Investigate the line emission spectra to examine the Balmer-Rydberg equation 1/lambda = R(1/n_f^2 - 1/n_i^2), and assess the limitations of the Bohr model of the hydrogen atom

What this dot point is asking

NESA wants you to state Niels Bohr's three postulates for the hydrogen atom, use the resulting energy levels $E_n = -13.6 \text{ eV} / n^2$ and the Balmer-Rydberg formula $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ to calculate transition wavelengths, identify the named spectral series, and assess the limitations of the model.

The answer

Why Bohr's model was needed

By 1911 Rutherford had established that the atom has a tiny dense nucleus surrounded by electrons. The classical problem: an orbiting electron is accelerating and should radiate electromagnetic waves continuously, losing energy and spiralling into the nucleus in about $10^{-11}$ s. Atoms are stable, so classical electromagnetism cannot be the whole story.

A second puzzle was that hot rarefied gases of hydrogen emit a discrete pattern of spectral lines, not a continuous spectrum. Balmer (1885) had found an empirical formula for the visible lines, generalised by Rydberg (1890) for all hydrogen lines:

with $n_i > n_f$ for emission. There was no underlying theory for this formula.

Bohr's postulates (1913)

Bohr postulated three rules to fix both problems.

Postulate 1: stationary orbits. The electron in a hydrogen atom occupies certain discrete circular orbits in which it does not radiate. Each such orbit is a stationary state with a well-defined energy.

Postulate 2: quantisation of angular momentum. The allowed orbits are those for which the orbital angular momentum is an integer multiple of $\hbar = h/(2\pi)$:

Postulate 3: photon emission. Radiation occurs only when the electron makes a transition between two stationary states. The photon energy equals the energy difference:

The first postulate rejects classical electromagnetism for bound electrons. The second is the new quantum rule. The third converts energy differences into spectral line wavelengths.

Energy levels

Combining quantised angular momentum with the Coulomb-centripetal force balance gives the allowed orbital radii and energies. The result for hydrogen:

Properties:

• The ground state ($n = 1$) has $E_1 = -13.6$ eV. This is the ionisation energy: removing the electron to infinity requires 13.6 eV.
• $E_n \to 0$ as $n \to \infty$, the unbound (ionised) limit.
• Spacing decreases as $n$ grows (the level "bunching" near zero).

Recovering the Rydberg formula

For a transition $n_i \to n_f$ with $n_i > n_f$:

Photon wavelength:

with $R = 13.6 \text{ eV} / (h c) = 1.097 \times 10^7$ m$^{-1}$. This is the Rydberg formula derived, not just postulated.

Named spectral series

The Balmer series is the visible band first discovered, which is why it has its own name. The full pattern lets astronomers identify hydrogen even from the most distant galaxies.

Worked example: Lyman alpha

Find the wavelength of the photon emitted when an electron drops from $n = 2$ to $n = 1$ in hydrogen.

$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 1.097 \times 10^7 \times (1 - 0.25) = 8.23 \times 10^6$ m$^{-1}$.

$\lambda = 1.22 \times 10^{-7}$ m = 122 nm (ultraviolet).

This is the Lyman alpha line, a powerful tracer of cold neutral hydrogen in the universe.

Try it: Rydberg spectrum calculator to compute wavelengths for arbitrary $n_i \to n_f$ transitions in hydrogen and hydrogen-like ions.

Limitations of the Bohr model

The Bohr model works astonishingly well for hydrogen (and for hydrogen-like ions such as He$^+$ and Li$^{2+}$, with $Z^2$ corrections), but it has clear limitations:

• Multi-electron atoms. The energy levels in helium, lithium and beyond are not predicted accurately. The model has no way to handle electron-electron repulsion.
• Definite orbits. The model puts the electron on a sharp circular path with a definite position and velocity, contrary to the uncertainty principle (which we now know to be exact).
• Fine structure. Spectral lines split into closely spaced sub-lines (fine structure) that the Bohr model does not predict. Relativistic and spin-orbit effects are needed.
• Zeeman and Stark effects. Splitting in external magnetic or electric fields is unexplained.
• Intensities. The model gives line positions but not their intensities. Selection rules and transition probabilities require the full quantum-mechanical treatment.
• No mechanism for quantisation. The angular-momentum rule is postulated, not derived. De Broglie later showed it can be motivated as a standing-wave condition; Schrödinger's wave equation gave it a proper derivation.

The Bohr model is best seen as a transitional model: not the final theory, but the bridge between the Rutherford atom and quantum mechanics.

Examples in context

Example 1. Sydney Observatory student measures the Balmer H-alpha line. Bohr predicts the transition $n_i = 3 \to n_f = 2$ in hydrogen produces a photon with $1/\lambda = R(1/n_f^2 - 1/n_i^2) = 1.097 \times 10^7 \times (1/4 - 1/9) = 1.097 \times 10^7 \times 0.1389 = 1.524 \times 10^6 \text{ m}^{-1}$, giving $\lambda = 656.3 \text{ nm}$ - the bright red H-alpha line. Using $E_n = -13.6/n^2 \text{ eV}$: $E_3 = -1.51 \text{ eV}$, $E_2 = -3.40 \text{ eV}$, so $\Delta E = 1.89 \text{ eV}$. Cross-check: $hc/\lambda = 1240 / 656.3 = 1.89 \text{ eV}$. Sydney Observatory's heritage spectrograph easily resolves H-alpha against the night sky.

Example 2. Lyman-alpha UV transition probed at the Australian Synchrotron. The $n = 2 \to n = 1$ transition in hydrogen produces $\Delta E = E_1 - E_2 = -13.6 - (-3.40) = -10.2 \text{ eV}$, so a photon of $E = 10.2 \text{ eV}$ is emitted. Its wavelength: $\lambda = hc/E = 1240 / 10.2 = 121.6 \text{ nm}$ (deep UV, Lyman-alpha). The same line is absorbed by hydrogen clouds along quasar sight-lines, producing the "Lyman alpha forest" that the Australian Synchrotron's UV beamline has been used to calibrate. Bohr's quantised orbits explain why this discrete wavelength dominates interstellar UV absorption.

Try this

Q1. State Bohr's two key postulates for the hydrogen atom. [2 marks]

• Cue. (1) Electron occupies stationary orbits without radiating; (2) Angular momentum is quantised: $L = n h / (2\pi)$.

Q2. Calculate the wavelength of the photon emitted when a hydrogen electron drops from $n = 4$ to $n = 2$. [3 marks]

• Cue. $1/\lambda = 1.097 \times 10^7 (1/4 - 1/16) = 1.097 \times 10^7 \times 0.1875 = 2.057 \times 10^6 \text{ m}^{-1}$; $\lambda = 486 \text{ nm}$ (H-beta blue-green).

Q3. An astronomer at Siding Spring Observatory observes hydrogen Balmer-series lines from a star. (a) Identify the energy of the $n = 5$ level in eV. (b) Calculate the wavelength of the transition $n = 5 \to n = 2$ (H-gamma). (c) State two limitations of the Bohr model. [2+3+2 marks]

• Cue. (a) $E_5 = -13.6/25 = -0.544 \text{ eV}$. (b) $\Delta E = 2.86 \text{ eV}$; $\lambda = 1240/2.86 = 434 \text{ nm}$. (c) Fails for multi-electron atoms; cannot predict line intensities or fine structure.