The atom that almost worked
Pass an electric spark through hydrogen gas and look at the glow through a prism. Instead of a smooth rainbow you see four thin lines: red, blue-green, and two in the violet, with darkness between them. Why does hydrogen emit only certain colours?
The previous two essays found energy exchanged in steps and light delivered in photons of energy $h\nu$. A sharp spectral line is a photon of one exact energy, so the question becomes: why can a hydrogen atom give out only certain amounts of energy? In 1913 Niels Bohr answered it with a model that reproduced the lines to a fraction of a percent. The model is not correct, and this essay will say where it breaks. But it computed, for the first time, the size and energy of an atom from fundamental constants, and its central idea survives: an atom exists only in a set of definite energy levels.
Why a classical atom cannot exist
By 1911 Ernest Rutherford's scattering experiments had shown that an atom has a tiny, heavy, positive nucleus with electrons somewhere around it. The natural picture is a miniature solar system: the electron circling the proton, held by the Coulomb force as a planet is held by gravity.
Classical physics rules this out twice over. First, a circling charge is an accelerating charge, and accelerating charges radiate. The electron should lose energy continuously and spiral inward. Run the calculation for an electron starting at hydrogen's actual size and it hits the nucleus in about $10^{-11}$ seconds: every atom in the universe would have collapsed at once. Second, as the electron spirals its orbital frequency changes smoothly, so the light it emits should sweep through a continuous range of colours. A hydrogen atom does neither. It is stable, and it emits sharp lines.
Bohr's two rules
Bohr kept the orbit and added two rules that classical physics could not justify.
Stationary states. Only certain orbits are allowed, and in them the electron does not radiate, whatever electromagnetism says. The allowed orbits are those whose angular momentum is a whole-number multiple of $h/2\pi$:
$$mvr = n\hbar,\qquad n = 1, 2, 3, \dots$$The combination $\hbar = h/2\pi = 1.055\times10^{-34}$ J s, read "h-bar", appears so often in quantum mechanics that it has its own symbol.
Jumps. Light is emitted or absorbed only when the electron jumps from one allowed orbit to another, and it goes as a single photon carrying the energy difference:
$$h\nu = E_{\text{upper}} - E_{\text{lower}} .$$The second rule is Einstein's photon applied to an atom, and it is the idea that lasts. The first rule was a guess, chosen because it gave the right answer. We will see shortly what it looks like from the far side of the theory.
From the rules to the levels
For a circular orbit the Coulomb attraction supplies the inward force:
$$\frac{mv^2}{r} = \frac{e^2}{4\pi\varepsilon_0\, r^2}.$$Bohr's condition gives $v = n\hbar/mr$. Substituting and solving for $r$:
$$r_n = n^2\,\frac{4\pi\varepsilon_0\,\hbar^2}{m e^2} = n^2 a_0,$$where $a_0$ is called the Bohr radius. The orbits grow as the square of $n$.
For the energy, the force balance tells us that $mv^2 = e^2/4\pi\varepsilon_0 r$, so the kinetic energy $\tfrac12 mv^2$ is exactly half the size of the potential energy $-e^2/4\pi\varepsilon_0 r$, with the opposite sign. The total is therefore minus the kinetic energy:
$$E_n = -\frac{e^2}{8\pi\varepsilon_0\, r_n} = -\frac{e^2}{8\pi\varepsilon_0\, a_0}\cdot\frac{1}{n^2}.$$Every quantity here is a fundamental constant except $n$. Putting in the numbers (below), the whole content of the model for hydrogen is two lines:
$$a_0 = 0.0529\ \text{nm},\qquad E_n = -\frac{13.6\ \text{eV}}{n^2}.$$Worked example: the red line
The constants. With $e = 1.602\times10^{-19}$ C and $\varepsilon_0 = 8.854\times10^{-12}$ F/m, the Coulomb combination is $e^2/4\pi\varepsilon_0 = 2.307\times10^{-28}$ J m, which is $1.440$ eV nm. With $m = 9.109\times10^{-31}$ kg,
$$\begin{aligned} a_0 &= \frac{\hbar^2}{m\,(e^2/4\pi\varepsilon_0)}\\[4pt] &= \frac{(1.055\times10^{-34})^2\ \text{m}}{(9.109\times10^{-31})(2.307\times10^{-28})}\\[4pt] &= 5.29\times10^{-11}\ \text{m},\\[4pt] E_1 &= -\frac{1.440\ \text{eV nm}}{2\times 0.0529\ \text{nm}}\\[4pt] &= -13.6\ \text{eV}. \end{aligned}$$The measured energy needed to ionise hydrogen from its ground state is 13.6 eV, and atoms are indeed about a tenth of a nanometre across.
The red line is the jump from $n = 3$ to $n = 2$:
$$\begin{aligned} E_3 - E_2 &= 13.6\ \text{eV}\left(\frac14 - \frac19\right)\\[4pt] &= 1.89\ \text{eV},\\[4pt] \lambda &= \frac{hc}{E_3 - E_2} = \frac{1240\ \text{eV nm}}{1.89\ \text{eV}}\\[4pt] &= 656\ \text{nm}. \end{aligned}$$This is the line spectroscopists call H-alpha, measured at 656.5 nm (in vacuum). The small remaining gap closes when one allows for the nucleus moving slightly as the electron circles it: the corrected value is 656.5 nm too, in agreement to better than one part in ten thousand.
