Chapter 43, Problem 22SP

To determine

The energies of the two longest wavelengths in Paschen series of the hydrogen atom and their corresponding wavelengths.

Answer to Problem 22SP

Solution:

$0.66\text{ eV}$, $0.97\text{ eV}$, $\text{1}\text{.9}\times {10}^{-6}\text{ m}$, and $1.3\times {10}^{-6}\text{ m}$.

Explanation of Solution

Given data:

Paschen series of the hydrogen atom.

Formula used:

The expression for the energy of the hydrogen atom in the $n^{\text{th}}$ state is written as

$E_n=\frac{\left(-13.6\text{ eV}\right)}{n^2}$

The expression for the difference between the two energy levels for Paschen series is written as

$\Delta E_{n,3}=E_n-E_3$

Here, $E_n$ is the energy of the $n^{\text{th}}$ state and $E_3$ is the energy of the ground state of Paschen series of the hydrogen atom.

The expression for the difference between the two energy levels is written as

$\Delta E=\frac{hc}{\lambda }$

Here, $h$ is the Planck’s constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the wave.

Explanation:

Recall the expression for the difference between the two energy levels is written as

$\Delta E=\frac{hc}{\lambda }$

Since, wavelength is inversely proportional to the difference in the energies. Hence, the longest wavelength will exist for the minimum energy difference between the states.

Thus, the two longest wavelengths will be obtained for Paschen series, when the transition of the electron will happen from $n=4$ to $n=3$ level and $n=5$ to $n=3$ level, respectively. So, the energies of these wavelengths will be the energies corresponding to the longest wavelengths.

Now, recall the expression for the energy of the hydrogen atom in the $n^{\text{th}}$ state:

$E_n=\frac{\left(-13.6\text{ eV}\right)}{n^2}$

Substitute $3$ for $n$

$\begin{array}{c}{E}_3=\frac{\left(-13.6\text{ eV}\right)}{3^2}\\ =-1.51\text{ eV}\end{array}$

Similarly, substitute $4$ for $n$ in the expression of $E_n$

$\begin{array}{c}{E}_4=\frac{\left(-13.6\text{ eV}\right)}{4^2}\\ =-0.85\text{ eV}\end{array}$

This is the first excited state energy of Paschen series of the hydrogen atom.

Again, substitute $5$ for $n$ in the expression of $E_n$

$\begin{array}{c}{E}_5=\frac{\left(-13.6\text{ eV}\right)}{5^2}\\ =-0.54\text{ eV}\end{array}$

This is the second excited state energy ofPaschen series of the hydrogen atom.

Write the expression for the difference between the energy for the transition of electron from $n=4$ to $n=3$ level of Paschen series:

$\Delta E_{4,3}=E_4-E_3$

Substitute $-0.85\text{ eV}$ for $E_4$ and $-1.51\text{ eV}$ for $E_3$

$\begin{array}{c}\Delta E_{4,3}=-0.85\text{ eV}-\left(-1.51\text{ eV}\right)\\ =0.66\text{ eV}\end{array}$

Or,

$\Delta E_{4,3}=\frac{hc}{{\lambda }_{4,3}}$

Here, ${\lambda }_{4,3}$ is the first longest wavelength of the photon emitted for the transition of electron from $n=4$ to $n=3$ level.

Rearrange the expression for ${\lambda }_{4,3}$

${\lambda }_{4,3}=\frac{hc}{\Delta E_{4,3}}$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $0.66\text{ eV}$ for $\Delta E_{4,3}$

$\begin{array}{c}{\lambda }_{4,3}=\frac{1240\text{ eV}\cdot \text{nm}}{0.66\text{ eV}}\\ =\left(1878.79\text{ nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\\ =1878.79\times {10}^{-9}\text{ m}\\ \simeq \text{1}\text{.9}\times {10}^{-6}\text{ m}\end{array}$

Similarly, write the expression for the difference between the energy for the transition of electron from $n=5$ to $n=3$ level of Paschen series.

$\Delta E_{5,3}=E_5-E_3$

Substitute $-0.54\text{ eV}$ for $E_3$ and $-1.51\text{ eV}$ for $E_3$

$\begin{array}{c}\Delta E_{5,3}=-0.54\text{ eV}-\left(-1.51\text{ eV}\right)\\ =0.97\text{ eV}\end{array}$

Or,

$\Delta E_{5,3}=\frac{hc}{{\lambda }_{5,3}}$

Here, ${\lambda }_{5,3}$ is the wavelength of the photon emitted for the transition of electron from $n=5$ to $n=3$ level.

Rearrange the expression for ${\lambda }_{5,3}$

${\lambda }_{5,3}=\frac{hc}{\Delta E_{5,3}}$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $0.97\text{ eV}$ for $\Delta E_{3,1}$

$\begin{array}{c}{\lambda }_{5,3}=\frac{1240\text{ eV}\cdot \text{nm}}{0.97\text{ eV}}\\ =\left(1278.35\text{ nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\\ =1278.35\times {10}^{-6}\text{ m}\\ \simeq 1.3\times {10}^{-6}\text{ m}\end{array}$

Conclusion:

Hence, the energies of the two longest wavelengths in Paschen series of the hydrogen atom are $0.66\text{ eV}$ and $0.97\text{ eV}$, and their corresponding wavelengths are $\text{1}\text{.9}\times {10}^{-6}\text{ m}$ and $1.3\times {10}^{-6}\text{ m}$.