Chapter 7, Problem 7.81P

(a)

Interpretation Introduction

Interpretation:

The highest frequency and minimum wavelength of radiation emitted when the electron makes a transition from $n=3$ is to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The difference between the energies of two energy levels is the amount of energy in a photon emitted or absorbed when an electron makes a transition. The energy of a photon emitted or absorbed by an electron is,

$\Delta E=h\nu$ (1)

Here,

$\Delta E$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency

Answer to Problem 7.81P

The highest frequency of radiation emitted when the electron makes a transition from $n=3$ is $8\times {10}^{14}{\text{ s}}^{-1}$. The wavelength of this radiation is $4\times {10}^{-7}\text{ m}$.

Explanation of Solution

The energies of the energy levels 3 and 1 are $-15\times {10}^{-19}\text{ J}$ and $-20\times {10}^{-19}\text{ J}$ respectively.

The formula to calculate the difference in the energies of energy level 1 and 2 is,

$\Delta E=E_3-E_1$ (2)

Substitute $-15\times {10}^{-19}\text{ J}$ for $E_3$ and $-20\times {10}^{-19}\text{ J}$ for $E_1$ in equation (2).

$\begin{array}{c}\Delta E=\left(-15\times {10}^{-19}\text{ J}\right)-\left(-20\times {10}^{-19}\text{ J}\right)\\ =-15\times {10}^{-19}\text{ J}+20\times {10}^{-19}\text{ J}\\ =5\times {10}^{-19}\text{ J}\end{array}$

Substitute $5\times {10}^{-19}\text{ J}$ for $\Delta E$ and $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ in equation (1).

$5\times {10}^{-19}\text{ J}=\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\nu$

Rearrange the above equation to calculate the value for $\nu$ as follows:

$\begin{array}{c}\nu =\frac{5\times {10}^{-19}\text{ J}}{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}\\ =7.546\times {10}^{14}{\text{ s}}^{-1}\\ =8\times {10}^{14}{\text{ s}}^{-1}\end{array}$

The equation to relate the frequency and wavelength of radiation is as follows:

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

Substitute $3\times {10}^8\text{ m/s}$ for $c$ and $8\times {10}^{14}{\text{ s}}^{-1}$ for $\nu$ in equation (3).

$8\times {10}^{14}{\text{ s}}^{-1}=\frac{3\times {10}^8\text{ m/s}}{\lambda }$

Rearrange the above equation to calculate the value of $\lambda$ as follows:

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{ m/s}}{8\times {10}^{14}{\text{ s}}^{-1}}\\ =3.75\times {10}^{-7}\text{ m}\\ =4\times {10}^{-7}\text{ m}\end{array}$

Conclusion

The highest frequency of radiation emitted when the electron makes a transition from $n=3$ is $8\times {10}^{14}{\text{ s}}^{-1}$. The wavelength of this radiation is $4\times {10}^{-7}\text{ m}$.

(b)

Interpretation Introduction

Interpretation:

The ionization energy of the atom in its ground state in $\text{kJ/mol}$ is to be determined.

Concept introduction:

Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom. The energy required to remove an electron from an atom depends on the position of the electron in the atom. The closer the electron is to the nucleus in the atom, the harder it is to pull it out of the atom. As the distance of an electron from the nucleus increases, the magnitude of the forces of attraction between the electron and the nucleus decreases. Thus it becomes easier to remove it from the atom.

Answer to Problem 7.81P

The ionization energy of the atom in its ground state in $\text{kJ/mol}$ is $1.2\times {10}^3$.

Explanation of Solution

The total number of atoms in one mole of a compound is $6.023\times {10}^{23}$.

The ionization energy required to remove an electron from the ground state is the same as the energy of the state.

Thus for one atom, the ionization energy for the ground state is $20\times {10}^{-19}\text{ J}$.

For 1 mole $\left(6.023\times {10}^{23}\text{ atoms}\right)$, the ionization energy is calculated as,

$\begin{array}{c}\text{Ionization energy}=\left(20\times {10}^{-19}\text{ J/atom}\right)\left(\frac{1\text{ kJ}}{{10}^3\text{ J}}\right)\left(6.023\times {10}^{23}\text{ atoms/mol}\right)\\ =1204.4\text{ kJ/mol}\\ =1.2\times {10}^3\text{ kJ/mol}\end{array}$

Conclusion

The ionization energy of the atom in its ground state in $\text{kJ/mol}$ is $1.2\times {10}^3$.

(c)

Interpretation Introduction

Interpretation:

The shortest wavelength of radiation that could be absorbed by the electron in the $n=4$ level without causing ionization is to be determined.

Concept introduction:

Electromagnetic waves are radiations that are formed by oscillating electric and magnetic fields. The electric and magnetic field components of an electromagnetic wave are perpendicular to each other.

The difference between the energies of two energy levels is the amount of energy in a photon emitted or absorbed when an electron makes a transition. The energy of a photon emitted or absorbed by an electron is,

$\Delta E=h\nu$ (1)

Here,

$\Delta E$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency

The equation to relate the frequency and wavelength of radiation is as follows:

$\nu =\frac{c}{\lambda }$

The above relation can be modified as follows:

$\Delta E=\frac{hc}{\lambda }$ (4)

Answer to Problem 7.81P

The shortest wavelength of radiation that could be absorbed by the electron in the $n=4$ level without causing ionization is $2\times {10}^2\text{ nm}$.

Explanation of Solution

Since the absorption of radiation occurs, hence the electron from $n=4$ will move to a level of higher energy.

The shortest wavelength of radiation will be absorbed for a transition from $n=4$ to $n=6$.

The energy of the energy level 4 and 6 are $-11\times {10}^{-19}\text{ J}$ and $-2\times {10}^{-19}\text{ J}$ respectively.

The formula to calculate the difference in the energies of energy level 4 and 6 is,

$\Delta E=E_6-E_4$ (5)

Substitute $-2\times {10}^{-19}\text{ J}$ for $E_6$ and $-11\times {10}^{-19}\text{ J}$ for $E_4$ in equation (5).

$\begin{array}{c}\Delta E=\left(-2\times {10}^{-19}\text{ J}\right)-\left(-11\times {10}^{-19}\text{ J}\right)\\ =-2\times {10}^{-19}\text{ J}+11\times {10}^{-19}\text{ J}\\ =9\times {10}^{-19}\text{ J}\end{array}$

Substitute $9\times {10}^{-19}\text{ J}$ for $\Delta E$, $3\times {10}^8\text{ m/s}$ for $c$ and $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ in equation (4).

$9\times {10}^{-19}\text{ J}=\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\lambda }$

Rearrange the above equation to calculate the value of $\lambda$ as follows:

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{9\times {10}^{-19}\text{ J}}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =2.21\text{ nm}\\ =2\times {10}^2\text{ nm}\end{array}$

Conclusion

The shortest wavelength of radiation that could be absorbed by the electron in the $n=4$ level without causing ionization is $2\times {10}^2\text{ nm}$.