Chapter 35Bohr Theory of Hydrogen

CHAPTER 35 BOHR THEORY OFHYDROGEN

The hydrogen atom played a special role in the historyof physics by providing the key that unlocked the newmechanics that replaced Newtonian mechanics. Itstarted with Johann Balmer's discovery in 1884 of amathematical formula for the wavelengths of some ofthe spectral lines emitted by hydrogen. The simplicityof the formula suggested that some understandablemechanisms were producing these lines.

The next step was Rutherford's discovery of the atomicnucleus in 1912. After that, one knew the basicstructure of atoms—a positive nucleus surrounded bynegative electrons. Within a year Neils Bohr had amodel of the hydrogen atom that "explained" thespectral lines. Bohr introduced a new concept, theenergy level. The electron in hydrogen had certainallowed energy levels, and the sharp spectral lineswere emitted when the electron jumped from oneenergy level to another. To explain the energy levels,Bohr developed a model in which the electron hadcertain allowed orbits and the jump between energylevels corresponded to the electron moving from oneallowed orbit to another.

Bohr's allowed orbits followed from Newtonian me-chanics and the Coulomb force law, with one small butcrucial modification of Newtonian mechanics. Theangular momentum of the electron could not vary

continuously, it had to have special values, be quan-tized in units of Planck's constant divided by 2π , h/2π .In Bohr's theory, the different allowed orbits corre-sponded to orbits with different allowed values ofangular momentum.

Again we see Planck's constant appearing at just thepoint where Newtonian mechanics is breaking down.There is no way one can explain from Newtonianmechanics why the electrons in the hydrogen atomcould have only specific quantized values of angularmomentum. While Bohr's model of hydrogen repre-sented only a slight modification of Newtonian me-chanics, it represented a major philosophical shift.Newtonian mechanics could no longer be consideredthe basic theory governing the behavior of particlesand matter. Something had to replace Newtonianmechanics, but from the time of Bohr's theory in 1913until 1924, no one knew what the new theory would be.

In 1924, a French graduate student, Louis de Broglie,made a crucial suggestion that was the key that led tothe new mechanics. This suggestion was quicklyfollowed up by Schrödinger and Heisenberg who de-veloped the new mechanics called quantum mechan-ics. In this chapter our focus will be on the develop-ments leading to de Broglie's idea.

35-2 Bohr Theory of Hydrogen

For an electron in a circular orbit, predicting the motionis quite easy. If an electron is in an orbit of radius r,moving at a speed v, then its acceleration a is directedtoward the center of the circle and has a magnitude

a = v2

r (2)

Using Equation 1 for the electric force and Equation 2for the acceleration, and noting that the force is in thesame direction as the acceleration, as indicated inFigure (2), Newton's second law gives

F = m a

e2

r2 = mv2

r (3)

One factor of r cancels and we can immediately solvefor the electron's speed v to get v2 = e2/mr, or

velectron = emr

(4)

The period of the electron's orbit should be the distance 2πr travelled, divided by the speed v, or 2πr/v sec-

onds per cycle, and the frequency should be the inverseof that, or v/2πr cycles per second. Using Equation 4for v, we get

frequency ofelectron in orbit

= v2πr

= e2πr mr

(5)

According to Maxwell's theory, this should also be thefrequency of the radiation emitted by the electron.

THE CLASSICAL HYDROGEN ATOMWith Rutherford's discovery of the atomic nucleus, itbecame clear that atoms consisted of a positivelycharged nucleus surrounded by negatively chargedelectrons that were held to the nucleus by an electricforce. The simplest atom would be hydrogen consist-ing of one proton and one electron held together by aCoulomb force of magnitude

Fe = e2

r2 p

r

FeFe e(1)

(For simplicity we will use CGS units in describing thehydrogen atom. We do not need the engineering units,and we avoid the complicating factor of 1/4πε0 in theelectric force formula.) As shown in Equation 1, both theproton and the electron attract each other, but since theproton is 1836 times more massive than the electron, theproton should sit nearly at rest while the electron orbitsaround it.

Thus the hydrogen atom is such a simple system, withknown masses and known forces, that it should be astraightforward matter to make detailed predictions aboutthe nature of the atom. We could use the orbit programof Chapter 8, replacing the gravitational force GMm/r2

by e2/r2 . We would predict that the electron moved inan elliptical orbit about the proton, obeying all of Kepler'slaws for orbital motion.

