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

Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
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Chapter42: Molecules And Solids
Section: Chapter Questions
Problem 47CP: As an alternative to Equation 42.1, another useful model for the potential energy of a diatomic...
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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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