Write an equation that shows the process corresponding to the first ionization energy of Hg. Express your answer as a chemical equation. Identify all of the phases in your answer. 0 - ΑΣΦ ? A chemical reaction does not occur for this question.
Write an equation that shows the process corresponding to the first ionization energy of Hg. Express your answer as a chemical equation. Identify all of the phases in your answer. 0 - ΑΣΦ ? A chemical reaction does not occur for this question.
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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![Write an equation that shows the process corresponding to the first ionization energy of Hg.
Express your answer as a chemical equation. Identify all of the phases in your answer.
0
ΑΣΦ
wwwwww
?
A chemical reaction does not occur for this question.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb45a7b7-d1aa-4ac7-ad98-90aa0325071f%2Fb2e2e6d1-8e98-40c7-aecf-8a4e5450e25f%2Fg4w9nwl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Write an equation that shows the process corresponding to the first ionization energy of Hg.
Express your answer as a chemical equation. Identify all of the phases in your answer.
0
ΑΣΦ
wwwwww
?
A chemical reaction does not occur for this question.
![One way to measure ionization energies is photoelectron
spectroscopy (PES), a technique based on the
photoelectric effect. In PES, monochromatic light is
directed onto a sample, causing electrons to be emitted.
The kinetic energy of the emitted electrons is measured.
The difference between the energy of the photons and the
kinetic energy of the electrons corresponds to the energy
needed to remove the electrons (that is, the ionization
energy). Suppose that a PES experiment is performed in
which mercury vapor is irradiated with ultraviolet light of
wavelength 58.4 nm. (Figure 1)
Figure
lonization energy (kJ/mol)
2500
2000
1500
1000
500
0
419
Rb
403
496
1A 2A
549
Ha
509
H
1312
$20 M
Na 738
Be
899
Ca
990
801
Al 786
Sa
709
558 PD
11 710
Ga 762
Ge 947 941
579
S 1012 1000
BE
703
Sb
834
O
1402 1314
Po
812
Increasing ionization energy
s 1251
Te 1008
SAY
1681
Kr
Se 1140
Be 1351
Xo
1170
Rn
1037
3A 4A SA 6A 7A SA
<
Ar
1521
Ne
2081
1 of 1 >
He
Increasing ionization energy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb45a7b7-d1aa-4ac7-ad98-90aa0325071f%2Fb2e2e6d1-8e98-40c7-aecf-8a4e5450e25f%2F3nvt2g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:One way to measure ionization energies is photoelectron
spectroscopy (PES), a technique based on the
photoelectric effect. In PES, monochromatic light is
directed onto a sample, causing electrons to be emitted.
The kinetic energy of the emitted electrons is measured.
The difference between the energy of the photons and the
kinetic energy of the electrons corresponds to the energy
needed to remove the electrons (that is, the ionization
energy). Suppose that a PES experiment is performed in
which mercury vapor is irradiated with ultraviolet light of
wavelength 58.4 nm. (Figure 1)
Figure
lonization energy (kJ/mol)
2500
2000
1500
1000
500
0
419
Rb
403
496
1A 2A
549
Ha
509
H
1312
$20 M
Na 738
Be
899
Ca
990
801
Al 786
Sa
709
558 PD
11 710
Ga 762
Ge 947 941
579
S 1012 1000
BE
703
Sb
834
O
1402 1314
Po
812
Increasing ionization energy
s 1251
Te 1008
SAY
1681
Kr
Se 1140
Be 1351
Xo
1170
Rn
1037
3A 4A SA 6A 7A SA
<
Ar
1521
Ne
2081
1 of 1 >
He
Increasing ionization energy
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