There is one important point we would have to take intoaccount in our analysis of the hydrogen atom that we didnot have to worry about in our study of satellite motion.The electron is a charged particle, and accelerated chargedparticles radiate electromagnetic waves. Suppose, forexample, that the electron were in a circular orbit movingat an angular velocity ω as shown in Figure (1a). If wewere looking at the orbit from the side, as shown in Figure(1b), we would see an electron oscillating up and downwith a velocity given by v = v0sin ωt .

In our discussion of radio antennas in Chapter 32, we sawthat radio waves could be produced by moving electronsup and down in an antenna wire. If electrons oscillated upand down at a frequency ω , they produced radio wavesof the same frequency. Thus it is a prediction of Maxwell'sequations that the electron in the hydrogen atom shouldemit electromagnetic radiation, and the frequency of theradiation should be the frequency at which the electronorbits the proton.

Figure 1The side view of circular motionis an up and down oscillation.

p

v0 v = v sin(ωt)

a) electron in circular orbit

b) side view of circular orbit

0e e

p

35-3

Electromagnetic radiation carries energy. Thus, to seewhat effect this has on the electron’s orbit, let us lookat the formula for the energy of an orbiting electron.

From Equation 3 we can immediately solve for theelectron's kinetic energy. The result is

12

mv2 = e2

2r

electronkineticenergy

(6)

The electron also has electric potential energy just as anearth satellite had gravitational potential energy. Theformula for the gravitational potential energy of asatellite was

potential energyof an earth satellite

= – GMmr (10-50a)

where M and m are the masses of the earth and thesatellite respectively. This is the result we used inChapter 8 to test for conservation of energy (Equations8-29 and 8-31) and in Chapter 10 where we calculatedthe potential energy (Equations 10-50a and 10-51).The minus sign indicated that the gravitational force isattractive, that the satellite starts with zero potentialenergy when r = ∞ and loses potential energy as it fallsin toward the earth.

We can convert the formula for gravitational potentialenergy to a formula for electrical potential energy bycomparing formulas for the gravitational and electricforces on the two orbiting objects. The forces are

Fgravity = GMmr2 ; Felectric = e2

r2

Since both are 1/r2 forces, we can go from the gravi-tational to the electric force formula by replacing the

constant GMm by e2. Making this same substitutionin the potential energy formula gives

PE = – e2

r

electrical potential energyof the electron in thehydrogenatom

(7)

Again the potential energy is zero when the particlesare infinitely far apart, and the electron loses potentialenergy as it falls toward the proton. (We used this resultin the analysis of the binding energy of the hydrogenmolecule ion, explicitly in Equation 18-15.)

The formula for the total energy Etotal of the electron inhydrogen should be the sum of the kinetic energy,Equation 6, and the potential energy, Equation 7.

Etotal = kinetic

energy +potentialenergy

= e2

2r – e2

r

Etotal = – e2

2rtotal energyof electron (8)

The significance of the minus (–) sign is that theelectron is bound. Energy is required to pull theelectron out, to ionize the atom. For an electron toescape, its total energy must be brought up to zero.

We are now ready to look at the predictions that followfrom Equations 5 and 8. As the electron radiates lightit must lose energy and its total energy must becomemore negative. From Equation 8 we see that for theelectron's energy to become more negative, the radiusr must become smaller. Then Equation 5 tells us thatas the radius becomes smaller, the frequency of theradiation increases. We are lead to the picture of theelectron spiraling in toward the proton, radiating evenhigher frequency light. There is nothing to stop theprocess until the electron crashes into the proton. It isan unambiguous prediction of Newtonian mechanicsand Maxwell's equations that the hydrogen atom isunstable. It should emit a continuously increasingfrequency of light until it collapses.

p

v

a

r

Fe

e

Figure 2For a circular orbit, both the acceleration a and theforce F point toward the center of the circle. Thus wecan equate the magnitudes of F and ma.

35-4 Bohr Theory of Hydrogen

Energy LevelsBy 1913, when Neils Bohr was trying to understand thebehavior of the electron in hydrogen, it was no surprisethat Maxwell's equations did not work at an atomicscale. To explain blackbody radiation and the photo-electric effect, Planck and Einstein were led to thepicture that light consists of photons rather thanMaxwell's waves of electric and magnetic force.

To construct a theory of hydrogen, Bohr knew thefollowing fact. Hydrogen gas at room temperatureemits no light. To get radiation, it has to be heated torather high temperatures. Then you get distinct spectrallines rather than the continuous radiation spectrumexpected classically. The visible spectral lines are the

Hα , Hβ and Hγ lines we saw in the hydrogen spec-trum experiment. These and many infra red lines wesaw in the spectrum of the hydrogen star, Figure (33-28) reproduced below, make up the Balmer series oflines. Something must be going on inside the hydrogenatom to produce these sharp spectral lines.

