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Quantum Theory of the Atoms slit 准直缝 electrons X 射线 crystal 晶体 fringes 劳厄斑 Electrons diffraction §20-1 Rutherford’s Experiment and the Nuclear Atom 卢瑟夫实验和原子有核模型 §20-2 Atomic Spectra Bohr Model of Hydrogen Atom 原子光谱 玻尔的氢原子模型 §20-3 De Broglie’s Postulates and Matter Waves 德布罗意假设和物质波 §20-4 The Uncertainty Principle 不确定关系 §20-5 Wave Function and Schrodinger Equation Born’s Interpretation 波函数和薛定谔方程 §20-6 The infinite Potential Well 一维无限深的势阱 §20-7 Hydrogen Atom and Electron Spin 氢原子和 电子自旋 §20-8 Multielectron Atom and Periodic Table 多电 子和周期表 §20-1 Rutherford’s Experiment and the Nuclear Atom 1. Thomson’s model electron +Ze of an atom distributes uniformly in sphere. Atom size (10-10m) electrons panel in the atom uniformly. Plum pudding model 2. Rutherford’s –particles scattering experiment (1907) – particles atom Some of the scattering angles are so large that it cannot be interpreted with “Plum pudding model” 3. Rutherford’s “nuclear atom model” (1911) All the positive charge and almost all the mass of the atom is concentrated in about 10-15~10-14 m the volume of the atom. 4. Limitation Rutherford’s model Cannot explain why the structure of atom is stable . Cannot explain why the radiating spectrum is discrete. §20-2 Atomic spectra and the Bohr model of hydrogen atom 1. The experimental laws of hydrogen atom ultraviolet 4102Å 4341Å H H H 4861Å 6563Å H H 3646Å visible found their regularities in 1885 : 1 1 ~ R( 2 2 ) -- Balmer formula 2 n 1 ~ Wave number J.Balmer J.R.Rydberg proposed a general formula in 1890 1 1 ~ R( 2 2 ) k n k 1,2, n are integers with n k -- Rydberg formula R=1.096776107m-1 -- Rydberg constant 1 1 ~ R( 2 2 ) k n Same k, different n – a series line (线系) (1)k=1,n=2,3,… Lyman series, ultraviolet (2)k=2,n=3,4,… Balmer series, visble (3)k=3,n=4,5,… Paschen series, infrared (4)k=4,n=5,6,… Brackeff series, infrared (5)k=5,n=6,7,… Pfund series, infrared ……………… 1 1 ~ R( 2 2 ) T ( k ) T ( n ) k n ----T(k) , T(n) terms of spectrum ~ kn T (k ) T (n) [T ( k ) T ( n)] [T ( n) T ( n)] ~kn ~nn ----Ritz combination principle 2.Bohr’s postulates In order to support the nuclear model, Bohr proposed three postulates for explaining the experimental regularities of hydrogen atom. (1) Stable states postulate: The motion of electron in a circular orbit is stable (2) Postulate about the quantization of orbit angular momentum : The orbit angular momentum L=mvr of an electron must satisfy , h L n n 2 n 1,2, n—quantum number h --simplified Plank’s constant 2 (3) radiation postulate: When an electron charges its orbit, a photon with frequency is emitted or absorbed. kn Ek En h En>Ek--- a photon is emitted En<Ek---a photon is absorbed Bohr got Noble Prize of physics in 1922. 3. The results of Bohr’s model (1) The radius of orbit According to Newton’s law e 2 2 v m 2 4 0 r r nh v 2mr h and L mvr n 2 0h n 1 , 2 , So rn n 2 me --quantization 2 2 For n=1, 0h rn n 2 me ----first Bohr’s radius 2 r1=0.5310-10 m 2 rn n r1 2 (2) energy The energy of hydrogen atom=potential+kinetic 2 1 e 2 E n mv n 2 4 0 rn 2 2 e v using m 2 4 0 r r 2 1 e 2 We get mv n 2 8 0 r 4 e 1 me En 2 2 2 8 0 rn n 8 0 h 2 ---quantization The quantized energy ---- energy-leverl n----quantum number discussion: 4 1 me En 2 2 2 n 8 0 h 2 h 2 0 rn n 2 me (1) For n=1, E1 13.6eV ----ground state Lowest energy, most stable state (2) For n>1 ----excited state (3) For n , rn , En 0 ---The ionization energy of hydrogen atom in ground state is 13.6eV E4 E3 E 2 n 1 n2 Excited