The Bohr Atom

Michael Fowler, University of Virginia

Bohr Comes to Cambridge

In 1911, the 26-year-old Niels Bohr earned a Ph. D. at the University of Copenhagen; his dissertation was titled "Studies on the Electron Theory of Metals". He was awarded a postdoctoral fellowship funded by the Carlsberg Brewery Foundation, which enabled him to go to Cambridge in September to study with J. J. Thomson. Bohr was a great admirer of Thomson's many achievements, both experimental and theoretical. In his thesis work, he had closely studied some of the problems covered in Thomson's bookConduction of Electricity through Gases.He had uncovered some apparent errors in Thomson's work, and looked forward to discussing these points with the great man. Unfortunately, by the time Bohr arrived, the Cavendish Laboratory had grown to the point where Thomson as director had more than he could manage. He had no spare time to think about electrons, and was not happy to hear from Bohr that some of his earlier work might be incorrect. In fact, Thomson went out of his way to avoid theoretical discussions with Bohr (Pais, page 195). He did assign Bohr an experiment on positive rays, but Bohr was not enthusiastic. (Rhodes, page 65) Bohr kept himself busy writing a paper on electrons in metals, reading Dickens to improve his English, and playing soccer.

In December, Rutherford came down from Manchester for the annual Cavendish dinner. Bohr later said that he was deeply impressed by Rutherford's charm, his force of personality, and his patience to listen to every young man who might have an idea—certainly a refreshing change after J. J.! A little later, Bohr met with Rutherford again when he visited one of his father's friends in Manchester, someone who also knew Rutherford. Although Rutherford was usually skeptical of theorists, he liked Bohr. For one thing, Rutherford was a soccer fan, and Bohr's brother Harald (only nineteen months younger than Bohr) was famous—he had played in the silver medal winning Danish soccer team at the 1908 Olympics in London.

After talking it over with Harald, who visited Cambridge in January, Bohr moved to Manchester in March, and took a six-week lab course, given by Geiger, Marsden and others. Really, though, his interests were theoretical, and he talked a lot with Charles Galton Darwin—"grandson of the real Darwin", as Bohr put it in a letter to Harald. Darwin had just completed a theoretical analysis of the loss of energy of an $α$ -particle going through matter—that is, an $α$ that doesn't get close enough to a nucleus to be scattered. Such $α$ 's gradually lose energy by churning through the electrons, and the rate of loss depends on how many electrons they encounter. In particular, Bohr concluded, after reviewing and improving on Darwin's work, it seemed clear that the hydrogen atom almost certainly had asingleelectron outside the nucleus.

What Determines Atomic Size?

A big problem with the nuclear hydrogen atom was: what determined its size? Classical mechanics gives a simple dynamical equation for circular orbits:

$\frac{m v^{2}}{r} = \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{r^{2}} .$

Now this equation is satisfied by any circular orbit centered at the nucleus, however large or small. (Note, by the way, that multiplying both sides by $r / 2$ gives that the magnitude of the kinetic energy in the circular orbit is just half the magnitude of the negative potential energy. We need this below.) There is no hint here that the atom in its "natural" ground state should have any particular radius. But it does! This means we're missing something. But what?

Bohr (and others) thought that Planck's constant must somehow play a role in determining the size of the orbit. After all, it did play a role in restricting allowed orbital changes in the oscillators in black body radiation—and these oscillators, although not very clearly understood, were of the same general size as atoms. So evidently the standard picture of how an oscillating charge radiated couldn't be right at the atomic level. Bohr concluded that in an atom in its natural rest state, the electron must be in a special orbit, he called it a "stationary state" to which the usual rules of electromagnetic radiation didn't apply. In this orbit, which determined the size of the atom, the electron, mysteriously, didn't radiate.

