Chapter 6, Problem 57GQ

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.

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.

  $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{where, c}\text{=}\text{speed}\text{of}\text{light}\\ \text{ν}=\text{frequency}\\ \text{λ}=\text{wavelength}\end{array}$

Answer to Problem 57GQ

The wavelength of cell phone signal is $0.35\text{m}$

Explanation of Solution

The wavelength of phone signal is calculated below.

Given,

The frequency of cell phone signal is $850\text{MHz}\text{=}850\text{ ×}{\text{10}}^6\text{Hz}\text{=}850\text{ ×}{\text{10}}^6{\text{s}}^{-1}$

  $\text{The speed of light,c}\text{=}\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}$

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

    $\begin{array}{c}\text{λ=}\frac{\text{c}}{\text{ν}}\\ =\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{850\text{ ×}{\text{10}}^6{\text{s}}^{-1}}\\ =0.35\text{m}\end{array}$

The wavelength of cell phone signal is $0.35m$

Interpretation Introduction

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

Concept introduction:

Planck’s equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \text{where, E}\text{=}\text{energy}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{ν}=\text{frequency}\end{array}$

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

Answer to Problem 57GQ

The energy per mole of photons of cell phone signal is $0.34\text{J/mol}$

Explanation of Solution

The energy per photon cell phone signal is calculated,

Given,

The frequency of cell phone signal is $850\text{MHz}\text{=}850\text{ ×}{\text{10}}^6\text{Hz}\text{=}850\text{ ×}{\text{10}}^6{\text{s}}^{-1}$

  $\begin{array}{c}\text{Planck's}\text{constant},\text{h}\text{=}\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\\ \text{The speed of light,c}\text{=}\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}\\ {\text{Avogadro's number,N}}_{\text{A}}\text{=}\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{photons/mol}\end{array}$

The energy per photon cell phone signal is calculated,

  $\text{E}\text{=}\text{hν}$

Substituting the values to the above equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \\ =\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\times 850\text{ ×}{\text{10}}^6{\text{s}}^{-1}\\ \\ =5.6321\times {10}^{-25}\text{J}\end{array}$

The energy per photon is $5.6321\times {10}^{-25}\text{J}$

The energy per mole of photons of cell phone signal is calculated,

The energy per mole of photons is the product of energy per photon and Avogadro’s number,${\text{N}}_{\text{A}}$.

Therefore,

The energy per mole of photons of cell phone signal is,

$\begin{array}{c}\text{Energy per mole of photon}={\text{N}}_{\text{A}}\text{hν}\\ \\ \text{=}\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{photons/mol}\times 5.6321\times {10}^{-25}\text{J/photon}\\ \\ =0.34\text{J/mol}\end{array}$

The energy per mole of photons of cell phone signal is $0.34J/mol$

Interpretation Introduction

Interpretation: The energy of violet light is to be compared with $850\text{MHz}$ cell phone signal

Concept introduction:

• Planck’s equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \text{where, E}\text{=}\text{energy}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{ν}=\text{frequency}\end{array}$

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.

  $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{where, c}\text{=}\text{speed}\text{of}\text{light}\\ \text{ν}=\text{frequency}\\ \text{λ}=\text{wavelength}\end{array}$

Answer to Problem 57GQ

The energy per mole of photons of cell phone signal is $0.34\text{J/mol}$ whereas the energy per mole of photons of violet light is $280\text{kJ/mol}$

Explanation of Solution

Given,

The wavelength violet light is $420\text{nm}\text{=}4.20\text{ ×}\text{10}{}^{-7}\text{m}$

  $\begin{array}{c}\text{Planck's}\text{constant},\text{h}\text{=}\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\\ \text{The speed of light,c}\text{=}\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}\\ {\text{Avogadro's number,N}}_{\text{A}}\text{=}\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{photons/mol}\end{array}$

The frequency of violet light is,

    $\text{ν}\text{=}\frac{\text{c}}{\text{λ}}$                                        (a)

The energy per photon of violet light is,

  $\text{E}\text{=}\text{hν}$                                        (b)

Combining (a) and (b)

    $\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

Substituting the values to the above equation,

  $\begin{array}{c}\text{E}\text{=}\frac{\text{hc}}{\text{λ}}\\ =\frac{\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\times \text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{4.20\text{ ×}\text{10}{}^{-7}\text{m}}\\ =4.72\times {10}^{-19}\text{J}\end{array}$

The energy per photon is $4.72\times {10}^{-19}\text{J}$

• The energy per mole of photons of violet light is calculated,

The energy per mole of photons is the product of energy per photon and Avogadro’s number,${\text{N}}_{\text{A}}$.

Therefore,

The energy per mole of photons of violet light is,

$\begin{array}{c}\text{Energy per mole of photon}={\text{N}}_{\text{A}}\text{hν}\\ \text{=}\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{photons/mol}\times 4.72\times {10}^{-19}\text{J/photons}\\ =280\text{kJ/mol}\end{array}$

The energy per mole of photons of violet light is $280\text{kJ/mol}$

The energy per mole of photons of cell phone signal is $0.34\text{J/mol}$ whereas the energy per mole of photons of violet light is $280\text{kJ/mol}$

Interpretation Introduction

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

Concept introduction:

Planck’s equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \text{where, E}\text{=}\text{energy}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{ν}=\text{frequency}\end{array}$

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

Answer to Problem 57GQ

The energy per mole of photons of violet light is $840000$ times greater than the energy per mole of photons of cell phone signal.

Explanation of Solution

The energy per mole of photons of cell phone signal is $0.34\text{J/mol}$ whereas the energy per mole of photons of violet light is $280\text{kJ/mol}$

Hence,

$\begin{array}{c}\frac{\text{Energy per mole of photons of violet light}}{\text{Energy per mole of photons of cell phone signal}}\text{=}\frac{\text{280}\text{×}{\text{10}}^{\text{3}}\text{J/mol}}{\text{0}\text{.34}\text{J/mol}}\\ \\ \text{=}\text{823529}\\ \\ \text{=}\text{82}0000\end{array}$

Therefore,

The energy per mole of photons of violet light is $\text{82}0000$ times greater than the energy per mole of photons of cell phone signal.