Chapter 7, Problem 7.29QP

(a)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$, the energy (in joules) of a photon to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$ and the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level should be calculated using the concept of Bohr’s theory.

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

${\text{E}}_{\text{n}}=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}^{\text{2}}}\right)$

Where, $\text{n}$ gets an integer values such as $\text{n }=\text{ 1, 2, 3}$ and so on.  This is the energy of electron in ${\text{n}}^{\text{th}}$ orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from ${\text{n}}_{\text{f}}$ (the $\text{i and f}$ subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

$\text{ΔE }={\text{ E}}_{\text{f}}-{\text{ E}}_{\text{i}}$

This transition results in the photon’s emission with frequency $\text{v}$ and energy $\text{hv}$.  The following equation is resulted.

$\text{ΔE }=\text{ hν }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)$

When, ${\text{n}}_{\text{i}}{\text{ > n}}_{\text{f}}$, a photon is emitted.  The term in parentheses is positive, making $\text{ΔE}$ negative.  As a result, energy is lost to the surroundings.  When ${\text{n}}_{\text{i}}{\text{ < n}}_{\text{f}}$, a photon is absorbed.  The term in parentheses is negative, so $\text{ΔE}$ is positive.  As a result, energy is absorbed from the surroundings.

Answer to Problem 7.29QP

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$ is $\text{140 nm}$

Explanation of Solution

To find: Calculate the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$

The given energy levels of a hypothetical atom are given as follows:

$\begin{array}{l}{\text{E}}_{\text{4}}=-\text{1}{\text{.0 × 10}}^{-\text{19}}\text{ J}\hfill \\ {\text{E}}_{\text{3}}=-\text{5}{\text{.0 × 10}}^{-\text{19}}\text{ J}\hfill \\ {\text{E}}_{\text{2}}=-{\text{10 × 10}}^{-\text{19}}\text{ J}\hfill \\ {\text{E}}_{\text{1}}=-{\text{15 × 10}}^{-\text{19}}\text{ J}\hfill \end{array}$

The energy difference ($\text{ΔE}$) between ${\text{E}}_{\text{4}}{\text{ and E}}_{\text{1}}$ is calculated using the formula:

$\begin{array}{l}{\text{ΔE = E}}_{\text{4}}-{\text{ E}}_{\text{1}}\\ \text{ΔE = (}-\text{1}{\text{.0 × 10}}^{-\text{19}}\text{ J) }-\text{ (}-{\text{15 × 10}}^{-\text{19}}\text{ J)}\\ {\text{ΔE = 14 × 10}}^{-\text{19}}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{E}}_{\text{4}}{\text{ and E}}_{\text{1}}$ is ${\text{14 × 10}}^{-\text{19}}\text{ J}$.  By Bohr’s theory,

$\begin{array}{l}\text{ΔE }=\text{ hν}\\ \text{ΔE }=\frac{\text{hc}}{\text{λ}}\because \text{ ν }=\frac{\text{c}}{\text{λ}}\\ \text{λ }=\frac{\text{hc}}{\text{ΔE}}\end{array}$

Planck’s constant, $\text{h }=\text{ 6}{\text{.63 × 10}}^{-\text{34}}\text{ Js}$ and the speed of light, $\text{c }=\text{ 3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$.  This formula is used to find the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$.  Substitute the given values in the formula:

$\begin{array}{l}\text{λ }=\frac{\text{(6}{\text{.63 × 10}}^{-\text{34}}\text{ Js)(3}{\text{.00 × 10}}^{\text{8}}\text{ m/s)}}{{\text{14 × 10}}^{-\text{19}}\text{ J}}\\ \text{λ }=\text{ 1}{\text{.4 × 10}}^{-\text{7}}\text{ m}\\ \text{λ }=\text{ 1}{\text{.4 × 10}}^{\text{2}}\text{ nm}\end{array}$

Therefore, the wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$ is $\text{140 nm}$.

(b)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$, the energy (in joules) of a photon to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$ and the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level should be calculated using the concept of Bohr’s theory.

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

${\text{E}}_{\text{n}}=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}^{\text{2}}}\right)$

Where, $\text{n}$ gets an integer values such as $\text{n }=\text{ 1, 2, 3}$ and so on.  This is the energy of electron in ${\text{n}}^{\text{th}}$ orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from ${\text{n}}_{\text{f}}$ (the $\text{i and f}$ subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

$\text{ΔE }={\text{ E}}_{\text{f}}-{\text{ E}}_{\text{i}}$

This transition results in the photon’s emission with frequency $\text{v}$ and energy $\text{hv}$.  The following equation is resulted.

$\text{ΔE }=\text{ hν }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)$

When, ${\text{n}}_{\text{i}}{\text{ > n}}_{\text{f}}$, a photon is emitted.  The term in parentheses is positive, making $\text{ΔE}$ negative.  As a result, energy is lost to the surroundings.  When ${\text{n}}_{\text{i}}{\text{ < n}}_{\text{f}}$, a photon is absorbed.  The term in parentheses is negative, so $\text{ΔE}$ is positive.  As a result, energy is absorbed from the surroundings.