Viewing the light radiated by hydrogen in terms ofEinstein's photon picture, we see that the hydrogenatom emits photons with certain precise energies. Asan exercise in the last chapter you were asked tocalculate, in eV, the energies of the photons in the Hα ,

Hβ and Hγ spectral lines. The answers are

EHα= 1.89 eV

EHβ= 2.55 eV

EHγ= 2.86 eV (9)

The question is, why does the electron in hydrogen emitonly certain energy photons? The answer is Bohr'smain contribution to physics. Bohr assumed that theelectron had, for some reason, only certain allowedenergies in the hydrogen atom. He called these allowedenergy levels. When an electron jumped from oneenergy level to another, it emitted a photon whoseenergy was equal to the difference in the energy of thetwo levels. The red 1.89 eV photon, for example, wasradiated when the electron fell from one energy level toanother level 1.89 eV lower. There was a bottom,lowest energy level below which the electron could notfall. In cold hydrogen, all the electrons were in thebottom energy level and therefore emitted no light.

When the hydrogen atom is viewed in terms of Bohr’senergy levels, the whole picture becomes extremelysimple. The lowest energy level is at -13.6 eV. This isthe total energy of the electron in any cold hydrogenatom. It requires 13.6 eV to ionize hydrogen to rip anelectron out.

Figure 33-28Spectrum of a hydrogen star

–13.6

–3.40

–1.51

–.850–.544 0

n = 1

n = 2

n = 3

n = 4n = 5

Hα Hβ Hγ

3.65 10 3.70 10 3.75 10 3.80 10

H9 H10 H11 H12 H13 H14 H15 H20 H30 H40

wavelength 3.85 10 –5 –5 –5 –5 –5

Figure 3Energy leveldiagram for thehydrogen atom.All the energylevels are given bythe simple formula

En = – 13.6/n2 eV.All Balmer serieslines result fromjumps down to then = 2 level. The 3jumps shown giverise to the threevisible hydrogenlines.

35-5

The first energy level above the bottom is at –3.40 eVwhich turns out to be (–13.6/4) eV. The next level is at–1.51 eV which is (–13.6/9) eV. All of the energylevels needed to explain every spectral line emitted byhydrogen are given by the formula

En = – 13.6 eVn2 (10)

where n takes on the integer values 1, 2, 3, .... Theseenergy levels are shown in Figure (3).

Exercise 1

Use Equation 10 to calculate the lowest 5 energy levelsand compare your answer with Figure 3.

Let us see explicitly how Bohr's energy level diagramexplains the spectrum of light emitted by hydrogen. If,for example, an electron fell from the n=3 to the n=2level, the amount of energy E3–2 it would lose andtherefore the energy it would radiate would be

E3–2 = E3 – E2

= – 1.51 eV – ( – 3.40 eV)

= 1.89 eV

=energy lost in fallingfrom n = 3 to n = 2 level

(11)

which is the energy of the red photons in the Hα line.

Exercise 2

Show that the Hβ and Hγ lines correspond to jumps tothe n = 2 level from the n = 4 and the n = 5 levelsrespectively.

From Exercise 2 we see that the first three lines in theBalmer series result from the electron falling from thethird, fourth and fifth levels down to the second level,as indicated by the arrows in Figure (3).

All of the lines in the Balmer result from jumps downto the second energy level. For historical interest, let ussee how Balmer's formula for the wavelengths in thisseries follows from Bohr's formula for the energylevels. For Balmer's formula, the lines we have beencalling Hα , Hβ and Hγ are H3 , H4 , H5 . An arbitraryline in the series is denoted by Hn , where n takes on thevalues starting from 3 on up. The Balmer formula forthe wavelength of the Hn line is from Equation 33-6

λn = 3.65 × 10– 5cm × n2

n2 – 4(33-6)

Referring to Bohr's energy level diagram in Figure (3),consider a drop from the nth energy level to the second.The energy lost by the electron is ( En – E2) which hasthe value

En – E 2 = 13.6 eV

n2 – 13.6 eV22

energy lost byelectron goingfrom nth tosecond level

This must be the energy E Hn carried out by thephoton in the Hn spectral line. Thus

E Hn = 13.6 eV

14

–1

n2

= 13.6 eVn2 – 4

4n2

(12)

We now use the formula

λ = 12.4 × 10– 5cm ⋅ eVEphoton in eV

(34-8)

relating the photon's energy to its wavelength. UsingEquation 12 for the photon energy gives

λn = 12.4 × 10– 5cm ⋅ eV

13.6 eV4n2

n2 – 4

λn = 3.65 × 10– 5cm n2

n2 – 4which is Balmer's formula.

