CHAPTER 6 SPECIAL TOPIC: PHOTOELECTRON SPECTROSCOPY (PES)

What is Photoelectron Spectroscopy?

Photoelectron spectroscopy (PES), also known as photoemission spectroscopy, involves the ionization of a sample by a source of photons (usually X-rays or UV light). This technique is based on the photoelectric effect. When an atom absorbs a photon with sufficient energy to cause ionization, electrons are emitted from the atom. The emitted electrons are called photoelectrons. The kinetic energy of the photoelectrons can be analyzed. The basic process can summarized as follows:

A + h A+ + e–

where A represents a neutral atom and h represents the radiation used to ionize the atom.

The basic requirements for a photoemission experiment are a source of fixed-energy radiation, an electron energy analyzer, and a high vacuum environment.

(http://commons.wikimedia.org/wiki/File:ARPESgeneral.png)

X-ray radiation (XPS) is used to study core electrons. Ultraviolet light (UPS) is used to study valence energy levels and chemical bonding.

The kinetic energy of the photoelectrons is measured in the photoelectron spectrometer. The difference between the energy of the photon (h) and the kinetic energy (K.E.) of the emitted electron is the binding energy (or ionization energy) of the electron.

Binding energy (or ionization energy) = h – K.E.

Why would different electrons in an atom have different binding energy values? Coulomb’s law is helpful to understand this point:

The magnitude of the electrostatic force of attraction (or repulsion) between two point charges(Q1 & Q2) is directly proportional to the product of their charges, and it is inversely proportional to the square of the distance between them (d2)

So as the distance between the electron and proton decreases, their coulombic attraction increases. Therefore it makes sense that electrons that are closer to the nucleus will have a higher binding energy. Electrons that are farther away from the nucleus will have a lower binding energy.

PES allows scientists to measure the energy needed to remove any electron on an atom. Measurements of the binding energies of the various electrons in an atom help to figure out the shell structure of the atom and the electron configuration. In multi-electron atoms, the electrons can be thought of as existing in shells and subshells. Core electrons are closer to the nucleus and are more difficult to remove. Valence electrons are farther away from the nucleus and are easier to remove.

What do photoelectron spectra look like?

Data from PES experiments are obtained as peaks in a spectrum. The intensity of the signal is plotted on the vertical axis, and the energy needed to remove an electron is plotted on the horizontal axis. By convention, the energy increases from right to left on the horizontal axis. The scale on the x-axis for the following diagrams is logarithmic, so that a wide range of values can be shown on one diagram.

Element Peak (MJ/mol)Hydroge

n1.31

Helium 2.37

Compare these two photoelectron spectra. Explain why the peak for Helium is twice as high as the peak for hydrogen.

Each spectrum shows binding energy values for electrons located in a 1s orbital. Explain why helium’s electrons have a higher binding energy (2.37 MJ/mol) than hydrogen’s electrons (1.31 MJ/mol).

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Photoelectron Spectrum for Hydrogen

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Photoelectron Spectrum for Helium

Element First Peak (MJ/mol) Second Peak (MJ/mol)Lithium 6.26 0.52Berylliu

m11.5 0.90

Which electrons have higher binding energy…1s electrons or 2s electrons? Explain why.

Element First Peak (MJ/mol) Second Peak (MJ/mol) Third Peak (MJ/mol)Boron 19.3 1.36 0.80

Carbon 28.6 1.72 1.09Nitroge

n 39.6 2.45 1.40

Oxygen 52.6 3.12 1.31Fluorine 67.2 3.88 1.68

Neon 84.0 4.68 2.08

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Photoelectron Spectrum for Lithium

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Photoelectron Spectrum for Beryllium

Label each peak in the following diagrams as 1s, 2s, or 2p.

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Photoelectron Spectrum for Boron

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Photoelectron Spectrum for Carbon

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Photoelectron Spectrum for Nitrogen

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Photoelectron Spectrum for Oxygen

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Photoelectron Spectrum for Fluorine

Identify each of the following elements from the PES diagram.

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Photoelectron Spectrum for Neon

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Photoelectron Spectrum

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Photoelectron Spectrum

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Photoelectron Spectrum

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Photoelectron Spectrum

The following questions are taken from AP Chemistry exams.

The photoelectron spectra at right show theenergy required to remove a 1s electron froma nitrogen atom and from an oxygen atom.Which of the following statements bestaccounts for the peak in the upper spectrumbeing to the right of the peak in the lowerspectrum?

(A) Nitrogen atoms have a half-filled p subshell.

(B) There are more electron-electron repulsionsin oxygen atoms than in nitrogen atoms.

(C) Electrons in the p subshell of oxygen atomsprovide more shielding than electrons in thep subshell of nitrogen atoms.

(D) Nitrogen atoms have a smaller nuclear chargethan oxygen atoms.

A sample containing atoms of C and F wasanalyzed using x-ray photoelectronspectroscopy. The portion of the spectrumshowing the 1s peaks for atoms of the twoelements is shown at right. Which of the following correctly identifies the 1s peak forthe F atoms and provides an appropriateexplanation?

(A) Peak X, because F has a smaller firstionization energy than C has.

(B) Peak X, because F has a greater nuclearcharge than C has.

(C) Peak Y, because F is more electronegativethan C is.

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Photoelectron Spectrum

(D) Peak Y, because F has a smaller atomic radiusthan C has.

The photoelectron spectra of the 1s electrons of two isoelectronic species, Ca2+ and Ar, are shown above.Which of the following correctly identifies the species associated with peak X and provides a valid justification?

(A) Ar, because it has completely filled energy levels

(B) Ar, because its radius is smaller than the radius of Ca2+

(C) Ca2+, because its nuclear mass is greater than that of Ar

(D) Ca2+, because its nucleus has two more protons than the nucleus of Ar has

The complete photoelectron spectra of neutral atoms of two unknown elements, X and Y, are shown above. Which of the following can be inferred from the data?

(A) Element X has a greater electronegativity than element Y does.

(B) Element X has a greater ionization energy than element Y does.

(C) Element Y has a greater nuclear charge than element X does.

(D) The isotopes of element Y are approximately equal in abundance, but those of element X are not.

The complete photoelectron spectrum of an unknown element is shown above. The frequency ranges of different regions of the electromagnetic spectrum are given in the table below.

(a) To generate the spectrum above, a source capable of producing electromagnetic radiation with an energy of 7 × 104 kJ per mole of photons was used. Such radiation is from which region of the electromagnetic spectrum? Justify your answer with a calculation.

(b) A student examines the spectrum and proposes that the second ionization energy of the element is3.88 × 103 kJ/mol. To refute the proposed interpretation of the spectrum, identify the following.

(i) The subshell from which an electron is removedin the second ionization of an atom of the element

(ii) The subshell that corresponds to the secondpeak of the photoelectron spectrum above