The Bohr Atom

Michael Fowler,University of Virginia

Bohr Comesto Cambridge

In 1911, the 26-year-oldNiels Bohr earned a Ph. D. at the University of Copenhagen; his dissertationwas titled "Studies on the Electron Theory of Metals". He was awardeda postdoctoral fellowship funded by the Carlsberg Brewery Foundation, whichenabled him to go to Cambridge in September to study with J. J. Thomson. Bohrwas a great admirer of Thomson's many achievements, both experimental andtheoretical. In his thesis work, he had closely studied some of the problemscovered in Thomson's bookConduction of Electricity through Gases.Hehad uncovered some apparent errors in Thomson's work, and looked forward todiscussing these points with the great man. Unfortunately, by the time Bohrarrived, the Cavendish Laboratory had grown to the point where Thomson asdirector had more than he could manage. He had no spare time to think aboutelectrons, and was not happy to hear from Bohr that some of his earlier workmight be incorrect. In fact, Thomson went out of his way to avoid theoreticaldiscussions with Bohr (Pais, page 195). He did assign Bohr an experiment onpositive rays, but Bohr was not enthusiastic. (Rhodes, page 65) Bohr kepthimself busy writing a paper on electrons in metals, reading Dickens to improvehis English, and playing soccer.

In December, Rutherfordcame down from Manchester for the annual Cavendish dinner. Bohr later said thathe was deeply impressed by Rutherford's charm, his force of personality, andhis 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 againwhen he visited one of his father's friends in Manchester, someone who alsoknew Rutherford. Although Rutherford was usually skeptical of theorists, heliked Bohr. For one thing, Rutherford was a soccer fan, and Bohr's brotherHarald (only nineteen months younger than Bohr) was famous—he had played in the silver medal winning Danishsoccer team at the 1908 Olympics in London.

After talking it overwith Harald, who visited Cambridge in January, Bohr moved to Manchester inMarch, and took a six-week lab course, given by Geiger, Marsden and others.Really, though, his interests were theoretical, and he talked a lot withCharles Galton Darwin—"grandson of the real Darwin", as Bohrput it in a letter to Harald. Darwin had just completed a theoretical analysisof 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 bescattered. Such $α$ 's gradually lose energy by churning through theelectrons, 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 asingleelectronoutside the nucleus.

WhatDetermines Atomic Size?

A big problem with thenuclear hydrogen atom was: what determined its size? Classical mechanics givesa 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 issatisfied byany 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 energyin the circular orbit is just half the magnitude of the negative potentialenergy. We need this below.) There is no hint here that the atom in its"natural" ground state should have any particular radius. But itdoes! This means we're missingsomething. But what?

Bohr (and others)thought that Planck's constant must somehow play a role in determining the sizeof the orbit. After all, itdid play a role in restrictingallowed orbital changes in the oscillators in black body radiation—and these oscillators, although not very clearlyunderstood, were of the same general size as atoms. So evidently the standardpicture of how an oscillating charge radiated couldn't be right at the atomiclevel. Bohr concluded that in an atom in its natural rest state, the electronmust be in a special orbit, he called it a "stationary state" towhich the usual rules of electromagnetic radiation didn't apply. In this orbit,which determined the size of the atom, the electron, mysteriously, didn'tradiate.

Just how to bringPlanck's constant into a discussion of the hydrogen atom was not so clear. For the black body oscillators, it related thefrequencyf of the oscillator with the allowed energychange $E$ by $E = h f .$ The obvious parallel approach for the hydrogenatom was to identify the frequency $f$ with the circular frequency of the electron inits orbit. However, in contrast to the simple harmonic oscillator this hydrogenatom frequencyvaries with the size of the orbit. Still, it was theonly frequency around, and, dimensionally, multiplying it by $h$ gave an energy. What energy could that beidentified with? Again, the choice was limited—the electron had a kinetic energy $E,$ the potential energy was $- 2 E$ and thetotal energy $- E .$ If ahydrogen nucleus captured a passing electron into its ground state, and emittedone quantum of electromagnetic radiation, that quantum would have energy $E,$ the same as the electron kinetic energy in thenatural stationary state (called theground state). Bohr suggestedin a note to Rutherford in the summer of 1912 that requiring this energy besome constant (assumed to be of order of magnitude one) multiplied by $h f$ would fix the size of the atom. Actually hisargument 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 atomicsize, it would be fixed uniquely if we knew the constant. If we assume theconstant 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 togetherwith the dynamic equation above determines the atomic radius $r .$ It is easy to check that it predicts a radiusof $4 π ε_{0} h^{2} / π^{2} m e^{2},$ which isjust four times the "right answer" defined as the Bohr radius (seebelow). The correct Bohr radius comes out if we choose the constant to beone-half, $E = \frac{1}{2} h f,$ which Bohr used later.

