Chapter 7, Problem 7.132AP

 (a)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of energies lost in the two-step process is same as the energy lost in the single transition from $n=3$ to $n=1$”.

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

Given

The transition occurs directly from $n=3$ to $n=1$.

The transition occurs first from energy level $n=3$ to $n=2$ and then from $n=2$ to $n=1$.

The energy difference between two energy levels is given as,

$\Delta E=R_H{Z}^2\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)$ (1)

Where,

• $R_H$ is the Rydberg constant $\left(-2.18\times {10}^{-18}\text{J}\right)$.
• $n_f$ is the final energy level.
• $n_i$ is the initial energy level.
• $\Delta E$ is the energy difference.
• $Z$ is the atomic number.

The energy lost in one step process is given as,

Substitute the value of $n_f=1$, $n_i=3$ and $R_H$ in the equation (1).

$\begin{array}{c}\Delta E=R_H{Z}^2\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(-2.18\times {10}^{-18}\text{J}\right){\left(2\right)}^2\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\\ =-7.75\times {10}^{-18}\text{J}\end{array}$

The negative sign indicates the loss of energy.

The energy lost in two step process is given as,

Substitute the value of $n_f=2$, $n_i=3$ , $Z$ and $R_H$ in the equation (1).

$\begin{array}{c}\Delta E_1=R_H{Z}^2\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(-2.18\times {10}^{-18}\text{J}\right){\left(2\right)}^2\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\\ =-1.21\times {10}^{-18}\text{J}\end{array}$

Substitute the value of $n_f=1$, $n_i=2$ , $Z$ and $R_H$ in the equation (1).

$\begin{array}{c}\Delta E_2=R_H{Z}^2\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(-2.18\times {10}^{-18}\text{J}\right){\left(2\right)}^2\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right)\\ =-6.54\times {10}^{-18}\text{J}\end{array}$

The total energy lost in two step process $\begin{array}{c}=\Delta E_1+\Delta E_2\\ =\left(\left(-1.21\times {10}^{-18}\right)-\left(6.54\times {10}^{-18}\right)\right)\text{J}\\ =-7.75\times {10}^{-18}\text{J}\end{array}$

Therefore, the value of energy lost is same whether the transition occurs in one step process or in two step process.

(b)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of wavelengths of the two photons emitted in the two-step process is equal to the wavelength of the single photon emitted in the transition from $n=3$ to $n=1$”.

Answer to Problem 7.132AP

Solution

The given statement is false.

Explanation of Solution

Explanation

The relation between energy, wavelength and speed of light is given as,

$\Delta E=h\frac{c}{\lambda }$ (2)

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{Js}\right)$.
• $\lambda$ is the wavelength of the light.
• $c$ is the speed of the light $\left(3\times {10}^8\text{m}/\text{s}\right)$.
• $\Delta E$ is the energy difference.

The wavelength of photon emitted in one step transition is given as,

Substitute the value of $h$, $c$ and $\Delta E$ in the equation (2),

$\begin{array}{c}\Delta E=h\frac{c}{\lambda }\\ 7.75\times {10}^{-18}\text{J}=\text{6}\text{.626}\times {\text{10}}^{-34}\text{Js}\times \frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\lambda }\\ 7.75\times {10}^{-18}\text{J}=\frac{19.878\times {10}^{-26}\text{J}\text{m}}{\lambda }\\ \lambda =2.56\times {10}^{-8}\text{m}\end{array}$

The wavelength that corresponds to given transition is $2.56\times {10}^{-8}\text{m}$.

The wavelength of photon emitted in two step transition is given as,

Substitute the value of $h$, $c$ and $\Delta E$ in the equation (2),

$\begin{array}{c}\Delta E=h\frac{c}{{\lambda }_1}\\ 1.21\times {10}^{-18}\text{J}=\text{6}\text{.626}\times {\text{10}}^{-34}\text{Js}\times \frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{{\lambda }_1}\\ 1.21\times {10}^{-18}\text{J}=\frac{19.878\times {10}^{-26}\text{J}\text{m}}{{\lambda }_1}\\ {\lambda }_1=16.43\times {10}^{-8}\text{m}\end{array}$

Substitute the value of $h$, $c$ and $\Delta E$ in the equation (2),

$\begin{array}{c}\Delta E=h\frac{c}{{\lambda }_2}\\ 6.54\times {10}^{-18}\text{J}=\text{6}\text{.626}\times {\text{10}}^{-34}\text{Js}\times \frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{{\lambda }_2}\\ 6.54\times {10}^{-18}\text{J}=\frac{19.878\times {10}^{-26}\text{J}\text{m}}{{\lambda }_2}\\ {\lambda }_2=3.04\times {10}^{-8}\text{m}\end{array}$

The total wavelength of light in two step process $\begin{array}{c}={\lambda }_1+{\lambda }_2\\ =\left(\left(16.43\times {10}^{-8}\right)+\left(3.04\times {10}^{-18}\right)\right)\text{J}\\ =19.47\times {10}^{-8}\text{m}\end{array}$

Therefore, the wavelengths that are emitted in two step process is not similar to wavelength of photons emitted in one step process.

