Rotation and vibration of Molecules

Transcription

1 Rotation and vibration of Molecules Overview of the two lectures... 2 General remarks on spectroscopy... 2 Beer-Lambert law for photoabsorption... 3 Einstein s coefficients... 4 Limits of resolution... 7 Instrumental... 7 Inhomogeneous broadening... 7 Thermal broadening (Doppler effect)... 8 Natural linewidth (Heisenberg s uncertainty)... 9 Molecular rotation Classical description Quantum-mechanical description: fixed orientation of the axis The potential Schrödinger equation Ansatz Boundary condition Normalization Familiarization Recapitulation Rotation of a diatomic molecule with a free orientation of the axis Term levels Angular momentum Wavefunctions

2 Overview of the two lectures - it is on internal motion of molecules - spectroscopies are experimental methods to learn about the quantized energies and how to derive chemically useful information from them examples: - photoabsorption (microwaves, IR, VIS, UV, VUV, X-ray; modern: terra-hertz spectroscopy) photoemission (luminescence, fluorescence, phosphorescence) more exotic spectroscopies: - neutron energy-loss (good for solids, expensive) - electron energy-loss (good for spin-forbidden transitions; done in FR) - photoelectron spectroscopy (good for cations and anions) - magnetic resonance (not treated in this course) General remarks on spectroscopy 2

3 Beer-Lambert law for photoabsorption - cross section is a (fictive) 100% opaque area which would give the same absorption probability as the molecule - A is area here, c is concentration in how many molecules in a m 3, c is concentration in mol/l - A is absorbance here, is molar absorption coefficient (formerly extinction coefficient ) - typical values of 3

4 - problem: peak absorbance depends on resolution take integrated absorbance A - A is linked to other quantities Einstein s coefficients Einstein s coefficient B for absorption and stimulated emission, A for spontaneous emission Einstein s coefficient B is also good for stimulated emission, A is for spontaneous emission 4

5 - stimulated emission is the basis of LASER - A is related to the radiative lifetime - A and B are related: - note: A is larger, the lifetime is shorter, for energetic radiation (UV) than for low-energy radiation (microwaves). Consequence: it is hard to make a UV or VUV (or even X-ray) laser than IR/VIS 5

6 - B (and consequently A) are related to the experimental value A and thus to radiative lifetime: - but caution there are generally also other, parallel, nonradiative, decay channels and the true lifetime of the excited state eff may be shorter than rad 6

7 - B is also related to the transition dipole moment tr, calculated from quantum theory (see later, the lecture on electronic structure) Limits of resolution Instrumental Inhomogeneous broadening 7

8 Thermal broadening (Doppler effect) - Doppler-free spectroscopy 8

9 Natural linewidth (Heisenberg s uncertainty) - limits? see - take a forbidden transition to make long for example the 1s 2s transition in the H- atom: transition energy measured to 15 significant digits!! useless? - the last digits permit to test exotic physics, or perhaps even to find new physics - relativistic correction, quantum electrodynamics (QED),... 9

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11 Molecular rotation Classical description - for stable rotation, find center-of-mass - how to calculate moment of inertia? 11

12 - how to find the principal axis? 12

13 Quantum-mechanical description: fixed orientation of the axis - a valuable illustrative case for how QM works you can solve everything yourself! 13

14 The potential (later we shall discuss a case where the molecule is immersed in an external electric field and V is not zero: the Stark effect) Schrödinger equation Ansatz - a differential equation is solved by guessing: look for an ansatz 14

15 Boundary condition - boundary condition leads to quantization 15

16 Normalization 16

17 so now we have the final form of the wavefunction: Let us recapitulate: 1. write the Schrödinger equation for our physical situation rotation with fixed orientation of the axis by inserting the right potential V and the right kinetic energy operator 2. solve the equation with an ansatz and respecting the boundary conditions 3. normalize the wavefunction 4. we received a set of quantized energies and corresponding wavefunctions 17

