內容文稿案例6

1、Brief ReviewThe most important properties of particle1 The quantization e.g quantization of energy energy levels2 Particle - Wave Duality h Ph /Planck-Eistain- de Broglie relationsParticleWaveInterference and Diffractionx Px h/4impossible to specify simultaneously the precise position and momentum.s

2、tate wavefunctionDynamic equationwave equationamplitude* the probability of finding the particleProbability wave Superposition Wavefunction：1 The state description2* Probability density3 The value of observable4 The average value of the observable The problem is How to get Wavefunction?The only way

3、is5 The motions of particle Translational motion Rotational motion Vibrational motion Electronic motion Nuclear motionThe Energy of the particle：6.6 Some Analytically Soluble Problems Translational motionRotational motion with translational motiononly rotational motiontranslationVibrational motionSt

4、retching of the bond between the atomsVibrational motionBending of the bond between the atomsDescribe the types of motionTranslationalRotationalVibrationalmotion that changes the shape of the molecule stretching, bending, and rotation of bondswhole atom or molecule changes its location in three dime

5、nsional space whole molecule spins around an axis in three dimensional space Motion of whole molecule Motion within molecule6.6.1 The Free Particle A free particle is one which moves through space without experiencing any forces. Hence it travels in a straight line. Its potential energy is everywher

6、e constant, and so can be assigned to be 0. The energy states are NOT quantized, but any value is allowed. 6.6.2 The Particle in a Box1. The 1-Dimensional Particle-in-a-Box(1) Schrdinger Equation The particle of mass m is confined between two walls:V(x) = 0 (0 xl)V(x) (x 0 and x 0)letBoundary condit

7、ions x =0, (0) A sin0B cos0 =0; B=0(x)Asinkxx=l, (l)A sinkl = 0; sinkl = 0, kl = nsquare, n = 1,2,3 quantum numberThe general solutions are (x)A sinkxB coskx n = 1,2,3 .(2) Properties of the solutions Therefore, the complete solution to the problem is(i) The quantization of energy n = 1,2,3 . quantu

8、m numberThis lowest, irremovable energy is called the zero-point energy. E = T+VThe 1-Dimensional Particle-in-a-Box , V = 0, E = T(a) Zero-point energy (b) E l or m，EClassical or free particle，E0.(ii) Wavefunction and quantum number nGround state and excitated state(iii)Probability distributions(iv)

9、 Applications1,3-butadieneb-carotenel = 210.140 nm = 3.08 nm.And the lowest 11 energy levels will be filled. Carrots are orange because the absorption of the short wavelength (blue) light leaves only the red-orange to reflect. Tunnelling If the potential energy of a particle does not rise to infinit

10、y when it is in the walls of the container, and E and |2,1 are degenerate.The wavefunctions for a particle confined to a rectangular surface depicted as contours of equal amplitude. (a) n1 = 1, n2 = 1, the state of lowest energy, (b) n1 = 1, n2 = 2, (c) n1 = 2, n2 = 1, and (d) n1 = 2, n2 = 2. 3. Mot

11、ion in three dimensions(1) Schrdinger EquationIn box，V0Separation of variablesX(x)Y(y)Z(z)E ExEyEz(2)Solution (3) DegeneracyCubic，a bc112121211E112 = E121 = E2116.6.3 Vibration motion3.4.1 The Harmonic Oscillator (1) Schrdinger Equation Consider a particle subject to a restoring force F = -kx, the p

12、otential is thenZero-point：(2)The solutions(i)The energy levelsv = 0, 1, 2, 3(ii)The wavefunctions3.5 Rotational Motion The rigid rotor is a simple model of a rotating diatomic molecule. We consider the diatomic to consist of two point masses at a fixed internuclear distance. Lets begin by reviewing

13、 what we know about the classical mechanics of rotation.3.5.1 The classical mechanics of rotation1 Moment of inertia- reduced mass Some coments:Erot is purely kinetic nergy; there is no potential energy in the problem.Erot increases as L increases.Erot decreases as or R increases. In other words, as

