Hydrogen without little planetary orbits
The Bohr model got hydrogen's colours right with orbits that turned out not to exist, and it got the angular momentum of the ground state wrong. Quantum mechanics has no orbits. So why does it produce exactly the same spectrum, and what does it put in place of the electron's path?
Hydrogen is the one real atom whose Schrödinger equation can be solved exactly, and it repays the effort. This essay solves it for the ground state by hand, states the general result, and then separates two things the Bohr model ran together: the energy levels, which decide the spectrum, and the spatial distributions of the electron, which decide the atom's size and shape. The two agree with Bohr in some places and part company in others, and the places tell us which parts of his picture were right.
Setting up the problem
The proton is about 1,800 times heavier than the electron, so to a good approximation it sits still at the origin, and the electron moves in its Coulomb potential. (Allowing for the proton's small motion replaces the electron mass by the slightly smaller reduced mass, met in the oscillator essay, and shifts every energy by about one part in 2,000.) The stationary states solve Schrödinger's equation in three dimensions,
$$-\frac{\hbar^2}{2m_e}\nabla^2\psi - \frac{e^2}{4\pi\varepsilon_0\, r}\,\psi = E\,\psi ,$$where $\nabla^2 = \partial^2/\partial x^2 + \partial^2/\partial y^2 + \partial^2/\partial z^2$ is the three-dimensional version of the curvature term. Here $m_e$ is the electron's mass, written with a subscript so that plain $m$ can keep its meaning from the angular momentum essay, and $\psi$ stands for the spatial shape of a stationary state, which earlier essays wrote $\varphi$. Because the potential depends only on the distance $r$, the problem has the symmetry of a sphere. That lets each solution be written as a function of $r$ times a function of direction, and the directional part is fixed by the angular momentum quantum numbers $l$ and $m$ of the previous essay. What remains is an equation for the radial part alone.
Worked example: the ground state by hand
The simplest guess is a state with no angular dependence at all, $l = 0$, falling off exponentially with distance:
$$\psi(r) = e^{-r/a}.$$For a function of $r$ alone, the curvature term takes the form $\nabla^2\psi = \psi'' + (2/r)\,\psi'$. (That is a standard formula, quoted rather than derived here.) With $\psi' = -\psi/a$ and $\psi'' = \psi/a^2$, Schrödinger's equation divided by $\psi$ becomes
$$-\frac{\hbar^2}{2m_e}\left(\frac{1}{a^2} - \frac{2}{a\,r}\right) - \frac{e^2}{4\pi\varepsilon_0\, r} = E .$$The right side is a constant, so the two terms in $1/r$ on the left must cancel:
$$\frac{\hbar^2}{m_e a} = \frac{e^2}{4\pi\varepsilon_0},\quad\text{so}\quad a = \frac{4\pi\varepsilon_0\,\hbar^2}{m_e e^2}.$$Putting in the numbers, $a = 0.0529$ nm.
That is Bohr's radius $a_0$, now as the decay length of a wavefunction. What is left gives the energy:
$$\begin{aligned} E &= -\frac{\hbar^2}{2m_e a_0^2} = -\frac{0.0381\ \text{eV nm}^2}{(0.0529\ \text{nm})^2}\\[4pt] &= -13.6\ \text{eV}. \end{aligned}$$Hydrogen's ionisation energy, with no orbit anywhere in the calculation. And the state has $l = 0$: zero angular momentum. This is where Bohr's model was wrong, and experiment sides with the wave equation.
Check: where is the electron? The probability of finding it at a distance between $r$ and $r + dr$ is $|\psi|^2$ times the volume of that thin shell, $4\pi r^2\,dr$. The factor $r^2 e^{-2r/a_0}$ rises and then falls, peaking exactly at $r = a_0$. The most probable distance is Bohr's radius; the average distance, $1.5\,a_0$, is a little further out, because the tail stretches a long way.
The general result
Solving the radial equation in general takes more work (the optional section says how), and the result is compact. The allowed energies are
$$E_n = -\frac{13.6\ \text{eV}}{n^2},\qquad n = 1, 2, 3, \dots,$$the same as Bohr's. For each $n$, the angular momentum can be $l = 0, 1, \dots, n-1$, and for each $l$ there are $2l + 1$ values of $m$. Counting them, level $n$ holds $1 + 3 + 5 + \dots + (2n-1) = n^2$ different states, all with the same energy. By tradition the values $l = 0, 1, 2, 3$ are labelled s, p, d, f, so the ground state is 1s and the first excited level contains 2s and three 2p states.
Now the spectrum. A photon is emitted when the atom drops from one level to another, carrying $E_{n_i} - E_{n_f}$. The energies are Bohr's, so the differences are Bohr's, so the lines are Bohr's: H-alpha from $n = 3$ to $n = 2$ at 656 nm, and the rest of the Balmer series. The spectrum only ever tested the energy levels. That is why a model with the wrong picture could still get every colour right.