35-6 Bohr Theory of Hydrogen

It does not take great intuition to suspect that there areother series of spectral lines beyond the Balmer series.The photons emitted when the electron falls down tothe lowest level, down to -13.6 eV as indicated inFigure (4), form what is called the Lyman series. In thisseries the least energy photon, resulting from a fall from-3.40 eV down to -13.6 eV, has an energy of 10.2 eV,well out in the ultraviolet part of the spectrum. All theother photons in the Lyman series have more energy,and therefore are farther out in the ultraviolet.

It is interesting to note that when you heat hydrogen andsee a Balmer series photon like Hα , Hβ or Hγ ,eventually a 10.2 eV Lyman series photon must beemitted before the hydrogen can get back down to itsground state. With telescopes on earth we see manyhydrogen stars radiating Balmer series lines. We do notsee the Lyman series lines because these ultravioletphotons do not make it down through the earth'satmosphere. But the Lyman series lines are all visibleusing orbiting telescopes like the Ultraviolet Explorerand the Hubble telescope.

Another series, all of whose lines lie in the infra red, isthe Paschen series, representing jumps down to then = 3 energy level at -1.55 eV, as indicated in Figure (5).There are other infra red series, representing jumpsdown to the n = 4 level, n = 5 level, etc. There are manyseries, each containing many spectral lines. And allthese lines are explained by Bohr's conjecture that thehydrogen atom has certain allowed energy levels, allgiven by the simple formula En = (– 13.6/n2) eV.This one simple formula explains a huge amount ofexperimental data on the spectrum of hydrogen.

Exercise 3Calculate the energies (in eV) and wavelengths of the 5longest wavelength lines in

(a) the Lyman series

(b) the Paschen series

On a Bohr energy level diagram show the electronjumps corresponding to each line.

Exercise 4

In Figure (33-28), repeated 2 pages back, we showedthe spectrum of light emitted by a hydrogen star. Thelines get closer and closer together as we get to H40 andjust beyond. Explain why the lines get closer togetherand calculate the limiting wavelength.

–13.6

–3.40

–1.51

–.850–.544 0

n = 1

n = 2

n = 3

n = 4n = 5

Figure 4The Lyman seriesconsists of all jumpsdown to the –13.6eVlevel. (Since this is asfar down as theelectron can go, thislevel is called the“ground state”.)

Figure 5The Paschen seriesconsists of all jumpsdown to the n = 3level. These are all inthe infra red. –1.51

–.850–.544

0

n = 3

n = 4n = 5

–.378 n = 6–.278 n = 7

35-7

Lyman seriesBalm

er series

Paschen

series

r2

r1

r3

THE BOHR MODELWhere do Bohr's energy levels come from? Certainlynot from Newtonian mechanics. There is no excuse inNewtonian mechanics for a set of allowed energylevels. But did Newtonian mechanics have to berejected altogether? Planck was able to explain theblackbody radiation formula by patching up classicalphysics, by assuming that, for some reason, light wasemitted and absorbed in quanta whose energy wasproportional to the light's frequency. The reason whyPlanck's trick worked was understood later, withEinstein's proposal that light actually consisted ofparticles whose energy was proportional to frequency.Blackbody radiation had to be emitted and absorbed inquanta because light itself was made up of these quanta.

By 1913 it had become respectable, frustrating per-haps, but respectable to modify classical physics inorder to explain atomic phenomena. The hope was thata deeper theory would come along and naturally ex-plain the modifications.

What kind of a theory do we construct to explain theallowed energy levels in hydrogen? In the classicalpicture we have a miniature solar system with theproton at the center and the electron in orbit. This canbe simplified by restricting the discussion to circularorbits. From our earlier work with the classical model

of hydrogen, we saw that an electron in an orbit ofradius r had a total energy E(r) given by

E(r) = – e2

2r

total energy ofan electron ina circular orbitof radius r

(8 repeated)

If the electron can have only certain allowed energies En = –13.6/n2 eV, then if Equation (8) holds, the

electron orbits can have only certain allowed orbits ofradius rn given by

En = – e2

2rn(13)

The rn are the radii of the famous Bohr orbits. Thisleads to the rather peculiar picture that the electron canexist in only certain allowed orbits, and when theelectron jumps from one allowed orbit to another, itemits a photon whose energy is equal to the differencein energy between the two orbits. This model isindicated schematically in Figure (6).

