1 1. The discovery of the electron Measure the charge to mass ratio of electron § 26.1 Some important discoveries at the end of the 20 th century 22 2 LB yE m q = 2 J. J. Thomson (1856--1940) § 26.1 Some important discoveries at the end of the 20 th century J. J. Thomson’s original tube § 26.1 Some important discoveries at the end of the 20 th century 3 a. Measured the charge of the cathode rays; b. Make a static electric deflection of the cathode rays; c. Measured the charge to mass ratio of the cathode rays; d. Prove the universal existence of the electron. § 26.1 Some important discoveries at the end of the 20 th century The charge to mass ratio of electron C/kg107588.1 11 ×= m q The mass of electron kg1011.9 31? ×=m The charge of electron C10602.1 19? ×?=?= eq The discovery of electron was the first clear evidence that the indivisible atoms of nature had structure. § 26.1 Some important discoveries at the end of the 20 th century The charge of electron is quantized. 4 2. The discovery of X-rays § 26.1 Some important discoveries at the end of the 20 th century The characteristics of X-rays: a. It is generated whenever high-energy cathode rays strike solid materials. b. Matter is more or less transparent to X-rays. § 26.1 Some important discoveries at the end of the 20 th century 5 c. Photographic film is affected by X-rays, so its use as a detector was assured from the beginning. d. The rays is not deviated by electric and magnetic fields, and so are uncharged. The wave nature of X-rays makes them useful tools for the study of the structure of crystals and molecules, where the atoms and molecules acts as three-dimensional diffraction gratings. The applications of X-rays: § 26.1 Some important discoveries at the end of the 20 th century § 26.1 Some important discoveries at the end of the 20 th century 6 3. The discovery of radioactivity Radioactivity occur naturally and have with us on the earth from the very beginning. Henri Becquerel discovered Uranium, Marie Curie discovered Polonium and Radium Ernest Rutherford found that the substances emit several distinct types of radiations. One is a penetrating radiation, dubbed α, that propagates through several centimeters in air and can even penetrate very thin metal foils. Another less penetrating radiation, dubbed β, is easily stopped by even a sheet of paper. Another type, called γ, was discovered in 1900 and is much more penetrating than even the α radiation. § 26.1 Some important discoveries at the end of the 20 th century § 26.2 The appearance of Plank’s constant h 1. Blackbody Radiation 1400K800K 1000K 1200K ?Thermal radiation Sr Rb Cu The color emitted by atoms after they have exited by heat is characteristic of the particular element they comprise. 7 ?Ideal blackbody A body that absorbs all radiation incident on it is called ideal blackbody. 1879 Josef Stefan found an empirical relation between the power per unit area radiated by a blackbody and the temperature. 4 TR σ= § 26.2 The appearance of Plank’s constant h § 26.2 The appearance of Plank’s constant h ?Wien’s displacement law Km1028978.0 2 max ?×= ? Tλ 8 0 1 2 3 4 5 6 7 8 9 m)(μλ ),( 0 Te λ ?Experimental law of blackbody radiation 4 ),( ? = λλ CTTe o Wien T C o eCTe λ λλ 2 5 1 ),( ? ? = Rayleigh-Jeans Ultraviolet catastrophe § 26.2 The appearance of Plank’s constant h 152 0 )1(2),( ?? ?= Tk hc ehcTe λ λπλ 0 λ (T,λ) 0 e 2. Plank’s law and Plank’s constant § 26.2 The appearance of Plank’s constant h Every attempt to explain the blackbody spectrum based on electromagnetic theory and thermodynamics failed to predict the shape of the blackbody spectrum. 9 § 26.2 The appearance of Plank’s constant h In 1901, Plank assumed that the energy E associated with the light inside the cavity was present only in finite packets (bundles ) proportional to the frequency ν. νhE = Where h was an unknown constant that he hoped to be able to set to zero after taking appropriate mathematical limits. sJ10626.6 34 ?