   Chapter 9.3, Problem 1.4ACP

Chapter
Section
Textbook Problem

The kinetic energy of an electron ejected from the σ*2s molecular orbital of N2 using 58.4 nm radiation is 4.23 × 10–19 J. What is the ionization energy, in both kJ/mol and eV, of an electron from this orbital?

Interpretation Introduction

Interpretation:

The ionization energy in both kJ/mol and eV should be determined for an electron in the given (σ*2s) orbital.

Concept Introduction:

Photoelectron spectroscopy: This is to energy measurement of electrons emitted from solids, gases or liquids by the photoelectric effect, in order to determine the binding energies of electrons in a substance.

The key to understanding PES is that all of the ejected electrons results from bombarding the atom with photons of the same energy (hν). The energy goes into ejecting the electrons from the atom corresponding to the ionization energy (IE) of the electron with the remainder going into kinetic energy of the electron. The formula of photoelectron spectroscopy E=hv= IE + KE (electron)

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

The energy E of electromagnetic radiation is directly proportional to the frequency ν of the radiation.

Therefore,

E=hνE=hcλ

Where, E is the energy, h is the Plank’s constant (6.626×1034Js), ν is the frequency of radiation, c is the velocity of light and λ is the wavelength of the radiation.

Explanation

The given photon has the wavelength of 58.4nm=58.4×10-9m thus the energy of corresponding photon can be determined as follows,

E=hνE=hcλ=6.63×10-34Js×3.0×108m/s58.4×10-9m=0.340×10-17J/photons=3.40×10-18J

Ionization energy in both kJ/mol and eV can be calculated using the equation,

IE=hνKE

=3.40×10-18JKE=4.23×10-19J

IE=3.40×10-18J-4

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