10^ eV_-À , ^7 ^ ự ( m „ ( ~ ) K ~ v'2(0.511 X lO'’ e V )p e V ) ” T his is ditTerent from the o b s e r\e d w avelenaih o f 1.65 À The kinelic energy coư e sp o n d in g to / = 1.65 Ả is found from v 2(m„c-)Ả’' 2(moC=);.- 2^0.5^ 10»eV)(1.65 A)- X Theretbre, the eíĩective accelerating potential o f Ihe nickel crystal is Fig 10-7 = 5 V - V = 1.3V The distance bctween adjacent fringes is then given by L l, T H E PR O B A B IL IT Y LNTERPRETATION O F DE B R O G L IE VVAVES 10.18 Determine the photon flux associated with a beam o f monochromatic light o f wavelength 0 Ả and iniensity X '''* w /m l Substituting Ihe values for our problem, wc have r- L Ac ( 6 x - ” J -s ) (3 x 10* m/s) £ = /iv = - ^ = i - ; - — À / X lO"''* J /s • ,y ss — = - - - I hv , = 6.63 X 10 ” J/p h o to n X -’ m 6.63 X 10“ '^ J /p h o to n p h o to n s ^ X 10 "■ '■ s • p h o to n s =: ' ' s-c m * On the average, 4.5 photons will strikc a í cm‘ area (of photographic film, say) đuring a pcriod of s Oi' course, only integral numbers of photons can be observcd Thus, for a given I cm^ arca, we might observ-e phoions or photons in a one-second intcrval, but never 4.5 photons Only if an average is taken over many intervals will the average number approach 4.5 photons Also, for a given one-second interval, the arriving photons may cluster wiihìn a fixed l cm" arca Only after a long period of lime will the photon positions approach a uniform đistribution 10.19 Suppose /1 = 6,625 X 10“ ’ J s instead o f 6.625 X Balls o f mass 66.25 grams are throvvn with a speed o f m /s into a house ứirough two tall,narrovv, parallel windows spaced 0.6 m apart, thechoice o f window as target being ranđom at each toss Determine the spacing between the friiiges that would be formed on a wall 12m behind the vvindovvs 0.6 m This problem illustrates the probabilistic interpreiation o f de Broglie waves Any sừigle ball will strike the wall at a speciíic, although undeterminable, position Although it cannot be predicted in advance where any ball vvill strike, each ball has a high probability o f aniving at a maximum and zero probability o f arriving at a minimum, o f the interíerence pattem The actual interference pattera is experimentally determined by counting the number o f balls that strike each part o f the wall In the beginning o f the experiment the balls will hit the wall in a more or less sporadic fashion Only after a large number o f balls have been thrown through Ihe windows will the interference pattem become disceraible, since the number o f hits at the eventual maxiina will increase, while the number o f hits at the minima will retnain zero 10.20 A particle o f mass m is coníìned to a one-dimensional line o f length L From arguments based on the wave interpretation o f matter, show that the energy o f the parlicle can have only discrete values and determine these values [CHAP 10 MATTER WAVES lf thc particic is confined ío a line scgment, say from x = lo X = L, the probability o f iìnding the particlc outsiđc this region m u st be zero Thercíore, thc wave íiinction ^ m usl bc zero for X < or ,r > L, sincc thc squarc of gives the probability for lìnding the particlc at a certain location Inside the limitcđ region, the waveiength o f ip must be such that fp vanishes at the boundaries X = and X = L, so that il can vary continuously 10 ứic ouiside region Hence only those wavcỉengihs will be possible f o r which an iniegral number of half-wavelengths fii bctween JT= and X = L, i.e., L = nÀ/2 where n is an inlcger cailed the quantum number, with values /I = ,3 From the de Broglie relationship À = /i/p we ứien find that the panicles momentum can havc only discrcte vaiues givcn by CHAP 10] MATTER WAVES 10.22 Wìiat is íhe unccrtainty in the location o f a photon o f vvavelcngth 3000 Ả if this wave!cngth is knovvn lo an accuracy o f one part in a miUion? Ans The momenrumof ỉhe photon is given by hc 12.40 X lO^eV-Ả ’’ ^ T c ~ {3 X 10^ A)c eV - ' ^ The unceriainty in the photon momentum is (vvorking wiih magnitudes only): A/J = — Since the panlcic Ì3 n o t acied upon by any íorces inside the region, iis polentiai energy vvill be a constant vvhich we set equal to zero Thereíore, thc cncrgy o f ửie body is entirely kinciic and will havc the discrete values obtained from A > 4n:Ap 10"^ = ^ 4/tcủp 4;rct4.13 X 10"^ eV/c) X = 239 10“* ^ X ,0« À = 23.9 mm 10.23 What is the minimum uncertainty in the energy State o f an atom if an elecưon remains in ứũs State for 10-* s? £ = n SmL^ Ans n = ,2 ,3 , r- >' t'c x l0 ^ e V Ả — = - - - ^ = 329 X 10" eV 4rtA/ 47tcA/ 47t(3 X ÍO*m/s)(lO“*s)(IO’'^À/m) \F > _ _ This very sim ple problem itlustrates One o f the basic íeatures o f the probability interpretation o f m atter; nam ely, that the energy o f a bound system ca n take on only discrete values, w ith zero energy The minimum uncertainty in ihe energy of a State, r = hỊụ,nz), where t is the mean lifctime of the excited State, is calleđ the