×= ? h It is a fundamental constant of nature, called by Plank the quantum of action. Now we called it Plank’s constant. § 26.3 The photoelectric effect 1. Apparatus and phenomenon A photocell is made by enclosing a metal plate and a collecting wire in an evacuated glass tube. EM radiation(visible or UVB) falls on the metal plate; some of the emitted electrons make their way to the collecting wire, which complete the circuit. An ammeter measures the current in the circuit. 10 2. Experimental results and the troubles of the classical theory ? Bright light causes an increase in current but does not cause the individual electrons to gain higher energies. The maximum kinetic energy of the electrons is independent of the intensity of the light. Classically, more intense light has larger amplitude EM field and thus delivers more energy. That should not only enable a larger number of electrons to escape from the metal, it should also enable the electrons emitted to have more kinetic energy. § 26.3 The photoelectric effect ? The maximum kinetic energy of emitted electrons does depend on the frequency of the incident radiation. Thus, if the incident light is very dim but high in frequency, electrons with large kinetic energies are released. Classically, there is no explanation for a frequency dependence. ? For a given metal, there is a threshold frequency ν c . If the frequency of the incident light is below the threshold, no electrons are emitted—no matter what the intensity of the incident light. Again, classical physics has no explanation of the frequency dependence. § 26.3 The photoelectric effect 11 ? When EM radiation falls on the metal, electrons are emitted virtually instantaneously; the time delay observed experimentally is about 10 -9 s, regardless of the light intensity. If the EM radiation behaves as a classical wave, when the intensity of the light is low, it should take some time for enough energy to accumulate on a particular spot to liberate an electron. ? For a given frequency of light, if we reverse the polarity of the potential difference and increase the value of the potential difference, the current in the circuit decreases to zero, i. e. there is a stopping potential V s . § 26.3 The photoelectric effect c ν (ν ν § 26.3 The photoelectric effect 12 max KEWh +=ν Slope h eV s ννν c hν W max KE -W 2. Einstein’s photon theory νhE = photon Einstein imagined light to consist of wavelike particles, pockets(bundles) of electromagnetic energy, called photon. s eVmvKE == 2 mmax 2 1 c hW ν= νhWeV s +?= § 26.3 The photoelectric effect Example 1: A low Helium-neon laser has a power output of 1.00 mW of light of wavelength 632,8 nm.(a) Calculate the energy of each photon, expressing your result in joules and electron –volts; (b) Determine the number of photons emitted by the laser each second. Solution: 1.96eVJ1013.3 108.632 )100.3)(10626.6( 19 9 834 =×= × ×× = == ? ? ? λ ν hc hE (a) § 26.3 The photoelectric effect 13 photons/s1018.3 1014.3 1000.1 15 19 3 ×= × × = ? ? N(b) Example 2: Ultraviolet light of wavelength 200 nm is incident on a freshly polished iron surface. Find (a) the stopping potential; (b)the maximum kinetic energy of the liberated electrons; and (c) the speed of these fastest electrons. Solution: (a) s eVWh +=ν From Table 26.1 J105.710602.17.4eV7.4 1919 ?? ×=××==W § 26.3 The photoelectric effect then V5.1 / = ? = ? = e Whc e Wh V s λν (b) The maximum kinetic energy J104.2 19 max ? ×=?== W hc eVKE s λ (c) The speed of fastest electrons m/s103.7 1011.9 104.222 J104.2 2 1 5 31 19 192 max ×= × ×× == ×== ? ? ? m eV v eVmv s s § 26.3 The photoelectric effect 14 § 26.4 The momentum of a photon—Compton effect 1. The phenomenon of experiment λ′ depend only on the scattering angle θ . 1923, Arthur H. Compton The spectrum of the scattered X- ray consisted of the incident wavelength as well as one shifted slightly to a longer wavelength. λλ > ′ θ graphite 2. The experimental results θ Target λ λλ ′ λλ ′ λλ ′ λλ ′ ? θ↑ , λ′ ↑ § 26.4 The momentum of a photon—Compton effect 15 § 26.4 The momentum of a photon—Compton effect o 0=θ o 45=θ o 90=θ o 135=θ ??