Chapter 7, Problem 7A.4BE

(i)

Interpretation Introduction

Interpretation: Under the given conditions, the energy of quantum has to be calculated.

Concept introduction: The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

Answer to Problem 7A.4BE

The energy of quantum in joule and kilo joule per mole are $\underset{\_}{2.65\times 10^{-19}J}$ and $\underset{\_}{159.6kJmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation

Therefore,

  $E=h\nu$                                                                                                            (1)

Where,

• $E$ is the energy of quantum.
• $\nu$ is the frequency of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$

The frequency in terms of time period is expressed as,

  $\nu =\frac{1}{T}$                                                                                                              (2)

Where,

• $T$ is the time period of  electronic oscillation.

Substitute the value of $\nu$ in equation (1).

  $E=h\times \frac{1}{T}$                                                                                                       (3)

It is given that,

The time period is $2.50\text{fs}$.

The conversion of $\text{fs}$ to $\text{s}$ is done as,

  $1\text{fs}={10}^{-15}\text{s}$

Therefore, the conversion of $2.50\text{fs}$ to $\text{s}$ is done as,

  $2.50\text{fs}=2.50\times {10}^{-15}\text{s}$

Substitute the value of $h$ and $T$ in equation (3).

  $\begin{array}{c}E=6.626\times {10}^{-34}\text{J}\text{s}\times \frac{1}{2.50\times {10}^{-15}\text{s}}\\ =\underset{\_}{2.65\times 10^{-19}J}\end{array}$

Per mole energy ($E_{{\text{N}}_{\text{A}}}$) is calculated by using the formula,

  $E_{{\text{N}}_{\text{A}}}=N_{\text{A}}\times E$                                                                                                 (4)

Where,

• ${\text{N}}_{\text{A}}$ is Avogadro’s number $\left(6.023\times \text{1}{0}^{23}\right)$.

Substitute the value of ${\text{N}}_{\text{A}}$ and $E$ in equation (4).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=6.023\times \text{1}{0}^{23}{\text{mol}}^{-1}\times \text{2}{\text{.65×10}}^{-\text{19}}\text{J}\\ =\text{15}\text{.96}\times \text{1}{0}^4\text{J}{\text{mol}}^{-1}\\ =\text{159}\text{.6}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}\end{array}$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{J}{\text{mol}}^{-1}={10}^{-3}\text{kJ}{\text{mol}}^{-1}$

Therefore, the conversion of $\text{159}\text{.6}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{159}\text{.6}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}=\underset{\_}{159.6kJmo{l}^{-1}}$

(ii)

Interpretation Introduction

Interpretation: Under the given conditions, the energy of quantum has to be calculated.

Concept introduction: The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

Answer to Problem 7A.4BE

The energy of quantum in joule and kilo joule per mole are $\underset{\_}{3\times 10^{-19}J}$ and $\underset{\_}{180.69kJmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation

Therefore,

  $E=h\nu$                                                                                                           (1)

Where,

• $E$ is the energy of quantum.
• $\nu$ is the frequency of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$

The frequency in terms of time period is expressed as,

  $\nu =\frac{1}{T}$                                                                                                              (2)

Where,

• $T$ is the time period of  electronic oscillation.

Substitute the value of $\nu$ in equation (1).

  $E=h\times \frac{1}{T}$                                                                                                       (3)

It is given that,

The time period is $2.21\text{fs}$.

The conversion of $\text{fs}$ to $\text{s}$ is done as,

  $1\text{fs}={10}^{-15}\text{s}$

Therefore, the conversion of $2.21\text{fs}$ to $\text{s}$ is done as,

  $2.21\text{fs}=2.21\times {10}^{-15}\text{s}$

Substitute the value of $h$ and $T$ in equation (3).

  $\begin{array}{c}E=6.626\times {10}^{-34}\text{J}\text{s}\times \frac{1}{2.21\times {10}^{-15}\text{s}}\\ =\underset{\_}{3\times 10^{-19}J}\end{array}$

Per mole energy is calculated by using the formula,

  $E_{{\text{N}}_{\text{A}}}=N_{\text{A}}\times E$                                                                                                (4)

Where,

• ${\text{N}}_{\text{A}}$ is Avogadro’s number $\left(6.023\times \text{1}{0}^{23}\right)$.

Substitute the value of ${\text{N}}_{\text{A}}$ and $E$ in equation (4).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=6.023\times \text{1}{0}^{23}{\text{mol}}^{-1}\times {\text{3×10}}^{-19}\text{J}\\ =\text{180}\text{.69}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}\end{array}$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{J}{\text{mol}}^{-1}={10}^{-3}\text{kJ}{\text{mol}}^{-1}$

Therefore, the conversion of $\text{180}\text{.69}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{180}\text{.69}\times \text{1}{0}^3\text{J}{\text{mol}}^{-1}=\underset{\_}{180.69kJmo{l}^{-1}}$

(iii)

Interpretation Introduction

Interpretation: Under the given conditions, the energy of quantum has to be calculated.

Concept introduction: The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

Answer to Problem 7A.4BE

The energy of quantum in joule and kilo joule per mole are $\underset{\_}{6.626\times 10^{-31}J}$ and $\underset{\_}{39.9\times 10^{-11}kJmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation

Therefore,

  $E=h\nu$                                                                                                          (1)

Where,

• $E$ is the energy of quantum.
• $\nu$ is the frequency of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$

The frequency in terms of time period is expressed as,

  $\nu =\frac{1}{T}$                                                                                                             (2)

Where,

• $T$ is the time period of  electronic oscillation.

Substitute the value of $\nu$ in equation (1).

  $E=h\times \frac{1}{T}$                                                                                                        (3)

It is given that,

The time period is $1.0\text{ms}$.

The conversion of $\text{ms}$ to s is done as,

  $1\text{ms}={10}^{-3}\text{s}$

Substitute the value of $h$ and $T$ in equation (3).

  $\begin{array}{c}E=6.626\times {10}^{-34}\text{J}\text{s}\times \frac{1}{{10}^{-3}\text{s}}\\ =\underset{\_}{6.626\times 10^{-31}J}\end{array}$

Per mole energy ($E_{{\text{N}}_{\text{A}}}$) is calculated by using the formula,

  $E_{{\text{N}}_{\text{A}}}=N_{\text{A}}\times E$                                                                                                 (4)

Where,

• ${\text{N}}_{\text{A}}$ is Avogadro’s number $\left(6.023\times \text{1}{0}^{23}\right)$.

Substitute the value of ${\text{N}}_{\text{A}}$ and $E$ in equation (4).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=6.023\times \text{1}{0}^{23}{\text{mol}}^{-1}\times \text{6}{\text{.626×10}}^{-31}\text{J}\\ =\text{39}\text{.9}\times \text{1}{0}^{-8}\text{J}{\text{mol}}^{-1}\end{array}$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{J}{\text{mol}}^{-1}={10}^{-3}\text{kJ}{\text{mol}}^{-1}$

Therefore, the conversion of $\text{39}\text{.9}\times \text{1}{0}^{-8}\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is done as,

  $\text{39}\text{.9}\times \text{1}{0}^{-8}\text{J}{\text{mol}}^{-1}=\underset{\_}{39.9\times 10^{-11}kJmo{l}^{-1}}$