n3 n4 Electronic orbits E1 Energy-level ground En/eV 0 -0.85 -1.51 Paschen -3.39 Balmer -13.6 Lyman n 4 3 2 1 1 me 4 En 2 2 2 n 8 0 h an electron jumps EnEk , the H-atom emits a photon. If En Ek me 1 1 2 2 3 2 h n 8 0 h k 4 Wave number is me 1 1 1 1 ~ 2 3 2 2 R 2 2 c 8 0 h c k n k n 4 4 me 7 1 Here R 2 3 1.0973731 10 m 8 0 h c ---agree perfectly with the experiment. Rexp=1.096776107m-1 Bohr’s theory made great success in hydrogen atoms and hydrogenlike ions (类氢离子) But this theory cannot explain the experimental results of more complex particles. More than one electron outside the nuclear 4. The limitation of Bohr’s model (1) using classical physics theory, but the electron has acceleration and no radiation is inexplicable. (2) No any theory can explain the angular momentum quantization. (3) Cannot get the intensity of the spectra from this theory. §20-3 De Broglie’s postulates and matter waves 1. De Broglie’s postulates(1924) Just as radiation has particle-like properties, electron and other material particles possess wave-like properties. E h h p ---De Broglie’s formulae The waves related to the material particles----matter waves De Broglie won Noble prize in physics in 1929. 2. Testing De Broglie hypothesis The wave property of electrons was confirmed by Davisson and Germer in experiment in 1927. observed that the diffraction patterns of the electrons are similar to with the ones of x-ray. E-gun detector They --electron have wave property. Acce. anode Nickle crystal Thomson’s experiment gave another method to confirm the wave property of electrons. Meanwhile, They won Noble prize in physics in 1937 x-ray electrons The results of slits diffraction experiments in 1960 single It double three four was confirmed by great amount of experiments that all micro-particles such as electrons, neutrons, proton, as well as atoms and molecules have wave properties. And their wave properties agree with De Broglie’s formulae. [Example] an electron is accelerated by electric field. If the Acc. potential is U, the electron is at rest before it is accelerated and its speed V<<C after it is accelerated. Find its wavelength Solution 1 2 eU m0 v 2 h m0v 2eU v m0 h 2em0U 12.2 o A U Scanning electron microscope---SEM ---observe the microscopic morphology. Mosquito Head with X 1,000 A shell of a radiolarian — a single-celled animal with X 2,000 Transmission electron microscope --TEM ---study the microstructure. a leaf of a green plant. §20-4 The Uncertainty Principle For microscopic particles :their position , velocity and others cannot be determined by actual experiment at same instant because of wave properties. a sin k The single slit diffraction of electrons A beams of electrons is incident on a slit normally, considering the central bright fringe only, 0 px p sin 1 px p sin1 The first dark fringe satisfies, x px py x sin 1 p x p x p x 1 y h p px h x xpx h p x p x Considering all diffraction fringes, xpx h --estimated result Hesbreg deduced the precise results in 1927. xp x , yp y , zpz 2 2 2 Hesbreg won Noble prize in physics in 1932. Any experiment cannot determine simultaneously the exact values of a component of momentum and its corresponding coordinate. xpx h ---Coordinate-momentum uncertainty relation Similarly, measuring the energy of a particle in a time interval t, the uncertainty E of the energy is tE h [Example] according to classical electromagnetism and mechanics, the speed of the electron moving around its nuclear is about 106 m/s. the size of an atom is 10-10m. Estimate the uncertainty of the speed. Solution the uncertainty of the position of the electron is x 10 10 m 34 p x h 6 . 