λ is independent of the scattering materials; the intensity of the scattering light is decreased as the atom ordinal number increases. Incident light λλ ′ λλ ′ § 26.4 The momentum of a photon—Compton effect 16 3. The trouble of the classical physics theory According to the electromagnetic theory and the forced oscillation mechanism of a particle, the wavelength of scattering light could not be longer than wavelength of incident light. 4. The momentum of photons 42222 cmcpE += From special relativity theory For photons, the rest mass is zero, therefore pcE = λ ν h c h c E p === § 26.4 The momentum of a photon—Compton effect Photon as a particle: λ ν h p hE = = 5. Compton’s explanation Compton assumed that an incoming X-ray photon collided elastically with an electron, conserving both energy and momentum. A recoiling electron meant the scattered X-ray photon would have less energy and, therefore, a longer wavelength than the incident X-ray. § 26.4 The momentum of a photon—Compton effect 17 For the elastic collision of two particles jmvimvj h i h i h mchmch ? sin ? cos ? sin ? cos ? 22 θγφγθ λ θ λλ γνν ?+ ′ + ′ = + ′ =+ i h p i ? λ = r j h i h p ? sin ? cos fp θ λ θ λ ′ + ′ = r jmvimvp ? sin ? cos fe φγφγ +=′ r φ θ x y φ νh νh m v r θ x y § 26.4 The momentum of a photon—Compton effect 2 sin2 2 θ λλλλ? c =?′= nm00243.0== mc h c λ --Compton wavelength 2 sin 2 )cos1( 2 θ θλλ mc h mc h = ?=? ′ φγθ λ φγθ λλ γ λλ sinsin0 coscos 22 mv h mv hh mc c hmc c h ? ′ = + ′ = + ′ =+ § 26.4 The momentum of a photon—Compton effect 18 Example 1: An X-ray photon of wavelength 0.01 nm is scattered through 110.0° by an electron. What is the kinetic energy of the recoiling electron? Solution: λλ νν ′ ?= ′ ?= hchc hhKE According to energy conservation nm00326.0 2 110 sin00243.02 2 sin2 2 2 =×= =? ′ = θ λλλλ? c § 26.4 The momentum of a photon—Compton effect nm01326.000326.001.0 =+=+= ′ λ?λλ KeV5.30= + ?= ′ ?= λ?λλλλ hchchchc KE Example 2: A X-ray photon with energy of 0.5MeV incident on a certain material and scattered. If the kinetic energy of the recoiling electron is 0.1 MeV, then what is the ratio of ?λ to λ. § 26.4 The momentum of a photon—Compton effect Solution:According to the energy conservation λλ νν ′ ?=′?= hchc hhKE 19 λ λλ λ λ 4 5 5.0 1.0 1 = ? = ? = ? = ′ E hc KEhc hc KE hc hc 25.0 4 5 = ? = λ λλ λ λ? § 26.4 The momentum of a photon—Compton effect 6. The radiation pressure i h p i ? λ = r i h p ? λ ?= r mirror p r i h pi h ?? mirror λλ =+? r if pp rr =Momentum conservation i h pp ? 2 mirrormirror λ ? == rr tFp ?? r r = t p F ? ? r r = § 26.4 The momentum of a photon—Compton effect 20 The comet and radiation pressure § 26.4 The momentum of a photon—Compton effect Planck Einstein Bohr § 26.5 The structure of atom and the Bohr model of a hydrogen atom 21 § 26.5 The structure of atom and the Bohr model of a hydrogen atom 1. The experimental results of atomic spectrum The spectrum of hydrogen atoms § 26.5 The structure of atom and the Bohr model of a hydrogen atom Empirical formula(Johann Balmer) nm)( 4 6.364 2 2 ? = n n n λ (n=3,4,5, …) Rydberg-Ritz formula ) 11 ( 1 22 mn R n ?= λ For m>n where 17 17 m10097373.1 m10096776.1 ? ∞ ? ×= ×= R R H For hydrogen For heavy elements 22 § 26.5 The structure of atom and the Bohr model of a hydrogen atom 2. Thomson’s atomic model—plum pudding This model explained: ?Atoms are electrically neutral; ?Atoms emitted electromagnetic waves duo to the oscillation of the electrons in the atom. electron Glob of positive charge J. J. Thomson (1856--1940) r E R 3. Rutherford’s atomic model —scattering of α particles 1 β 12 ββ > 2 β We expect 2 0 ex 4 r ze E πε = 3 0 in 4 R zer E πε = 2 0 2 ex 4 2 2 r Ze eEF πε == R r R Ze eEF 2 0 2 in 4 2 2 πε == § 26.5 The structure of atom and the Bohr model of a hydrogen atom 23 The experiments revealed that a very small number of the α particles were scattered through quite a large angle, even in the backward direction. Experimental results The experimental results forced Rutherford to conclude that atom