## Calculating the Energy and Wavelength of Electron Transitions in a One–electron (Bohr) System

What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with *n* = 4 to the orbit with *n* = 6? In what part of the electromagnetic spectrum do we find this radiation?

### Solution

In this case, the electron starts out with *n* = 4, so *n*_{1} = 4. It comes to rest in the *n* = 6 orbit, so *n*_{2} = 6. The difference in energy between the two states is given by this expression:

\(\text{Δ}E={E}_{1}-{E}_{2}=2.179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{n}_{1}^{2}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{n}_{2}^{2}}\phantom{\rule{0.2em}{0ex}}\right)\)

\(\text{Δ}E=7.566\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}\phantom{\rule{0.2em}{0ex}}\text{J}\)

This energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron from the *n* = 4 orbit up to the *n* = 6 orbit. The wavelength of a photon with this energy is found by the expression \(E\text{=}\phantom{\rule{0.2em}{0ex}}\cfrac{hc}{\lambda }.\) Rearrangement gives:

\(\lambda =\phantom{\rule{0.2em}{0ex}}\cfrac{hc}{E}\)

\(\require{cancel}\begin{array}{}=\left(6.626\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\phantom{\rule{0.2em}{0ex}}\cancel{\text{J}}\phantom{\rule{0.2em}{0ex}}\cancel{\text{s}}\right)\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.3em}{0ex}}\cfrac{2.998\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}{\cancel{\text{s}}}^{-1}}{7.566\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}\phantom{\rule{0.2em}{0ex}}\cancel{\text{J}}}\phantom{\rule{0.2em}{0ex}}\\ =2.626\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}\end{array}\)

From a previous lesson, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum.

\(\text{Δ}E=2.179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{4}^{2}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{6}^{2}}\phantom{\rule{0.2em}{0ex}}\right)\phantom{\rule{0.2em}{0ex}}\text{J}\)

\(\text{Δ}E=2.179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-18}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{1}{16}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{36}\phantom{\rule{0.2em}{0ex}}\right)\phantom{\rule{0.2em}{0ex}}\text{J}\)