Chapter 28.1, Problem 5PP

To determine

To Find: The emitted wavelengths and the lines to which it corresponds.

Explanation of Solution

Given:

Transition between energy levels in the hydrogen atom: $E_3$ and $E_2$ ; $E_4$ and $E_2$

Formula Used:

Wavelength of emitted light when hydrogen atom makes a transition from one level to another can be obtained by:

  $\frac{1}{\lambda }=R\left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)$

Here, R is the Rydberg’s constant.

Calculations:

Wavelength of emitted light when hydrogen atom makes a transition from $E_3$ to $E_2$ can be calculated as follows:

  $\begin{array}{l}\frac{1}{{\lambda }_{32}}=-109677\left(\frac{1}{3^2}-\frac{1}{2^2}\right){\text{cm}}^{\text{-1}}\\ \frac{1}{{\lambda }_{32}}=15232.9{\text{cm}}^{\text{-1}}\\ {\lambda }_{32}=6.56\times {10}^{-5}\text{cm}\text{= 656 nm}\end{array}$

This corresponds to lowest transition energy (red line).

Wavelength of emitted light when hydrogen atom makes a transition from $E_4$ to $E_2$ can be calculated as follows:

  $\begin{array}{l}\frac{1}{{\lambda }_{32}}=-109677\left(\frac{1}{4^2}-\frac{1}{2^2}\right){\text{cm}}^{\text{-1}}\\ \frac{1}{{\lambda }_{32}}=20564.44{\text{cm}}^{\text{-1}}\\ {\lambda }_{32}=4.86\times {10}^{-5}\text{cm}\text{= 486 nm}\end{array}$

This corresponds to second transition energy (green line).