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One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. (Section 6.2) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (a) What is the energy of a photon of this light, in eV? (b) Write an equation that shows the process corresponding to the first ionization energy of Hg. (c) The kinetic energy of the emitted electrons is measured to be 10.75 eV. What is the first ionization energy of Hg, in kJ/mol? (d) Using Figure 7.10, determine which of the halogen elements has a first ionization energy closest to that of mercury.

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Jamar Ferry
Jamar FerryLv2
20 May 2020

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