naturaỉ width o f the State For this problem the mean lifetime is 10“®s and the natural width is 0.329 X 10"’ eV not being a possible value T H E H E ISE N B ER G U N C ER TA IN TY PR IN C IPL E 10.21 Suppose that ứie m o m e n tu r a o f a c e r ta in p a r tic le c a n be m e a s u re d to a n a c c u c y o f One p a rt in a thousand Determine the minimum uncertainty in the position o f the particle if ứie particle is (a) a X 10“^ kg mass moving with a speed o f m /s , (6) an elecơon moving with a speed o f 1.8 X 10*m /s — = 10 ’ Then, from The time available for measuring the energy is 10“* s Therctbre, Irom AE Ai > h/4iỉ, or 10.24 The width o f a spectral line o f wave!engứĩ 4000 Ả is measured as 10“**Ả What is the average từne th a t th e atomic s y s te m remains in th e corresponding e n e rg y S tate? Ans r = A£ is the cnergy spread correspondữig to From Problem 10.23, z — h /(4 n r ỵ where AÀ = lO"'* Ả From E = hc/À, |A£| = Ap = 10"’p = 10 ^mv A/ > A/4ít, Áx > 4itẠp 4ílO -% t; ’'4ìtc& x ( 1) 4;i(3 X 10»m/s)(10-''‘ m) Note that Planck's constant does not enter into ứie íìnai exprcssion A > - s 10->(5 X 10->kg)(2m /s) The minimum uncertainty is 5.28 (b) X = 5-28 X - » m = 5.28 X -“ Ả 10"^“ Ả, a value that b clearly unmeasuiable The rclativistic raass o f the electron, m = ntỊiy/] -( v ^ /c ^ ), must be used in Ụ) ^ ( 6 x l - » * J s ) y i- ( ) ^ ;rÌ0-^m oiJ 4iitO~H9.U X 10"^‘ kgXỈ.8 X 10* m/s) The minimum uncertainty is 2.57 Ả 10.25 Suppose the uncertainty in the momentum o f a paiticle is equal to the particle’s momentum How is thc minimum uncertainty in the particle’s location related to its de Broglie wavelength? Ans We are givcn that p = p ,s o that /ir > h .= — Aĩtâíp h ss — Aitp X An since the đe Đrogiie wavelengih o f a paiticle is ^ =: hỊp Thus Ihc minimum unccrtainty in thc position is XỊAn [CHAP 10 MATTER WAVES CHAP 10] MATTER \VAVES T h u s , th e m in im u m m a g n itu đ c o f p , IS h / ^ ĩĩL , a n d 10.26 Proin the rclation A/ỉA-t > h ị ị i t , show that for a particle moving in a circle, A i Aỡ > A /4n The qiiantity A/ is thc unccrtainty in the angular momt-nmm and Aiỉ is the uncertainty in the angle Since ihe panỉcle movcs in a cìrcle, Ihe uncertainty principle will apply to directions tangent to the circlc Thus This value compares rcasonably we!i, consỉdcring the cnideness o f our argumenl, with the valuc h k■SmLfrom Probiem 10.20 The resuh further iliustrates ihat, under the uncertainty principle, bound systems cannot have zero energy vvhere í is nicasured along the circumference of the circle The angular momentum is relaied to the linear momcnUim by L = m v R ^ p ,R thercfore \ p , = AL/R The angular displacement is related to the arc lengứi by = s /R ; therefore \ ỉ = R SO Hence - ịùJ./R)(R ủfl) = A L A S > 10.29 Calculate the minimum kinetic enerey o f a neutron in a nucleus o f diameter 10 '■‘ Ans The situation is ihat of Problem 10.28, with I equal to the nuclear diameter Thus, /i/ it / A y ' m \i:tL j For State of fixcd angular momentum (e.g an electron in a Bohr orbit, which will be discussed in Chapter 11), the uncertainty in the angular momcntum, A I, is zero Therefore, the uncertainty in ihe angular position AO, is intìnite, so thai the position of the particle in the orbit is indcterrninate ị" y { m í~ )\ỉĩĩL j ■ ' 12.4 X 10-’ MeV Ả ' 2(940McV) 811( 10-* Ả) 10.30 If an elecữon were in the nucleus o f Problem 10.29, what would be its minimum kinetic energy? 10.27 11' we assume that E — km v^ for a partĩcle moving in a straight line, show that A £ A/ > A /4n, where Ai = ^ 2m Ans For an electron, a rcialivistic calculation is necessaiy As in Problem 10.28, the minimmn magnitude o f the momentum is Ip U = ĩ8Z7 n l = 8,t(10-‘ À) 2m Taking diATerentials of both sides o f this expression, we obtain + £ o f = ( ỉ p L ^ c f +Eo^ A £ = P ^ = ' ĩ ỉ ^ = vAp ( K , + 0.51IMeV)= = Then, from &p Ax > A/4it, Since the particle must be somewhere in the given segment, ứie uncertainty in its position, A* cannol be grealer ứian L If &x is set equal to i , the uncertainty relatíon A i/ip i — i" implies that the momentum must b« uncertaứi by the amount Ap, > h/AnL We are looklng for the smallest possible value of the energy and hence, since K = p Ĩ Ị l m , the smallest possible lp,| We identiíy ứie uncertainty in Ịp^ị wỉth that in and assume that the unccrtainty interval is symmetrical about |pjjị Then (see Fig 10-8), lp ,l = ị A l p J > - lí.l Fig 10-8 ỉ - ( 41x1) 8nl X 10-’ McV8^(10-* À) + (0.511M eVr Solving, = 4.45 MeV When the emission o f electtons (^-rays) ftom nuclei was first observed, it was believed that ứie electrons must resiđe inside the nucleus The energies o f the emitted electrons, however, were often a few hundred keV and not the minimum MeV predicted by the íoregoing calculation It was therefore concỉuded that clecưons are not nuclear building blockỉ (See also Problem ỉ 7.