# Electron Transitions in a One-Electron System

## 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 n1 = 4. It comes to rest in the n = 6 orbit, so n2 = 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.

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

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

Bohr’s model of the hydrogen atom provides insight into the behavior of matter at the microscopic level, but it is does not account for electron–electron interactions in atoms with more than one electron. It does introduce several important features of all models used to describe the distribution of electrons in an atom. These features include the following:

• The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed value and used to characterize the arrangement of electrons in an atom.
• An electron’s energy increases with increasing distance from the nucleus.
• The discrete energies (lines) in the spectra of the elements result from quantized electronic energies.

Of these features, the most important is the postulate of quantized energy levels for an electron in an atom. As a consequence, the model laid the foundation for the quantum mechanical model of the atom. Bohr won a Nobel Prize in Physics for his contributions to our understanding of the structure of atoms and how that is related to line spectra emissions.

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