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PROGRAM OF “PHYSICS2B” Lecturer: Dr. DO Xuan Hoi Room A1.413 E-mail : [email protected] ANALYTICAL PHYSICS 2B 03 credits (45 periods) Chapter 1 Geometric Optics Chapter 2 Wave Optics Chapter 3 Relativity Chapter 4 Quantum Physics Chapter 5 Nuclear Physics Chapter 6 The Standard Model of Particle Physics References : Young and Freedman, University Physics, Volume 2, 12th Edition, Pearson/Addison Wesley, San Francisco, 2007 Halliday D., Resnick R. and Merrill, J. (1988), Fundamentals of Physics, Extended third edition. John Willey and Sons, Inc. Alonso M. and Finn E.J. (1992), Physics, Addison-Wesley Publishing Company Hecht, E. (2000), Physics. Calculus, Second Edition. Brooks/Cole. Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole. Roger Muncaster (1994), A-Level Physics, Stanley Thornes. http://ocw.mit.edu/OcwWeb/Physics/index.htm http://www.opensourcephysics.org/index.html http://hyperphysics.phyastr.gsu.edu/hbase/HFrame.html http://www.practicalphysics.org/go/Default.ht ml http://www.msm.cam.ac.uk/ http://www.iop.org/index.html . . . PHYSICS 2B Chapter 4 Quantum Physics & Quantum Mechanics Nature of light - Matter Wave The Schrödinger’s Equation The Heisenberg’s uncertainty principle The Bohr Atom The Schrödinger Equation for the Hydrogen Atom The Zeeman Effect The Dirac Equation and antimatter 1. Nature of light - Matter Wave Diffraction, Interference phenomena light has wave nature Photoelectric, Compton’s effects light has particle nature What is the nature of light? ANSWER: From a modern viewpoint, the light has both wave and particle characteristics That is The Wave-Particle Duality of Light “… the wave and corpuscular descriptions are only to be regarded as complementary ways of viewing one and the same objective process…” (Pauli, physicist) 1. 1 Continuous and line spectra Prism (or a diffraction grating) to separate the various wavelengths in a beam of light into a spectrum Light source: Hot solid (such as a light-bulb filament) or liquid continuous spectrum Source: Gas line spectrum (isolated sharp parallel lines) Each line: a definite wavelength and frequency. Continuous spectrum White light: 380 nm (violet) 760 nm (red) Line spectrum Explain? 1. 2 Rutherford's gold foil experiment The scattering of alpha particles () by a thin metal foil: The source of alpha particles: a radioactive element (Ra) alpha particles are scattered by the gold foil. Result: The atom has a nucleus; a very small, very dense structure (10-14 m in diameter; 99.95% of the total mass of the atom) 1. 3 The photoelectric effect The photoelectric effect: light strikes a metal surface emission of electrons electron light _ _ _ _ ++ + + metal Einstein's Photon Explanation: a beam of light consists of small packages of energy called photons or quanta. The energy of a photon: hc E hf f, : frequency, wavelength of the light Planck's constant h = 6.626069310-34J.s The photoelectric effect: hc E hf K max : work function (minimum energy needed to remove an electron from the surface) Maximum kinetic energy : K max 1 mv 2max 2 V0: The stopping potential Theorem of variation of kinetic energy: K max K 'max eV0 K 'max 0 K max 1 2 mv max eV0 2 EXAMPLE While conducting a photoelectric-effect experiment with light of a certain frequency, you find that a reverse potential difference of 1.25 V is required to reduce the current to zero. Find (a) the maximum kinetic energy and (b) the maximum speed of the emitted photoelectrons. (a) (b) EXAMPLE When ultraviolet light with a wavelength of 254 nm falls on a clean copper surface, the stopping potential necessary to stop emission of photoelectrons is 0.181 V. What is the photoelectric threshold wavelength for this copper surface? Photon Momentum: E hf h p c c Photons have zero rest mass. The direction of the photon's momentum is the direction in which the electromagnetic wave is. Compton Scattering When X rays strike matter, some of the radiation is scattered: Some of the scattered radiation has smaller frequency (longer wavelength) than the incident radiation and that the change in wavelength depends on the angle through which the radiation is scattered. Compton Scattering 2 Energy conservation: pc mc p'c Ee Momentum conservation: p p ' Pe Pe2 p2 p'2 2pp'cos h h With p and p ' : ' h ' 1 cos mc (Compton shift) EXAMPLE You use the x-ray photons = 0.124 nm in a Compton scattering experiment. (a) At what angle is the wavelength of the scattered X rays 1.0% longer than that of the incident X rays? (b) At what angle is it 0.050% longer? (a) (b) 1.4 The Davisson - Germer experiment Electron diffraction experiment (Davisson, Germer, Thomson, 1927) A beam of either X rays (wave) or electrons (particle) is directed onto a target which is a gold foil The scatter of X rays or electrons by the gold crystal produces a circular diffraction pattern on a photographic film The pattern of the two experiments are the same Particle could act like a wave; both X rays and electrons are WAVE Discussion Similar diffraction and interference experiments have been realized with protons, neutrons, and various atoms (see “Atomic Interferometer”, Xuan Hoi DO, report of practice of Bachelor degree of Physics, University Paris-North, 1993) In 1994, it was demonstrated with iodine molecules I2 Wave interference patterns of atoms Small objects like electrons, protons, neutrons, atoms, and molecules travel as waves - we call these waves “matter waves” 1.5 De Broglie’s Theory - Matter Wave De Broglie’s relationships We recall that for a photon (E, p) associated to an electromagnetic wave (f, ): h f 1 p h E particle wave De Broglie’s hypothesis: To a particle (E, p) is associated a matter wave which has a frequency f and a wavelength E f h h p E h From f and h p if we put: h 2 E p K E 2f 2 p Planck-Einstein’s relation is called de Broglie wavelength Conclusion. Necessity of a new science When we deal with a wave, there is always some quantity that varies (the coordinate u, the electricity field E) For a matter wave, what is varying? ANSWER: That is the WAVE FUNCTION New brand of physics: THE QUANTUM MECHANICS PROBLEM 1 In a research laboratory, electrons are accelerated to speed of 6.0 106 m/s. Nearby, a 1.0 10-9 kg speck of dust falls through the air at a speed of 0.020m/s. Calculate the de Broglie wavelength in both case For the electron: SOLUTION h h 6.625 1034 J .s p mv 9.111031 kg 6.0 106 m / s 1.2 10 10 m For the dust speck: h h 6.625 1034 J .s d pd mvd 1.0 109 kg 6.0 0.020m / s d 3.3 10 23 m DISCUSSION: The de Broglie wavelength of the dust speck is so small that we do not observe its wavelike behavior PROBLEM 2 An electron microscope uses 40-keV electrons. Find the wavelength of this electron. SOLUTION The velocity of this electron: v 2K / m v 2 40 103 1.6 1019 1.2 108 m / s 9.1 1031 The wavelength of this electron: h mv o 6.63 1034 10 m 6.1 A 31 8 6.1 10 9.110 1.2 10 2 The Schrödinger’s Equation 2.1 Wave Function and Probability Density Matter waves: A moving particle (electron, photon) with momentum p is described by a matter wave; its wavelength is h / p A matter wave is described by a wave function: ( x, y, z; t ) (called uppercase psi) ( x, y, z; t ) is a complex number(a + ib with i2 = -1, a, b: real numbers) ( x, y, z; t ) depends on the space (x, y, z) and on the time (t) The space and the time can be grouped separately: ( x, y, z; t ) ( x, y, z ) e it ( x, y, z ) : space- dependentpart (lower casepsi) i t : time - dependentpart ( : angular frequency) e • The meaning of the wave function: ( x, y, z ) The function ( x, y, z ) has no meaning 2 Only has a physical meaning. That is: The probability per unit time of detecting a particle in a small volume centered on a given point in the matter 2wave is proportional to the value at that point of 2 greater it is easier to find the particle * * N.B.: with is the complex conjugate of If we write a ib * a ib (a, b: real numbers) 2 • How can we find the wave equation? Like sound waves described by Newtonian mechanics, or electromagnetic waves by Maxwell’s equation, matter waves are described by an equation called Schrödinger’s equation (1926) 2.2 The Schrödinger’s equation For the case of one-dimensional motion, when a particle with the mass m has a potential energy U(x) Schrödinger’s equation is d 2 dx 2 8 2 m h 2 [ E U ( x)] 0 where E is total mechanical energy (potential energy plus kinetic energy) Schrödinger’s equation is the basic principle (we cannot derive it from more basic principles) EXAMPLE: Waves of a free particle For a free particle, there is no net force acting on it, so 1 2 U ( x) 0 and E mv 2 d 2 8 2 m 1 2 Schrödinger’s equation becomes: 2 2 2 mv 0 dx h d 2 2 p 2 By replacing: mv p 2 h 0 dx 1 p With the de Broglie wavelength: h 2 and the wave number: K we have the Schrödinger’s equation for free particle: d 2 K 2 0 dx 2 This differential equation has the most general solution: ( x) AeiKx Be iKx (A and B are arbitrary constants) The time-dependent wave function: ( x, t ) ( x) ei t ( AeiKx Be iKx ) ei t ( x, t ) Aei ( Kx t ) Be i ( Kx t ) ( x, t ) Aei ( Kx t ) Be i ( Kx t ) : traveling waves Aei ( Kx t ) : wave traveling in the direction of increasing x Be i (kx t ) : wave traveling in the negative direction of x Probability density: Assume that the free particle travels only in the positive direction Relabel the constant A as 0 : ( x) 0eiKx The probability density is: 2 0e iKx 2 2 iKx 2 (0 ) e Because: e iKx 2 (eiKx )(eiKx )* (eiKx )(eiKx ) e0 1 we have: 2 (0 ) 2 const What is the meaning of a constant probability? 2 (0 ) 2 const 2 The plot of (0 ) 2 const is a straight line parallel to the x axis: 2 0 2 x O The probability density is the same for all values of x The particle has equal probabilities of being anywhere along the x axis: all positions are equally likely expected 3 The Heisenberg’s uncertainty principle • In the example of a free particle, we see that if its momentum is completely specified, then its position is completely unspecified • When the momentum p is completely specified we write: p 0 (because: p p1 p2 0) and when the position x is completely unspecified we write: x • In general, we always have: x p a constant This constant is known as: (called h-bar) h 2 h is the Planck’s constant (h 6.625 1034 J .s) So we can write: x p That is the Heisenberg’s uncertainty principle “ it is impossible to know simultaneously and with exactness both the position and the momentum of the fundamental particles” N.B.: • We also have for the particle moving in three dimensions x px y p y z pz • With the definition of the constant : p h / hK / 2 • Uncertainty for energy : E t p K PROBLEM 3 An electron is moving along x axis with the speed of 2×106 m/s (known with a precision of 0.50%). What is the minimum uncertainty with which we can simultaneously measure the position of the electron along the x axis? Given the mass of an electron 9.1×10-31 kg SOLUTION From the uncertainty principle: x p if we want to have the minimum uncertainty: x p We evaluate the momentum: p mv (9.11031) (2.05 106 ) p 9.35 1027 kg.m / s The uncertainty of the momentum is: p 0.5% p 0.5 / 100 1.87 1024 9.35 1027 kg.m / s 6.635 1034 / 2 8 1 . 13 10 m 11nm x 27 p 9.35 10 PROBLEM 4 In an experiment, an electron is determined to be within 0.1mm of a particular point. If we try to measure the electron’s velocity, what will be the minimum uncertainty? SOLUTION p v m mx 6.63 1034 J .s v 9.11031 kg 1.0 104 m 2 v 1.2m / s Observation: We can predict the velocity of the electron to within 1.2m/s. Locating the electron at one position affects our ability to know where it will be at later times PROBLEM 5 A grain of sand with the mass of 1.00 mg appears to be at rest on a smooth surface. We locate its position to within 0.01mm. What velocity limit is implied by our measurement of its position? SOLUTION p v m mx 6.63 1034 J .s v 1106 kg 1.0 105 m 2 v 1.11023 m / s Observation: The uncertainty of velocity of the grain is so small that we do not observe it: The grain of sand may still be considered at rest, as our experience says it should PROBLEM 6 An electron is confined within a region of width 1.010- 10 m. (a) Estimate the minimum uncertainty in the x-component of the electron's momentum. (b) If the electron has momentum with magnitude equal to the uncertainty found in part (a), what is its kinetic energy? Express the result in jou1es and in electron volts. (a) (b) SOLUTION PROBLEM 7 A sodium atom is in one of the states labeled ''Lowest excited levels". It remains in that state for an average time of 1.610-8 s before it makes a transition back to a ground state, emitting a photon with wavelength 589.0 nm and energy 2.105 eV. What is the uncertainty in energy of that excited state? What is the wavelength spread of the corresponding spectrum line? SOLUTION The fractional uncertainty of the photon energy is 4 The Bohr Atom 4.1 The energy levels The hydrogen atom consists of a single electron (charge – e) bound to its central nucleus, a single proton (charge + e), by an attractive Coulomb force 1 (e)(e) e2 The electric potential energy is : U 40 r 40 r We can demonstrate that the energies of the quantum electron is me4 1 En 2 2 2 8 0 h n En 13.6eV n2 where n is an integer, that is the principal quantum number The energies of the hydrogen atom is quantized 4.2 Spectral emission lines When the electron jumps down from an energy level Em to a lower one En , the hydrogen atom emits a photon of energy: hf mn • If • If • If • If . . . hc mn Em En E E4 E3 En E1 : Lyman series En E2 : Balmer series En E3 : Paschen series En E4 : Backett series Paschen E2 Balmer E1 Lyman PROBLEM 8 1/ What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines? 2/ What is the wavelength of the line H in the Balmer series? SOLUTION 1/ For the Lyman series: hf m1 hc m1 Em E1 The least energetic photon is the transition between E1 and the level immediately above it; that is E2 . The energy difference is: 13.6eV 13.6eV 10.2eV E E2 E1 22 12 hc (6.63 1034 ) (3 108 ) 7 1 . 22 10 m 0.122 m 122nm 19 E 10.2 1.6 10 (in the ultraviolet range) 2/ The line H in the Balmer series corresponds to the transition between E3 and E2 . The energy difference is: 13.6eV 13.6eV 1.89eV E E3 E2 2 2 3 2 hc (6.63 1034 ) (3 108 ) 7 6 . 