energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm. (Figure 1)

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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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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
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
What is the energy of a photon of this light, in eV?
VE ΑΣΦ
ΕΞ
H
?
eV
Transcribed Image Text:What is the energy of a photon of this light, in eV? VE ΑΣΦ ΕΞ H ? eV
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