ĩ.) 10.28 A particle o f mass m is conAned to a one-dim ensional line o f length L From arguments based upon the uncertainty principle, estimate the value o f the sm allest energy that ứie body can have, Am 12.4 10.31 The position o f a particle is measured by passing it through a slit o f width d Fứid the corresponding uncertainty induceđ m the particle’s momentum Arts When monochromatic waves of wavelength A pass through a slit o f widlh J, a diữraction panem will be produced on a screen as shơwn in Fig 10-9 The location o f the fữst point o f zero intensity is found from dỉíĨTactỉon thcory to bc sin 3t = Ằ/íỉ Because of its associated đe Broglie wave, whose vravelength is Ẳ = A/p,- the particle will be diHracted as it passes through the slit and hence will acquire some unknoivn momentum in the Xdirection Although we đo not know exactly where the particle will strike Ihe screen, Ihe most probable place for it to hit will be somewhere within Ihe C entral region o f the diAiacúoa pattern Therefore we can be reasonably certain that the *-component o f tìie patticle's momentum has a magnitude between and p sina; i.e., /1 Ẳ h [CHAP 10 MATTER WAVES MATTER WAVES CHAP 10] F i g -1 momentum Hence, when we (inally measure the momentum o f the electron, the value o f its moraentum will be uncertaữi by This uncertaínty can be made as smalỉ as desỉređ by increasỉng d However, sínce d = Ax, the = TSỈnot u n c e r ta i n ty in th e p a n i c l e ’s x - p o s itio n , w e s e e th a t Ai = h We can make ố p , as small as desired by making /l suíRciently large, but then Ax becomes correspondingly laiger Taking the product o f the tv,'0 uncertainties, we obtain in agreement with Heisenberg's imcertainty principle AxA p^ = h in agreement with Heisenberg's uncertainty principle It is desired to measure the position and m om entum o f an electron by observing it ửưough a m icroscope A nalyze the observation p rocess m detail to sh ow that rcsults consistent with the unccrtainty principle are obtained Ans When lig h t is sc a tte re d fro m th e e le c tro n d u rin g th e o b se rv a tio n process, ứ ie m om entum o f the elecơon, which we are trying to mcasure, will be affected because the úicidem light itseir carries momentum Hence, we will consider the experiment to be períbrmed with the smallest amount o f light possible, namely vvith a single photon When light reAecting from a particle passes through ữie objective lens o f a microscope, a difÍT3 Ction p a n e rn is p ro d u c e d at th e lo c atio n o f ứ ie eye (o r photographic film ) T hus, a “ fu 2z y ” panem rather than a ptecise S h a rp point will be observed with normally intense light consisting o f many photons DiRìaction theory o f light shows that the diameter o f the centtal disk o f the dif&acrion pattem is given approximately by S u p p le m e n ta ry P rob lem s D E B R O G L IE W AV ES 10J3 Calculate the de Broglie wavelength o f a 2kg mass w h o se 10J Calculate the de Broglie wavelength o f a 0.08 eV neutron velocity is 25m /s Ans Ans 1.33 X 10~“ A 1.01 A sin vvhere X is the wavelength o f thc light and 2ct is the angle subtended at the particle by the mieroscope objective, as shown in Fig 10-10 When we obsetve the single photon in our experiment, we can be rea so n a b ly c e rtain o n ly th a t it w ill h av e a rriv e d so m e w h e re in ứ ie Central disk o f th e diỉ& action panem Hence, Ihe uncenaĩnty in the electron’s position can be taken as A r = í/ = sin The uncertainty in the position can be made as small as desired by using a suHìcietitly small wavelength In the scattering process some o f the photon’s momentum will be transfened to the electron If the momentum o f the scattered photon were known exactly, it would be a relatìvely easy matter to 10 J Calculate the kinetic energy o f a neuttxỉn whose de Broglie wavelength is 0.7 Ằ I0 J What is the minừnum-energy electron needed to observe a À object? Ảns Ảns 0,167eV 6.02 eV 10J7 For the object o f Problem 10.36, whflt is the iránimum-ínergy proton that can be used? Ans 3.28 X 1Q-’ eV J A proton u accelerated ftom rest through a potentíal o f kV Find its de BroglỊe wavelengứL Ans 9.05 X 10 ’ Ẳ w ork backward to determ ine h o w the original m om entum o f the electton was alĩected However, since all w e lcnow ỉs th a t ie s c a tte re d p h o to n e n te re d th e o b je c tỉv e ỈCIỈS som ew here, ỉts jCH:omponent o f momentum could have a magnitude anywhere betwéen and p sin a, where p = h /x is the photon^ X- J9 Find the de BrogUe vravelength o f a keV a-particle {mo = 3728 MeV) Ans 4.54 X 10"^ Ằ (CHAP, 10 MATTER WAVES 10.40 At 'A h a t kinetic cncrgy %vill ihc nonrclativisiic calculalion o f thc de Broglic \vavelcngth of a proion be in ciT or by 5%? Ans 192 MeV CIỈAP 10] MATTER VVAVES 10.54 Rcpeai Problem 10.53 for an cioctron Ans 29.Om 10.55 Repeat Problem 10.22 fbr a gamma-niy photon o f vvavelengih 10”’ Ả 10.41 Wh u is the ratio of a particles Compton and de Broglie wavelcngths? /Í/tí 0.796Ả -Íní 