Chapter 7, Problem 7.82P

(a)

Interpretation Introduction

Interpretation:

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 3 to level 2 are 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,

$\Delta E=h\nu$

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:

$\Delta E=\frac{hc}{\lambda }$ (1)

Answer to Problem 7.82P

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 3 to level 2 are $7.56\times {10}^{-18}\text{ J}$ and $2.63\times {10}^{-8}\text{ m}$ respectively.

Explanation of Solution

The energies of the electronic transitions $E_{31}$ and $E_{21}$ are $4.854\times {10}^{-17}\text{ J}$ and $4.098\times {10}^{-17}\text{ J}$ respectively.

The formula to calculate the difference in the energies of energy levels 3 and 2 is,

$\Delta E=E_{31}-E_{21}$ (2)

Substitute $4.854\times {10}^{-17}\text{ J}$ for $E_{31}$ and $4.098\times {10}^{-17}\text{ J}$ for $E_{21}$ in equation (2).

$\begin{array}{c}\Delta E=\left(4.854\times {10}^{-17}\text{ J}\right)-\left(4.098\times {10}^{-17}\text{ J}\right)\\ =0.756\times {10}^{-17}\text{ J}\\ =7.56\times {10}^{-18}\text{ J}\end{array}$

Substitute $7.56\times {10}^{-18}\text{ J}$ 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 (1) to calculate $\lambda$ for the photon emitted in the transition.

$7.56\times {10}^{-18}\text{ J}=\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\lambda }$

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

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{7.56\times {10}^{-18}\text{ J}}\\ =2.6309\times {10}^{-8}\text{ m}\\ =2.63\times {10}^{-8}\text{ m}\end{array}$

Conclusion

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 3 to level 2 are $7.56\times {10}^{-18}\text{ J}$ and $2.63\times {10}^{-8}\text{ m}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 4 to level 1 are 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,

$\Delta E=h\nu$

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:

$\Delta E=\frac{hc}{\lambda }$ (1)

Answer to Problem 7.82P

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 4 to level 1 are $5.122\times {10}^{-17}\text{ J}$ and $3.88\times {10}^{-8}\text{ m}$ respectively.

Explanation of Solution

The energies of the electronic transitions $E_{42}$ and $E_{21}$ are, $1.024\times {10}^{-17}\text{ J}$ and $4.098\times {10}^{-17}\text{ J}$ respectively.

The formula to calculate the difference in the energies of energy levels 4 and 1 is,

$\Delta E=E_{42}+E_{21}$ (3)

Substitute $1.024\times {10}^{-17}\text{ J}$ for $E_{42}$ and $4.098\times {10}^{-17}\text{ J}$ for $E_{21}$ in equation (3).

$\begin{array}{c}\Delta E=\left(1.024\times {10}^{-17}\text{ J}\right)+\left(4.098\times {10}^{-17}\text{ J}\right)\\ =5.122\times {10}^{-17}\text{ J}\end{array}$

Substitute $5.122\times {10}^{-17}\text{ J}$ 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 (1) to calculate $\lambda$ for the photon emitted in the transition.

$5.122\times {10}^{-17}\text{ J}=\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\lambda }$

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

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{5.122\times {10}^{-17}\text{ J}}\\ =3.883\times {10}^{-8}\text{ m}\\ =3.88\times {10}^{-8}\text{ m}\end{array}$

Conclusion

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 4 to level 1 are $5.122\times {10}^{-17}\text{ J}$ and $3.88\times {10}^{-8}\text{ m}$ respectively.

(c)

Interpretation Introduction

Interpretation:

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 5 to level 4 are 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,

$\Delta E=h\nu$

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:

$\Delta E=\frac{hc}{\lambda }$ (1)

Answer to Problem 7.82P

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 5 to level 4 are $1.2\times {10}^{-18}\text{ J}$ and $1.66\times {10}^{-7}\text{ m}$ respectively.

Explanation of Solution

The energies of the electronic transitions $E_{51}$ and $E_{41}$ are, $5.242\times {10}^{-17}\text{ J}$ and $5.122\times {10}^{-17}\text{ J}$ respectively.

The formula to calculate the difference in the energies of energy levels 5 and 4 is,

$\Delta E=E_{51}-E_{41}$ (4)

Substitute $5.242\times {10}^{-17}\text{ J}$ for $E_{51}$ and $5.122\times {10}^{-17}\text{ J}$ for $E_{41}$ in equation (4).

$\begin{array}{c}\Delta E=\left(5.242\times {10}^{-17}\text{ J}\right)-\left(5.122\times {10}^{-17}\text{ J}\right)\\ =0.12\times {10}^{-17}\text{ J}\\ =1.2\times {10}^{-18}\text{ J}\end{array}$

Substitute $1.2\times {10}^{-18}\text{ J}$ 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 (1) to calculate $\lambda$ for the photon emitted in the transition.

$1.2\times {10}^{-18}\text{ J}=\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\lambda }$

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

$\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{1.2\times {10}^{-18}\text{ J}}\\ =16.575\times {10}^{-8}\text{ m}\\ =1.66\times {10}^{-7}\text{ m}\end{array}$

Conclusion

The values of $\Delta E$ and $\lambda$ for the photon emitted in the transition of an electron from energy level 5 to level 4 are $1.2\times {10}^{-18}\text{ J}$ and $1.66\times {10}^{-7}\text{ m}$ respectively.