What replaces the orbit
Each stationary state of the electron is called an orbital. Its $|\psi|^2$ is a probability distribution for where the electron will be found, fixed and unchanging in time. It is not a path the electron travels, and it is not a smeared-out electron: a position measurement always finds the electron whole, at one point, and the orbital's distribution gives the odds.
Three words that the Bohr model used interchangeably now mean different things. An energy level is a value of $E$, such as $-3.4$ eV for $n = 2$. An orbital is one stationary state with that energy, such as 2p with $m = 0$; a level can contain several. A trajectory, a definite path in time, does not exist for the electron at all, since position and momentum are never both sharp.
Where Bohr was right, the figure shows it. For the states with the largest angular momentum at each level, $l = n - 1$ (1s, 2p, 3d, and so on), the most probable distance is exactly $n^2a_0$, Bohr's radius. As $n$ grows these states concentrate more and more tightly around that distance, and the classical orbit emerges as a limit: the correspondence principle again.
Optional: how the general solution goes
Write $\psi = R(r)\,Y_{lm}(\theta, \phi)$, where $Y_{lm}$ is the angular function with angular momentum quantum numbers $l$ and $m$. Substituting and using $\hat L^2 Y_{lm} = l(l+1)\hbar^2 Y_{lm}$ leaves an equation for $R$ alone, in which the angular motion appears as an extra repulsive term $l(l+1)\hbar^2/2m_er^2$ added to the Coulomb potential: the energy cost of going round, which keeps states with $l > 0$ away from the nucleus.
Far from the nucleus the solution must die away as $e^{-r/na_0}$, and near it it must behave as $r^l$. Writing $R = r^l e^{-r/na_0}$ times a power series and substituting, the series terminates and gives a normalisable wavefunction only when $n$ is a whole number greater than $l$. That condition is the quantisation of energy, and the terminating series are polynomials, the reason the 2s and 3s curves in the figure have nodes.
Two questions
1. Inside the Bohr radius. In the ground state, the most probable distance of the electron is $a_0$. What is the probability of finding it closer to the nucleus than $a_0$?
Worked answer
The normalised radial probability is $P(r) = (4r^2/a_0^3)\,e^{-2r/a_0}$. Integrating from $0$ to $a_0$ (by parts, twice) gives $1 - 5e^{-2} = 0.32$. About a third of the time the electron is found inside the Bohr radius, and two-thirds of the time outside it.
The tempting slip is to read "most probable distance" as a half-way point, expecting 50 percent. The peak of a lopsided distribution is not its middle. This one has a long tail outwards, which is also why the average distance, $1.5a_0$, exceeds the peak.
2. Level, orbital, trajectory. Which of these statements about a hydrogen atom are meaningful, and which are not? (a) "The electron has energy $-3.4$ eV." (b) "The electron is in a 2p orbital." (c) "The electron is moving round a circle of radius $4a_0$."
Worked answer
(a) is meaningful: it says the atom is in the $n = 2$ energy level, and an energy measurement would confirm it. (b) is meaningful and more specific: of the four states at that energy, the atom is in one of the three with $l = 1$. It says what a position measurement's odds would be: most likely near $4a_0$, but spread over a range. (c) is not meaningful. It describes a trajectory, which would need sharp position and momentum at every moment.
The tempting wrong turn is to treat (c) as a vivid version of (b), because the 2p distribution does peak at $4a_0$. The peak is where a measurement most often finds the electron. Nothing in the theory says the electron is travelling round that radius between measurements.
Further reading
- Feynman, Lectures on Physics, Vol. III, Ch. 19 "The Hydrogen Atom and The Periodic Table", §19-1 to §19-5: the solution built up from the spherically symmetric states, with pictures of the wavefunctions.
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §36 "Motion in a Coulomb field (spherical polar co-ordinates)": the full series solution.
- Born, Atomic Physics (Dover, 8th ed.), Ch. V §4 "Wave Mechanics", with the three-dimensional solution in Appendix XVIII.
Where this leaves us
Quantum mechanics produces Bohr's spectrum because it produces Bohr's energies, and the spectrum tests nothing else. In place of orbits it gives orbitals: fixed probability distributions for where the electron will be found, sized by $a_0$, with the ground state carrying no angular momentum at all. Energy levels, orbitals and trajectories are three different things, and only the first two exist.
The count of states is not quite finished. Real hydrogen lines, seen at high resolution, are split into close pairs, and atoms in a magnetic field split in ways that orbital angular momentum alone cannot produce. Something else about the electron is quantised, with only two possible values. The next essay meets it directly, in a beam of silver atoms that splits into exactly two.