Exercise 5

From Equation 13 and the fact that E1 = – 13.6 eV,calculate the radius of the first Bohr orbit r1. [Hint: firstconvert eV to ergs.] This is known as the Bohr radiusand is in fact a good measure of the actual radius of acold hydrogen atom. [The answer is

r1 = .529 × 10– 8cm= .529A° .] Then show that rn = n2r1 .

Figure 6The Bohr orbits are determined byequating the allowed energy

En = – 13.6 n2– 13.6 n2 to the energy En = – e2 2rn– e2 2rnfor an electron in an orbit of radius rn.The Lyman series represents all jumpsdown to the smallest orbit, the Balmerseries to the second orbit, the Paschenseries to the third orbit, etc. (The radii inthis diagram are not to scale, the radii rnincrease in size as n2, as you can easilyshow by equating the two values for En.)

35-8 Bohr Theory of Hydrogen

Angular Momentum in the Bohr ModelNothing in Newtonian mechanics gives the slightesthint as to why the electron in hydrogen should haveonly certain allowed orbits. In the classical picturethere is nothing special about these particular radii.

But ever since the time of Max Planck, there was aspecial unit of angular momentum, the amount givenby Planck's constant h. Since Planck's constant keepsappearing whenever Newtonian mechanics fails, andsince Planck's constant has the dimensions of angularmomentum, perhaps there was something special aboutthe electron's angular momentum when it was in one ofthe allowed orbits.

We can check this idea by re expressing the electron'stotal energy not in terms of the orbital radius r, but interms of its angular momentum L.

We first need the formula for the electron's angularmomentum when in a circular orbit of radius r. Backin Equation 4, we found that the speed v of the electronwas given by

v = emr (4 repeated)

Multiplying this through by m gives us the electron'slinear momentum mv

mv = memr = e m

r (14)

The electron's angular momentum about the center ofthe circle is its linear momentum mv times the leverarm r, as indicated in the sketch of Figure (7). The resultis

L = mv r = e mr r

= e mr(15)

where we used Equation 14 for mv.

The next step is to express r in terms of the angularmomentum L. Squaring Equation 13 gives

L2 = e2mror

r = L2

e2m(16)

Finally we can eliminate the variable r in favor of theangular momentum L in our formula for the electron'stotal energy. We get

total energyof the electron E = – e2

2r

= – e2

2 L2 e2mL2 e2m

= – e2

2e2mL2

= – e4m2L2

(17)

In the formula – e4m/2L2 for the electron's energy,only the angular momentum L changes from one orbitto another. If the energy of the nth orbit is En , thenthere must be a corresponding value Ln for the angularmomentum of the orbit. Thus we should write

En = – e4m2Ln

2(18)

v

L = mvr

r

m

Figure 7Angular momentum of a particlemoving in a circle of radius r.

35-9

At this point, Bohr had the clue as to how to modifyNewtonian mechanics in order to get his allowedenergy levels. Suppose that angular momentum isquantized in units of some quantity we will call L0. Inthe smallest orbit, suppose it has one unit, i.e.,

L1 = 1 × L0. In the second orbit assume it has twice asmuch angular momentum, L2 = 2 L0 . In the nth orbitit would have n units

Ln = nL0

quantizationof angularmomentum

(19)

Substituting Equation 19 into Equation 18 gives

En = – e4m

2L20

1n2 (20)

as the total energy of an electron with n units of angularmomentum. Comparing Equation 20 with Bohr'senergy level formula

En = –13.6 eV 1n2 (10 repeated)

we see that we can explain the energy levels byassuming that the electron in the nth energy level has nunits of quantized angular momentum L0. We can alsoevaluate the size of L0 by equating the constant factorsin Equations 10 and 20. We get

e4m2L0

2 = 13.6 eV (21)

Converting 13.6 eV to ergs, and solving for L0 gives

e4m2L0

2 = 13.6 eV × 1.6 × 10– 12ergseV

With e = 4.8 × 10– 10esu and m = .911 × 10– 27gmin CGS units, we get

L0 = 1.05 × 10– 27gm cm2

sec (22)

which turns out to be Planck's constant divided by 2π .

L0 = h

2π =6.63 × 10– 27gm cm/sec

2π

= 1.05 × 10– 27gm cmsec

This quantity, Planck's constant divided by 2π , ap-pears so often in physics and chemistry that it is giventhe special name “h bar” and is written h

h ≡ h2π "h bar" (23)

Using h for L0 in the formula for En , we get Bohr'sformula

En = – e4m

2h21n2 (24)

where e4m/2h2, expressed in electron volts, is 13.6 eV.This quantity is known as the Rydberg constant.[Remember that we are using CGS units, where e is inesu, m in grams, and h is erg-sec.]