58 10 m 0.658 m 658nm 19 E 1.89 1.6 10 (red color) • The line H : E3 • The line H : E4 • The line H : E5 • The line H : E6 E2 E2 E2 E2 ( = 658 nm) ( = 486 nm) ( = 434 nm) ( = 410 nm) PROBLEM 9 An atom can be viewed as a numbers of electrons moving around a positively charged nucleus. Assume that these electrons are in a box with length that is the diameter of the atom (0.2 nm). Estimate the energy (in eV) required to raised an electron from the ground state to the first excited state and the wavelength that can cause this transition. SOLUTION En h2 8ma E2 2 n 2 h2 8ma E1 h2 8ma 2 1 2 (6.625 1034 ) 2 8 9.11031 (0.2 109 ) 2 1.6 10 9.40eV 2 2 4E1 37.6eV 2 2 1 19 Energy required: E E2 E1 37.6 9.40 28.2eV The wavelength that can cause this transition: hc / E 44.0nm PROBLEM 10 According to the basic assumptions of the Bohr theory applied to the hydrogen atom, the size of the allowed electron orbits is determined by a condition imposed on the electron’s orbital angular momentum: this quantity must be an integral multiple of : mvr n ; n 1,2,3,... 1/ Demonstrate that the electron can exist only in certain allowed orbit determined by the integer n 2/ Find the formula for the wavelength of the emission spectra. SOLUTION 1 2 e2 1/ E KE PE mv k 2 r e2 v2 Newton’s second law: F ma; k 2 m r r e2 The energy of the atom: E k 2r KE ke2 2r 2 n2 2 2 v With: v2 2 2 : k 2 m r mr r We have: rn With: a0 e2 n2 mke2 r2 2 2 m n2 2 n ; ke2 r m2r 2 mr 2 2 mke 2 ke2 (the electronic orbits are quantized) 0.0529nm (Bohr radius), rn a0n2 2 4 mk e 1 2/ We have : E 2 2 n2 En 13.6eV n2 Ei E f The frequency of the emitted photon is given by: f h 2 4 mk e 1 f mk2e4 1 1 7 1 R 1.097 10 m 2 With: H 3 3 2 4 c c 4 c n f ni (Rydberg constant) 1 1 1 RH 2 2 n f ni PROBLEM 11 The result of the Bohr theory of the hydrogen atom can be extended to hydrogen-like atoms by substituting Ze2 for e2 in the hydrogen equations. Find the energy of the singly ionized helium He+ in the ground state in eV and the radius of the ground-state orbit SOLUTION For He+: Z = 2 2 mk2 Z 2e4 1 Z (13.6eV ) E 2 En 2 2 n n2 The ground state energy: E1 4(13.6eV ) 54.4eV • The radius of the ground state : rn n2 2 mkZe2 n2 rn a0 Z 12 r1 a0 0.0265nm (The atom is smaller; the electron is more 2 tightly bound than in hydrogen atom) 5 The Schrödinger Equation for the Hydrogen Atom Electron cloud In spherical coordinates (r, , ), the hydrogen-atom problem is formulated as : (r ) (r , , ) R (r )Y ( , ) ; Y ( ,) ( )() 2 e The potential energy is : U (r ) 4 0 r 1 The Schrödinger equation in three dimensions : 2 x 2 2 x 2 2 y 2 2 y 2 2 z 2 2 z 2 2m 2 [E U (r )] 0 2m 1 e2 2 E 0 4 0 r 5.1 Radial function of the hydrogen atom If depends only on r : (r) r r x 2x r x y z ; 2r x x r 2 d r x d x d ; x r dr x 2 x x x dr x r dr 2 2 2 2 2 x r d x d 2 r 1 d x 2 d x 2 d 2 2 3 2 2 2 r dr r x dr r dr x r dr x r dr r dr 2 1 d 2 x 2 d x 2 d 2 3 2 2 r dr x r dr r dr 2 1 d The same result for y and z : 2 1 d y 2 d y 2 d 2 3 2 2 r dr y r dr r dr 2 2 z 2 d z 2 d 2 3 2 2 r dr z r dr r dr 2 2 x 2 1 df 2 y 2 2 z 2 d 2 2 d 2 r dr dr The Schrödinger equation : d 2 2 d 2m 2 2 r dr dr 1 e2 E 0 4 0 r Wave function for the ground state The wave function for the ground state of the hydrogen atom: (r ) 1 a 3/2 e r / a 10 a 0 . 529 10 m 52.9nm Where a is the Bohr radius: Wave function for the first excited state (r ) Ce ar / 2 (2 ar ) PROBLEM 12 Knowing that the wave function for the ground state of the hydrogen atom is r / a (r ) Ae Where a is the Borh radius: a 0.529 1010 m 52.9nm 1/ What is the value of the normalization constant A ? 2/ What is the value of x at which the radial probability density has a maximum? SOLUTION 1/ Because the electron moves in the three dimensional space, the probability of finding the electron in a volume dV is written: dV A2e2r / a dV 2 where: dV 4 r 2dr (r ) Normalization condition: (1) We put: I dV 1 0 4 A2 e 2r / a r 2 dr 1 0 2 e2r / a r 2dr a 3 z 2 I e z dz By changing variable: z 2r / a 2 0 By integration by parts, we can demonstrate the general formula: 0 e z z n dz n ! 