18 - There is one point to add: spectroscopists have the tradition of using the wavenumber ~ (in cm -1 ), called the term instead of the energy. How to calculate the term? - B is called the rotational constant. Each molecule has a different value, depending on its moment of inertia I. Familiarization - Now we have all the solutions, we want to get an intuitive feeling about what they mean to become familiar with them. - the first thing is the term diagram: 18

19 - then the wavefunction for k = 0. It is just a constant value. - Note that it is in line with the DeBroglie relation. Constant value means an infinite wavelength, that is, zero velocity and thus zero kinetic energy. - Now the k = 1 wavefunction. The DeBroglie wavelength is shorter, the kinetic energy higher. - Philosophically it is perturbing that we need imaginary numbers to describe real world. 19

20 Recapitulation - Subject: rotation of a diatomic molecule rigid molecule bond length is fixed orientation of the axis is fixed - How did we proceed? - find center of inertia (center of mass) - calculate moment of inertia I - set up and solve the Schrödinger equation - the result is a set of wavefunctions (WF) and corresponding (quantized) energies, indexed by a quantum number k - the energies are traditionally expressed as wavenumber and called the term F the term diagram, with (very) schematic drawings of the wavefunctions: 20

21 - note that wavefunctions with a negative k have the same real part, but a reversed sign of the imaginary part. The relative phase of the real and the imaginary parts expresses the sense (right or left) of the rotation. - the Schrödinger equation is an eigenvalue equation of the Hamilton operator, the operator for energy. The solutions are therefore eigenvalues and eigenfunctions of energy. The states described by these eigenfunctions have well-defined energies (see course on elementary Quantum Mechanics). - but we have another important quantity describing the rotation : the angular momentum J. - the question now arises whether our WF are also eigenfunctions of the angular momentum operator Ĵ. Let s test it by applying the operator to our WFs. - the answer is yes, and the eigenvalues are J k Rotation of a diatomic molecule with a free orientation of the axis - applies to molecules in the gas phase - we have operators for the three components of the angular momentum and for the total value of the angular momentum. The Hamiltonian can be derived from the operator for the total angular momentum. 21

22 - the angular position of the molecule is now expressed by two angles in the spherical coordinates Term levels - solving the Schrödinger equation gives the quantized energies (here expressed as wavenumbers the rotational term, as before). J is the quantum number - and the term diagram (energy diagram): (note that for large J the formula becomes 2 F J BJ, like with the fixed axis) 22

23 - one more time, compared to the scheme with fixed axis orientation: Angular momentum - what about the angular momentum? - Nature does not want to let us know all the details about the angular momentum. We are allowed to know the magnitude of the total angular momentum (the length of the arrow of the angular momentum vector) and only partial information about its orientation in space, namely one component (traditionally named J z ). Nature will not let us know the direction in the xy plane, i.e., the J x and J y components. 23

24 - Mathematically, this is expressed as commuting properties of the operators (see lecture Introduction to Quantum Mechanics): - This is what we obtain for the angular momentum and its orientation: - A side remark: which is the correct direction for z? Is it vertical? Or tilted? In the microscopic world measurement is linked to what the values of quantities are. Imagine that you have a machine which measures one component of angular momentum. Nature will always quantize the angular momentum respective to the axis along which you measure. If you measure along vertical it will be quantized along the vertical. If you tilt the machine and measure along a tilted axis, it will be quantized along the tilted axis. The correct z axis is the axis of measurement. 24

25 - Why is J z always smaller than J? If J z were equal to J then the J vector would be exactly vertical and J x and J y were equal to zero. But then we would know exactly the values of all three components, J x J y J z and that would be against the Heisenberg requirement. - Note that for large J the J is nearly vertical and J z is nearly equal to J this is the transition toward the macroscopic world. - The situation for J = 2: Wavefunctions 25

26 - A picture of J = 1, M J = 1 : - and a picture of J = 2, M J = 2 : Remember the models: 26

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