The whole series. Jumps from $n = 4, 5, 6$ down to $2$ give 486, 434 and 410 nm: the other three visible lines. In general a jump from $n_i$ to $n_f$ gives
$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right),$$with $R = 13.6\ \text{eV}/hc = 1.097\times10^{7}$ m−1.
Johann Balmer had found the $n_f = 2$ case of this pattern by fitting the measured lines in 1885, with no idea why it held. Bohr's model derived it and predicted the value of $R$ from $e$, $m$ and $h$.
Check: the electron's speed. In the $n = 1$ orbit, $v = \hbar/ma_0 = 2.19\times10^{6}$ m/s, less than one percent of the speed of light. Ignoring relativity was a fair approximation.
Bohr held his rules to one more test. Where classical physics is known to work, for large systems, quantum rules must give the same answers; for hydrogen that means large orbits, high $n$. This requirement is called the correspondence principle, and the model passes it: for large $n$ the frequency of the light from a jump to the next level down approaches the orbit's own frequency, which is what a classical circling charge would emit. The optional section shows the numbers. The principle will serve as a check throughout the book.
Optional: large orbits behave classically
Bohr insisted that his rules must agree with classical physics where classical physics is known to work: for large orbits. There, a classical electron circling at frequency $f$ would radiate light of frequency $f$. Does the quantum jump give the same?
The orbital frequency in orbit $n$ is $v/2\pi r_n$, which works out to $f_n = 2|E_1|/hn^3$. The photon from the jump $n \to n-1$ has frequency $\nu = (|E_1|/h)\left[1/(n-1)^2 - 1/n^2\right]$. For large $n$, the bracket is close to $2/n^3$, so $\nu \approx f_n$. Numerically, the ratio $\nu/f_n$ is 3.0 at $n = 2$, 1.17 at $n = 10$, 1.015 at $n = 100$, and 1.0015 at $n = 1000$.
For small orbits the two disagree completely; for large ones they merge. This requirement, that quantum rules must go over into classical ones for large systems, is called the correspondence principle. It will serve as a check throughout the book.
What the model could not do
The success was real, and so were the failures. Helium, with two electrons, could not be made to work: no choice of orbits gave its measured energy. The model said nothing about how bright each line should be, or why some jumps never occur. When a magnetic field splits the lines, the pattern is often more intricate than the model allows. And its prediction for the ground state is wrong in a way that matters: it gives the electron one unit of angular momentum, $\hbar$, where the correct theory, and experiment, give zero. An orbit with no angular momentum is not an orbit at all.
Most basically, the model never said why. Why should $mvr$ be a multiple of $\hbar$, and why should an orbiting charge stop radiating? It was a scaffold, not a building: essential for getting up, but not the thing itself.
One hint of the answer came a decade later, from the relation at the end of the essay on photons. If an electron has a wavelength $\lambda = h/mv$, then Bohr's condition $mvr = n\hbar$ can be rewritten as $2\pi r = n\lambda$: a whole number of wavelengths fits around the orbit, like a standing wave on a closed loop. Orbits that do not close on themselves are forbidden because the wave would interfere with itself. This picture is still not right, and the real theory has no orbits, but it shows where to look. The allowed energies come from waves that fit.
Two questions
1. The invisible line. Find the wavelength of the photon emitted when a hydrogen atom drops from $n = 2$ to $n = 1$. Why is this line, the strongest in hydrogen's spectrum, missing from what you see through the prism?
Worked answer
$E_2 - E_1 = 13.6\ \text{eV}\,(1 - \tfrac14) = 10.2$ eV, and $\lambda = 1240/10.2 = 122$ nm. That is far into the ultraviolet, which the eye cannot see and ordinary glass absorbs. It is called Lyman alpha. The visible lines all end on $n = 2$ because only those jumps fall between about 1.8 and 3.1 eV, the energy range of visible photons.
2. Does 656 nm prove the orbits? The Bohr model predicts H-alpha correctly to a fraction of a percent. Does that show the electron really travels in a circle of radius $9a_0$ in the $n = 3$ state?
Worked answer
No. The prediction uses only the energies of the levels: the wavelength depends on $E_3 - E_2$ and nothing else. Any theory that produces the same energies, $-13.6\ \text{eV}/n^2$, predicts the same line, whatever picture of the electron it uses. Quantum mechanics does exactly that, with no orbits at all.
The tempting wrong turn is to treat an accurate prediction as confirming every part of the model behind it. A result confirms only the parts it depends on. Here the energy levels were confirmed; the circular paths were never tested by the spectrum, and the ground state's angular momentum shows the orbit picture is wrong.
Further reading
- Born, Atomic Physics (Dover, 8th ed.), Ch. IV §2–§3, on why a classical atom fails and Bohr's stationary states, and Ch. V §1 "The Bohr Atom; Stationary Orbits for Simply Periodic Motions", which works through the hydrogen levels, the Balmer formula and the correspondence principle at greater length.
Where this leaves us
Hydrogen emits only certain colours because the atom can only exist at certain energies, $-13.6\ \text{eV}/n^2$, and each photon carries away the difference between two of them. Bohr's model found those energies. It did so with orbits that turned out not to exist, and with a rule it could not explain.
The question the model leaves is what replaces a trajectory. If an electron in an atom does not follow a path, what does quantum mechanics describe instead, and how does it still produce definite predictions? The clearest place to see the answer is not inside an atom but in a simpler experiment, where electrons are fired at a wall with two slits and asked to choose a route. That is where the next essay begins.