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requnc TonnEution step that is, Hg9+(g) Hg80+ (g) e 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 to be ejected 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 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- tron from potassium is found to be 5.34 x 10-19 J using 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)? 11.37 The energy needed for the following process is 1.96 x 104 kJ mol-1: Li(g) → Li3+(g) + 3e- If the first ionization of lithium is 520 kJ mol-1, calcu- late the second ionization of lithium; that is, calculate the energy required for the process

requnc TonnEution step that is, Hg9+(g) Hg80+ (g) e 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 to be ejected 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 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- tron from potassium is found to be 5.34 x 10-19 J using 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)? 11.37 The energy needed for the following process is 1.96 x 104 kJ mol-1: Li(g) → Li3+(g) + 3e- If the first ionization of lithium is 520 kJ mol-1, calcu- late the second ionization of lithium; that is, calculate the energy required for the process

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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11.36

requnc TonnEution step that is, Hg9+(g) Hg80+ (g) e 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 to be ejected 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 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- tron from potassium is found to be 5.34 x 10-19 J using 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)? 11.37 The energy needed for the following process is 1.96 x 104 kJ mol-1: Li(g) → Li3+(g) + 3e- If the first ionization of lithium is 520 kJ mol-1, calcu- late the second ionization of lithium; that is, calculate the energy required for the process
Transcribed Image Text:requnc TonnEution step that is, Hg9+(g) Hg80+ (g) e 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 to be ejected 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 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- tron from potassium is found to be 5.34 x 10-19 J using 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)? 11.37 The energy needed for the following process is 1.96 x 104 kJ mol-1: Li(g) → Li3+(g) + 3e- If the first ionization of lithium is 520 kJ mol-1, calcu- late the second ionization of lithium; that is, calculate the energy required for the process
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