Chapter 6, Problem 56GQ

### 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

# Radiation in the ultraviolet region of the electro-magnetic spectrum is quite energetic. It is this radiation that causes dyes to fade and your skin to develop a sunburn. If you are bombarded with 1.00 mol of photons with a wavelength of 375 nm, what amount of energy, in kilojoules per mole of photons, are you being subjected to?

Interpretation Introduction

Interpretation: The energy per mole of photons of ultraviolet light is to be calculated.

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 ultraviolet radiation lies in the wavelength of 100nm and 400nm
• 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 energy per mole of photons is the product of energy per photon and Avogadro’s number,NA.

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

ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Explanation

The energy per mole of photons of ultraviolet light is calculated

Given,

The wavelength ultraviolet light is 375â€‰nmâ€‰=â€‰3.75Â 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 ultraviolet light is,

Â Â Â Â Î½â€‰=â€‰cÎ»Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (a)

The energy per photon of ultraviolet light is,

â€‚Â Eâ€‰=â€‰hÎ½Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b)

Combining (a) and (b)

Â Â Â Â Eâ€‰=â€‰â€‰hcÎ»

Substituting the values to the above equation,

â€‚Â Eâ€‰=â€‰â€‰hcÎ»=â€‰6

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