1 2 ... (n 1) n 0 I a / 2 3 e z z 2 dz 2 a / 2 3 a 3 / 4 0 Substituting the value of I into (1): 4 A a / 4 1 2 3 (r ) 1 a3 / 2 e r / a A 1 a3 / 2 2/ Because: 2 2r / a 2 dV A e dV 1 a 3e 2r / a 4 r dr 2 4 a 2r / a 2 e r dr 3 the radial probability density is 4 2r / a 2 P(r) P( r ) 3 e r a dP (r ) When P (r) has a maximum: 0 dr dP(r ) 4 d 2r / a 2 3 e r dr a dr O 4 2 2 2r / a 3 r ( )e 2re 2r / a a a 8 2r / a r ( a r ) e 0 4 a ra a (The value r = 0 corresponds to a minimum of P( r) ) Physical meaning: The position r = a is the most probable for the electron electronic clouds r PROBLEM 13 Calculate the probability that the electron in the ground state of the hydrogen atom will be found outside the Bohr radius SOLUTION 1 With: (r ) 3 / 2 e r / a a The probability is found by: P (r ) dV where: dV 4 r 2dr 2 a P 4 a 1 P 2 3 r 2e2r / a dr a By changing variables: z 2r / a , 1 z 2e z dz 2 {z 2 2 z 2}e z 5e2 0.677 P 67.7% 2 2 PROBLEM 14 The wave function of a particle is given as: (r ) Ce x / a 1/ Find C in terms of a such that the wave function is normalized in all space 2/ Calculate the probability that the particle will be found in the interval -a x a SOLUTION 2C 2 e 2 x / a dx 1 1/ With: C 2 e 2 x / a dx 1 2 a 1 C a 2C 1 2 a 2/ P ( x) a 0 2 a a dx 2 ( x) dx 2C 2 e 2 x / a dx 0 2 2 0 P 2C 2 (a / 2)(1 e ) 2(1 / a )2 (a / 2)(1 e2 ) 86.5% 5.2 Quantization of Orbital Angular Momentum NOTES : In classical mechanics : Force and linear momentum in translational motion : dv d (mv ) dp F ma m dt dt dt Torque and angular momentum in rotational motion Fd F r L pr L r p L mvr m (r )r (mr 2 ) I In quantum mechanics : In the wave function of electron in hydrogen atom : (r ) (r , , ) R (r )Y ( , ) ; Y ( ,) ( )() The requirement that the () function must be finite at = 0 and = gives the result : L can take some possible values : L l (l 1) ( l 0 ; 1 ; 2 ; 3 ;...; n 1 ) The number l is called : the orbital angular-momentum quantum number or the orbital quantum number for short. In quantum mechanics : L l (l 1) ( l 0 ; 1 ; 2 ; 3 ;...; n 1 ) On the other hand, the permitted values of the component of L in a given direction, say the z-component Lz are determined by the requirement that the () function must equal ( + 2). The possible values of Lz are Lz ml ( ml 0 ; 1 ; 2 ; 3 ;...; l ) ( ml l ; l 1 ; l 2 ; ... 1 ; 0 ; 1 ;...; l 1 ; l ) We call ml the orbital magnetic quantum number ( l 0 ; 1 ; 2 ; 3 ;...; n 1 ) L l (l 1) Lz ml ( ml 0 ; 1 ; 2 ; 3 ;...; l ) ( ml l ; l 1 ; l 2 ; ... 1 ; 0 ; 1 ;...; l 1 ; l ) EXAMPLE : For n = 3; the possible values of l are : 0 ; 1 ; 2 With l = 2; the possible values of ml are : -2 ; -1 ; 0 ; +1 ; +2 L 2(2 1) 6 Lz 0 ; 1 ; 2 2.45 PROBLEM 15 How many distinct (n, l, ml ) states of the hydrogen atom with n = 3 are there? Find the energy of these states. SOLUTION For n = 3; the possible values of l are : 0 ; 1 ; 2 With l = 0 ; the possible value of ml is : 0 With l = 1 ; the possible values of ml are : -1 ; 0 ; +1 With l = 2 ; the possible values of ml are : -2 ; -1 ; 0 ; +1 ; +2 The total number of (n , l , m, ) states with n = 3 is therefore 1 + 3 + 5 = 9. En 13.6eV n 2 E3 13.6 eV 2 3 1.51 eV PROBLEM 16 Consider the n = 4 states of hydrogen. (a) What is the maximum magnitude L of the orbital angular momentum? (b) What is the maximum value of Lz? (c) What is the minimum angle between L and the z-axis? SOLUTION (a) When n = 4, the maximum value of the orbital angularmomentum quantum number l is (n 1) (4 1) 3 L l (l 1) 3(3 1) 12 (b) For l = 3 the maximum value of the magnetic quantum number ml is 3 : Lz ml 3 PROBLEM 17 Consider the n = 4 states of hydrogen. (a) What is the maximum magnitude L of the orbital angular momentum? (b) What is the maximum value of Lz? (c) What is the minimum angle between L and the z-axis? SOLUTION (c) The minimum allowed angle between L and the z-axis corresponds to the maximum allowed values of Lz and ml cos min (Lz )max L min 300 3 12 PROBLEM 18 Represent all the possible orientations of the angular momentum with the value l = 0 ; 1 ; 2 ; 3 SOLUTION l = 0 : L = 0 ; ml = 0 2 Lz l=1 L l (l 1) l 1 l 2 L 2 L 6 0 2 1(1 1) ml = -1 ; 0 ; +1 l 0 L 0 2 Lz ml 0 ; 1 l=2 L l (l 1) 2(2 1) ml = -2 ; -1 ; 0 ; +1 ; +2 6 Lz ml 0 ; 1 ; 2 PROBLEM 19 (a) If the value of Lz is known, we cannot know either Lx or Ly precisely. But we can know the value of the quantity L2x L2y . Write an expression for this quantity in terms of l and ml . (b) What is the meaning of L2x L2y ? (c) For a state of nonzero orbital angular momentum, find the maximum and minimum values of L2x L2y . Explain your results. SOLUTION (a) L2x L2y L2 L2z l (l 1) ml2 2 2 l (l 1) ml2 (b) This is the magnitude of the component of angular momentum perpendicular to the z-axis (c) The maximum value : L2x L2y l (l 1) when ml = 0 The minimum value : L2x L2y MAX MIN l when ml = l 5.3 The spectroscopic notation and the shell notation The existence of more than one distinct state with the same energy is called degeneracy Example : n = 2 4 states : degeneracy g = 4 5.4 Electron Spin Analogy : The earth travels in a nearly circular orbit around the sun, and at the same time it rotates on its axis. Each motion has its associated angular momentum. which we call the orbital and spin angular momentum, respectively. L S Each electron possesses an intrinsic angular momentum called its spin. Like orbital angular momentum. the spin angular momentum of an electron (denoted by S) is found to be quantized. 1 S s (s 1) ; s 2 1 1 S 1 22 3 2 The projection of the spin on z-axis is called Sz S z ms 2 1 ( ms : magnetic spin number ) 2 The spin angular momentum vector S can have only two orientations in space relative to the z-axis: "spin up" with a z-component of 2 and "spin down" with a z-component of 2 CONCLUSION : State of an electron is defined by 5 quantum numbers : n : the principal quantum number l : the orbital quantum number ml : the orbital magnetic quantum number s : the spin number ms : the magnetic spin number Wave function of an electron is denoted as : n ,l ,ml ,s 1 / 2,ms 1 / 2 PROBLEM 20 (a) Show that the total number of atomic states (including different spin states) in a shell of principal quantum number n is 2n2 . (b) Which shell has 50 states ? SOLUTION n 1 n 1 n 1 n 1 n 1 l 0 l 0 l 0 l 0 l 0 (a) N 2 (2l 1) 4 l 2 1 4 l 2 1 (n 1)n 4 2(n ) 2n 2 2n 2n 2 N 2n 2 (b) The n = 5 shell (O - shell) has 50 states 6. The Zeeman Effect The Zeeman effect is the splitting of atomic energy levels and the associated spectrum lines when the atoms are placed in a magnetic field The interaction energy of an electron (mass m) with magnetic quantum number ml , in a magnetic field B along the + z-direction : ` U ml B B (ml 0; 1; 2; 3;...) e B : the Bohr magneton 2m PROBLEM 21 An atom in a state with l = 1 emits a photon with wavelength 600.000 nm as it decays to a state with l = 0. If the atom is placed in a magnetic field with magnitude B = 2.00 T, determine the shifts in the energy levels and in the wavelength resulting from the interaction of the magnetic field and the atom's orbital magnetic moment. SOLUTION 7. The Dirac Equation and antimatter Paul Dirac had developed a relativistic generalization of the Schrödinger equation for the electron (1928). The Dirac Equation explain the spin magnetic moment of the electron mechanism for the creation of positrons like photons, electrons can be created and destroyed Antiparticle: positron (e+), the electron's antiparticle, which is identical to the electron but with positive charge