Chapter 43, Problem 26SP

To determine

The three longest wavelengths of the photon that the singly ionized helium atoms will absorb strongly.

Answer to Problem 26SP

Solution:

$30.4\text{ nm}$, $25.6\text{ nm}$, and $24.3\text{ nm}$.

Explanation of Solution

Given data:

The helium atom is singly ionized.

Formula used:

The expression for the energy of an atom having a single electron in the $n^{\text{th}}$ state is written as

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

Here, $Z$ is the atomic number of the atom.

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

$\Delta E_{n,1}=E_n-E_1$

Here, $E_n$ is the energy of the $n^{\text{th}}$ state and $E_1$ is the energy of the ground state.

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:

The minimum energies of the photon will be observed for the transition of electron from $n=1$ to $n=2$ level, $n=1$ to $n=3$ level, and $n=1$ to $n=4$ level. So, the corresponding values of the wavelengths will be the maximum values for the wavelength of the photon. The minimum amount of energy of the photon will give the maximum wavelength of the photon as energy is inversely proportional to the wavelength.

Now, the singly ionized helium atom behaves like a hydrogen atom. And, the wavelength of the photon emitted during the transition from higher to lower state will be equal to the wavelength of the photon absorbed during the transition from lower to higher state.

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

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

Substitute $1$ for $n$ and $2$ for $Z$

$\begin{array}{c}{E}_1=\frac{\left(-13.6\text{ eV}\right)\left(2^2\right)}{1^2}\\ =-54.4\text{ eV}\end{array}$

This is the ground state energy of the singly ionized helium atom.

Similarly, substitute $2$ for $n$ and $2$ for $Z$ in the expression of $E_n$

$\begin{array}{c}{E}_2=\frac{\left(-13.6\text{ eV}\right)\left(2^2\right)}{2^2}\\ =-13.6\text{ eV}\end{array}$

This is the first excited state energy of the singly ionized helium atom.

Similarly, substitute $3$ for $n$ and $2$ for $Z$ in the expression of $E_n$

$\begin{array}{c}{E}_3=\frac{\left(-13.6\text{ eV}\right)\left(2^2\right)}{3^2}\\ =-6.04\text{ eV}\end{array}$

This is the second excited state energy of the singly ionized helium atom.

Again, substitute $4$ for $n$ and $2$ for $Z$ in the expression of $E_n$

$\begin{array}{c}{E}_4=\frac{\left(-13.6\text{ eV}\right)\left(2^2\right)}{4^2}\\ =-3.4\text{ eV}\end{array}$

This is the third excited state energy of the singly ionized helium atom.

Write the expression for the difference of energy for the transition of electron from $n=2$ to $n=1$ level:

$\Delta E_{2,1}=E_2-E_1$

Substitute $-13.6\text{ eV}$ for $E_2$ and $-54.4\text{ eV}$ for $E_1$

$\begin{array}{c}\Delta E_{2,1}=-13.6\text{ eV}-\left(-54.4\text{ eV}\right)\\ =40.8\text{ eV}\end{array}$

Write the expression for energy of the photon emitted for the transition of electron from $n=2$ to $n=1$ level:

$\Delta E_{2,1}=\frac{hc}{{\lambda }_{2,1}}$

Here, ${\lambda }_{2,1}$ is the wavelength of the photon emitted for the transition of electron from $n=2$ to $n=1$ level.

Rearrange the expression for ${\lambda }_{2,1}$

${\lambda }_{2,1}=\frac{hc}{\Delta E_{2,1}}$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $40.8\text{ eV}$ for $\Delta E_{2,1}$

$\begin{array}{c}{\lambda }_{2,1}=\frac{1240\text{ eV}\cdot \text{nm}}{40.8\text{ eV}}\\ =30.4\text{ nm}\end{array}$

Write the expression for the difference of energy for the transition of electron from $n=3$ to $n=1$ level:

$\Delta E_{3,1}=E_3-E_1$

Substitute $-6.04\text{ eV}$ for $E_3$ and $-54.4\text{ eV}$ for $E_1$

$\begin{array}{c}\Delta E_{3,1}=-6.04\text{ eV}-\left(-54.4\text{ eV}\right)\\ =48.4\text{ eV}\end{array}$

Write the expression for energy of the photon emitted for the transition of electron from $n=3$ to $n=1$ level:

$\Delta E_{3,1}=\frac{hc}{{\lambda }_{3,1}}$

Here, ${\lambda }_{3,1}$ is the wavelength of the photon emitted for the transition of electron from $n=3$ to $n=1$ level.

Rearrange the expression for ${\lambda }_{3,1}$

${\lambda }_{3,1}=\frac{hc}{\Delta E_{3,1}}$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $48.4\text{ eV}$ for $\Delta E_{3,1}$

$\begin{array}{c}{\lambda }_{3,1}=\frac{1240\text{ eV}\cdot \text{nm}}{48.4\text{ eV}}\\ =25.6\text{ nm}\end{array}$

Write the expression for the difference of energy for the transition of electron from $n=4$ to $n=1$ level:

$\Delta E_{4,1}=E_4-E_1$

Substitute $-3.4\text{ eV}$ for $E_4$ and $-54.4\text{ eV}$ for $E_1$

$\begin{array}{c}\Delta E_{4,1}=-3.4\text{ eV}-\left(-54.4\text{ eV}\right)\\ =51\text{ eV}\end{array}$

Write the expression for energy of the photon emitted for the transition of electron from $n=4$ to $n=1$ level:

$\Delta E_{4,1}=\frac{hc}{{\lambda }_{4,1}}$

Here, ${\lambda }_{4,1}$ is the wavelength of the photon emitted for the transition of electron from $n=4$ to $n=1$ level.

Rearrange the expression for ${\lambda }_{4,1}$

${\lambda }_{4,1}=\frac{hc}{\Delta E_{4,1}}$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $51\text{ eV}$ for $\Delta E_{4,1}$

$\begin{array}{c}{\lambda }_{4,1}=\frac{1240\text{ eV}\cdot \text{nm}}{51\text{ eV}}\\ =24.3\text{ nm}\end{array}$

Conclusion:

Since the wavelength of the photon emitted during the transition from higher to lower state will be equal to the wavelength of the photon absorbed during the transition from lower to higher state, the three longest wavelengths of the photon, that the singly ionized helium atoms will absorb strongly, are $30.4\text{ nm}$, $25.6\text{ nm}$, and $24.3\text{ nm}$.