Successes and Failures of the Bohr Model
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You should be comfortable with the derivation of the Bohr energy levels for hydrogen and understand the physical meaning of the quantum number \(n\).
Why this matters
The Bohr model was a triumph, but its limitations define the boundary of classical physics.
While the Bohr model accurately predicts the spectral lines of hydrogen, it fails completely for multi-electron atoms. Understanding exactly where it breaks down is crucial because it highlights the necessity of the full quantum mechanical model, which treats electrons as wavefunctions rather than particles in fixed orbits. This lecture dissects the specific successes and failures to prepare you for the transition to wave-particle duality.
The Triumph: Hydrogen and Hydrogen-like Ions
The most significant success of the Bohr model is its quantitative agreement with experimental data for one-electron systems. By postulating that angular momentum is quantised in integer multiples of \(\hbar\), Bohr derived a set of discrete energy levels that perfectly match the observed emission and absorption spectra of hydrogen. This was not a qualitative fit; it was a precise numerical match. The model successfully explained why atoms emit light at specific, discrete frequencies rather than a continuous spectrum, resolving a major crisis in early 20th-century physics. It also provided a theoretical basis for the Rydberg constant, linking atomic structure directly to the geometry of the atom through the Bohr radius \(a_0\). For any system with a single electron, such as \(\mathrm{He^+}\) or \(\mathrm{Li^{2+}}\), the model remains valid, scaling the energy levels with the square of the nuclear charge \(Z\).
The Failure: Multi-Electron Atoms
The model’s primary failure is its inability to describe atoms with more than one electron. In a multi-electron atom, the electron-electron repulsion forces are significant and complex. The Bohr model assumes a central potential where the electron moves in a perfect circle under the influence of the nucleus alone. It does not account for the shielding effect, where inner electrons reduce the effective nuclear charge felt by outer electrons. Consequently, the simple formula \(E_n = -\frac{13.6\,Z^2}{n^2}\text{ eV}\) does not hold for elements like helium or lithium. The energy levels in multi-electron atoms depend on both the principal quantum number \(n\) and the orbital angular momentum quantum number \(l\), a distinction the Bohr model cannot make because it only allows circular orbits. This failure necessitates the introduction of the full set of quantum numbers to describe electron configurations accurately.
The Failure: Fine Structure and Zeeman Effect
Even for hydrogen, the Bohr model is incomplete. It predicts that all states with the same principal quantum number \(n\) are degenerate, meaning they have the same energy. However, high-resolution spectroscopy reveals that these levels are actually split into closely spaced sub-levels, a phenomenon known as fine structure. This splitting arises from relativistic corrections to the electron's kinetic energy and the spin-orbit interaction, neither of which are included in the non-relativistic, spinless Bohr model. Furthermore, the model fails to explain the anomalous Zeeman effect, where spectral lines split in the presence of an external magnetic field. The Bohr model cannot account for the magnetic moment associated with the electron's spin, which is a fundamental property revealed by quantum mechanics. These failures demonstrate that the classical concept of a particle in a defined orbit is insufficient to describe the true behaviour of electrons in atoms.
Quantitative Comparison: Energy Scaling
To illustrate the success for hydrogen-like ions, we can compare the ground state energies of hydrogen (\(Z=1\)) and helium ion (\(\mathrm{He^+}\), \(Z=2\)). the energy scales with \(Z^2\). For hydrogen, \(E_1 = -13.6\ \mathrm{eV}\). For \(\mathrm{He^+}\), the energy is \(E_1 = -13.6 \times 2^2 = -54.4\ \mathrm{eV}\). This factor of four increase is experimentally verified. However, if we attempt to apply this to neutral helium (\(Z=2\), two electrons), the model would incorrectly predict a ground state energy of \(-54.4\ \mathrm{eV}\) for the first electron and a similar magnitude for the second, ignoring the repulsion. The actual ionisation energy of helium is \(24.6\ \mathrm{eV}\), significantly lower than the Bohr prediction for a hydrogen-like ion, highlighting the critical role of electron-electron interactions that the model neglects.
The result
The Bohr model of hydrogen
For a one-electron atom or ion with nuclear charge \(+Ze\) (hydrogen has \(Z=1\)), postulate that the electron moves in a circular orbit under the Coulomb attraction of the nucleus, and that its orbital angular momentum is quantised in integer multiples of \(\hbar\): \(L=n\hbar\), \(n=1,2,3,\dots\). Then only discrete orbital radii and energies are allowed, with the electron's total energy given by \(E_n=-\dfrac{Z^2}{n^2}\cdot\dfrac{m_ee^4}{8\varepsilon_0^2h^2}=-\dfrac{13.6\,Z^2}{n^2}\text{ eV}\), and orbital radius \(r_n=\dfrac{n^2}{Z}a_0\), where \(a_0=\dfrac{\varepsilon_0h^2}{\pi m_ee^2}\approx0.529\,\text{Å}\) is the Bohr radius.
A common misconception is that the Bohr model is "wrong" because it is replaced by quantum mechanics. In reality, it is an approximation that works perfectly for one-electron systems. The failure lies in its classical assumptions about trajectories and angular momentum quantisation, not in the energy formula itself for \(Z\)-scaled systems. Do not discard the Bohr radius \(a_0\) or the \(1/n^2\) energy dependence; they remain fundamental constants and scaling laws in modern quantum chemistry. C
Worked example
Problem. Calculate the wavelength of the photon emitted when an electron in a \(\mathrm{Li^{2+}}\) ion (hydrogen-like, \(Z=3\)) transitions from the \(n=3\) orbit to the \(n=1\) orbit. Use \(E_n = -\frac{13.6\,Z^2}{n^2}\text{ eV}\) and \(hc = 1240\ \mathrm{eV\,nm}\).
- Determine the energy of the initial state (\(n=3\)) for \(Z=3\): \[ E_3 = -\frac{13.6 \times 3^2}{3^2} = -13.6\ \mathrm{eV} \]
- Determine the energy of the final state (\(n=1\)) for \(Z=3\): \[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -13.6 \times 9 = -122.4\ \mathrm{eV} \]
- Calculate the energy difference \(\Delta E\) emitted: \[ \Delta E = E_3 - E_1 = -13.6 - (-122.4) = 108.8\ \mathrm{eV} \]
- Convert energy to wavelength using \(\lambda = \frac{hc}{\Delta E}\): \[ \lambda = \frac{1240\ \mathrm{eV\,nm}}{108.8\ \mathrm{eV}} \approx 11.397\ \mathrm{nm} \]
Units check. The ratio of \(\mathrm{eV\,nm}\) to \(\mathrm{eV}\) yields \(\mathrm{nm}\), which is a valid unit of length for wavelength.
Three quick tests
- Why does the Bohr model fail for helium?
Answer
It neglects electron-electron repulsion and shielding, which are significant in multi-electron atoms. - How does the ground state energy of \(\mathrm{He^+}\) compare to hydrogen?
Answer
It is four times more negative (\(-54.4\ \mathrm{eV}\) vs \(-13.6\ \mathrm{eV}\)) because energy scales with \(Z^2\). - What phenomenon reveals the failure of the Bohr model even for hydrogen?
Answer
Fine structure splitting, which arises from relativistic and spin-orbit effects not included in the model.
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.