Chapter 7, Problem 7A.5BE

(i)

Interpretation Introduction

Interpretation:

The energy of photon and the energy per mole of photon for the radiation of wavelength $200\text{nm}$ have 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.5BE

The energy of photon and the energy per mole of photon are $\underset{\_}{9.93\times 10^{-19}J}$ and $\underset{\_}{59.86\times 10^{4}Jmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of photon is inversely proportional to the wavelength of radiation.

Therefore,

  $E=\frac{hc}{\lambda }$                                                                                                          (1)

Where,

• $E$ is the energy of photon.
• $c$ is the velocity of light $\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)$.
• $\lambda$ is the wavelength of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$.

It is given that,

The wavelength of radiation is $200\text{nm}$.

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $200\text{nm}$ to $\text{m}$ is done as,

  $200\text{nm}=200\times {10}^{-9}\text{m}$

Substitute the values of $h,c$ and $\lambda$ in equation (1).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)}{200\times {10}^{-9}\text{m}}\\ =\underset{\_}{9.93\times 10^{-19}J}\end{array}$

Therefore, the energy of photon is $\underset{\_}{9.93\times 10^{-19}J}$.

The energy of photon per mole is calculated by using the relation,

  $\text{Energy of photon per mole}=E\times N_{\text{A}}$                                                            (2)

Where,

• $N_{\text{A}}$ is Avogadro’s number $6.023\times {10}^{23}\text{per}\text{mol}$.

Substitute the value of $N_{\text{A}}$ and $E$ in equation (2).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=9.93{\text{×10}}^{-\text{19}}\text{J}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ \text{=}\underset{\_}{59.86\times 10^{4}Jmo{l}^{-1}}\end{array}$

Therefore the energy of photon per mole for the radiation is $\underset{\_}{59.86\times 10^{4}Jmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation:

The energy of photon and the energy per mole of photon for the radiation of wavelength $150\text{pm}$ have 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.5BE

The energy of photon and the energy per mole of photon are $\underset{\_}{3.6\times 10^{-19}J}$ and $\underset{\_}{7.98\times 10^{8}Jmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of photon is inversely proportional to the wavelength of radiation.

Therefore,

  $E=\frac{hc}{\lambda }$                                                                                                          (1)

Where,

• $E$ is the energy of photon.
• $c$ is the velocity of light $\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)$.
• $\lambda$ is the wavelength of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$

It is given that,

The wavelength of radiation is $150\text{pm}$.

The conversion of $\text{pm}$ to $\text{m}$ is done as,

  $\text{1}\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $150\text{pm}$ to $\text{m}$ is done as,

  $150\text{pm}=150\times {10}^{-12}\text{m}$

Substitute the values of $h,c$ and $\lambda$ in equation (1).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)}{150\times {10}^{-12}\text{m}}\\ =\underset{\_}{1.3\times 10^{-15}J}\end{array}$

Therefore, the energy of photon is $\underset{\_}{1.3\times 10^{-15}J}$.

The energy of photon per mole is calculated by using the relation,

The energy of photon per mole is calculated by using the relation,

  $\text{Energy of photon per mole}=E\times N_{\text{A}}$                                                            (2)

Where,

• $N_{\text{A}}$ is Avogadro’s number $6.023\times {10}^{23}\text{per}\text{mole}$.

Substitute the value of $N_{\text{A}}$ and $E$ in equation (2).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=\text{1}{\text{.3×10}}^{-\text{15}}\text{J}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ \text{=}\underset{\_}{7.98\times 10^{8}Jmo{l}^{-1}}\end{array}$

Therefore the energy of photon per mole for the radiation is $\underset{\_}{7.98\times 10^{8}Jmo{l}^{-1}}$.

(iii)

Interpretation Introduction

Interpretation:

The energy of photon and the energy per mole of photon for the radiation of wavelength $\text{1}\text{.0}\text{cm}$ have 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.5BE

The energy of photon and the energy per mole of photon are $\underset{\_}{19.87\times 10^{-24}J}$ and $\underset{\_}{11.97Jmo{l}^{-1}}$ respectively.

Explanation of Solution

The energy of photon is inversely proportional to the wavelength of radiation.

Therefore,

  $E=\frac{hc}{\lambda }$                                                                                                          (1)

Where,

• $E$ is the energy of photon.
• $c$ is the velocity of light $\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)$.
• $\lambda$ is the wavelength of radiation.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{s}\right)$

It is given that,

The wavelength of radiation is $\text{1}\text{.0}\text{cm}$.

The conversion of $\text{cm}$ to $\text{m}$ is done as,

  $\text{1}\text{cm}={10}^{-2}\text{m}$

Substitute the values of $h,c$ and $\lambda$ in equation (1).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\left(3\times {\text{10}}^{\text{8}}\text{m}{\text{s}}^{-1}\right)}{{10}^{-2}\text{m}}\\ =\underset{\_}{19.87\times 10^{-24}J}\end{array}$

Therefore, the energy of photon is $\underset{\_}{19.87\times 10^{-24}J}$.

The energy of photon per mole is calculated by using the relation,

  $\text{Energy of photon per mole}=E\times N_{\text{A}}$                                                            (2)

Where,

• $N_{\text{A}}$ is Avogadro’s number $6.023\times {10}^{23}\text{per}\text{mole}$.

Substitute the value of $N_{\text{A}}$ and $E$ in equation (2).

  $\begin{array}{c}{E}_{{\text{N}}_{\text{A}}}=\text{19}{\text{.87×10}}^{-24}\text{J}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ \text{=}\underset{\_}{11.97Jmo{l}^{-1}}\end{array}$

Therefore the energy of photon per mole for the radiation is $\underset{\_}{11.97Jmo{l}^{-1}}$.