Chapter 28, Problem 40P

To determine

The wavelength of $\text{He-Ne}$ laser.

Answer to Problem 40P

  $\text{633}\text{nm}$

Explanation of Solution

Given info:

The laser of $\text{He-Ne}$ .

Formula used:

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}$

Where, $\text{λ}$ is a wavelength and $\text{E}$ is energy in eV.

Calculation:

Transition from the ${\text{E}}_3^{\text{'}}$ state to the ${\text{E}}_2^{\text{'}}$ state releases photons with energy $\text{1}\text{.96}\text{eV}$ .

The wavelength is calculated by

  $\begin{array}{c}\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\Delta \text{E}}\\ =\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left(\text{1}\text{.96}\text{eV}\right)}\\ =\text{633}\text{nm}\end{array}$

Conclusion:

The required value is $\text{633}\text{nm}$ .