Chapter 7, Problem 7.34P

(a)

Interpretation Introduction

Interpretation:

The absorption and emission transitions in the image are to be determined.

Concept introduction:

Atomic spectrum is a series of electromagnetic radiations absorbed or emitted when electrons in an atom undergo transitions between different energy levels.

Absorption spectra - When an atom is subjected to energy in the form of heat or light, the electrons absorb the energy. If an electron in a lower energy level absorbs a photon whose energy is equal to the difference in the energies of the lower energy level and a higher energy level, the electron jumps to the higher energy level. The absorption spectra are characterized by the presence of a series of dark lines separated by colored bands.

Emission spectra – In the emission spectra, an electron in the higher energy level jumps to a lower energy level by releasing energy. The emission spectra are characterized by the presence of a series of fine lines at specific wavelengths separated by black spaces.

Answer to Problem 7.34P

The absorption transitions are A, C and D. The emission transitions are B, E, and F.

Explanation of Solution

The absorption of radiation by an electron takes place when it jumps from a lower energy level to a higher energy level. By the absorption of radiation, the electron gains energy and jumps to an energy level with a higher value of principal quantum number $\left(n\right)$. The orbits in an atom are specified by principal quantum numbers which represent their energies and location. An orbit with a higher value of principal quantum number will be further from the nucleus and higher in energy compared to an orbit with a lower value of principal quantum number.

The emission of radiation by an electron occurs when an electron loses energy and jumps from a higher to a lower energy level. Thus, by the emission of radiation, an electron makes a transition from an orbit with a higher principal quantum number to an orbit with a lower principal quantum number.

In the transition A, the electron jumps from $n=2$ to $n=5$. Since the transition is occurring from an orbit of lower energy to an orbit of higher energy, hence absorption of radiation takes place. Therefore, transition A is absorption.

In the transition B, the electron jumps from $n=6$ to $n=1$. Since the transition is occurring from an orbit of higher energy to an orbit of lower energy, hence emission of radiation takes place. Therefore, transition B is emission.

In the transition C, the electron jumps from $n=3$ to $n=6$. Since the transition is occurring from an orbit of lower energy to an orbit of higher energy, hence absorption of radiation takes place. Therefore, transition C is absorption.

In the transition D, the electron jumps from $n=1$ to $n=5$. Since the transition is occurring from an orbit of lower energy to an orbit of higher energy, hence absorption of radiation takes place. Therefore, transition D is absorption.

In the transition E, the electron jumps from $n=5$ to $n=3$. Since the transition is occurring from an orbit of higher energy to an orbit of lower energy, hence emission of radiation takes place. Therefore, transition E is emission.

In the transition F, the electron jumps from $n=4$ to $n=1$. Since the transition is occurring from an orbit of higher energy to an orbit of lower energy, hence emission of radiation takes place. Therefore, transition F is emission.

Conclusion

The absorption transitions are A, C and D. The emission transitions are B, E, and F.

(b)

Interpretation Introduction

Interpretation:

The increasing order of energy of emissions is to be determined.

Concept introduction:

Atomic spectrum is a series of electromagnetic radiations absorbed or emitted when electrons in an atom undergo transitions between different energy levels.

Emission spectra – In the emission spectra, an electron in the higher energy level jumps to a lower energy level by releasing energy. The emission spectra are characterized by the presence of a series of fine lines at specific wavelengths separated by black spaces.

The equation to find the difference in the energy between the two levels in hydrogen-like atoms is,

$\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{n_{\text{final}}^2}-\frac{1}{n_{\text{initial}}^2}\right)$        (1)

Here,

$\Delta E$ is the difference in the energy between two levels.

$n_{\text{final}}$ is the lower energy level.

$n_{\text{initial}}$ is the higher energy level.

Answer to Problem 7.34P

The order of increasing energy of emissions is $\text{E}<\text{F}<\text{B}$.

Explanation of Solution

In the case of B, the transition of the electron has occurred from $n=6$ energy level to $n=1$ energy level.

Substitute 1 for $n_{\text{final}}$ and 6 for $n_{\text{initial}}$ in equation (1).

$\begin{array}{c}\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{1^2}-\frac{1}{6^2}\right)\\ =-2.18\times {10}^{-18}\text{ J}\left(1-\frac{1}{36}\right)\\ =-2.11\times {10}^{-18}\text{ J}\end{array}$

In the case of E, the transition of the electron has occurred from $n=5$ energy level to $n=3$ energy level.

