Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. 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 28.0 nm.

(a) What is the energy of a photon of this light in eV (1 eV = 1.602e-19J)?
___eV

(b) Write an equation that shows the process corresponding to the first ionization energy of Hg. (Use the lowest possible whole number coefficients. Omit states-of-matter in your answer.)

Answer 1