10.42 A( 'Ahat encrgy wili (he nonrelativistic calcation o f ỉhe đe Brogỉie wavelength of an clectron be in eưor by O.OỈ5MeV 10.56 What is the minimum uncertainty in ihe energy of an excited that State for lO"" s? Ans 3.29 X 10"^ eV State of a system if on the avcrage, ii remains in 10.57 Refer to Problem 10.23 l f ih e tr a n s iiio n fro m th e e n c r g y State in q u e s tio n to the g ro u n d 3.39 eV, determine the mìnimum u n c e n a in ty in the vvaveiengih of the emitted pholon Ans 3.55 X 10-5 A EXPERIMENTALVERIPICATION F DE BROGLIE'S HYPOTHESIS 10.43 The spacing between ihe nuclei in a certain crystal is 1.2 Ả- At what angỉe will tìrst-order Bragg reAeciion occur for neutrons with kinetic cnergy o f 0.020 eV? Ans 57.4'’ State c o rr e s p o n d s to If the energy w idth o f an excited State o f a system is 1.1 eV what is the average [ifetime o f that State? 2.99 X 10 ‘^ s A n s Ỉ0.44 A 3,1 eV neutroii bcam cr:\tters from w hat IS the B racg plane spacinc? an unknoNvn sample If a first-orderBraggreữection isfound at 28’ , A n s 0.96"^ Ả ỉf the State o f Problem 10.58 is located an excitation energy of 1.6 keV, vvhai is the minimum uncertainty ìn A n s the w avelength o f the photon em itteđ w hen the exciied State decays? 10.45 5.33 X 10"^ A Thíirmal neutrons are incident on a crystal whose interatomic spacing is 1.8 Á If ìrst-order Bragg reAection ÍTom the neutrons? p nncipal Ans Bragg planes is found at 22"*, w hat is the kinetic energy of the therm al x l " - e v If the A ns 10.46 For the cr>stal of Problem 10.11 what wouỉd be the energy o f thcrmaỉ ncuưons observeđ at 30'" if ihis were a second-order B ragg rertection? Ans 4.14 X " ’ eV 10.47 Re fer to Problem 10.12 Determine the rađius of the second-order diffraction pattem from ứìe principal Bragg planes 9.6 mm u n c e n a in ty in th e e n e r g y o f a n u c le a r S tate is 33 keV, what is iis a v erag e lifeiime? 9.97 X 10“ - ' s I f th e u n c e r ta in ry in a p h o t o n ’s w a v e le n g ih is o n e p a rt in a m illio n , fin đ th e rainimum v a lu e o f th e u n c e r ta in ty in its position if the ph o to n ’s wavlength is (ư) 3000 A, (b) 0.5 A, and (c) X 10“** A A ns (fl)2 c m ;( ) X IQ-* Ả: (c) 15.9 A For an object o f size 0.5 Á, vvhat is the longest-vvavelengih photon with which il can be observed? Ans 0.5 Á 10.48 A bcam of neutrons with kineỉic cnergy 0.020eV is incident on KCl povvder The lattice spacing o f KCl is ỉ Ả NVTiat is the nidius o f the circie on a flat photo^raphic plaie placed cm behind the from íìrstorđer relìections from the Bragg pianes that are 3.14 A apart? Ans 3.85cm For the object o f Problem 10.62, measurement? Ans 602 eV 10.49 Refer to Problem 10.48 What is the radius of the circle due second-order rcAections from the same Bragg planes? Ans 28.9 cm For the object o f Problem 10.62, Nvhat is the smallest-energy proton which can be used to make the mcasurement? Ans 0.328 eV X 10"’^ W/m^ the smallest-energy eleciT on which can be If a photon were in the nucleus of Problem iO.29, vvhat would be its minimum energy? THE PROBABILITY INTERPRETATION OF DE BROGLIE WAVES 10.50 Do Problem 10.18 for > = 4000Ả and an ịntensity o f w h a t is Ans X 10"* photons/(s • m') 10.51 Rer to Problem 10.20 Suppose that the particle is an electron coníìned on a Hnc of length L = A(which is o f atomic dũnensions) Determine the lowest energy Ans 1.5 eV 10.52 Refer lo Problem 10.20 Supposc Ihat the particle is a smatl but macroscopic body o f mass 0.1 miUigram c o n íìn e d to a length L = 0.1 mm Determine its lowest cnergy Ans 3.43 X 10“^ eV TH E H E ISE N B E R G Ư N C ER TA IN TY P R IN C IP L E 10,53 Suppose that the x-componenl o f ihe velocity o f a X I0~^kg mass is measurcd to an accuracy o f ilO "^m /s What then is the limit o f the accuracy with vvhich we can ỉocate the particỉe along thc Xaxis? Ans 1.32 X 10"^^ m Ans hc _ ■= 4.9 MeV %Tzd ' u s e d to m a k e the The Bohr Atom Í 1 ĩiT H E H Y D R O G E N S PEC TR U M ''- v B ý 'th e e n d o f th e n in e te e n th c e n tu r y m u c h e x p e n m c n ta l vvork h a d b e e n d o n e o n th e a n a lv s is o f th e discrete spèctnim o f radiation emitted when electrical discharges were produced in gases The lightest and sunplest 01 all atoms is hydroeen, being composed o f a nucleus and one electron It was perhaps not suipnsing, ịthen, that very precise spectroscopic measurements showed that hydrogen had the simplest spéctnũìi o f all the elemcnts It was tbund that the various lines in the optical and nonoptical regions were systematicaỉly spaced ìn various series .Ajnazinaly, it tumed out that aỉl the wavelengihs o f atomic hyđrogen were aiven bv a single empirical relation the Rvciherg/ornnilu' R = 1.096 7758 / V '/ X 10“^ Ả ' «íi/ where ni — ^and«„ = 3, ves the ivm anienei-(ultraviolet region) = 2an d«„ = 4, giveslheBu/m tTieTíVííoptical reaion) ^ = 3andn„ = givesth eP ứ ic/ie/u ericiíin ữ ared reaion ) ' " / 1, = 4ÌndH „ = 5, 6, g iv e s th e S r o c í-« fíe n fj(fa r infrareilregion) 'ĩ and 50 onVor wher series lying in the tarther infrared U " T H E B Ỏ H R T H E O R Y O F T H E H Y D R O G E N ATOM eĩectron m'tts ó itit is the attractive Coulomb force: F = k ^ i- = 9.0 X lO’ N ■ mVc’ z w tó fe'ỉo r nlrogen = A straightfor\vard classical calculation (Problem 11.14) then shovvs that the orbit velocity o fI h e electron is relãted to the radius o f its o rb it, assumed by Bohr to be circular, by kzè- (11 I) whcrc m is the mass ot’ thc elcctron, and thc total energy o f the electron (klnetic encrgy Is givcn by E = - THE BOHR ATOM [CMAP II THE DOHR ATOM kzỷ ■potential cnergy) Substituting the values o f the orbital encrgies given by (11.