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In a particular photoelectron spectrum using 21.21eV photons, electrons were ejected with kinetic energies of 11.01 eV, 8.23 eV, and 5.22 eV. Sketch the molecular orbital energy level diagram for the species, showing the ionization energies of the three identifiable orbitals.

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Jean Keeling
Jean KeelingLv2
6 Nov 2019
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Related questions


Mercury in the environment can exist in oxidation states 0, +1,
and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury’s 4f orbitals at 105 eV, from an X-ray source that provided 1253.6 eV of energy. The oxygen on the mineral surface gave emitted electron energies at 531 eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and –2 for O. (a) Calculate the wavelength of the X rays used in this experiment. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground-state electron configurations for Hg2+ and O2-; which electrons are the valence electrons in each case? (d) Use Slater’s rules to estimate Zeff for the 4f and valence electrons of Hg2+ and O2-; assume for this purpose that all the inner electrons with (n – 3) or less screen a full +1.

A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with ultraviolet (UV) light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write hv=IE+1/2mv^2, in which v is the frequency of the UV light, and m and v are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be 5.34x10^-19 J using a UV source of wavelength 162 nm. Calculate the ionization energy of potassium (in what units is it usually reported?). Answer: 420 kj/mol Please help with the steps and how they got this answer? Thank you!

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