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A cell phone sends signals at about 850 MHz (where 1 MHz = 1 × 10 6 Hz or cycles per second). (a) What is the wavelength of this radiation? (b) What is the energy of 1.0 mol of photons with a frequency of 850 MHz? (c) Compare the energy in part (b) with the energy of a mole of photons of violet light (420 nm). (d) Comment on the difference in energy between 850 MHz radiation and violet light.

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Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

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Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 6, Problem 57GQ
Textbook Problem
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A cell phone sends signals at about 850 MHz (where 1 MHz = 1 × 106 Hz or cycles per second). (a) What is the wavelength of this radiation? (b) What is the energy of 1.0 mol of photons with a frequency of 850 MHz? (c) Compare the energy in part (b) with the energy of a mole of photons of violet light (420 nm). (d) Comment on the difference in energy between 850 MHz radiation and violet light.

(a)

Interpretation Introduction

Interpretation: The wavelength of cell phone signal has to be calculated.

Concept introduction:

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

  ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Explanation of Solution

The wavelength of phone signal is calculated below.

Given,

The frequency of cell phone signal is 850MHz=850 ×106Hz=850 ×106s1

  The speed of light,c=2.998×108m/s

The wavelength of cell phone signal is calculated by using the equation,

    <

(b)

Interpretation Introduction

Interpretation: The energy per mole of photons of cell phone signal has to be calculated.

Concept introduction:

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.

(c)

Interpretation Introduction

Interpretation: The energy of violet light is to be compared with 850MHz cell phone signal

Concept introduction:

  • 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

(d)

Interpretation Introduction

Interpretation: The energy difference in violet light and cell phone signal with 850MHz has to be compared.

Concept introduction:

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.

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Chapter 6 Solutions

Chemistry & Chemical Reactivity
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