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PES

PES. Photoelectron Spectroscopy (a simplified introduction). Photoelectron Spectroscopy. PES uses high-energy electromagnetic radiation (X-ray or UV) to probe the electronic structure of an atom. When sufficient energy is provided, electrons will be emitted. X+ h ν → X + + e -.

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PES

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  1. PES Photoelectron Spectroscopy (a simplified introduction)

  2. Photoelectron Spectroscopy PES uses high-energy electromagnetic radiation (X-ray or UV) to probe the electronic structure of an atom. When sufficient energy is provided, electrons will be emitted. X+ hν → X+ + e-

  3. Energy of the Photon The energy of the photon is consumed in two ways: • Break the force of attraction between the electron and the protons in the nucleus. • This is called the binding energy (BE), measured in Joules (J), megajoules/mole (MJ/mole) or electron volts (eV). 1 J = 6.24 x 1018eV

  4. Energy of the Photon • Remember coulomb’s law: As the distance between the electron and proton DECREASES, the attraction INCREASES and the amount of energy necessary to break the attraction (the binding energy) INCREASES • The BINDING ENERGY is thus related to the IONIZATION ENERGY. If the sample is in the gaseous phase, they are typically considered as identical. For our purposes we will treat them as synonyms.

  5. Energy of the Photon • Increase the kinetic energy of the electron. The energy of the impinging photons is KNOWN the kinetic energy values of emitted electrons are MEASURED, allowing the binding energy to be GRAPHED. Ephoton= Kinetic energy + Binding energy OR Ephoton= Kinetic energy + Ionization energy

  6. Photoelectron Spectroscopy • We will be exploring the binding energy for the removal of electrons in different sublevels as the FIRST electron removed. • We will not be exploring the CHANGES in binding energy that occur with the removal of SEQUENTIAL electrons. • Each peak in the spectrum will represent the binding energy for an electron when it is the first electron removed.

  7. PES Spectra PES spectra will thus provide us with two key pieces of information: • A spectrum will show the binding energy for each sublevel in an atom and hence links experimental evidence to the electron configuration model. • The peak height shows the relative number of electrons in the sublevel. • The spectrum provides evidence for the claim that a 2s electron is at a little higher energy than the 2p electron.

  8. PES Spectra-Fluorine

  9. PES Spectra-Fluorine CAUTION: evaluate the x-axis carefully. Sometimes energy is graphed low to high and others high to low.

  10. PES Spectra-Fluorine 2p5 2s2 1s2

  11. PES Spectra-Fluorine Note that the IE of the 1s electron is higher than the 2s & 2p. It is closer to the nucleus so experiences greater attraction. 2p5 2s2 1s2

  12. Note (1) the ratio of peak height is 5:2 representing the electron ratio of 5:2 (2) the IE for the “p” electrons is less than the “s” electrons PES Spectra-Fluorine 2p5 2s2

  13. PES Spectra PES spectra will thus provide us with two key pieces of information: • Comparison of spectra beautifully demonstrates how the actual energy of energy levels varies as the nuclear charge, Z, increases. This in turn illustrates why each element has a unique atomic spectrum!

  14. PES Spectra-Nitrogen vs. Fluorine

  15. PES Spectra-Nitrogen vs. Fluorine F 2p5 N 2p3 N 1s2 F 1s2 F 2s2 N 2s2

  16. PES Spectra-Nitrogen vs. Fluorine Note that the IE for a Fluorine 1s electron is greater than for nitrogen. This is due to the differences in nuclear charge, Z. REMEMBER COULOMB’S LAW! ZF = 9 and ZN= 7 F 2p5 N 2p3 N 1s2 F 1s2 F 2s2 N 2s2

  17. PRACTICE Q’s

  18. Identify the element and write its electron configuration. 2p6 1s2 2s2 2p6 3s2 3p1 Aluminum 2s2 1s2 3s2 3p1

  19. Sketch the expected spectrum of phosphorus on the graph.

  20. How many peaks would you anticipate for copper? 1s22s2 2p6 3s2 3p6 4s13d10 7 peaks The relative intensity of peaks for fluorine is 2:2:5. What is the relative intensity of peaks for copper? 2:2:6:2:6:1:10

  21. A student makes the following claim regarding the PES spectra of Mg2+ and Ne. Is the statement true or false? Justify your answer. Since Mg2+ is isoelectronic with Ne, the PES spectra will be identical. FALSE-although isoelectronic, Mg has a greater nuclear charge, and therefore to remove electrons will require more energy than for Ne.

  22. A sample of an unknown element was studies using photons with energy of 198.2 MJ/mole. The kinetic energy of the ejected electrons was determined to be 38.6 MJ/mole. What is the binding energy for these electrons? BE = KEphoton + KEelectron BE = 198.2 MJ/mol - 38.7 MJ/mol = 159.5 MJ/mol

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