Substitute 3 for $n_{\text{final}}$ and 5 for $n_{\text{initial}}$ in equation (1).

$\begin{array}{c}\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{3^2}-\frac{1}{5^2}\right)\\ =-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{9}-\frac{1}{25}\right)\\ =-0.154\times {10}^{-18}\text{ J}\end{array}$

In the case of F, the transition of the electron has occurred from $n=4$ energy level to $n=1$ energy level.

Substitute 1 for $n_{\text{final}}$ and 4 for $n_{\text{initial}}$ in equation (1).

$\begin{array}{c}\Delta E=-2.18\times {10}^{-18}\text{ J}\left(\frac{1}{1^2}-\frac{1}{4^2}\right)\\ =-2.18\times {10}^{-18}\text{ J}\left(1-\frac{1}{16}\right)\\ =-2.04\times {10}^{-18}\text{ J}\end{array}$

Conclusion

The order of increasing energy of emissions is $\text{E}<\text{F}<\text{B}$.

(c)

Interpretation Introduction

Interpretation:

The increasing order of the wavelengths for absorption transitions is to be determined.

Concept introduction:

Atomic spectrum is a series of electromagnetic radiations absorbed or emitted when electrons in an atom undergo transitions between different energy levels.

Absorption spectra - When an atom is subjected to energy in the form of heat or light, the electrons absorb the energy. If an electron in a lower energy level absorbs a photon whose energy is equal to the difference in the energies of the lower energy level and a higher energy level, the electron jumps to the higher energy level. The absorption spectra are characterized by the presence of a series of dark lines separated by colored bands.

The equation used to predict the position and wavelength of any line in a given series is called the Rydberg’s equation.

Rydberg’s equation is as follows:

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$        (2)

Here,

$\lambda$ is the wavelength of the line.

$n_1$ and $n_2$ are positive integers, with ${\text{n}}_2>n_1$.

$R$ is the Rydberg’s constant.

The conversion factor to convert wavelength from $\text{nm}$ to $\text{m}$ is,

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

Answer to Problem 7.34P

The order of the increasing wavelength of absorption is $\text{D}<\text{A}<\text{C}$.

Explanation of Solution

The value of the Rydberg’s constant is $1.096776\times {10}^7{\text{ m}}^{-1}$.

In the case of A, the transition of the electron has occurred from $n=2$ energy level to $n=5$ energy level.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 2 for $n_1$ and 5 for $n_2$ in equation (2).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{4}-\frac{1}{25}\right)\\ =2,303,229.6{\text{ m}}^{-1}\end{array}$

Convert the value of $\lambda$ from $\text{m}$ to $\text{nm}$ as follows;

$\begin{array}{c}\lambda \left(\text{nm}\right)=\left(\frac{1}{2,303,229.6{\text{ m}}^{-1}}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =434.17\text{ nm}\\ =434.2\text{ nm}\end{array}$

In the case of C, the transition of the electron has occurred from $n=3$ energy level to $n=6$ energy level.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 3 for $n_1$ and 6 for $n_2$ in equation (2).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{3^2}-\frac{1}{6^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{9}-\frac{1}{36}\right)\\ =914,711.184{\text{ m}}^{-1}\end{array}$

Convert the value of $\lambda$ from $\text{m}$ to $\text{nm}$ as follows;

$\begin{array}{c}\lambda \left(\text{nm}\right)=\left(\frac{1}{914,711.184{\text{ m}}^{-1}}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =1093.24\text{ nm}\\ =1093.2\text{ nm}\end{array}$

In the case of D, the transition of the electron has occurred from $n=1$ energy level to $n=5$ energy level.

Substitute $1.096776\times {10}^7{\text{ m}}^{-1}$ for $R$, 1 for $n_1$ and 5 for $n_2$ in equation (2).

$\begin{array}{c}\frac{1}{\lambda }=\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(\frac{1}{1^2}-\frac{1}{5^2}\right)\\ =\left(1.096776\times {10}^7{\text{ m}}^{-1}\right)\left(1-\frac{1}{25}\right)\\ =10,529,049.6{\text{ m}}^{-1}\end{array}$

Convert the value of $\lambda$ from $\text{m}$ to $\text{nm}$ as follows;

$\begin{array}{c}\lambda \left(\text{nm}\right)=\left(\frac{1}{10,529,049.6{\text{ m}}^{-1}}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =94.97\text{ nm}\\ =94.10\text{ nm}\end{array}$

Conclusion

The order of the increasing wavelength of absorption is $\text{D}<\text{A}<\text{C}$.