63 10 v x 31 10 m m x 9.1 10 10 6 7.3 10 m/s v v. the uncertainty of the speed of the electron in atoms is very apparent. §20-5 The wave function and Schrodinger equation Born’s interpretation of wave function h E h p 1.wave function A classical plane wave function travels in +x axis, y y0 cos 2 (t x ) Write it in complex form, y y0 e i 2 (t x ) Similarly, for a free particle with energy E and momentum p, its wave function, x i i 2 (t ) Et px x , t 0e 0e 2. Schrodinger equation Free x , t 0 e i Et px particle:moves in +x axis p 2 2 x 2 p 2 2m x 2m 2 2 i E t 2 p E 2m 2 2 i E t 2 2 i 2 2m x t ---S-Equation of a free particle The particle has potential energy V(x,t), its total energy is E Ek V ( x, t ) p 2m V ( x, t ) 2 i p i V ( x , t )] E [ 2m t 2 p i.e. i [ V ( x , t )] 2 t 2m 2 2 p2 2 2m x 2m V ( x , t ) i 2 2m x t 2 2 If a particle moves in three dimension space, 2 2 2 2 2 x x y z 2 2 2 2 2 ( r , t ) ( r , t ) V ( r , t ) i 2m t 2 Introducing energy operator, 2 2 ˆ H V ( r , t ) --Hamiltonian operator 2m ---S-Equation of a then Hˆ i particle in a t potential field V ( x , t ) i 2 2m x t 2 Stationary time Let 2 state: potential is independent on V V ( x) ( x, t ) ( x ) f (t ) ( x) f (t ) V ( x ) ( x ) f ( t ) 2 2m x 2 2 f ( t ) i ( x ) t Divided by ( x ) f ( t ) , we get 1 ( x) 1 f (t ) V ( x ) i 2 2m ( x ) x f (t ) t 2 2 Left side of the equation is the function of x, and the right side is the function of t, If it is useful for any t or any x, it must satisfies, left side = right side = a constant i.e. 1 f ( t ) E (1) i f ( t ) t 1 ( x) (2) V ( x ) E 2 2m ( x ) x 2 2 the solution of Eq. (1) is f ( t ) ~ e E has the demission of energy. i Et Rewriting (2), ( x) V ( x ) ( x ) E ( x ) 2 2m x 2 or 2 ( x ) 2m 2 ( E V ) ( x ) 0 2 x 2 ----Schordinger’s equation in stationary state The wave function of the particle is x , t x e i Et 3.The physical meaning of wave function ---Born’s interpretation of wave function --- complex function no physical meaning M.Born postulated ( r , t ) ( r , t ) * 2 (r , t ) represents probability density, i.e. the probability of finding the particle per unit volume about the point r at time t. The probability of finding the particle within the volume element dv =dxdydz about the point r at time t is dw dV 2 The wave function ( r , t ) should satisfy continuous, single-valued and finite. ---Standard conditions And V 2 dV 1 ---Normalized condition The space of the particle moving in The superposition principle i.e If the 1 and 2 are two solutions of the Schrodinger equation, then c11 c22 is also the solutions of the S- equation. Here, c1 and c2 are normalized factor. The conservation of probability 2 ( r , t ) ---probability density * * j [ ] 2mi ---probability current density From Sch. Eq., we can get j 0 t ---probability is conservative. 4. Operators and physical observables The value of a physical observable= (its operator )dV * V Such as 2 2 ˆ Energy operator is H V (r , t ) 2m ˆ dV So E * H V ˆ momentum operator is P i * ˆ So P P dV V If the particle is in stationary state, 2 ( r , t ) ( r )e i Et 2 2 (r ) -- probability density has nothing to do with t. Schrodinger won Noble prize in 1933. § 20-6 The infinite potential well A particle with mass m moves along x axis. It potential is V ( x) 0 0 x a( in well ) x 0, x a(out well ) Out well V ( x ) 0 In well: V ( x ) 0 d ( x) E ( x ) 2 2m dx 2 0 a x Let k 2mE 2 2 d ( x) 2 