Ch 35 Bohr Theory of Hydrogen

8/12/2019 Ch 35 Bohr Theory of Hydrogen

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Chapter 35

Bohr Theory of Hydrogen

CHAPTER 35 BOHR THEORY OF

HYDROGEN

The hydrogen atom played a special role in the historyof physics by providing the key that unlocked the new

mechanics that replaced Newtonian mechanics. It

started with Johann Balmer's discovery in 1884 of a

mathematical formula for the wavelengths of some of

the spectral lines emitted by hydrogen. The simplicity

of the formula suggested that some understandable

mechanisms were producing these lines.

The next step was Rutherford's discovery of the atomic

nucleus in 1912. After that, one knew the basic

structure of atomsa positive nucleus surrounded by

negative electrons. Within a year Neils Bohr had a

model of the hydrogen atom that "explained" the

spectral lines. Bohr introduced a new concept, the

energy level. The electron in hydrogen had certain

allowed energy levels, and the sharp spectral lines

were emitted when the electron jumped from one

energy level to another. To explain the energy levels,

Bohr developed a model in which the electron had

certain allowed orbits and the jump between energy

levels corresponded to the electron moving from one

allowed orbit to another.

Bohr's allowed orbits followed from Newtonian me-

chanics and the Coulomb force law, with one small but

crucial modification of Newtonian mechanics. The

angular 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 2, h/2 .In Bohr's theory, the different allowed orbits corre-

sponded to orbits with different allowed values of

angular momentum.

Again we see Planck's constant appearing at just the

point where Newtonian mechanics is breaking down.

There is no way one can explain from Newtonian

mechanics why the electrons in the hydrogen atom

could have only specific quantized values of angular

momentum. 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 considered

the basic theory governing the behavior of particles

and matter. Something had to replace Newtonian

mechanics, but from the time of Bohr's theory in 1913

until 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 to

the new mechanics. This suggestion was quickly

followed up by Schrdinger 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.


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35-2 Bohr Theory of Hydrogen

For an electron in a circular orbit, predicting the motion

is quite easy. If an electron is in an orbit of radius r,

moving at a speed v, then its acceleration a is directed

toward the center of the circle and has a magnitude

a =

v2

r(2)

Using Equation 1 for the electric force and Equation 2

for the acceleration, and noting that the force is in the

same direction as the acceleration, as indicated in

Figure (2), Newton's second law gives

F = m a

e2

r2= m

v2

r(3)

One factor of r cancels and we can immediately solve

for the electron's speed v to get v2 = e2/mr, or

velectron =e

mr(4)

The period of the electron's orbit should be the distance

2r travelled, divided by the speed v, or 2r/v sec-onds per cycle, and the frequency should be the inverse

of that, or v/2r cycles per second. Using Equation 4for v, we get

frequency of

electron in orbit

= v

2r

= e

2r mr

(5)

According to Maxwell's theory, this should also be the

frequency of the radiation emitted by the electron.

THE CLASSICAL HYDROGEN ATOM

With Rutherford's discovery of the atomic nucleus, it

became clear that atoms consisted of a positively

charged nucleus surrounded by negatively charged

electrons that were held to the nucleus by an electric

force. The simplest atom would be hydrogen consist-ing of one proton and one electron held together by a

Coulomb force of magnitude

Fe =

e2

r2

p

r

FeFe e (1)

(For simplicity we will use CGS units in describing the

hydrogen atom. We do not need the engineering units,

and we avoid the complicating factor of 1/40 in theelectric force formula.) As shown in Equation 1, both the

proton and the electron attract each other, but since the

proton is 1836 times more massive than the electron, the

proton should sit nearly at rest while the electron orbitsaround it.

Thus the hydrogen atom is such a simple system, with

known masses and known forces, that it should be a

straightforward matter to make detailed predictions about

the nature of the atom. We could use the orbit program

of Chapter 8, replacing the gravitational force GMm/r2

by e2/r2 . We would predict that the electron moved in

an elliptical orbit about the proton, obeying all of Kepler's

laws for orbital motion.