Exercise 6Use Equation 21 to evaluate L0 .

Exercise 7What is the formula for the first Bohr radius in terms of theelectron mass m, charge e, and Planck's constant h.Evaluate your result and show that

r1 = .51 × 10– 8cm= .51A° . (Answer: r1 = h2/e2m.)

Exercise 8Starting from Newtonian mechanics and the Coulombforce law F = e2/r2, write out a clear and concise deriva-tion of the formula

En = – e4m2h2

1n2

Explain the crucial steps of the derivation.

A day or so later, on an empty piece of paper and a cleandesk, see if you can repeat the derivation withoutlooking at notes. When you can, you have a secureknowledge of the Bohr theory.

35-10 Bohr Theory of Hydrogen

Exercise 9An ionized helium atom consists of a single electronorbiting a nucleus containing two protons as shown inFigure (8). Thus the Coulomb force on the electron hasa magnitude

Fe =

e 2e

r2 = 2e2

r2

–e

2e

a) Using Newtonian mechanics, calculate the totalenergy of the electron. (Your answer should be – e2/r .Note that the r is not squared.)

b) Express this energy in terms of the electron's angularmomentum L. (First calculate L in terms of r, solve forr, and substitute as we did in going from Equations 16to 17.)

c) Find the formula for the energy levels of the electronin ionized helium, assuming that the electron's angularmomentum is quantized in units of h.

d) Figure out whether ionized helium emits any visiblespectral lines (lines with photon energies between 1.8eV and 3.1 eV.) How many visible lines are there andwhat are their wavelengths?)

Exercise 10

You can handle all single electron atoms in one calcu-lation by assuming that there are z protons in thenucleus. (z = 1 for hydrogen, z = 2 for ionized helium,z = 3 for doubly ionized lithium, etc.) Repeat parts a), b),and c) of Exercise 9 for a single electron atom with zprotons in the nucleus. (There is no simple formula formulti electron atoms because of the repulsive forcebetween the electrons.)

DE BROGLIE'S HYPOTHESISDespite its spectacular success describing the spectraof hydrogen and other one-electron atoms, Bohr'stheory represented more of a problem than a solution.It worked only for one electron atoms, and it pointed toan explicit failure of Newtonian mechanics. The ideaof correcting Newtonian mechanics by requiring theangular momentum of the electron be quantized inunits of h , while successful, represented a bandaidtreatment. It simply covered a deeper wound in thetheory. For two centuries Newtonian mechanics hadrepresented a complete, consistent scheme, applicablewithout exception. Special relativity did not harm theintegrity of Newtonian mechanics—relativistic New-tonian mechanics is a consistent theory compatiblewith the principle of relativity. Even general relativity,with its concepts of curved space, left Newtonianmechanics intact, and consistent, in a slightly alteredform.

The framework of Newtonian mechanics could not bealtered to include the concept of quantized angularmomentum. Bohr, Sommerfield, and others triedduring the decade following the introduction of Bohr'smodel, but there was little success.

In Paris, in 1923, a graduate student Louis de Broglie,had an idea. He noted that light had a wave nature, seenin the 2-slit experiment and Maxwell's theory, and aparticle nature seen in Einstein's explanation of thephotoelectric effect. Physicists could not explain howlight could behave as a particle in some experiments,and a wave in others. This problem seemed so incon-gruous that it was put on the back burner, more or lessignored for nearly 20 years.

De Broglie's idea was that, if light can have both aparticle and a wave nature, perhaps electrons can too!Perhaps the quantization of the angular momentum ofan electron in the hydrogen atom was due to the wavenature of the electron.

The main question de Broglie had to answer was howdo you determine the wavelength of an electron wave?

Figure 8Ionized helium has anucleus with two protons,surrounded by one electron.

35-11

An analogy with photons might help. There is, how-ever, a significant difference between electrons andphotons. Electrons have a rest mass energy and pho-tons do not, thus there can be no direct analogy betweenthe total energies of the two particles. But both particleshave mass and carry linear momentum, and the amountof momentum can vary from zero on up for bothparticles. Thus photons and electrons could havesimilar formulas for linear momentum.

