Chem 3240 · Lecture 1.4
Balmer (1885) fit part of the hydrogen spectrum. Rydberg generalized it:
\[\tilde{\nu} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\]
Fit an integer number of de Broglie waves around the orbit:
\[2\pi r = n \lambda_e, \qquad \lambda_e = \frac{h}{m_e v}\]
Substituting gives the quantization condition:
\[m_e v r = \frac{n h}{2\pi} = n \hbar\]
Electrostatic pull balances the centrifugal force:
\[\frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{m_e v^2}{r}\]
Combined with \(m_e v r = n\hbar\), the allowed radii are:
\[r = n^2 a_0, \qquad n = 1,2,3,...\]
\[a_0 = \frac{4\pi \varepsilon_0 \hbar^2}{m_e e^2} \approx 0.529 \,\text{Å}\]
Total energy (kinetic plus Coulomb) at quantized radius:
\[E_n = -\frac{m_e e^4}{8 \varepsilon_0^2 h^2}\cdot\frac{1}{n^2}\]
The practical form for problem solving:
\[E_n = -13.6\,\frac{1}{n^2}\,\,\,[\text{eV}]\]
Photon energy of a transition, \(\tilde{\nu} = \nu/c\):
\[\tilde{\nu} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]
Now \(R_H\) is derived, not fitted:
\[R_H = \frac{m_e e^4}{8 \varepsilon_0^2 c h^3}\]
For one-electron ions (\(He^+\), \(Li^{2+}\)), add nuclear charge \(Z\):
\[E_n = -13.6\,\frac{Z^2}{n^2}\,\,\,[\text{eV}]\]
Atomic spectra are discrete because energy is quantized: Bohr’s standing-wave condition fixes the orbits and yields \(E_n = -13.6\,/\,n^2\) eV, deriving the empirical Rydberg formula from first principles.