ups-紫外光电子发射谱课件.pdf

Ultraviolet Photoelectron Spectroscopy (UPS)-1 Louis ScudieroLouis Scudiero http://www.wsu.edu/~pchemlabhttp://www.wsu.edu/~pchemlab; 5; 5- -26692669 scudiero@wsu.eduscudiero@wsu.edu In photoelectron spectroscopy techniques such UPS, the photon energies range from 10-50 eV which is greater than any typical work function values (2-5 eV). So, electrons can be ejected from surfaces due to the photoelectric effect. UPS uses vacuum Ultraviolet radiation with energy 10 - 50 eV to determine work functions, ionization energies and examine valence levels. CasaXPS (Washington State University, Pullman, Wa) Vb/7 Au 111 -film x 103 10 15 20 25 30 35 CPS 46810121416182022 Kinetic Energy (eV) Koopmans’ Theorem and the Photoemission Process The binding energy of an electron in state i is equal to the negative of the orbital energy of the ithstate. (the ion is represented by (N-1) frozen orbitals) The ionization energy (Ii) for the removal of electrons from different orbitals in a molecule is given by the energy difference between the initial state of the neutral molecule (in the ground state) and the final state that is the state of the ionized molecule. Koopmans’ theorem makes possible the identification of calculated orbital energies with ionization potentials. But it does not allow for electronic relaxation ii I 1. Intra-molecular relaxation (relaxation energy for a free molecule) The N-1 electrons are rearranged around the hole, leading to lowering of the energy. 2. Extra-molecular Relaxation When a gas is chemisorbed on a surface the energy levels of the chemisorbed molecule are shifted relative to those of the free gas. Effects: Bonding (initial state) Relaxation (final state) or (polarization screening) The measured binding energy is always lower than the one calculated from Koopmans’ theorem. Example: O 1s in COO 2s BE (expt) = 532.3 eVBE (expt) = 38.9 eV BE (Koopmans) = 562.4 eV BE (Koopmans) = 41.1 eV Ionization sources are Ne I (16.6 eV), Ne II (26.8 eV) and He I (21.2 eV), He II (40.8 eV). These lines are produced by cold cathode capillary discharge.They represent resonance fluorescence produced when the gas is excited in the discharge and then decays back to its ground state. I -- light emitted from neutral atoms II -- light emitted by singly ionized atoms The resonance line produced by transition from the first excited state to the G.S is usually the most intense (called raie ultime). He I line is at 584Å or 21.22 eV and He II line at 304 Å or 40.8 eV. Turner and Jobory (J. Chem. Phys. 1962, 37, 3007) introduced the He I resonance line in 1962. He II line was first used by Price and co-workers in 1970 (Potts, A.W. Lenpka, H.J Streets D.G. and price, W.C. Phil. Trans Roy. Soc. London A, 1970, 268, 59). UV Commonly Used Lines (16.6 to 40.8 eV) He I and He II Grotian diagram for He I and He II showing the strongest resonance lines (584 Å, 98% of the emission intensity, the other lines present in a He discharge are 537 Å, 522 Å and 304 Å which can have intensity of the order of 2 % of the 584 Å line. Rabalais J. Wayne “Principle of Ultraviolet Photoelectron Spectroscopy”, John Wiley the boundary point at 2/a (1/2,1/2,1/2) is L, the line D runs between G and X. In (b) the corresponding symbols are H, P and D. Symmetry points and axes of the Brillouin zones of the fcc (a) and bcc (b) lattices. The zone centers are G . Photoionization Process In general, however the photoionization process is described by the following expression: ii KhI Case I: Metals UPS could be used to determine the work function of metals.By measuring the width of the emitted electrons (W) from the onset of the secondary electrons up to the Fermi edge and subtracting W from the energy of the incident UV light h, the work function mis then given by Wh m  I is the ionization energy Example: Au (111) single crystal W = 15.95 eV, Photon energy h = 21.2 eV  m= 21.2 –15.95 = 5.25 eV   0.15eV (resolution of the instrument) Literature value 5.3 eV (CRC) Washington State University--Pullman, WA Single Crystal Au111 15.95 UPS --Au (111) surface He I (21.2 eV) x 10 2 2 4 6 8 10 12 14 16 18 16 14 12 10 8 6 4 2 0 Binding Energy (eV) Case II: Molecular systems (molecular film adsorbed on metal) Not so simple in case of moleculesWhy? 1. Charge transfer across the interface (expected for the combination of strong acceptor-low work function or strong donor-high work function). 2. Redistribution of electron cloud (polarization of the electron cloud attracted by the image charge formed in the metal). Seki et al. Adv. Mater. 1999, 11, No. 8 4. Strong chemical interaction between the surface and the adsorbate leading to the rearrangement of the electronic cloud and also the molecular and surface geometries (both directions of dipoles possible), Others 5. Existence of interface state serving as a buffer of charge carriers 6. Orientation of polar molecules or functional groups. 3. Interfacial chemical reactions (well known case for small mol。