Chapter 7, Problem 7.115QP

(a)

Interpretation Introduction

Interpretation:

The ionization energy for the hydrogen atom and the ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom 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.

To calculate: The ionization energy for the hydrogen atom

Answer to Problem 7.115QP

The ionization energy for the hydrogen atom is $\text{1}{\text{.31 × 10}}^{\text{3}}\text{ kJ/mol}$

Explanation of Solution

Ionization energy is the minimum energy required to remove an electron from an atom. It is usually expressed in units of $\text{kJ/mol}$, that is, the energy in kilojoules required to remove one mole of electrons from one mole of atoms.  Consider this as an absorption process.  The energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 1 and n}}_{\text{f}}=\infty$ is calculated using the formula:

$\begin{array}{l}\text{ΔE }=\text{ (}-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J)}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ \text{ΔE }=\text{ (}-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J)}\left(\frac{\text{1}}{{\infty }^{\text{2}}}-\frac{\text{1}}{{\text{1}}^{\text{2}}}\right)\\ \text{ΔE }=2.18{\text{ × 10}}^{-18}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 1 and n}}_{\text{f}}=\infty$ is $2.18{\text{ × 10}}^{-18}\text{ J}$.  The ionization energy for the hydrogen atom is calculated as

$\begin{array}{l}\text{ionization energy }=\frac{\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J}}{\text{1 atom}}\times \frac{\text{6}{\text{.022 × 10}}^{\text{23}}\text{ atoms}}{\text{1 mol}}\\ \text{ionization energy }=\text{ 1}{\text{.31 × 10}}^{\text{6}}\text{ J/mol}\\ \text{ionization energy }=\text{ 1}{\text{.31 × 10}}^{\text{3}}\text{ kJ/mol}\end{array}$

Therefore, the ionization energy for the hydrogen atom is $\text{1}{\text{.31 × 10}}^{\text{3}}\text{ kJ/mol}$.

(b)

Interpretation Introduction

Interpretation:

The ionization energy for the hydrogen atom and the ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom 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.

To calculate: The ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom

Answer to Problem 7.115QP

The ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom is $\text{328 kJ/mol}$

Explanation of Solution

The energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 2 and n}}_{\text{f}}=\infty$ is calculated using the formula:

$\begin{array}{l}\text{ΔE }=\text{ (}-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J)}\left(\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\right)\\ \text{ΔE }=\text{ (}-\text{2}{\text{.18 × 10}}^{-\text{18}}\text{ J)}\left(\frac{\text{1}}{{\infty }^{\text{2}}}-\frac{\text{1}}{2^{\text{2}}}\right)\\ \text{ΔE }=5.45{\text{ × 10}}^{-19}\text{ J}\end{array}$

Therefore, the energy difference ($\text{ΔE}$) between ${\text{n}}_{\text{i}}={\text{ 2 and n}}_{\text{f}}=\infty$ is $5.45{\text{ × 10}}^{-19}\text{ J}$.  The ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom is calculated as

$\begin{array}{l}\text{ionization energy }=\frac{\text{5}{\text{.45 × 10}}^{-\text{19}}\text{ J}}{\text{1 atom}}\times \frac{\text{6}{\text{.022 × 10}}^{\text{23}}\text{ atoms}}{\text{1 mol}}\\ \text{ionization energy }=\text{ 3}{\text{.28 × 10}}^{\text{5}}\text{ J/mol}\\ \text{ionization energy }=\text{ 328 kJ/mol}\end{array}$

Therefore, the ionization energy in which the electrons are removed from the $\text{n }=\text{ 2}$ state instead of from the ground state in the hydrogen atom is $\text{328 kJ/mol}$.  It takes considerably less energy to remove the electron from an excited state.