近代物理

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1 Ch6. Quantum Theory of the Hydrogen Atom Introduction 1945 Schrodinger Schrodinger equation : If only I knew more mathematics wave function analytical solution Schrodinger Equation for the Hydrogen Atom Symmetry suggests spherical polar coordinate. We shall consider the proton to be stationary ( the correction for proton motion is simply a matter of replacing the electron mass m by the reduced mass m ). 1

2 The Schrodinger equation for the electron in three dimension of the hydrogen atom is 2 Ψ 2 Ψ 2 Ψ 2 x 2 y 2 z 2 m ( E U) Ψ = 0 where U is the electric potential energy and 2 e e U = qv = ( e) = 4πε0r 4πε 0r The Spherical Coordinate System 2

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4 2 2 2 r x y z, 0 r = + + z z θ = = x y z r y o φ = tan, 0 φ 360 x 1 1 o cos cos, 0 θ or x= rsin( θ )cos( φ) y = rsin( θ )sin( φ) z = rcos( θ ) Example Given point P(-2,6,3), express P in spherical coordinates. Sol: 4

5 In Spherical polar coordinates Schrodinger equation is written m r Ψ sinθ Ψ Ψ ( E U) Ψ = 0 r r r r sin( θ) θ θ r sin θ φ Substituting the above equation for the potential energy U and multiplying the entire equation by r 2 sin 2 θ, we obtain Ψ Ψ Ψ 2mr sin sin θ r + sin( θ) sinθ + + E Ψ = r r θ θ φ 4πε0r θ separation of variables ( )! e 5

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8 ( m ) m l = magnetic quantum number l l 8

9 l = orbital quantum number Θ R 9

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11 RΘΦ quantum n, l, ml = n principal quantum number, n= 1,2,3,... n 1 l = orbital quantum number l = 0,1,2,...( n 1) l 0 m = magnetic quantum number m = 0, ± 1, ± 2,... ± l, l m + l l l Rr () = R () r Θ () θ =Θ () θ Φ () φ =Φ () φ nl lm m l l l numbers. Principal Quantum Number n: Quantization of energy Classically the total energy can have any value, but in the Hydrogen, the electron can only have discrete value: E n me 1 E = =, n= 1,2, πε n n where E 1 = -13.6eV. The quantization of the electron energy in the hydrogen atom is described 11

12 by the principal quantum number n. Orbital Quantum Number l: Quantization of angular momentum magnitude Like total energy E, angular momentum is both conserved and quantized. And For example: L= l( l+ 1), l = 0,1,2,...( n 1) If 34 l = 2, L= l( l+ 1) = 2(2 + 1) = 6 = J. s By contrast the orbital angular momentum of the earth is 2.7x10 40 J.s, so large that the separation into discrete angular momentum state cannot be experimentally observed. Designation of Angular Momentum States 12

13 It is customary to specify electron angular states by a letter: s: sharp, p: principal, d: diffuse, f: fundamental 13

14 For example: A state in which n= 2, l = 0 is a 2s state 14

15 A state in which n= 3, l = 2 is a 3d state! Magnetic Quantum Number m l : Quantization of angular momentum direction L= l( l+ 1) L L L An electron revolving about a nucleus is a minute current loop and has a magnetic field like that of a magnetic dipole. 15

16 Hence an atomic electron that possesses angular momentum interacts with an external magnetic field B. m l The magnetic quantum number specifies the direction of L by determining the component of L in the field direction. This phenomenon is often referred to as space quantization. 16

17 Space quantization: Lz = ml, ml = 0, ± 1, ± 2,... ± l or l ml + l The number of possible orientations of L in a magnetic field is 2l + 1. For example: When l = 2, L z = 2, 1, 0, 1, 2 17

18 **In the absence of an external magnetic field, the direction of the z axis is arbitrary. What must be true is that the component of L in any direction we choose is m. What an external magnetic field does is l 18