Just how to bring Planck's constant into a discussion of the hydrogen atom was not so clear. For the black body oscillators, it related the frequency f of the oscillator with the allowed energy change $E$ by $E = h f .$ The obvious parallel approach for the hydrogen atom was to identify the frequency $f$ with the circular frequency of the electron in its orbit. However, in contrast to the simple harmonic oscillator this hydrogen atom frequency varies with the size of the orbit. Still, it was the only frequency around, and, dimensionally, multiplying it by $h$ gave an energy. What energy could that be identified with? Again, the choice was limited—the electron had a kinetic energy $E,$ the potential energy was $- 2 E$ and the total energy $- E .$ If a hydrogen nucleus captured a passing electron into its ground state, and emitted one quantum of electromagnetic radiation, that quantum would have energy $E,$ the same as the electron kinetic energy in the natural stationary state (called the ground state). Bohr suggested in a note to Rutherford in the summer of 1912 that requiring this energy be some constant (assumed to be of order of magnitude one) multiplied by $h f$ would fix the size of the atom. Actually his argument was a bit more complicated, he considered the several electron atom, and took the electrons to form rings. However, the basic point is the same—a condition like this constrains the atomic size, it would be fixed uniquely if we knew the constant. If we assume the constant is 1, for example, we have

$f = \frac{v}{2 π r}, \frac{1}{2} m v^{2} = \frac{h v}{2 π r} .$

Putting this together with the dynamic equation above determines the atomic radius $r .$ It is easy to check that it predicts a radius of $4 π ε_{0} h^{2} / π^{2} m e^{2},$ which is just four times the "right answer" defined as the Bohr radius (see below). The correct Bohr radius comes out if we choose the constant to be one-half, $E = \frac{1}{2} h f,$ which Bohr used later.

Hence the approximate size of the atom follows from dimensional arguments alone once one assumes that Planck's constant plays a role! Of course, the nucleus is irrelevant in determining the atomic size—it just provides a fixed center of electrostatic attraction. The relevant electronic parameters are the mass $m$ and the strength of attraction $e^{2} / 4 π ε_{0} .$ Together with $h,$ these parameters determine a length.

It should be mentioned that this assumption explained more than the size of the hydrogen atom. It was believed at the time that in the higher atoms, the electrons formed rings, thought to lie one outside the other, and various stability arguments indicated that there couldn't be more than seven electrons in a ring. The length scale above, $4 π ε_{0} h^{2} / π^{2} m e^{2},$ would decrease for larger atoms, with $e^{2}$ replaced by $Z e^{2}$ essentially, for nuclear charge $Z .$ Thus as the number of rings increased, the size of the rings would decrease, explaining the observed approximate periodicity in atomic volume with atomic number. (Note: this model isn’t right, but it’s on the right track in some ways: we now know the electrons don’t form rings, one inside the next, but they form shells, again, more or less, one inside the other, and these shells do indeed decrease in size as more electrons are added (because the corresponding increase in nuclear charge draws them in closer), and the first electron in a new shell gives a substantially bigger atom, hence the almost-periodicity.)

Also, in 1911 Richard Whiddington in Cambridge had found that to cause a substance having atomic number A to emit characteristic x-rays by bombarding it with electrons, it was necessary to use electrons of speed approximately Ax 10⁶ meters per second. Any substance on being bombarded with sufficiently fast electrons emits a continuum of x-ray frequencies up to a maximum frequency $f$ given by $h f =$ kinetic energy of electron; plus some sharply defined lines—x-rays at a particular frequency, which does not change as the electron speed is further increased. The frequency corresponding to these lines was found to increase with atomic number. Applying his length scale argument to the innermost ring of an atom, Bohr found that an electron in that ring would have a speed proportional to the nuclear charge, and hence, at least approximately, to the atomic number. Furthermore, the predicted speed in orbit was of the same order as that of Whiddington's electrons.

Exercise : assuming the length scale given above, check Bohr's prediction that the speed in orbit is proportional to the nuclear charge.

