the energy nano ther required for the last ionization step-that is, 11.42 are a 79+ Hg(g)Hg0(g)e 80+ 7.7 peric 11.36 A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sam- ple is irradiated with UV light, which causes electrons Add 11.4 be eiected from the valence shell. The kinetic energies of the ejected electrons are measured. Knowing the en- ergy of the UV photon and the kinetic energy of the ejected electron, we can write ener or abc tra enc hv -ΙE +5mv to kir (b) where v is the frequency of the UV light, and me and v are the mass and velocity of the electron, respectively. In one experiment, the kinetic energy of the ejected elec- an it lew tron from potassium is found to be 5.34 x 10-19 J using 11 a UV source of wavelength 162 nm. Calculate the ion- ization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)? h 11.37 The energy needed for the following process is 1.96 x 104 kJ mol: V 1 Li(g)Li (g)+3e If the first ionization of lithium is 520 kJ mol1, calcu- late the second ionization of lithium; that is, calculate

the energy nano ther required for the last ionization step-that is, 11.42 are a 79+ Hg(g)Hg0(g)e 80+ 7.7 peric 11.36 A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sam- ple is irradiated with UV light, which causes electrons Add 11.4 be eiected from the valence shell. The kinetic energies of the ejected electrons are measured. Knowing the en- ergy of the UV photon and the kinetic energy of the ejected electron, we can write ener or abc tra enc hv -ΙE +5mv to kir (b) where v is the frequency of the UV light, and me and v are the mass and velocity of the electron, respectively. In one experiment, the kinetic energy of the ejected elec- an it lew tron from potassium is found to be 5.34 x 10-19 J using 11 a UV source of wavelength 162 nm. Calculate the ion- ization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)? h 11.37 The energy needed for the following process is 1.96 x 104 kJ mol: V 1 Li(g)Li (g)+3e If the first ionization of lithium is 520 kJ mol1, calcu- late the second ionization of lithium; that is, calculate

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter6: Quantum Mechanics And Molecular Structure

Section: Chapter Questions

Problem 39P: The photoelectron spectrum of HBr has two main groups of peaks. The first has ionization energy...

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