k ( x) 0 2 dx 2 Its general solution is ( x ) C sinkx 0 C、 ---integral constants 0 0 a Using the continuous condition of wave function, ( 0) 0 ( a) 0 C sin 0 0 C sin ka 0 n k n 1,2, a ka n n ( x ) C sin x a x Using normalized condition of wave function, 2 2 2 a a a n ( x, t ) dx ( x ) dx C sin x dx 1 0 0 0 a We get C 2 a 0 x 0, x 0 n ( x ) 2 a sin n x 0 x a a 2mE k 2 2 n En 2ma 2 2 2 2 n k a n 1,2, --quantized energy n:quantum number Notes: n=1: E1 2 2 2ma 2 -- zero point energy -- the lowest energy of a particle cannot be zero in quantum mechanics. En n2 --- the interval of energies between two adjacent states is not uniform. E n 2 2 2ma 2 2n 1 For an eigen state n with quantum number n, there are n+1 nodes and n antinodes in the well. Minimum probability n (x ) Maximum probability n4 n (x ) n3 n2 0 n 1 a x 0 a x 2 [Example] A particle with mass m locates in a infinite potential well with length a. Calculate the probability of finding the particle in the range of 0xa/4 for the two different states with n=1 and n=. the positions of maximum probability for the state n=2. 2 n sin x Solution As ( x ) a a the probability of finding the particle in the range of 0xa/4 is a 2 4 2 n 4 dx w ( x ) dx sin 0 a 0 a a 2 1 1 n sin 4 2n 2 n 1 1 1 9% For n 1 w 4 2 For n 1 w 4 n 2 2 2 ( x ) sin x a a 2 2 2 2 2 ( x ) sin x a a d 2 2 ( x) 0 Let maximum dx We get a a 3a ,a , , x0 , 4 2 4 minimum Tunneling effect The potential distribution is U0 U ( x) 0 0 xa x 0, x a A particle move along x axis, U U0 1 2 3 For classical particle, If E U 0 , the particle can pass through area 2 and arrive to area 3. If E U0 , the particle cannot pass through area 2 and cannot be found in area 3. x For macroscopic particle, Using Schrodinger equation, d 1 Area 1 E 1 2 2m dx 2 2 d 2 Area 2 U0 2 E 2 2 2m dx 2 2 d 3 Area 3 E 3 2 2m dx 2 2 The solutions of S.Eq. are Area 1:incident waves +reflected waves Area 2:transmitted waves + reflected waves Area 3: transmitted waves No matter E>U0 or E<U0 E ---Tunneling effect 0 a x Application Scanning Tunnel Microscope (STM) 扫描隧道显 微镜(STM) 硅表面图像 癌细胞表面图像 “扫描隧道绘画” 一氧化碳“分子人” “原子和分子的观察与操纵” -- 白春礼 1993年人们用STM所做的量子围栏第一次看到波 函数 操纵48个铁原子在铜表 面围成半径为 71.3Å的 圆圈。表面电子在铁原 子上强烈反射,被禁锢 在该围栏中,它们的波 函数形成同心圆形驻波 量子围栏 ----实现波函数的测量 §20-7 The Hydrogen atom and Electron spin 1.The Schrodinger equation of hydrogen atom In the hydrogen atom system, the potential energy is U (r ) Schrodinger equation e 2 4 0 r is 2m e ( E ) 0 2 2 2 2 x y z 4 0 r 2 2 2 2 introducing the transitions, x r sin cos y r sin sin z r cos S.Eq. changes into 1 2 1 (r ) 2 (sin ) 2 r r r r sin 1 2m e 2 2 2 (E ) 0 2 r sin 4 0 r 2 2 Let (r , , ) R(r ) ( ) ( ) d 2 ml 0 2 d 2 We get (1) ml 1 d d (sin ) ( 2 ) 0 (2) sin d d sin 2 1 d 2 dR 2m e (3) ( r ) ( E ) R 0 2 2 2 r dr dr 40 r r 2 2.the results of the solution of Schrodinger equation for H-atom Quantized energy and principle quantum number n. The energy of H-atom system can be gotten from Eq. (3) 4 4 me 1 me 1 En 2 2 2 2 2 2 2 32 0 n 8 0 h n n 1,2,---- principle quantum number n. It is agrees with Bohr’s result. Quantized angular momentum and azimuthal quantum number l. The magnitude of orbital angular momentum about the electron moving around the nuclear can be gotten from Eq.(2) L l (l 1) l 0,1,2,( n 1) --azimuthal quantum number(角量子数) l has n possible values for a given value of n. The direction of orbital angular momentum about the electron moving around the nuclear can be gotten from Eq.