Nicolson: a Clever Idea about a Wrong Model

Meanwhile, in Cambridge one J. W. Nicolson was struggling to incorporate Planck's ideas in a model of the atom, in an attempt to understand some strange sets of spectral lines observed in nebulae and in the sun's corona. He conceived a rather exotic (and quite wrong!) model, in which, again, rings of electrons, like a necklace, orbited the nucleus. (Actually, many people, including Bohr himself, investigated models like this. The reason was that the classical radiation from a ring of electrons is a lot less that that from a single orbiting electron, the fields tend to cancel each other.) Oscillations of electrons in this ring gave the spectra. Nicolson predicted the frequencies emitted by a straightforward classical analysis of these oscillation frequencies, in the spirit of earlier work on the plum pudding model. He did bring in Planck's constant, though. He knew that dimensionally it was a unit of angular momentum, and he suggested that the atom could only lose angular momentum in discrete amounts—presumably constant multiples of $h .$ Nicolson felt that, given the dimensionality of Planck's constant, quantization of angular momentum was more plausible than quantization of energy. Of course, for the simple harmonic oscillator they amounted to the same thing, but not for any other system, in particular, not for a nuclear atom.

Bohr Returns to Denmark

Bohr left Manchester in July 1912 and was married on the first of August. In the fall, he began work at the University of Copenhagen, and gave a course of lectures. At the same time, he began setting down on paper some of his Manchester ideas about atoms. He read Nicolson's work. As he wrote to Rutherford at the end of January 1913, he and Nicolson were really looking at different things—Nicolson was considering atoms in a very hot environment (like the sun's corona, or an electrical discharge tube) and the spectra gave information about how energy was emitted as the atom settled into its ground state. Bohr himself was only interested in the state in which the system possessed the smallest amount of energy. He went on: "I do not at all deal with the question of calculation of the frequencies corresponding to the visible part of the spectrum". At that time, Bohr thought of spectra as pretty but peripheral, having as little to do with basic physics as the colors of a butterfly's wings had to do with basic biology.

Bohr Changes his Mind about Spectra

In February 1913, Bohr was surprised to find out in a casual conversation with the spectroscopist H. R. Hansen that some patterns had been discerned in the apparent chaos of spectral lines. In particular, Hansen (a colleague and former classmate of Bohr) showed him Balmer's formula for hydrogen. They had very likely seen this in class together, but, given Bohr's opinion of the value of spectra, he probably hadn't paid much attention.

Balmer's formula is:

$\frac{1}{λ} = R_{H} (\frac{1}{4} - \frac{1}{n^{2}})$

for the sequence of wavelengths of light emitted, with $n = 3, 4, 5, 6$ being in the visible, the lines used by Balmer in finding the formula. Hansen would doubtless have informed Bohr that the 1/4 could be replaced by $1 / m^{2},$ with $m$ another integer. The constant appearing on the right hand side is called the Rydberg constant, R_H = 109,737 cm^-1. This is the modern value—Balmer got it right to one part in 10,000, about the limit of spectral measurements at the time. (Note: the subscript $H$ denotes hydrogen, and depends slightly on the finite proton mass. Taking that to be infinity relative to the electron, an excellent approximation for heavy ions, gives $R_{\infty} .$ )

Bohr said later: "As soon as I saw Balmer's formula, the whole thing was immediately clear to me." What he saw was that the set of allowed frequencies (proportional to inverse wavelengths) emitted by the hydrogen atom could all be expressed as differences. This immediately suggested to him a generalization of his idea of a "stationary state" lowest energy level, in which the electron did not radiate. There must be a whole sequence of these stationary states, with radiation only taking place as the atom jumps from one to another of lower energy, emitting a single quantum of frequency $f$ such that

$h f = E_{n} - E_{m},$

the difference between the energies of the two states. Evidently, from the Balmer formula and its extension to general integers $m, n,$ these allowed non-radiating orbits, the stationary states, could be labeled 1, 2, 3, ... ,n, ... and had energies $- 1, - 1 / 4, - 1 / 9, \dots, - 1 / n^{2}, \dots$ in units of $h c R_{H}$ (using $λ f = c$ and the Balmer equation above). The energies are of course negative, because these are bound states, and we count energy zero from where the two particles are infinitely far apart.