ỉ ) , we then find iI Ì - ) wiih Now we comc 10 ứie point where B ohrs model diRers radically from a classical picture As the electron moves in ts orbit with linear momentum mu, it will have a de Broglie wavelength associated with it, given «^ = ^ Ĩ Ỉ Í ^ )W ) = (hcÝ bv Á = h, m r N ow , a w a v e c a n b e a s s o c ia te d w ith a g iv e n c ir c u la r o r b it o n ly if th e c irc u m te r e n c e o f th e orbit is an intcaral number ot' vvavplengths so that nÁ = — = 27tr mv L = m vr = n — 2rt or (11.3) where n = 1, 2, The quantity L = m vr is the anaular momentum o f the eleetron moving in its circular orbit, so it Is seen ứiat in the Bohr theory the ele ctro n s angiilar mom enium is quantized The i n te g e r n IS called th e p r i n c i p a l q u a n t u m n u m h e r Solvirm ( / / / ) , (11.2), and ( I I ỉ ) for the three unkno\vns r ,E , and u, one obtains the following quantized quantities; h._ m ” 27tA-w E N E R G Y L E V E L D ÍA G R ^\1S A convenient vvay to describe the transitions betvveen allowed states is in terms o f energy leveì diagrams In these, the allowed energy levels, given by (11.5), are plotted, as shovvn in Fig 11-1 for z = The tro n s itio n s are ữien in d ic a te d by a rro w s r u n n m g from the in itia l e n e r g y S tate, d e s ig n a te d b y n „ , to ứ ie íĩnal energy State, designated by í)( Thus, for example, the transitions that give rise to the Baliner series are indicated by the arrows ứiat end on rt, = The lines in the Balmer series are called H „ Hy, etc., as indicated in Fig 11-1 11.5 H Y D R O G E N IC A TO M S A hydrogenic atom is an atom that is stripped o f all but one o f its elecưons Thus, hydrogenic atoms are singly ionized helium (H e*, z = 2), double ionized lithium (Li-'*', z = 3), triply ionized beryllium (B e’’*', = 4) and so forth T hese atoms behave in all respects like hydrogen excep! that the nucleus has a positive charge o f Ze, where z is the atomic number o f the atom Equations ( / / / ) through {1 T) hold for hydrogenic atoms, provided the appropriate value o f z is used Pigure 11-2 shows the energy levels for H, H e*, and z S olved P roblem s 11.1 Determine, in angstroms, the shortest and longest wavelengths o f the Lyman series o f hydrogen Ans Wavelengths in the Lyman series are given by n, = 1: I ^ = (1.097 = ,3 , X The longest wavelengih corresponds lo - ^ = (1.097 X = 2; or 10 = 1215 Ả The shortest Nvavelength corresponds to5 Hy = co: Ạ -= ^in 11.2 1.097 X 10-’ Â- ■ *-'('- i ) or ^ = 912 Determine the wavelength o f the second line o f the Paschen series for hydrogen Ans ị = (1.097 X The Paschen series is deíined by n, = 3, and the second line coriesponds to ị = (1.097 11.3 X or = Hence, Ẳ = 12.820Ả The longest wavelength in tiie Lyman series for hydrogen is 1215 Ả Calculate the Rydbcrg constant For the Lyman series, n, = 1; ứie longest wavelength will correspond to tíie value 1215Ả Fig 11-2 R = 1.097 X 10-" A ' = 11.4 [CHAP 11 THE BOHR ATOM lOS THE BOHR ATOM This probỉem can also be solved using Rvdberg’s íormula, Detcrmine ữic wavclengths o f hydrogen that lie in the optical specưiim (3S00 A to 7700 A) The vvavelcngtlis for hydrogcn are given by ị = (1.097 X 10 Solving, À = 43-40 Ả In Problcm 11.1 it was found thM when n, = the wavelengths rangc from 12 À to 1215Ă,sothat none of ihese lie in the optical region For n, = the longest vvavdength coưcsponds to n, = 3, 11.7 giving Determine the ionization energy o f hydrogen if the shortest wavelength in the Balmer series is foand to be 3650 Ả Ans ị = (1.097 X or / = 6563Ả The Balmer series is given by n, = The shonest wavelength will correspond to n„ = 00 Consequently, from = -£7/n" where £J Ì5 the ionization cnergy, we have 4/ic ana iK n, is any pair o f unequal integers in the range to n, it is c le a r th a t th e re is a t le a í t One ro u te fro m State n d o w n to th e g r o u n d S tate th a t in c lu d e s th e tr a n s itio n - = (1.097 X or 10 = 9.9 n„ ->• n, Thus, the number of photons is equal to the number o f such paits, which is 3.S X lO ^ Ả Q Therefore, the lines in the optical region are given by ■ Ị = (1.097 n(n - 1) For n = 5, there are 5(4)/2 = 10 photons The above reasoning fails if there is “degeneracy,” i.e., if