Chapter 9.3, Problem 1.2ACP

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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.Â  When electrical excited, He atoms emit nearly monochromic radiation at a wavelength of 58.4 nm.

The given photon has the wavelength of 58.4â€‰nmâ€‰â€‰=â€‰â€‰58.4â€‰Ã—â€‰10-9â€‰m thus the energy of corresponding photon can be determined as follows,

Â  E=hÎ½Eâ€‰=â€‰hcÎ»=â€‰6.63â€‰Ã—â€‰10-34â€‰Jsâ€‰â€‰Ã—â€‰3.0â€‰Ã—â€‰108â€‰m/s58.4â€‰Ã—â€‰10-9â€‰m=â€‰0

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