Answer to Problem 7.29QP

The energy of a photon to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$ is ${\text{5 × 10}}^{-\text{19}}\text{ J}$

Explanation of Solution

To find: Calculate the energy (in joules) a photon must have in order to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$

The energy difference ($\text{ΔE}$) between ${\text{E}}_{\text{3}}{\text{ and E}}_{\text{2}}$ is calculated using the formula:

${\text{ΔE = E}}_{\text{3}}-{\text{ E}}_{\text{2}}$

Substitute the given values in the formula:

$\begin{array}{l}\text{ΔE = (}-\text{5}{\text{.0 × 10}}^{-\text{19}}\text{ J) }-\text{ (}-{\text{10 × 10}}^{-\text{19}}\text{ J)}\\ {\text{ΔE = 5 × 10}}^{-\text{19}}\text{ J}\end{array}$

Therefore, the energy of a photon to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$ is ${\text{5 × 10}}^{-\text{19}}\text{ J}$.

(c)

Interpretation Introduction

Interpretation:

The wavelength of the photon needed to excite an electron from the given energy levels of a hypothetical atom ${\text{E}}_{\text{1}}{\text{ to E}}_{\text{4}}$, the energy (in joules) of a photon to excite an electron from ${\text{E}}_{\text{2}}{\text{ to E}}_{\text{3}}$ and the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level should be calculated using the concept of Bohr’s theory.

Concept Introduction:

The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light.  Based on electrostatic interaction and law of motion, Bohr derived the following equation.

${\text{E}}_{\text{n}}=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}^{\text{2}}}\right)$

Where, $\text{n}$ gets an integer values such as $\text{n }=\text{ 1, 2, 3}$ and so on.  This is the energy of electron in ${\text{n}}^{\text{th}}$ orbital.

The electrons are excited thermally when the light is used by an object.  As a result, an emission spectrum comes.  Line spectra consist of light only at specific, discrete wavelengths.  In emission, the electron returns to a lower energy state from ${\text{n}}_{\text{f}}$ (the $\text{i and f}$ subscripts denote the initial and final energy states).  In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state.  The difference in the energies between the initial and final states is

$\text{ΔE }={\text{ E}}_{\text{f}}-{\text{ E}}_{\text{i}}$

This transition results in the photon’s emission with frequency $\text{v}$ and energy $\text{hv}$.  The following equation is resulted.

$\text{ΔE }=\text{ hν }=-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J }\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)$

When, ${\text{n}}_{\text{i}}{\text{ > n}}_{\text{f}}$, a photon is emitted.  The term in parentheses is positive, making $\text{ΔE}$ negative.  As a result, energy is lost to the surroundings.  When ${\text{n}}_{\text{i}}{\text{ < n}}_{\text{f}}$, a photon is absorbed.  The term in parentheses is negative, so $\text{ΔE}$ is positive.  As a result, energy is absorbed from the surroundings.

Answer to Problem 7.29QP

The wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$  level is $\text{200nm}$

Explanation of Solution

To find: Calculate the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level

The energy difference ($\text{ΔE}$) between ${\text{E}}_{\text{1}}{\text{ and E}}_{\text{3}}$ is calculated using the formula:

$\begin{array}{l}{\text{ΔE = E}}_{\text{1 }}-{\text{ E}}_{\text{3}}\\ \text{ΔE = (}-{\text{15 × 10}}^{-\text{19}}\text{ J) }-\text{ (}-\text{5}{\text{.0 × 10}}^{-\text{19}}\text{ J)}\\ \text{ΔE = }-{\text{10 × 10}}^{-\text{19}}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{E}}_{\text{1}}{\text{ and E}}_{\text{3}}$ is ${\text{10 × 10}}^{-\text{19}}\text{ J}$ where negative sign is ignored.  By Bohr’s theory,

$\begin{array}{l}\text{ΔE }=\text{ hν}\\ \text{ΔE }=\frac{\text{hc}}{\text{λ}}\because \text{ ν }=\frac{\text{c}}{\text{λ}}\\ \text{λ }=\frac{\text{hc}}{\text{ΔE}}\end{array}$

Planck’s constant, $\text{h }=\text{ 6}{\text{.63 × 10}}^{-\text{34}}\text{ Js}$ and the speed of light, $\text{c }=\text{ 3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$.  This formula is used to find the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level.  Substitute the given values in the formula:

$\begin{array}{l}\text{λ }=\frac{\text{(6}{\text{.63 × 10}}^{\text{-34}}\text{ Js)(3}{\text{.00 × 10}}^{\text{8}}\text{ m/s)}}{{\text{10 × 10}}^{-\text{19}}\text{ J}}\\ \text{λ }=\text{ 2}{\text{.0 × 10}}^{-\text{7}}\text{ m}\\ \text{λ }=\text{ 2}{\text{.0 × 10}}^{\text{2}}\text{ nm}\end{array}$

Therefore, the wavelength of the photon emitted when an electron drops from the ${\text{E}}_{\text{3}}$ level to the ${\text{E}}_{\text{1}}$ level is $\text{200 nm}$.