One way to measure ionization energies is ultraviolet photoelectron
spectroscopy (PES), a technique based on the photoelectric efect. (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
diference 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 joules? (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 \(1.72 \times
10^{-18} \mathrm{J} .\) What is the first ionization energy of \(\mathrm{Hg},\)
in \(\mathrm{kJ} / \mathrm{mol} ?(\mathbf{d})\) Using Figure \(7.10,\) determine
which of the halogen elements has a first ionization energy closest to that of
mercury.