Chapter 7, Problem 7.30QE

(a)

Interpretation Introduction

Interpretation:

The wavelength of light of frequency $9.83\times {10}^{14}\text{ Hz}$ has to be determined.

Concept Introduction:

The wave nature of any light can be described by its frequency, wavelength, and amplitude. The wavelength $\left(\text{λ, lambda}\right)$ of the light is defined as the distance between two successive peaks. Its SI unit is meter. The frequency $\left(v\right)$ is defined as the number of waves that can pass through the one point in 1 second. Its SI unit is ${\text{s}}^{-1}$. The amplitude $\left(A\right)$ of the light wave is maximum height of a wave.

The relation between frequency $\left(v\right)$ and wavelength $\left(\text{λ}\right)$ of the light is as follows:

  $v=\frac{c}{\text{λ}}$        (1)

Here, $c$ is the speed of light and its value is $3.00\times {10}^8\text{ m/s}$.

The relation between the energy $\left(E\right)$ of a photon and frequency $\left(v\right)$ is as follows:

  $E=hv$        (2)

Here, $h$ is known as Plank’s constant and its value is $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$.

Answer to Problem 7.30QE

The wavelength of light is $3.05\times {10}^{-7}\text{ m }$.

Explanation of Solution

Rearrange equation (1) to calculate $\text{λ}$ as follows:

  $\text{λ}=\frac{c}{v}$        (3)

Substitute $9.83\times {10}^{14}\text{ Hz}$ for $v$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (3)

  $\begin{array}{c}\text{λ}=\left(\frac{3.00\times {10}^8\text{ m/s}}{9.83\times {10}^{14}\text{ Hz}}\right)\left(\frac{1\text{ Hz}}{1{\text{ s}}^{-1}}\right)\\ =3.05\times {10}^{-7}\text{ m }\end{array}$

(b)

Interpretation Introduction

Interpretation:

The energy of one photon for light of frequency $9.83\times {10}^{14}\text{ Hz}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.30QE

The energy of one photon for light is $6.51\times {10}^{-19}\text{ J}$.

Explanation of Solution

Substitute $9.83\times {10}^{14}\text{ Hz}$ for $v$ and $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$ for $h$ in equation (2).

  $\begin{array}{c}E=\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(9.83\times {10}^{14}\text{ Hz}\right)\left(\frac{1{\text{ s}}^{-1}}{1\text{ Hz}}\right)\\ =6.51\times {10}^{-19}\text{ J}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The energy of one mole photon in $\text{kJ}$ for light of frequency $9.83\times {10}^{14}\text{ Hz}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.30QE

The energy of one mole photon for light is $3.92\times {10}^2\text{ kJ}$.

Explanation of Solution

Substitute $9.83\times {10}^{14}\text{ Hz}$ for $v$ and $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$ for $h$ in equation (2).

  $\begin{array}{c}E=\left(\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}\right)\left(9.83\times {10}^{14}\text{ Hz}\right)\left(\frac{1{\text{ s}}^{-1}}{1\text{ Hz}}\right)\\ =6.51\times {10}^{-19}\text{ J}\end{array}$

The energy per phone is $6.51\times {10}^{-19}\text{ J/photon}$.

The expression to calculate the energy of 1 mole photon is as follows:

  $E\left(\text{J/mol}\right)=E\left(\text{J/photon}\right)\times {\text{N}}_{\text{A}}$        (4)

Here, ${\text{N}}_{\text{A}}$ is the Avogadro number and its value is $6.022\times {10}^{23}{\text{ mol}}^{-1}$.

Substitute $6.51\times {10}^{-19}\text{ J/photon}$ for $E\left(\text{J/photon}\right)$ and $6.022\times {10}^{23}{\text{ mol}}^{-1}$ for ${\text{N}}_{\text{A}}$ in equation (4)

  $\begin{array}{c}E\left(\text{J/mol}\right)=\left(6.51\times {10}^{-19}\text{ J/photon}\right)\left(6.022\times {10}^{23}{\text{ mol}}^{-1}\right)\\ =\left(3.92\times {10}^5\text{ J/mol}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\\ =3.92\times {10}^2\text{ kJ/mol}\end{array}$