19 to provide an experimentally meaningful reference direction ( z ) Why is only one component of L ( L z ) quantized? The answer is related to the fact that L can never point to any specific direction but instead is somewhere on a cone in space such that it projection Lz is m. Were this not so, the uncertainty principle would be violated. l 19

20 (a) L z xy z = 0 z pz pz = ( ) 2 20

21 L = m L processes constantly about the z axis. z l Electron Probability Density: no definite orbit 2 Ψ Bohr model orbit. definite 21

22 orbit. : 22

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25 R () r nl 25

26 1. The most probable value of r for 1s electron turn out to be exactly a 0, the orbital radius of a ground-state electron in Bohr model. 2. < r >. 1s = 1.5a0 e U =, U < > 1s = It makes sense! 4πε r r r a 0 0 Example 26

27 Verify that the average value of 1/r for a 1s electron in the hydrogen exp( r/ a0 ) atom is 1/a 0. Ψ 1s = 3/2 π a0 Sol: Example How much more likely is a 1s electron in a hydrogen atom to be at the distance a 0 from the nucleus than at the distance a 0 /2? Sol: 27

28 The electron is 47% more likely to be a 0 from the nucleus than half that distance. 2 Ψ 28

29 ( r, θ, φ or x, y, z ) Ψ 3D 2 29

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35 The pronounced lobe patterns characteristic of many of the states turn out to be significant in chemistry since these patterns help determine the manner in which adjacent atoms in a molecules interact. 35

36 Selection Rule: (or quantum electrodynamics ), it is found that only transitions between states of different n that can occur are those in which the orbital quantum number l change by +1 or -1 and the m l magnetic quantum number does not change or changes by +1 or

37 n n ( n 0) i f ( ) i f l =± 1 l l = 1 ( ) m = 0, ± 1 m m = 0,1 l i f Besides, quantum electrodynamics (QED) shows that the radiation emitted during a transition from state m to state n in the form of a single proton. Selection rule line 37

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40 Zeeman Effect: How atoms interact with a magnetic field Zeeman B line 3 : 1. Magnetic moment µ : current loop orbiting electron µ = IA 40

41 2. µ L : L = angular momentum = µ = ( ef )( πr ) = ef πr 2 2 rp = rmv = rm( f 2 π r) = 2π mfr 2 e 2 e µ = ef πr = ( 2πmfr ) = L 2m 2m e = constant =gyromagnetic ratio 2 m 3. Torque: τ µ magnetic dipole( orbiting electron) B torque : 2 41

42 τ = µ Bsinθ 4. Magnetic potential energy U m Magnetic dipole B ( /2 θ = π = 90 o U m 0 ) U = τ dθ = µ B sinθdθ = µ Bcosθ m θ π /2 π /2 e e = LBcosθ = LBcosθ 2m 2m θ So, when µ points in the same direction as B, then U m minimum value. = µ B, its 42

43 L l 5. cos z m θ = = L L e e ml e 6. Um = LBcosθ = LB = ml B 2m 2m L 2m -24 = constant = Bohr magneton = 9.274x10 J/T = 5.788x10 2 em -5 ev/t E1 13.6eV 1. B E = n 2 n = n E n 2 2. B µ B U m e E = En + Um = En + ml B 2m 3. l m + l transition selection rule: = 0, ± 1 l spectral line B 3 normal Zeeman effect. m l 43

44 e E0 +, ml =+ 1 2m En = E0, ml = 0 e E0, ml = 1 2m Example A sample of a certain element is placed in a 0.3T magnetic field and suitably excited. How far apart are the Zeeman components of the 450-nm spectral line of this element? Sol: 44

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Untitled-3 SEC.. Separable Equations In each of problems 1 through 8 solve the given differential equation : ü 1. y ' x y x y, y 0 fl y - x 0 fl y - x 0 fl y - x3 3 c, y 0 ü. y ' x ^ y 1 + x 3 x y 1 + x 3, y 0 fl