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
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question

Transcribed Image Text: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
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