(c)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The sum of frequencies of the two photons emitted in the two-step process is equal to the frequency of the single photon emitted in the transition from $n=3$ to $n=1$”.

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

The relation between frequency, wavelength and speed of light is given as,

$\nu =\frac{c}{\lambda }$ (3)

Where,

• $\nu$ is frequency of the light.
• $\lambda$ is the wavelength of the light.
• $c$ is the speed of the light $\left(3\times {10}^8\text{m}/\text{s}\right)$.

The frequency of photon emitted in one step transition is given as,

Substitute the value of $\lambda$, $c$ in the equation (2),

$\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{3\times {10}^8\text{m}/\text{s}}{2.56\times {10}^{-8}\text{m}}\\ =1.17\times {10}^{16}{\text{s}}^{-1}\end{array}$

The frequency that corresponds to given transition is $1.17\times {10}^{16}{\text{s}}^{-1}$.

The frequency of photon emitted in two step transition is given as,

Substitute the value of $\lambda$, $c$ in the equation (2),

$\begin{array}{c}{\nu }_1=\frac{c}{\lambda }\\ =\frac{3\times {10}^8\text{m}/\text{s}}{16.43\times {10}^{-8}\text{m}}\\ =0.18\times {10}^{16}{\text{s}}^{-1}\end{array}$

Substitute the value of $\lambda$, $c$ in the equation (2),

$\begin{array}{c}{\nu }_2=\frac{c}{\lambda }\\ =\frac{3\times {10}^8\text{m}/\text{s}}{3.04\times {10}^{-8}\text{m}}\\ =0.99\times {10}^{16}{\text{s}}^{-1}\end{array}$

The total frequency of photons in two step process $\begin{array}{c}={\nu }_1+{\nu }_2\\ =\left(\left(0.18\times {10}^{16}\right)+\left(0.99\times {10}^{16}\right)\right)\text{J}\\ =1.17\times {10}^{16}{\text{s}}^{-1}\end{array}$

Therefore, the frequencies that are emitted in two step process are similar to wavelength of photon emitted in one step process.

(d)

Interpretation Introduction

Interpretation: The given statements are to be authenticated.

Concept introduction: The Rydberg formula is used for calculating the wavelength of light involved in the emission spectrum of Hydrogen. The transition occurs between two energy levels that involves either the gain or loss of energy depending on the initial and final level of transition.

To determine: The authenticity of the given statement, “The wavelength of photon emitted by the ${\text{He}}^{+}$ ion in the $n=3$ to $n=1$ transition is shorter than the wavelength of the photon emitted by $\text{H}$ atom in an $n=3$ to $n=1$ transition”.

Answer to Problem 7.132AP

Solution

The given statement is true.

Explanation of Solution

Explanation

The wavelength of photon emitted by ${\text{He}}^{+}$ ion during transition from $n=3$ to $n=2$ is calculated as $2.56\times {10}^{-8}\text{m}$.

The energy of Hydrogen atom is calculated by substituting the value of $n_f=1$, $n_i=3$, $Z=1$ and $R_H$ in the equation (1).

$\begin{array}{c}\Delta E=R_H{Z}^2\left(\frac{1}{n_f{}^2}-\frac{1}{n_i{}^2}\right)\\ =\left(-2.18\times {10}^{-18}\text{J}\right){\left(1\right)}^2\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\\ =-1.94\times {10}^{-18}\text{J}\end{array}$

Substitute this value of $h$, $c$ and $\Delta E$ in the equation (2),

$\begin{array}{c}\Delta E=h\frac{c}{\lambda }\\ 1.94\times {10}^{-18}\text{J}=\text{6}\text{.626}\times {\text{10}}^{-34}\text{Js}\times \frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\lambda }\\ 1.94\times {10}^{-18}\text{J}=\frac{19.878\times {10}^{-26}\text{J}\text{m}}{\lambda }\\ \lambda =10.24\times {10}^{-8}\text{m}\end{array}$

The wavelength of photon emitted by the ${\text{He}}^{+}$ ion in the $n=3$ to $n=1$ transition is shorter than the wavelength of the photon emitted by $\text{H}$ atom in an $n=3$ to $n=1$ transition.