has a minute massive center carrying charge called nucleus. From analysis, Rutherford estimated the diameter of the nucleus to be 10 -15 m. the diameter of atoms was known to be 10 -10 m. Thus most of an atom is empty space. § 26.5 The structure of atom and the Bohr model of a hydrogen atom rm e rr v mv e r 0 2 2 0 2 42 1 2 , 4 πεππ ν πε === 4. The Bohr model of a hydrogenic atom Bohr model nucleus Hydrogen atoms consiste of a nucleus with positive charge +e and a single electron. r mv r ee F 2 2 0 4 == πε Atoms should emit continuous spectrum. § 26.5 The structure of atom and the Bohr model of a hydrogen atom r e mvE 0 2 2 42 1 πε ?= 24 α H β H γ H δ H 5 5 6 2 . 8 4 8 6 1 . 3 4 3 4 0 . 5 4 1 0 1 . 7 3 9 7 0 . 1 The spectrum of hydrogen atoms The problems are 1the atoms emit discrete spectrum! 2the atoms are stable. § 26.5 The structure of atom and the Bohr model of a hydrogen atom Bohr’s postulates about the Hydrogen atoms: 1stationary state: the electron in an particular orbit was in a stable state, and simply would not radiate energy in spite of classical electrodynamics. r Ze mv PEmvE 2 0 2 elec 2 4 1 2 1 2 1 πε ?= += 2the quantized angular momentum: the magnitude of the orbital angular momentum of the electron was restricted to only certain values. § 26.5 The structure of atom and the Bohr model of a hydrogen atom 25 π2 orbit h nL = or hnL = orbit sJ10055.1 34 ?×= ? h According to the definition of the orbital angular momentum of a particle h rr r nrmvrp prL == ×= orbit mr n v h = Substitute for v in Newton’s second law equation 2 2 2 0 )( 4 1 mr n r m r Ze h = πε 2 0 22 )4( mZe n r πεh = n is integer, the electron orbits is only permitted at some specific radii, called allowed orbits. § 26.5 The structure of atom and the Bohr model of a hydrogen atom 222 0 42 )4(2 1 hn emZ E πε ?= r Ze mvE 2 0 2 4 1 2 1 πε ?= Only certain values of the energy are permitted. eV 6.13 )eV6.13( 22 2 nn Z E ?=?= n=1 is called ground state; the other values of n correspond to excited states. 3transition: when electrons on high energy state cascade down to the lower energy state, atoms emit photons or light wave § 26.5 The structure of atom and the Bohr model of a hydrogen atom 26 n E 6.13? 40.3? 51.1? 85.0? energy null eV null 1=n 2=n 3=n 4=n 5=n ∞=n 0 连续 Energy level diagram Bohr’s objective was to explain the discrete spectrum of hydrogen, the fact that only specific wavelengths of light are emitted by a collection of hydrogen atoms. If n >1, then photonemitted EEE fi += fi EEE =+ absorbedphoton § 26.5 The structure of atom and the Bohr model of a hydrogen atom eV 6.13 )eV6.13( 22 2 nn Z E ?=?= According to the transition hypothesis ) 11 (6.13 22 photonemitted nm EEhE if ??=?== ν 17 834 19 m10097.1 109979.210626.6 10602.16.13 ? ? ? ×= ××× ×× =R It is agreement with the experiment results. § 26.5 The structure of atom and the Bohr model of a hydrogen atom ) 11 (6.13 22 mn hc n ?= λ ) 11 ( 1 22 mn R n ?= λ 27 § 26.5 The structure of atom and the Bohr model of a hydrogen atom § 26.6 The de Broglie hypothesis 1. The de Broglie hypothesis and matter waves The true nature of light is difficult to assess. Light exhibits nature of wave showed by diffraction and interference. The photoelectric and Compton effects indicate that light has nature of particle. Light exhibits a wave-like duality. λ ν h mcp hmcE == == 2 photons The wave-particle duality was extended to particles by Louis de Broglie in the early 1920s. 28 λ ν h mvp hmcE == == 2 particles Louis de Broglie won the Nobel prize in physics in 1929. § 26.6 The de Broglie hypothesis Matter wave De Broglie wavelength of matter waves 1a Ping-pong Ball of mass 2.0 g with a speed of 5 m/s m106.6 5100.2 1063.6 32 3 34 ? ? ? ×= ×× × == mv h λ 2The Earth kg1098.5 24 ×=m 1 orbit skm8.29 ? ?=v mv h =λ (m)1072.3 1098.21098.5 1063.6 63 424 34 ? ? ×= ××× × = § 26.6 The de Broglie hypothesis 29 3electrons accelerated by an electric field § 26.6 The de Broglie hypothesis 2 0 2 0 2 0 2 )( 11 EKEE c EE c p ?+=?