two diRerent paứs o f quantum numbere = X ~ co espond to the sam e ẽnergy difference In that casc the num ber o f distincl photons is sm aller than Since the shortest wavelength o f ửie Paschen-series (n, = 3) is ị = (1.097 X or n(n - l)/2 A = 8200Ả 1 all other series will give rise to lines lying outside the optical region In a transition to a State o f e x citation en erg y 1 eV, a h yd rogcn a to m eiĩiits a Ả p hoton Determine the binding energy o f the ừiitial State Ans 11.5 Evalute ứie ionization potential o f hydrogen, The energy o f the emìtted photon is hc 12.40 /iv = ^ = — in units o f eV X Ans In-lce^m _ 2nHke^Ý(.mc^) h- 11.6 ~ ụ,cÝ (12.40 X 10^ eV A )^ E „ = E ị + E ^ = - e V + 10.19eV = -3 e V Find the vvavelength o f the photon that is emÌRed when a hydrogen atom undergoes a transition from n„ = to n, = Ans The photon arises from the transition betwecn energy states such that E ^ - E , = hv, hence £ ^ _ (_ e V ) = 2.54eV From the Bohr model, the cnergy kvels are £„ = (-1 eV )/n ^ Hence, 2^ - — - 10’ eV • Ả , „ , ■ = 2.54 eV The excitation energy {£,) is the energy to excite ưie atom to a level above the ground State Therefore, the energy o f the level is 2it-( 14.40eV Ã)-(0.5I1 X 10‘ eV) _ ” X 89 X 10^ Ằ Therefore, the binding eneigy o f an electron in the Note that ưie transition corresponds to £ , = _ i ỉ l 52 £ X = _0.544eV £ , = -0 e V or State is 0.87 eV - - l ĩ, /l3.6eV , , From the Bohr postulates the energy o f the emitted photon is Ị ĩi / 6•6eV ev ' E; = -0 4 eV - (-3 eV) = 2.86 eV The wavelength of this photon is given by 1 E lectron s o f c n er g y 12 e V are fired at h yd rogen a tom s in a g as d isch a rg e tube D e te n n ũ ie the w a v e le n g th s o f the lin e s that can be em itted by the hyd rogen Ảns The maximum enetgy that can be absorb«d by a hydrogen atom is equal to the electron energy 12 cY Absorption of th is e n c r g y w o u ld e x c iie thc atom was in itia lỉy in thc g r o u n d State) THE ĐOHR ATOM [CHAP lỉ THE BOHR ATOM th e a to m in to a n e n e r g y State g iv e n by (a s s u m in g À-à q Ằ (h -/l\fX ^) hc ụ ~ ~ “ 2{Mc^)À £ , = £, + 12.2eV = - i3 e V + 12.2eV = -1 e V Tnc v a lu e of n c o rr e s p o n d in g to ứ iis State is f o u n d from E„ — l3.6eV ứ iu s Since ứie wavelenglhs are o f the order Á ứiereíore negligible 12.40 X 10’ eV Ả 2(939 X eV)À ' 10^ Ả the ữactional changc is of the ordcr 10"’ and is n = 3.l2 Since n musi btí an integer, the highest State ihai can be reached corresponđs to n = Hence (Probleni ỉ l.S) there are three possible vvavelengihs that will be emitted as ửie atom retums to ừic ground State, corresponding to th e transitions -*■ 2 — and -* l Thesc wavelengths are found trom = (1.097 X 10' Ẳ = 6563 Ả = (1.097 X 10' /1= 1215Ả = (1.097 X 10 = 1026 Ả 11.13 For hydrogen, shovv that when /1 » the ữequency oí' the emitted photon in a transition from n to n — equals ứie rotational frequcncy Ans The r o ta t io n a l ír e q u e n c y in State w 0J 2k n is Inke^/nh _^4ĩĩ‘ k^me* 2nn‘ h^/4n-kme^ V, 2nr^ The frequency o f the emitted photon Ì5 v = c \^ c R ^ À For By {11.4) and ( U S ) , /I » ° n H n - \Ý in -\r 1, 2n ^^=«„2^2 11.11 According to the Bohr th e o ry , h o w m a n y re v o lu tio n s w ill a n e le c tro n m a k e in ứ ie first e x c ite d State o f hydroeen if the lifetime in that State is 10” * s? Ans 6 x l0 -^ Ả X ^ 2:t"k^e*m _ 47i‘ k^me* fịì which is the samc as the rotationai frequency given above This problem illustraies Bohr’s corresponding principie, vvhich states that for large n a quantum equation should go over into the corresponđing classical equation According to classical theory, rađiation emitted from a rotating charge will have a írequency equal lo the rotationai ửequency the radius and orbital velocity for the State Í1 = are given by rj = r ỉ = 4(0.529 Ả) = 2.12 Ằ = 2.12 X 10-'° m y? c X 10* m /s 11.14 An eiecưon rolates in a circle around a nucỉeus with positive charge Ze How is the electron’s velocity related 10 the radius o f its orbit? The angular velocity is then v-y l.lO x I0®m/s ,, ^ Ans Equating ứic Coulomb forcc to ihe (elecữon rnass)x(cenmpetal acceleralion), k{e){Ze) and the total number o f revolutions is ^ = ÍÍỊ = 2n mv^ = X ,0 « V 6.28rad/rev 11.15 How is the total energy o f the electron in Problem 11.14 related to the radius o f its orbit? Ans The electrical potcntial energy o f the elcctron is 1 D e te r m in e the correctio n to the w a v e le n g th o f an e m itte d p h o to n w h e n the re co il k in etic en erg y o f the hydrogen nucleus is taken into account Mns A ssum ing that the atom isinitially at rest, conscrvaiion o f energy gives The kinetic eneigy o f the electton is found by using the resull o f Problem 11.14; + or ^ hc = f hc where K is the nuclear kinctic cnergy The fìi3t term on ứie lcft is 1/-Ỉ0 Xq and A are the uncorrected and actual wavelengths Thus 1 K Ảq À /ÍC Xq hc the second is \/X , wherc ^ nir The total energy