14、 more mass is placed further away from the center of mass, the rotational energy ( for given L ) es smaller.2 Rotational energies for rigid rotorL - angular momentum 3 Angular momentum (the vector product)3.5.2 Schrdinger EquationFor a rigid rotorsoHow do we approach the quantum mechanics of this pr

15、oblem?The square of the angular momentum operator takes the form of a Laplacian and the Schrodinger equation takes the formR=ra+rbxyzrarbBAOTo treat the more general three-dimensional rotation of a rigid molecule, it is necessary to formulate the Hamiltonian in three dimensions. It is appropriate to

16、 formulate it in spherical polar coordinates, and this adds significantly to the mathematical complexity. 1 The solutions The approach to the solution of this differential equation is to separate the variables in the form (, )Substitution into the Schrodinger equation gives the two separated equatio

17、nswhere M is inserted as an arbitrary separation constant at this point, but will be found to represent Lz, the ponent of the angular momentum. Solutions to these equations which meet the necessary constraints on the wavefunction, i.e. single valued and normalizable, can be obtained only whenJ 0, 1,

18、 2, 3B rotational constant After a little effort, the eigenfunctions can be shown to be the spherical harmonics (, ) = Y (, )The wavefunctions associated with these allowed solutions are the association M=MJ is made and the projections of the rotational angular momentum along the polar axis (z-axis)

19、 can be expressed as 2 DiscusionThe quantization of energyJ 0, 1, 2, 3degeneracy g = 2J + 1The quantization of angular momentumJ - angular momentum quantum numberThe quantization of the projections of the rotational angular momentum along the polar axis (z-axis) MJ JHomework(1). An electron in a one

20、-dimensional box undergoes a transition from the n=3 level to the n=6 level by absorbing a photon of wavelength 500 nm. What is the width of the box?(2). What is the average location of a particle in a box of length l in the n=3 quantum state?(3). Calculate the lowest energy transition in the butadi

21、ene molecule.6.7 Molecular Spectroscopy6.7.1 General feature of spectroscopyEnergy of molecules：Emission spectroscopyAbsorption spectroscopy6.7.2 Experimental techniquesRaman spectroscopyThe rotational energy levels of a linear rotor, the transitions allowed by the selection rule (J = 1), and a typi

22、cal pure rotational absorption spectrum (displayed here in terms of the radiation transmitted through the sample). The intensities reflect the populations of the initial level in each case and the strengths of the transition dipole moments. 6.7.3 Intensities of spectral linesTransmittanceAbsorbance3

23、.6.4 LinewidthsLifetime broadeningBoltzmann distributionDoppler effect Doppler broadening6.7.4 Selection Rules For Infrared Activity: A polyatomic molecule will possess several vibrations and these may be Infrared active or inactive according to the symmetry of the vibrational mode. The fact that th

24、is behavior is related to the symmetrical structure of the molecule is one reason for the importance of I.R. spectroscopy in chemistry. A vibrational mode can give rise to an absorption of infrared radiation only if the vibration involves a change in the electric dipole moment of the molecule. So, i

25、t should be obvious that the vibration of a homonuclear diatomic molecule (e.g. O2, N2, etc.) does not result in infrared absorption. For Raman Activity: A molecule will only display a Raman spectrum if there is a change in the polarizability of the molecule during the vibration. One consequence of

26、this is that although diatomic molecules do not give i.r. spectra they do result in Raman spectra. According to the Rule of Mutual Exclusion, molecules with a centre of symmetry have no normal modes that are active in both the infrared and the Raman spectrum.If the molecule has N atoms then we need

27、3N displacement coordinates to fully describe all possible motions, i.e. each atoms movement may be described by giving the components of its displacement (from the stable geometry) along the x, y and z directions. 6.7.5 Vibrational modes for a non-linear molecule: of N atoms there will be 3N-6 vibrational m