= 2 0 2 1 KEKEE c += VeVemc hc p h ?? λ )2( 2 + == 222 0 2 pcEE += According to special relativity VeKE ?=from CWE theorem V100=V? nm1225.0=λ rayX~ ?For instance Two special cases: )nm( 225.1 2 VVme h ?? λ =≈ 0 EKE << VMe51.0<<Ve? a. b. 0 EKE>> nm 1024.1 3 VVe hc ?? λ × =≈ V10 6 =V? nm1024.1 3? ×=λFor instance Stiff X-ray or γ-ray § 26.6 The de Broglie hypothesis 30 2. The experimental verifications 1Clinton Davisson and Lester Germer nm167.0 54 225.1225.1 === V? λ According to de Broglie hypothesis § 26.6 The de Broglie hypothesis o 50=φ o 90=φ o 0=φ Nickel crystal λ? mdx == o 65sin2 nm165.0=λ 1=m λθ? mdx == sin2 o 50=φ According to X-ray diffraction and Bragg’s formula § 26.6 The de Broglie hypothesis 31 2 George Thomson’s experiment Clinton Davisson and George Thomson were awarded Nobel prize in physics in 1937. § 26.6 The de Broglie hypothesis 3diffraction of electrons through an aperture with straight edge Visible light § 26.6 The de Broglie hypothesis 38-keV electron beam 32 4In 1989, Tonomura double slits diffraction of electrons 100 3000 20000 70000 § 26.6 The de Broglie hypothesis ⊕ Electron beam Thin thread with positive charge 50V 50V -20 -10 0 10 20 angle intensity ×0.1 4The other experiments In 1929, Stern accomplished the diffraction experiment of molecules. § 26.6 The de Broglie hypothesis 33 In 1936, diffraction of neutrons Particles have the wave-particle duality. § 26.6 The de Broglie hypothesis Example: suppose the de Broglie wavelengths of photons and electrons are same, are their momentum and energy same? Solution: pho 22 2 e e 2 pho pho E v chc v c v cp v vcm cmE h vmp hc hmcE h mcp ee e e ===== == === == λ λ λ ν λ c v c h cm vm h u e e >=== 22 λνPhase speed § 26.6 The de Broglie hypothesis 34 3. The allowed orbits of electron λππ π nr p h nr h nmvrnL =?= =?= 22 2 h The allowed orbits of electron are precisely those for which an integral number of electron de Broglie wavelengths fit along the circumference. § 26.6 The de Broglie hypothesis 4. The radius of the orbit r m r e 2 2 0 2 4 υ πε =and According to r nhh p πλ 2 == r e mr hn m p mvE k 0 2 22 222 2 8822 1 πεπ ==== 2 2 2 0 n me h r n ?= π ε o A053.0 2 2 0 1 == me h r π ε The orbits of the electron is not continuous because of the wave nature of the electrons. Bohr radius § 26.6 The de Broglie hypothesis 35 § 26.7 Probability waves Single photon Semitransparent mirror Receiver A R T Receiver B 1. Nonclassical particle ?Single photon interference experiment 1 Statistical conclusion: the number of photons by transmission and reflection is 50% , respectively. Reflecting mirror Receiver A Receiver B Reflecting mirror Semitransparent mirror Single photon ?Single photon interference experiment 2 Results: when the two paths of the photon have definite difference, only receiver A detects photons; change the path difference, only receiver B detects the photons. The results indicate that the interference come from single photon, wave concomitant photon . § 26.7 Probability waves 36 NNhIhE ∝== νν, photon current: § 26.7 Probability waves Photons: Electrons: Bright: more electrons --large probability Dark: no electron --zero probability Bright: more photons --large probability Dark: no photons --zero probability microscopic particles are different from neither the classical particle nor the classical waves § 26.7 Probability waves 37 Microscopic particles (Photons and electrons …) are not classical particle! 1 Microscopic particles is a wave with a phase factor; 2 Microscopic particles have no definite moving paths or trajectory; 3 Microscopic particles can be here and there simultaneously before measurement. We must reconstruct the picture of particles ! § 26.7 Probability waves ),,,(),( tzyxtr ΨΨ = r 2. Probability waves For classical waves, the intensity of wave 2 ),,,(),,,( tzyxtzyxI Ψ∝ For particles, the intensity of wave ),,,(),,,( tzyxPNtzyxI ∝∝ 1926, Born suppose（ 1954 Nobel prize in physics） VtzyxVtzyxP d),,,(d),,,( 2 Ψ= Probability distribution function or Probability density Unify the natures of particle and wave. § 26.7 Probability waves