Bohr was very familiar with the dynamics of simple circular orbits in an inverse square field. (We spell out the details in the later sections below.)

He knew that if the energy of the orbit was $- h c R_{H} / n^{2},$ that meant the kinetic energy of the electron, $\frac{1}{2} m v^{2} = h c R_{H} / n^{2},$ and the potential energy would be

$- \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{r} = - \frac{2 h c R_{H}}{n^{2}} .$

It immediately follows that the radius of the $n$ th orbit is proportional to $n^{2},$ and the speed in that orbit is proportional to $1 / n .$

It then follows that the angular momentum of the $n$ th orbit is just proportional to $n .$

Evidently, then the angular momentum in the $n$ th orbit was $n K h,$ where $h$ is Planck's constant and $K$ is some multiplying factor, the same for all the orbits, still to be determined.

In fact, the value of $K$ follows from the results above. $R_{H}, m, h,$ and $c$ are all known, (being measured experimentally by observing the lines in the Balmer series) so the above formulas immediately give the electron's speed and distance from the nucleus in the $n$ th orbit, and hence its angular momentum. Therefore, by putting in these experimentally determined quantities, we can find $K .$

That is to say, the spectroscopic experiment that measured the Balmer frequencies, and hence the Rydberg constant, determines the value of $K :$ it turns out to be $1 / 2 π .$

Bohr Finds the Rydberg Constant without Doing an Experiment

Bohr gave a very clever argument to find $K$ without doing any experiment. First, think about how the size of $K$ affects the physical properties of the hydrogen atom. How would the atom be different for $K = 10$ compared with $K = 1$ ?

Exercise: what would change? Size? Spectral lines?

For $K = 1,$ the allowed orbits would be those having angular momentum $h, 2 h, 3 h, \dots$ .

For $K = 10,$ the only allowed orbits would be those having angular momentum $10 h, 20 h, \dots$ Evidently, for $K = 10$ there will be a lot fewer spectral lines, and the average spacing between them will begreaterthat for $K = 1.$ This will increase the energy differences, and also the spectral line frequencies.

The Correspondence Principle

Next, Bohr imagined a really immense hydrogen atom, an electron going around a proton in a circle of one meter radius, say. This would have to be done in the depths of space, but really this is just a thought experiment in the spirit of Einstein. The point is that for this very large atom, the electron is moving rather slowly over a distance scale we are familiar with. We know from many experiments that charges moving at these slow speeds over ordinary (human size) distances emit radiation according to Maxwell's equations. Or, more simply, if it's going round the circle at frequency $f$ revolutions per second, it will be emitting radiation at that frequency $f$ —because its electric field, as seen from some fixed point a few meters or so away, say, will be rotating $f$ times per second.

Exercise: give a ballpark estimate of the frequency in orbit for one meter proton-electron separation.

Bohr knew that if his theory was correct, it would have to give a frequency for this large atom the same as that given by classical physics (and well-verified). This became known as the Correspondence Principle, and is a useful constraint on any microscopic theories that have classical (macroscopic) limits. They had better agree with the well-established classical results.

Bohr's theory stipulates that the angular momentum quantization condition must be true for all circular orbits of the electron around the proton, including this very large atom. Furthermore, the radiation emitted must still be given by the difference in energies of neighboring orbits,

$h f = E_{n + 1} - E_{n} .$

But $E_{n + 1} - E_{n},$ the energy spacing between neighboring orbits,depends on $K .$

Therefore, Bohr concluded $K$ is fixed by requiring that the frequency of radiation emitted by a really large atom be correctly given by ordinary common sense—that is, the frequency of the radiation (the energy level separation divided by $h$ must be the same as the orbital frequency of the electron, the number of cycles a second.