is then XK hc The recoil momenrum o f the nuclcus is p s= y/ĩM K Then, by conservation o f momentum 11.16 Assuming that all transitions are possible, will ứie optical spectmm (3 0 Ả to 7 0 Ả) o f hydrogen have morê or fewer lines than ứie optical spectrum o f doubly ionized liứiium? [CHAP 11 TỈỈE BOHR ATOM CHAP 11| THE BOHR ATOM For hydrogen : Z-£? £ ỈS - £L ~ {n /ìÝ ForU^" : Hcnce the energy ievei diagram for contains all the energy levels of hydrogen, plus two extra levels for each hvdrogen level Since there are more levcls availablc, there will be more lines in the oplical spectrum of Li^'^ than in ữie opiical specưum of hydrogen 11.17 Deicrmine the mass raiio o f deuterium and hydrogen if, respectively, ứieir H , lines have wavelengths o f 6561.01 Ả and 6562 Ả, (It was through measurements o f this type that deuterium was discovered.) Ans In terms of the reduced mass o f the atom, the Rydberg formula is vvhere m and are the elecưonic and nuciear masses For a fìxed ữansìtion and íìxed that À is proponional to + (m /M ), so that z, this impĩies _ ịo '*H 1+ -^ Fig n - Àp - Ah M ọ M ị, -ỈH “ + “ - m m m m A/„ - ^Mọ Mh M „ /M M ọd /1 I = « p (ị-4 ) Substituting the daa and " i /M ịị = 1/1836, 1\ ” A = The ionization energy is the energy required to excite positronium ftom its ground S tate State = 00 Thus -1  l ả í ® - ) («( = 1) to the £ ,„ = hcRf = (12.40 X 10’ eV ■Ằ)(0.3485 X 10'^Ả"') = 6.8eV Solving, 11.20 Electrons are acceleratcd between a íìlament and a griđ through mercury vapor by a variable 11.18 Determine ứie Rydberg constant for posiưonium (a bound system com posed o f a positron and an electron) Ans The mass o f a position is the same as the mass o f an electron, so that « pp = = - ^2 = \ +{ m/ M) l+ (m /m ) Í2£ = 0.5485 X 10 “ ’ Á " ' p o ten tial K, as sh o w n in F ig l l - ( a ) A sm a ll retarding p o ten tia l, Vị i ^ O S V is m a in ta m ed b e tw e e n the grid and ứie c o lle c to r plate W h en th e c iư v e / in the c o lle c to r is m ea su r e d a s a ttm ction o f th e accelẽra tin g v o lta g e , th e cũ rv e o f F ig 1 -3 (6 ) is o b tain ed D e te r m in e th e fiis t ex c ita tio n energy o f mercury and the vvavelength o f the light em ined by mercury in the experiinent Ans In order to reach the collector the electrons must have a kìnetic energy greater than ữie retaiding potentiỉl energy o f about 0.5 eV bctwcen the grid and the collector As the accelerating potential is increased, the electrons acquire larger and laiger kinetic energies and hence more and mote reach the collector, resulting in an increasing cuưent Eventually, however, the electrons acquiie an cncrgy equal to the first excited S tate o f the m eroư y atoms A t this point thc electrons can excite the m ereuiy atoms into this State, thereby losing kinetic energy T h u s few er electrons w ill have suíB cient eneig y to 11.19 Rcfer to Problem 11.18 Find the ionization potential o f positronium.- overcome the retarding potential y^Ị resulting in the observed dip in the collector current In addition, thc mcrcuiy vapor, previously dark, wiíl emit radỉatỉon as the atoms rctura to Iheữ ground State THE BOHR ATOM THE BOilR ATOM [CHAP 11 Upon íurther increase in y , the current will again begin to increase because the electrons can acquirc additional kinetic cnergy alíer they excite a mercury actom At still grealer accclerating poicntials electrons will have suHìcicnt energy to excite two mercury atoms, resulting in second dip in ! and 30 on ("VVe are neglecting the possibility Ihat an elecư on m ight put a m ercury atom into a highcr excited staie This could happen, but special potential variations across the vapor tube would be rcquired.) The voltage dinerence benveen the vanous current peaks is thus secn to comspond to the energy required to cxcịte mcrcury into its íìrsT excited State, so that 11.24 Refer to Problems 11.21 through 11.23 For a -°"Pb /1-m esonic atom, what is the cnergy o f the photon given o ff in the first Lyiĩian transition (n„ = to n, = 1)? Ans £ , = A £ = £„ - E, = - Z = £ ^ i _ = 14.25 M = v A£ = eá K = 4.88eV (The potential of ihe lirst peak cannot be used because of the existence o ỉ Vg and various conlact potentials.) The W3velengths of the photon emined when the excited mercury atoms retum to their ground State is Su p p lem en tary Problem s 11.25 Repeat Problem 11.1 for ĩhe Balmer senes ■ _ £ _ ^ = jÍ!£ = l r i Ì Ì ! i X l A = 2540Ả ''” v “ A v~A £“ 4.88eV This experiment was íìrst pertbrmed by J Fnmck and G Hertz in 1914, and was the íirst expenment to demonsưate the existence o f stationary states in atoms, ÍUrther coníìnning Bohr's emerging quantum hvpothesis In addition, it shovved thai atoms can be excited by interacting with energetic eiectrons Ans 3646.À: 6563Ả 11.26 Determine in angstroms the vvavelength of the phoion emined in the transition third tnm sition in ửie Paschen series.) 