Back in Equation 34-13 we saw that the linear momen-tum p of a photon was related to its wavelength λ by thesimple equation

λ = hp

de Brogliewavelength (34-13)

De Broglie assumed that this same relationship alsoapplied to electrons. An electron with a linear momen-tum p would have a wavelength λ = h/p . This is nowcalled the de Broglie wavelength. This relationshipapplies not only to photons and electrons, but as far aswe know, to all particles!

With a formula for the electron wavelength, de Brogliewas able to construct a simple model explaining thequantization of angular momentum in the hydrogenatom. In de Broglie's model, one pictures an electronwave chasing itself around a circle in the hydrogenatom. If the circumference of the circle, 2πr did nothave an exact integral number of wavelengths, then thewave, after going around many times, would eventu-ally cancel itself out as illustrated in Figure (9).

But if the circumference of the circle were an exactintegral number of wavelengths as illustrated in Figure(10), there would be no cancellation. This wouldtherefore be one of Bohr's allowed orbits shown inFigure (6).

Suppose (n) wavelengths fit around a particular circleof radius rn . Then we have

nλ = 2πrn (25)

Using the de Broglie formula λ = h/p for the electronwavelength, we get

n hp = 2πrn (26)

Multiplying both sides by p and dividing through by 2π gives

n h2π = prn (27)

Now h/2π is just h , and prn is the angular momentum Ln (momentum times lever arm) of the electron. Thus

Equation 27 gives

nh = prn = Ln (28)

Equation 28 tells us that for the allowed orbits, theorbits in which the electron wave does not cancel, theangular momentum comes in integer amounts of theangular momentum h . The quantization of angularmomentum is thus due to the wave nature of theelectron, a concept completely foreign to Newtonianmechanics.

r

Figure 9De Broglie picture of anelectron wave cancellingitself out.

Figure 10If the circumference of the orbit isan integer number of wavelengths,the electron wave will go aroundwithout any cancellation.

Figure 10a--MovieThe standing waves on acircular metal band nicelyillustrate de Broglie’s waves

35-12 Bohr Theory of Hydrogen

When a graduate student does a thesis project, typicallythe student does a lot of work under the supervision ofa thesis advisor, and comes up with some new, hope-fully verifiable, results. What do you do with a studentthat comes up with a strange idea, completely unveri-fied, that can be explained in a few pages of algebra?Einstein happened to be passing through Paris in thesummer of 1924 and was asked if de Broglie's thesisshould be accepted. Although doubtful himself abouta wave nature of the electron, Einstein recommendedthat the thesis be accepted, for de Broglie just might beright.

In 1925, two physicists at Bell Telephone Laboratories,C. J. Davisson and L. H. Germer were studying thesurface of nickel by scattering electrons from thesurface. The point of the research was to learn moreabout metal surfaces in order to improve the quality ofswitches used in telephone communication. Afterheating the metal target to remove an oxide layer thataccumulated following a break in the vacuum line, theydiscovered that the electrons scattered differently. The

metal had crystallized during the heating, and thepeculiar scattering had occurred as a result of thecrystallization. Davisson and Germer then prepared atarget consisting of a single crystal, and studied thepeculiar scattering phenomena extensively. Their ap-paratus is illustrated schematically in Figure (11), andtheir experimental results are shown in Figure (12). Fortheir experiment, there was a marked peak in thescattering when the detector was located at an angle of50° from the incident beam.

Davisson presented these results at a meeting in Lon-don in the summer of 1927. At that time there was aconsiderable discussion about de Broglie's hypothesisthat electrons have a wave nature. Hearing of this idea,Davisson recognized the reason for the scattering peak.The atoms of the crystal were diffracting electronwaves. The enhanced scattering at 50° was a diffrac-tion peak, a maximum similar to the reflected maximawe saw back in Figure (33-19) when light goes througha diffraction grating. Davisson had the experimentalevidence that de Broglie's idea about electron waveswas correct after all.

electron gun

detector

θ

nickel crystal

electron beam

θ = 50°

Figure 11Scattering electrons from thesurface of a nickel crystal.

Figure 12Plot of intensity vs. angle for electrons scattered by anickel crystal, as measured by Davisson and Germer.The peak in intensity at 50° was a diffraction peaklike the ones produced by diffraction gratings. (Theintensity is proportional to the distance out from theorigin.)

Reflectedmaximum

transmittedmaximum

Figure 33-19Laser beamimpinging ona diffractiongrating.

