Chapter 6, Problem 6PS

### 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

# Violet light has wavelength of about 410 nm. What is its frequency? Calculate the energy of one photon of violet light. What is the energy of 1.0 mol of violet photons? Compare the energy of photons of violet light with those of red light. Which is more energetic?

Interpretation Introduction

Interpretation: The frequency, the energy of one photon and the energy of 1.0mol of photons of violet light have to be calculated. The energy of violet 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.
• The violet and red light lies in the visible light where violet light lies between 450 nm400 nm whereas red light is lies in 700nm635 nm.
• 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, the energy per photon and the energy per mole of photons of violet light is calculated below.

Given,

The wavelength violet light is 410â€‰nmâ€‰=â€‰4.10Â 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 violet light is calculated by using the equation,

Â Â Â Â Î½â€‰=â€‰cÎ»=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sâ€‰4.10Â Ã—â€‰10â€‰âˆ’7mâ€‰=â€‰7.31â€‰Ã—â€‰10â€‰14â€‰sâˆ’1

The frequency of violet light is 7.31â€‰Ã—â€‰10â€‰14â€‰sâˆ’1

• The energy per photon of violet light is calculated,

â€‚Â Eâ€‰=â€‰hÎ½

Substituting the values to the above equation,

â€‚Â Eâ€‰=â€‰hÎ½=â€‰6.626â€‰Ã—â€‰10â€‰âˆ’34â€‰J.sâ€‰Ã—â€‰7.31â€‰Ã—â€‰10â€‰14â€‰sâˆ’1=â€‰4.844â€‰Ã—â€‰10â€‰âˆ’19â€‰J

The energy per photon is 4

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