Chapter 28, Problem 35A

To determine

To rank: The given energy transitions of hydrogen atoms according to the energy of the released photon from least to greatest.

Answer to Problem 35A

The energy transition from least to greatest is as follows:

  $\text{B}<\text{A}<\text{D}<\text{C}<\text{E}$

Explanation of Solution

Given:

The following energy transitions:

From $n=5$ to $n=3$

From $n=5$ to $n=4$

From $n=4$ to $n=2$

From $n=3$ to $n=2$

From $n=2$ to $n=1$

Formula used:

Energy of a hydrogen atom is given by,

  $E_n=-13.6\text{ eV }\times \frac{1}{n^2}$

Where $n$ is the principal quantum number.

Energy of the emitted photon is given by,

  $E_{\text{photon}}=E_f-E_i$

Where $E_i$ is the initial energy level and $E_f$ is the final energy level.

Calculation:

To calculate the energy released by the photon in the given energy transitions, first find the energy of the hydrogen atom at different energy levels

The energies associated with the different energy levels of a hydrogen atom is calculated as:

  $E_1=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(1)}^2}=-13.6\text{ eV}$

  $E_2=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(2)}^2}=-3.40\text{ eV}$

  $E_3=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(3)}^2}=-1.51\text{ eV}$

  $E_4=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(4)}^2}=-0.85\text{ eV}$

  $E_5=-13.6\text{ eV }\times \frac{1}{n^2}=\frac{-13.6}{{(5)}^2}=-0.54\text{ eV}$

A. For energy transition from $n=5$ to $n=3$ :

Energy of the released photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=E_f-E_i\\ \text{ =}-1.51-(-0.54)\\ =-0.97\text{ eV}\end{array}$

B. For energy transition from $n=5$ to $n=4$ :

Energy of the released photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=E_f-E_i\\ \text{ =}-0.85-(-0.54)\\ =-0.31\text{ eV}\end{array}$

C. For energy transition from $n=4$ to $n=2$ :

Energy of the released photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=E_f-E_i\\ \text{ =}-3.40-(-0.85)\\ =-2.55\text{ eV}\end{array}$

D. For energy transition from $n=3$ to $n=2$ :

Energy of the released photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=E_f-E_i\\ \text{ =}-3.40-(-1.51)\\ =-1.89\text{ eV}\end{array}$

E. For energy transition from $n=2$ to $n=1$ :

Energy of the released photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=E_f-E_i\\ \text{ =}-13.6-(-3.40)\\ =-10.2\text{ eV}\end{array}$

Conclusion:

On comparing the calculated energy transition, the following order from the least to the greatest energy transition is obtained,

  $-0.31<-0.97<-1.89<-2.55<-10.2$

Or

  $\text{B}<\text{A}<\text{D}<\text{C}<\text{E}$