Chapter 28, Problem 40A

To determine

To calculate: Difference between the two lasing energy levels.

Answer to Problem 40A

Difference between the two lasing energy levels = $\underset{\_}{\text{1}\text{.94 eV}}$

Explanation of Solution

Given:

Wavelength = $640\text{ nm}$

Formula used:

Frequency is given by,

  $f=\frac{c}{\lambda }$

Where $c$ is the speed of light and $\lambda$ is the wavelength.

Energy difference is given by,

  $E=hf$

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

Calculation:

Frequency is calculated as:

  $\begin{array}{l}f=\frac{c}{\lambda }\\ \text{ =}\frac{3\times {10}^8\text{ m/s}}{640\times {10}^{-9}\text{ m}}\\ \text{ =4}\text{.69}\times {\text{10}}^{14}\text{ Hz}\end{array}$

Energy difference is calculated as:

  $\begin{array}{l}E=hf\\ \text{ =}\frac{\text{(6}\text{.626}\times {\text{10}}^{-34}\text{ Js)(4}\text{.69}\times {\text{10}}^{14}\text{ Hz)}}{1.60\times {10}^{-19}\text{ J/eV}}\\ \text{ =1}\text{.94 eV}\end{array}$

Conclusion:

Difference between the two lasing energy levels = $\underset{\_}{\text{1}\text{.94 eV}}$