35-13

IndexSymbols13.6 eV, hydrogen spectrum 35-4

AAllowed orbits, Bohr theory 35-1Angular momentum

Bohr model 35-1, 35-8Planck's constant 35-8

AtomsClassical hydrogen atom 35-2

BBalmer series

Energy level diagram for 35-6Formula from Bohr theory 35-5Hydrogen spectrum 35-4

Bell Telephone Lab, electron waves 35-12Bohr model

Allowed orbits 35-1Angular momentum 35-1, 35-8Chapter on 35-1De Broglie explanation 35-1Derivation of 35-8Energy levels 35-4Planck's constant 35-1, 35-8Quantum mechanics 35-1Rydberg constant 35-9

Bohr orbits, radii of 35-7

CCGS units

Classical hydrogen atom 35-2Circular orbit, classical hydrogen atom 35-2Classical hydrogen atom 35-2Coulomb's law

Classical hydrogen atom 35-2

DDavisson & Germer, electron waves 35-12De Broglie

Electron waves 35-11Formula for momentum 35-11Hypothesis 35-10Key to quantum mechanics 35-1Wavelength, formula for 35-11Waves, movie of standing wave model 35-11

EElectromagnetic radiation

Energy radiated by classical H atom 35-3Electron

In classical hydrogen atom 35-2Electron scattering

First experiment on wave nature 35-12Electron waves

Davisson & Germer experiment 35-12De Broglie picture 35-11Scattering of 35-12

EnergyElectric potential energy

In classical hydrogen atom 35-3Energy level 35-1Kinetic energy

Bohr model of hydrogen 35-3Classical hydrogen atom 35-3

Total energyClassical H atom 35-3

Energy level diagramBalmer series 35-6Bohr theory 35-4Lyman series 35-6Paschen series 35-6

FForce

Electric forceClassical hydrogen atom 35-2

Hh bar, Planck's constant 35-9Hydrogen atom

Bohr theory 35-1Classical 35-2

Hydrogen atom, classicalFailure of Newtonian mechanics 35-3

Hydrogen spectrumBalmer series 35-4Lyman series 35-6Of star 35-4Paschen series 35-6

IInfrared light

Paschen series, hydrogen spectra 35-6

KKinetic energy

Bohr model of hydrogen 35-3Classical hydrogen atom 35-3

LLight

Hydrogen spectrumBalmer formula 35-5

Spectral lines, hydrogenBohr theory 35-4

Lyman series, energy level diagram 35-6

35-14 Bohr Theory of Hydrogen

MMaxwell's equations

Failure ofIn classical hydrogen atom 35-2

MechanicsNewtonian

Classical H atom 35-3Momentum

De Broglie formula for momentum 35-11Movie

Standing De Broglie like waves 35-11

NNewtonian mechanics

Classical H atom 35-3Failure of

In the classical hydrogen atom 35-3Nucleus

Discovery of, Rutherford 35-1

OOrbits

Bohr, radii of 35-7Classical hydrogen atom 35-2

PParticle-wave nature

De Broglie picture 35-10Of electrons

Davisson and Germer experiment 35-12De Broglie picture 35-10

Paschen seriesEnergy level diagram 35-6Hydrogen spectra 35-6

PhotonHydrogen spectrum 35-5

Planck's constantAngular momentum, Bohr model 35-8Bohr theory 35-1In de Broglie wavelength formula 35-11

Potential energyElectric potential energy

In classical hydrogen atom 35-3

QQuantized angular momentum

In Bohr theory 35-9In de Broglie's hypothesis 35-10

Quantum mechanicsBohr theory of hydrogen 35-1

RRadiation

Radiated energy and the classical H atom 35-3Radio waves

Predicted from the classical hydrogen atom 35-2

Rutherford and the nucleus 35-1Rydberg constant, in Bohr theory 35-9

SSatellite motion

Classical hydrogen atom 35-2Scattering of waves

Davisson-Germer experiment 35-12Spectral lines

HydrogenBohr theory of 35-4

SpectrumHydrogen

Bohr theory of 35-4Lyman series, ultraviolet 35-6Paschen series, infrared 35-6

Hydrogen star 35-4Standing waves

De Broglie wavesMovie 35-11

StarHydrogen spectrum of 35-4

TTotal energy

Classical hydrogen atom 35-3

UUnit of angular momentum

In Bohr theory 35-9

WWave

De Broglie, standing wave movie 35-11Electron waves, de Broglie picture 35-11

WavelengthDe Broglie 35-11

Xx-Ch35

Exercise 1 35-5Exercise 2 35-5Exercise 3 35-6Exercise 4 35-6Exercise 5 35-7Exercise 6 35-9Exercise 7 35-9Exercise 8 35-9Exercise 9 35-10Exercise 10 35-10