That is to say, for a large orbit we must have

$E_{n + 1} - E_{n} = h f = h \frac{v}{2 π r} .$

Here v is the speed of the electron in the orbit, and the orbit radius is r.

Finding the Energy Levels in Terms of K

In a large $n$ orbit, angular momentum quantization gives

$L_{n} = m v_{n} r_{n} = n K h .$

Briefly reviewing Newtonian mechanics for circular orbits in an inverse-square force, $\vec{F} = m \vec{a}$ becomes

$m \frac{v_{n}^{2}}{r_{n}} = \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{r_{n}^{2}} .$

It easily follows that the total energy in the orbit

$K . E . + P . E . = E_{n} = - \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{2 r_{n}} .$

So for a given radius $r_{n}$ we know the energy $E_{n},$ and, from the angular momentum quantization equation above, we know $v_{n} .$

Therefore, we also know $L_{n} = m v_{n} r_{n},$ and now requiring Bohr's quantization of angular momentum, $L_{n} = n K h,$ gives the allowed values of $r_{n} .$

It's slightly easier if we square both sides of the quantization equation:

$\begin{matrix} n^{2} K^{2} h^{2} = m (m v_{n}^{2}) r_{n}^{2}, \\ = m (\frac{1}{4 π ε_{0}}) \cdot \frac{e^{2}}{r_{n}} \cdot r_{n}^{2}, \\ = \frac{1}{4 π ε_{0}} m e^{2} r_{n} . \end{matrix}$

So the radius of the $n$ th orbit is

$r_{n} = \frac{4 π ε_{0}}{m e^{2}} {(n K h)}^{2},$

and the radii follow the pattern 1, 4, 9, ... .

The velocity in the $n$ th orbit is (from quantization)

$v_{n} = \frac{n K h}{m r_{n}},$

so the classical orbital frequency is

$f_{n} = \frac{v_{n}}{2 π r_{n}} = \frac{n K h / m r_{n}}{2 π r_{n}} = \frac{n K h}{2 π m r_{n}^{2}},$

and writing $r_{n}$ in terms of $n$ (see equation for $r_{n}$ above)

$f_{n} = \frac{n K h}{2 π m} \cdot {(\frac{m e^{2}}{4 π ε_{0}})}^{2} \cdot \frac{1}{{(n K h)}^{4}} = {(\frac{1}{4 π ε_{0}})}^{2} \cdot \frac{m e^{4}}{2 π} \cdot \frac{1}{{(n K h)}^{3}} .$

This is the frequency that Bohr matched to $(E_{n + 1} - E_{n}) / h$ to determine the constant $K .$

The energy for $r_{n}$ is

$\begin{matrix} E_{n} = - \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{2 r_{n}}, \\ = - \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{2} \cdot \frac{1}{4 π ε_{0}} \cdot \frac{m e^{2}}{{(n K h)}^{2}}, \\ = - {(\frac{1}{4 π ε_{0}})}^{2} \cdot \frac{m e^{4}}{2 {(n K h)}^{2}}, \end{matrix}$

$\begin{matrix} \frac{E_{n + 1} - E_{n}}{h} = - {(\frac{1}{4 π ε_{0}})}^{2} \cdot \frac{m e^{4}}{2 K^{2} h^{3}} [\frac{1}{{(n + 1)}^{2}} - \frac{1}{n^{2}}], \\ ≅ {(\frac{1}{4 π ε_{0}})}^{2} \cdot \frac{m e^{4}}{K^{2} h^{3}} \cdot \frac{1}{n^{3}}, \end{matrix}$

where in the last line we've dropped terms which were smaller by a factor of order $1 / n$ .

Comparing the two expressions, we find they are identical provided $K = 1 / 2 π .$

Hence, the orbital angular momentum in the $n$ th orbit

$L_{n} = n h / 2 π = n ℏ,$

where the $ℏ$ is the standard notation.

Of course, Bohr already knew this was the answer from the spectroscopic experiment, but it's satisfying to see a number come from an abstract argument!

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