---

Example 6.5: 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 \(n = 4\) and comes to rest in the \(n = 6\) orbit, so

\(n_{\mathrm{1}}\) \(= 4\)


\(n_{\mathrm{2}}\) \(= 6\)


The difference in energy between the two states is given by this expression:


\(ΔE_{\mathrm{electron}}\) \(= 2.179\times 10^{-18}\ \mathrm{J} \cdot (\dfrac{1}{{n_{\mathrm{1}}}^{2}} - \dfrac{1}{{n_{\mathrm{2}}}^{2}})\)

\(\ \ \ =2.179\times 10^{-18}\ \mathrm{J} \cdot (\dfrac{1}{{4}^{2}} - \dfrac{1}{{6}^{2}})\)

\(\ \ \ =2.179\times 10^{-18}\ \mathrm{J} \cdot (\dfrac{1}{16} - \dfrac{1}{36})\)

\(\ \ \ =2.179\times 10^{-18}\ \mathrm{J} \cdot (\frac{1 }{ 16} - \frac{1 }{ 36})\)

\(\ \ \ =2.179\times 10^{-18}\ \mathrm{J} \cdot \frac{5 }{ 144}\)

\(\ \ \ =7.566\times 10^{-20}\ \mathrm{J}\)


This energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron from then \( = 4\) orbit up to the \(n = 6\) orbit. The wavelength of a photon with this energy is found by the expression


\(E_{\mathrm{photon}}\) \(= ΔE_{\mathrm{electron}}\)

\(\ \ \ =7.566\times 10^{-20}\ \mathrm{J}\)

\(\ \ \ =7.566\times 10^{-20}\ \mathrm{J}\)


\(c_{\mathrm{0}}\) \(= 2.998\times 10^{8}\ \frac{\mathrm{m}}{\mathrm{s}}\)


\(h_{\mathrm{Planck}}\) \(= 6.626\times 10^{-34}\ \mathrm{J}\ \mathrm{s}\)


Rearrangement gives:

\(λ\) \(= \dfrac{h_{\mathrm{Planck}} \cdot c_{\mathrm{0}}}{E_{\mathrm{photon}}}\)

\(\ \ \ =\dfrac{6.626\times 10^{-34}\ \mathrm{J}\ \mathrm{s} \cdot 2.998\times 10^{8}\ \frac{\mathrm{m}}{\mathrm{s}}}{7.566\times 10^{-20}\ \mathrm{J}}\)

\(\ \ \ =\dfrac{1.98647\times 10^{-25}\ \mathrm{J}\ \mathrm{m}}{7.566\times 10^{-20}\ \mathrm{J}}\)

\(\ \ \ =2.626\times 10^{-6}\ \mathrm{m}\)


\(λ\) \(= 2.626\times 10^{-6}\ \mathrm{m}\)

\(\ \ \ =2.626\ \mathrm{μm}\)


From Figure 3, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum.