Chapter 28, Problem 31A

To determine

To calculate: Kinetic energy of the electron as it is ejected from the atom.

Answer to Problem 31A

Kinetic energy of the electron = $\underset{\_}{\text{1}\text{.31 eV}}$

Explanation of Solution

Given:

Wavelength of photon = $6.00\times {10}^2\text{ nm}$

Energy level of the calcium atom = $E_{\text{8}}$

Formula used:

Energy of an emitted photon is given by,

  $E_{\text{photon}}=\frac{hc}{\lambda }$

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

Energy needed to ionize is given by,

  $E_{\text{ionization}}=E_f-E_i$

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

Kinetic energy of the electron is given by:

  $E_f=E_{\text{photon}}-E_{\text{ionization}}$

Where $E_{\text{photon}}$ is the energy emitted by the photon and $E_{\text{ionization}}$ is the ionization energy.

Calculation:

Energy of an emitted photon is calculated as:

  $\begin{array}{l}{E}_{\text{photon}}=\frac{hc}{\lambda }\\ \text{ =}\frac{1240\text{ eV-nm}}{6.00\times {10}^2\text{ m}}\text{ [}hc=1240\text{ eV-nm}]\\ \text{ =2}\text{.07 eV}\end{array}$

Energy needed to ionize = $6.08-5.32=0.76\text{ eV}$

Kinetic energy of the electron is calculated as:

  $\begin{array}{l}{E}_f=E_{\text{photon}}-E_{\text{ionization}}\\ \text{ =2}\text{.07}-0.76\\ \text{ =1}\text{.31 eV}\end{array}$

Conclusion:

Kinetic energy of the electron = $\underset{\_}{\text{1}\text{.31 eV}}$