12.3 The Line Spectra of Hydrogen and de Broglie's Explanation

Solar astronomers photograph the Sun through filters that pass only a band about $0.05$ nm wide centred on $656.3$ nm. In that red light the smooth disc becomes a seething surface of filaments, flares and prominences — gas whose hydrogen atoms are absorbing and re-emitting exactly that one colour. Space telescopes map clouds of hydrogen around distant galaxies in the ultraviolet line at $121.6$ nm, and radio telescopes trace the gas of the Milky Way in lines from atoms with enormous orbits. In every case the light has one precise wavelength because it is produced by an electron jumping between two precisely fixed energies.

In Section 12.2 we found the energies of Bohr's stationary orbits, $E_n=-13.6/n^2$ eV. Now we put them on an energy-level diagram and use Bohr's third postulate to explain the hydrogen spectrum line by line — even calculating the Rydberg constant from $m$, $e$, $h$ and $\varepsilon_0$. We then see how ionisation and excitation energies, absorption spectra and the number of lines emitted all follow from the same diagram. Finally we answer the question Bohr could not: why is the angular momentum quantised? The answer, due to Louis de Broglie, is that the electron behaves as a wave that must fit round its orbit.

Energy Levels and the Energy-Level Diagram

The energies $E_n$ are best shown as a ladder of horizontal lines, one for each value of $n$, drawn at heights proportional to energy. The zero of energy is an electron at rest far from the proton; every bound state lies below it.

For hydrogen the levels are at $-13.6$, $-3.40$, $-1.51$, $-0.85$, $-0.54$, … eV. They crowd together as $n$ grows and approach $E=0$, exactly as the lines of a series crowd towards the series limit.

An atom can be excited in two ways. It can absorb a photon, but only one whose energy exactly matches the energy difference between two levels (or exceeds the energy needed to ionise it). Or it can be struck by a fast electron or another atom in a collision, which can hand over any part of its kinetic energy — so a $12.5$ eV electron can give up $10.2$ eV and move on with $2.3$ eV. This is how the atoms in a discharge tube or a fluorescent lamp are excited.

Emission and Absorption by Hydrogen

By Bohr's third postulate, when the electron drops from a level $n_i$ to a lower level $n_f$ the atom emits one photon of energy $$h\nu=E_{n_i}-E_{n_f}=\frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).$$ Since $\nu=c/\lambda$, dividing by $hc$ gives the wavenumber of the line: $$\frac1\lambda=\frac{me^4}{8\varepsilon_0^2h^3c}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right).$$ This is exactly the Rydberg formula of Section 12.2.

With the course constants, $R=\dfrac{(9.1\times10^{-31})(1.6\times10^{-19})^4}{8(8.85\times10^{-12})^2(6.63\times10^{-34})^3(3.0\times10^8)}$ $\approx1.09\times10^7\ \text{m}^{-1}$; precise constants give $1.097\times10^7\ \text{m}^{-1}$, in agreement with the value measured from spectra to better than $0.1\%$. A constant that spectroscopists had fitted to their data for thirty years turned out to be built from the charge and mass of the electron, Planck's constant and the speed of light.

How many lines? An atom excited to level $n$ can reach the ground state by many routes: directly, or in steps through any of the levels in between. Each pair of levels gives one line.

Absorption spectra. At room temperature practically every hydrogen atom is in its ground state, so a beam of white light (or ultraviolet) passing through cold hydrogen can only lift atoms from $n=1$: cold hydrogen absorbs only the Lyman wavelengths. Balmer lines appear in absorption only where many atoms are already in $n=2$ — in the atmospheres of hot stars, at around $10\,000$ K. A photon with more than $13.6$ eV can always be absorbed by a ground-state atom: it ionises it, and the extra energy becomes the kinetic energy of the freed electron.

De Broglie's Explanation of Bohr's Second Postulate

Bohr's rule $mvr=nh/2\pi$ worked, but Bohr could give no reason for it. In 1924 Louis de Broglie proposed that every moving particle has a wave associated with it, of wavelength $\lambda=h/p=h/mv$ (Chapter 11 — Dual Nature of Radiation and Matter). He applied the idea to the electron in an atom.

Think of a wave travelling round and round a circular orbit. After one trip it meets itself. If the circumference is not a whole number of wavelengths, the wave arriving back is out of step with the wave leaving, and after many trips the contributions cancel: no steady wave can exist. Only if the circumference holds a whole number of wavelengths does the wave join smoothly onto itself and form a standing wave, like the standing waves on a stretched string fixed at both ends (Class 11 Physics, Chapter 14 — Waves) — but closed into a ring, so there are no ends. De Broglie identified these standing waves with Bohr's stationary orbits.

The quantisation of angular momentum is therefore not an arbitrary rule: it is what you get when you treat the electron as a wave confined to a closed path, just as the frequencies of a guitar string are quantised because the string's waves are confined between two fixed ends. This idea was taken much further by Erwin Schrödinger in 1926, whose wave mechanics replaced the circular orbits by three-dimensional standing waves and is the basis of all modern atomic physics.

Limitations of Bohr's model. It works only for atoms or ions with a single electron (hydrogen-like), and even there it cannot explain the relative intensities of the lines or their fine splitting into close pairs. For many-electron atoms, chemical bonding and line intensities one needs the full wave (quantum) mechanics that grew out of de Broglie's idea.

Exercises