(3) ml 0,1,2, l ---the orientation (取向) of L is quantized ---magnetic quantum number(磁量子数) The component Lz of L along z axis is Lz ml ml has (2 l+1) possible values for a given value of l . 3. The “electron cloud” The distribution of the probability density of finding the electron. 4. Electron spin Unlenbeck and Goudsmit arranged a experiment to check whether the orientation (取向) of L is quantized (1921) N Atom source No magnet S In magnet The results of the experiment: No magnet:there is one track of the atoms on the film. With magnet: there are two tracks of the atoms on the film. 无磁场 有磁场 contradiction:when the azimuthal quantum number is l , the orientation of L in the space should have (2l+1) possible values. So the number of the tracks of the atoms should be a odd number. . In order to explain the experimental results, Unlenbeck and Goudsmit proposed a postulation in 1925 : An electron not only revolves around a nucleus but has a spin. This is analogous that the earth revolves around the sun and meanwhile rotating about its own axis. The spin angular momentum S of an electron is quantized S ss 1 S z ms here 1 s 2 1 ms 2 ----spin quantum number 5. Four quantum numbers (n, l, ml, ms) ---determine the state of an electron. (1) Principle quantum number n(n=1,2,)--determine the energy of the atom system. (2) azimuthal l(l=0,1,2,,n-1)---determine the magnitude of orbital angular momentum of an electron. (3) magnetic ml(ml=0,1,2,,l)--determine the orientation of orbital angular momentum of an electron. (4) Spin ms(ms=1/2)---determine the orientation of spin angular momentum of an electron. §20-8 Multi-electron atoms and The Periodic table Multi-electron: “Average field” theory +Ze +Ze protons+ (Z-1) electrons forms a uniform field. +Ze moving uniform field Hydrogenlike (类氢) motion Conclusions The state of each electron can be specified by n, l , ml, ms . Pauli exclusion principle(泡利不相容原理):no two electrons in any one atom can have the same four quantum numbers n, l , ml , ms . Corresponding same n, l , ml , 1 ms has two values m s 2 i.e. only 2 electrons with same n, l , ml . Corresponding same n, l , ml has the values ml=0,1,2,,l i.e. 2(2l+1) electrons with same n, l . The electrons which have same l are said to belong to a subshell (次壳层). Corresponding l= 1, 2, 3, 4, 5, 6 With code letters s, p, d, e, f, g, to specify. So the maximum number of the electrons with same n is n 1 N e 22l 1 2n 2 l 0 The electrons which have same n are said to belong to a shell(壳层). Corresponding n= 1, 2, 3, 4, 5, 6 With code letters K, L, M, N, O, P to specify. minimum energy principle:If an atom is in ground state, the electrons must fill in such way as to minimize the total energy of the atom. Minimum energy corresponds the minimum values of n, l . the configuration of an atom is represented by symbol x1 x2 ( n1l1 ) ( n2 l 2 ) x i -- the number of the electrons with same ni , li For example, H-atom has one electron, its configuration is represented by [H] 1s1 Helium-atom has 2 electrons, its configuration is represented by [He] 1s2 Lithium-atom has 3 electrons, its configuration is represented by [Li] 1s2 2s1 Beryllium-atom has 4 electrons, its configuration is represented by [Be] 1s2 2s2 Boron-atom has 5 electrons, its configuration is represented by [B] 1s2 2s2 2p1 Neon-atom has 10 electrons, its configuration is represented by [Ne] 1s2 2s2 2p6 Sodium-atom has 11 electrons, its configuration is represented by [Na] 1s2 2s2 2p6 It is a great triumph for atomic theory to understand The periodic Table of the Elements in terms of atomic levels (or quantum states).