Hence the approximatesize of the atom follows fromdimensional arguments alone onceone assumes that Planck's constant plays a role! Of course, the nucleus isirrelevant in determining the atomic size—it just provides a fixed center of electrostaticattraction. The relevant electronic parameters are the mass $m$ and the strength of attraction $e^{2} / 4 π ε_{0} .$ Togetherwith $h,$ these parameters determine a length.

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

Also, in 1911Richard Whiddington in Cambridge had found that to cause a substance havingatomic numberA to emit characteristic x-rays by bombarding it withelectrons, it was necessary to use electrons of speed approximatelyAx10⁶ meters per second. Any substance on being bombarded withsufficiently fast electrons emits a continuum of x-ray frequencies up to amaximum frequency $f$ given by $h f =$ kinetic energy of electron;plus somesharply defined lines—x-rays at a particular frequency, which does notchange as the electron speed is further increased. The frequency correspondingto these lines was found to increase with atomic number. Applying his lengthscale argument to the innermost ring of an atom, Bohr found that an electron inthat ring would have a speed proportional to the nuclear charge, and hence, atleast approximately, to the atomic number. Furthermore, the predicted speed inorbit was of the same order as that of Whiddington's electrons.

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

Nicolson: aClever Idea about a Wrong Model

Meanwhile, in Cambridgeone J. W. Nicolson was struggling to incorporate Planck's ideas in a model ofthe atom, in an attempt to understand some strange sets of spectral linesobserved in nebulae and in the sun's corona. He conceived a rather exotic (andquite 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 fromaring of electrons is a lot lessthat that from a single orbiting electron, the fields tend to cancel eachother.) Oscillations of electronsin thisringgave the spectra. Nicolson predicted the frequencies emitted by astraightforward classical analysis of these oscillation frequencies, in thespirit 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 hesuggested that the atom could only lose angular momentum in discrete amounts—presumably constant multiples of $h .$ Nicolson felt that, given the dimensionalityof Planck's constant, quantization of angular momentum was more plausible thanquantization of energy. Of course, for the simple harmonic oscillator theyamounted to the same thing, butnotfor any other system, in particular, not for a nuclear atom.

BohrReturns to Denmark

Bohr left Manchester inJuly 1912 and was married on the first of August. In the fall, he began work atthe 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. Heread Nicolson's work. As he wrote to Rutherford at the end of January 1913, heand Nicolson were really looking at different things—Nicolson was considering atoms in a very hotenvironment (like the sun's corona, or an electrical discharge tube) and thespectra gave information about how energy was emitted as the atom settled intoits ground state. Bohr himself was only interested in the state in which thesystem possessed the smallest amount of energy. He went on: "I do not atall deal with the question of calculation of the frequencies corresponding tothe visible part of the spectrum". At that time, Bohr thought of spectraas pretty but peripheral, having as little to do with basic physics as thecolors of a butterfly's wings had to do with basic biology.

