Chapter 28, Problem 33A

To determine

To calculate: Energy and frequency of the photon that is absorbed.

Answer to Problem 33A

Energy of the photon = $\underset{\_}{\text{1}\text{.24 eV}}$

Frequency of the photon = $\underset{\_}{\text{2}\text{.99}\times {\text{10}}^{14}\text{ Hz}}$

Explanation of Solution

Given:

Energy of the mercury atom = $-4.98\text{ eV}$

The mercury atom rises to the next higher energy level after absorbing the photon.

Formula used:

Energy absorbed by an atom is given by,

  $\Delta E=E_f-E_i$

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

Energy of an emitted photon is given by,

  $E_{\text{photon}}=hf$

Where $h$ is Planck’s constant and $f$ is frequency.

Calculation:

Energy of the photon is calculated as:

  $\begin{array}{l}\Delta E=E_f-E_i\\ \text{ =}{E}_5-E_4\\ \text{ =}-3.74-(-4.98)\\ \text{ =1}\text{.24 eV}\end{array}$

Frequency of the photon is calculated as:

  $\begin{array}{l}f=\frac{E}{h}\\ \text{ =}\frac{1.24\text{ eV}\times 1.60\times {10}^{-19}\text{ J/eV}}{6.626\times {10}^{-34}\text{ Js}}\text{ [Charge of an electron = }1.60\times {10}^{-19}\text{ C, }h=6.626\times {10}^{-34}\text{ Js}]\\ \text{ =2}\text{.99}\times {\text{10}}^{14}\text{ Hz}\end{array}$

Conclusion:

Energy of the photon = $\underset{\_}{\text{1}\text{.24 eV}}$

Frequency of the photon = $\underset{\_}{\text{2}\text{.99}\times {\text{10}}^{14}\text{ Hz}}$