Chapter 6, Problem 70IL

### 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

# A solution of KMnO4 absorbs light at 540 nm (page 206). What is the frequency of the light absorbed? What is the energy of one mole of photons with λ = 540 nm?

Interpretation Introduction

Interpretation: The frequency, the energy per photon and the energy per mole of photons of absorbed light has to be calculated. The energy of absorbed light has to be compared with red light

Concept introduction:

• Electromagnetic radiations are a type of energy surrounding us. They are of different types like radio waves, IR, UV, X-ray etc.
• Planck’s equation,

E=where, E=energyh=Planck'sconstantν=frequency

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• The frequency of the light is inversely proportional to its wavelength.

ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Explanation

The frequency and the energy per mole of photons of absorbed light is calculated below.

Given,

The wavelength absorbed light is 540â€‰nmâ€‰=â€‰5.40Â Ã—â€‰10â€‰âˆ’7m

Â Â Planck'sâ€‰constant,hâ€‰=â€‰6.626â€‰Ã—â€‰10â€‰âˆ’34â€‰J.sTheÂ speedÂ ofÂ light,câ€‰=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sAvogadro'sÂ number,NAâ€‰=â€‰6.022â€‰Ã—â€‰1023â€‰photons/mol

• The frequency of absorbed light is calculated by using the equation,

Â Â Â Â Î½â€‰=â€‰cÎ»=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sâ€‰5.40Â Ã—â€‰10â€‰âˆ’7mâ€‰=â€‰5.552â€‰Ã—â€‰10â€‰14â€‰sâˆ’1

The frequency of absorbed light is 5.552â€‰Ã—â€‰10â€‰14â€‰sâˆ’1

• The energy per photon of absorbed light is calculated,

â€‚Â Eâ€‰=â€‰hÎ½

Substituting the values to the above equation,

â€‚Â Eâ€‰=â€‰(6

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