BohrChanges his Mind about Spectra

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

Balmer's formula is:

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

for the sequence ofwavelengths of light emitted, with $n = 3, 4, 5, 6$ being in the visible, the lines used by Balmerin finding the formula. Hansen would doubtless have informed Bohr that the 1/4could be replaced by $1 / m^{2},$ with $m$ another integer. The constant appearing on theright hand side is called theRydberg constant, R_H = 109,737 cm^-1. This is the modern value—Balmer got it right to one part in 10,000, aboutthe limit of spectral measurements at the time. (Note: the subscript $H$ denoteshydrogen, and depends slightly on the finite proton mass. Taking that to beinfinity 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 clearto me." What he saw was that theset of allowedfrequencies (proportional to inversewavelengths) emitted by the hydrogen atom could all be expressed asdifferences.This immediately suggested to him a generalization of his idea of a"stationary state" lowest energy level, in which the electron did notradiate. There must be awhole sequence of these stationarystates, with radiation only taking place as the atom jumps from one to anotherof lower energy, emitting a single quantum of frequency $f$ such that

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

the difference betweenthe energies of the two states. Evidently, from the Balmer formula and itsextension to general integers $m, n,$ these allowed non-radiating orbits, thestationary states, could be labeled 1, 2, 3, ... ,n, ... andhad 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 energiesare of course negative, because these are bound states, and we count energyzero from where the two particles are infinitely far apart.

Bohr was very familiarwith the dynamics of simple circular orbits in an inverse square field. (Wespell out the details in the later sections below.)

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

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

It immediately followsthat theradius of the $n$ th orbit is proportional to $n^{2},$ and thespeed in that orbit isproportional to $1 / n .$

It then follows that the angularmomentum of the $n$ th orbitis just proportional to $n .$

Evidently, then theangular 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 allthe orbits, still to be determined.

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

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

Bohr Finds the Rydberg Constant without Doing an Experiment

Bohr gave a very cleverargument to find $K$ without doing any experiment. First, think about how the size of $K$ affectsthephysical properties of the hydrogen atom. How would theatom 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 havingangular momentum $h, 2 h, 3 h, \dots$ .

For $K = 10,$ the onlyallowed orbits would be those having angular momentum $10 h, 20 h, \dots$ Evidently, for $K = 10$ there will be a lot fewer spectral lines, andthe averagespacing between them will begreaterthatfor $K = 1.$ Thiswill increase the energy differences, and also the spectral line frequencies.

The Correspondence Principle

Next, Bohr imagined areallyimmense hydrogen atom, an electron going around aproton in a circle of one meter radius, say. This would have to be done in thedepths of space, but really this is just a thought experiment in the spirit ofEinstein. The point is that for this very large atom, the electron is movingrather slowly over a distance scale we are familiar with. We know from manyexperiments that charges moving at these slow speeds over ordinary (human size)distances emit radiation according to Maxwell's equations. Or, more simply, ifit's going round the circle at frequency $f$ revolutions per second, it will be emittingradiation at that frequency $f$ —because its electric field, as seen from somefixed 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 meterproton-electron separation.

Bohr knew that if histheory was correct, it would have to give a frequency for this large atom thesame as that given by classical physics (and well-verified). This became known as theCorrespondence Principle, and is a useful constraint on anymicroscopic theories that have classical (macroscopic) limits. They had betteragree with the well-established classical results.

Bohr's theory stipulatesthat the angular momentum quantizationcondition must be true forall circular orbits of the electron aroundthe proton, including this very large atom. Furthermore, the radiation emitted must stillbe given by the difference in energies of neighboring orbits,

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

But $E_{n + 1} - E_{n},$ the energyspacing betweenneighboring orbits,depends on $K .$

Therefore, Bohrconcluded $K$ is fixedby requiring that thefrequency of radiation emitted by a really large atom be correctly given byordinary common sense—that is, the frequency of the radiation (theenergy level separation divided by $h$ must bethe same as the orbital frequency of the electron, the number ofcycles a second.

That is to say, for alarge orbit we must have

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

Herev isthe speed of the electron in the orbit, and the orbit radius isr.

Finding the Energy Levels in Terms ofK

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 inan 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 quantizationequation above, we know $v_{n} .$

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

It's slightly easier if we square both sides of thequantization 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 oforder $1 / n$ .

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

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 thespectroscopic experiment, but it's satisfying to see a number come from anabstract argument!

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