   Chapter 9.3, Problem 1.2ACP

Chapter
Section
Textbook Problem

What is the energy of a photon with a wavelength of 58.4 nm in kilojoules/mole?

Interpretation Introduction

Interpretation:

The energy of photon with the given wavelength of 58.4nm in kJ/mol has to be determined.

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 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.

1J=0.001kJ

Explanation

The Helium (He) gas is the most common UV radiation source used for the photoelectron spectroscopy of molecules.  When electrical excited, He atoms emit nearly monochromic radiation at a wavelength of 58.4 nm.

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

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