11.27 /1/Í5 = to n, = (Tliis is the 1.094 X 10'*Ầ Calculate ứie shortesi-vvavelength photon in ứie series o f transitions wiih rt/= {the Bnickett series) Ans 1.459 x 10'* Ả 11.28 The shortesi vvaveicngth in the Balmer series for hyđrogen is 3646 Ả Determine the Rydberg constant from this vaỉue 1.097 X 10"^ Ả“' J1-M ES0NIC A N D TT-MESONĨC ATOM S In Yukawa’s explanation o f nuclear binding forces (i.e., strong interactions), the existence o f a panicle called a mesonic, o f rest mass 264 times the rest mass o f an elecơon, was predicted Two years after this prediction in 1937, a panicle with a rest mass o f 207 electron masses was discovered However, in 1946, it was show.li that this Ịi-m eson was not the predicted particle, and a short time after, ứie Yukawa particle, called a ji-meson, was found, Both 7t- and /j-m esons can be found with negative charge, and can thereíore form hydrogenic atoms The Bohr orbits o f ứiese particles, due to their large m asses, are much smaller than the electron’s orbits 11.29 Find the value of in the series that gives rise to the line in ihe hydrogen spectrum 1026 Ả (Note: this is in the Lyman series.) Ans 11.30 Repcat Problem 11.29 for the series.) Ans hyđrogen spectraỉ line at 486ÌẢ (Noie; this is in the Balmer 1131 Evaluate ke^ in uniis o f eV • Ả 1U Show that v ự c = 2ĩi ke^/hc = ũt is 1/137 The dimensionỉess qiumtity a is called the Jỉne structure consianĩ Aris 14.40 eV • Ả 11.21 Determine the ionization energy o f a íi-m esonic atom that is formed when a n-m eso n is captured by a proton Arts The analysis is identical in all respects to that of a hydrogen atom, with the mass, m, o f the electron replaccd by 207 m U From (//.^ ) evaluale the radius o f the íìrsi Bohr orbit o f hyđrogen in angstroms Ans 0.529 Ả £ ,„ = 207(13.6eV) = 2.82 keV 11.34 Determinc the ratio o f the Compton wavclcngth o f an clectron (Chapter 9) to ứic radius o f the íìrst Bohr orbii oíhyđrogen Ans 21.8 11.22 Refer to Probíem 11.21 Calculate the radius o f the íìrst Bohr orbit in “ *Pb (Z = 82) for a ^•mesonic atom Ans By ụ ! 4), rj varies inveiseiy with Zm Hence, 1U What is the minimum accelerating potential that will enable an electron to excite a hydrogen alom out o f its ground State? Ans 10.2 V ỉ ỉ J Determine the minimum energy ứiat must bc given to a hydrogen atom so ihat it can cmit ứie lữie corKsponds to a —►2 transitíon.) Ans 2.55 eV 11.23 For Problem 11.22, calculate the energy o f the íirst Bohr orbit A/ĩS U Determine the binding energy o f an clectron in the third cxciteđ S tate of hydrogen line (The Ans 0.85 cV By ( / / i), £ , varies directly with z^m Hence, £ , , = (82)^(207)(-13.58eV ) = -19.0M eV U What accelcrating potential wiU enable an electron to ionize a hydrogen atom? Ans 13.6V TIÍE BQMR ATOM 11.39 (CHAR l W h.it IS tho liig h ciil S tate ih a t u n c x c ite d h y đ r o g c n a to m s c a n r e a c h w h e n th c y artì b o m b a rd c d w iih e V n = eỉccirons? 1J.4() Find the rccoil cnerey o f a hydrogen atom when a phoion is em itted in a tninsition from n, , A n s- = 10 to x l O “ ''e V 11.41 Calcuiatc thc ín c tio n a l changc in the w avelength o f a spectral line that aríses ÍTom a sm all change in the reduccd mass o f ihc atom Ans AÂ/à = ~ ầ u /ị.i 11.42 D eterm inc the radius o f thtí second B ohr orbii tbr doubly ionized liihium .-Iní 0.705 Ả 11.43 For tnpiy ionized ber>'llium (Z = 4) d eterm ine the tirst Bohr o rbit radius Ảns Ầ 11.44 D eiem iine ihe vvavelcnỵth o f the ^ l-0*^"07 X 10~^ A''’) Ans line o f đeuterium if the Electron Orbìtal Motion line o f hydrogen is 4862.6 Ả 4861.3Ả 11.45 CaỉcuUte the first and second Bohr radii for positronium .-Im 1.06 Ả; 4.23 Ả 11.46 (ư) C alculaie the first thrce energy levels for positronium (ò) Finđ the w aveỉenẹth o f the H , line ( - ^ transition) o f positroniurTì (a) - e V , - e V , - e V ; (b) Ỉ3 A showĩj inTPíg is^ìhe';ỉ)ẽiT;)^< speL J h e direction o f L is given by the usual right-hand rule, as shown in the íìgure From N ew ton’s ị» n đ ] ,la w lith e n e t to rq u e T o n th e p a rtic le w ill b e equal to th e rate at w h ic h its angular m o m e n tu m c Ì ẩ ĩ{ ^ :Ì T ^ t/L /r fr Howcver, since the force on the panicle is a cenưal force, the torque exerted will be ạẹrt.'-Hencc w e particle’s angular momentum L will have a constant magnitude and dircction at every point Btical tnýectory, that is, angular momentum is conserved with the C entral force Fig 12-1 10WS various possible elUptical m otions, nm ging &om a circle to nearly a straight lúie, all Ịjor axis, 2a It can be shown that the total energy E (kinetic and potential) depends only major axis and hence will have the sa m e value (e.g., E = - k ê - Ị l a for the Coulomb 1 ) for all th e se e llip se s T h e o ib ita i an gu lar m o m en tu m , hovvever, w ill v ary from ranging continuously from a maximum value o f a ^ / —2m E for the circle to nearly zero for ly straight line (A straight-line ellipse has zero angular momentum because d = 'ng the trajectory.)