There is one important point we would have to take into

account in our analysis of the hydrogen atom that we did

not have to worry about in our study of satellite motion.

The electron is a charged particle, and accelerated charged

particles radiate electromagnetic waves. Suppose, for

example, that the electron were in a circular orbit moving

at an angular velocity 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 down

with a velocity given by v = v0sin t .

In our discussion of radio antennas in Chapter 32, we sawthat radio waves could be produced by moving electrons

up and down in an antenna wire. If electrons oscillated up

and down at a frequency , they produced radio wavesof the same frequency. Thus it is a prediction of Maxwell's

equations that the electron in the hydrogen atom should

emit electromagnetic radiation, and the frequency of the

radiation should be the frequency at which the electron

orbits the proton.

Figure 1

The side view of circular motion

is an up and down oscillation.

p

v0 v = v sin(t)

a) electron incircular orbit

b) side view ofcircular orbit

0e e

p


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35-3

Electromagnetic radiation carries energy. Thus, to see

what effect this has on the electrons orbit, let us look

at the formula for the energy of an orbiting electron.

From Equation 3 we can immediately solve for the

electron's kinetic energy. The result is

12

mv2 =e2

2r

electronkineticenergy

(6)

The electron also has electric potential energy just as an

earth satellite had gravitational potential energy. The

formula for the gravitational potential energy of a

satellite was

potential energy

of an earth satellite= GMmr (10-50a)

where M and m are the masses of the earth and the

satellite respectively. This is the result we used in

Chapter 8 to test for conservation of energy (Equations

8-29 and 8-31) and in Chapter 10 where we calculated

the potential energy (Equations 10-50a and 10-51).

The minus sign indicated that the gravitational force is

attractive, that the satellite starts with zero potential

energy when r = and loses potential energy as it fallsin toward the earth.

We can convert the formula for gravitational potential

energy to a formula for electrical potential energy by

comparing 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 substitution

in the potential energy formula gives

PE =

e2

r

electrical potential energy

of theelectronin the

hydrogenatom(7)

Again the potential energy is zero when the particlesare infinitely far apart, and the electron loses potential

energy as it falls toward the proton. (We used this result

in the analysis of the binding energy of the hydrogen

molecule ion, explicitly in Equation 18-15.)

The formula for the total energy Etotalof the electron in

hydrogen should be the sum of the kinetic energy,

Equation 6, and the potential energy, Equation 7.

Etotal =

kinetic

energy

+potential

energy

=e2

2r

e2r

Etotal =

e2

2r

total energy

of electron(8)

The significance of the minus () sign is that the

electron is bound. Energy is required to pull the

electron out, to ionize the atom. For an electron to

escape, its total energy must be brought up to zero.

We are now ready to look at the predictions that follow

from Equations 5 and 8. As the electron radiates light

it must lose energy and its total energy must become

more negative. From Equation 8 we see that for the

electron's energy to become more negative, the radius

r must become smaller. Then Equation 5 tells us that

as the radius becomes smaller, the frequency of the

radiation increases. We are lead to the picture of the

electron spiraling in toward the proton, radiating even

higher frequency light. There is nothing to stop the

process until the electron crashes into the proton. It isan unambiguous prediction of Newtonian mechanics

and Maxwell's equations that the hydrogen atom is

unstable. It should emit a continuously increasing

frequency of light until it collapses.

p

v

a

r

Fe

e

Figure 2

For a circular orbit, both the acceleration a and the

force F point toward the center of the circle. Thus wecan equate the magnitudes of F and ma.


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Energy Levels

By 1913, when Neils Bohr was trying to understand the

behavior of the electron in hydrogen, it was no surprise

that Maxwell's equations did not work at an atomic

scale. To explain blackbody radiation and the photo-

electric effect, Planck and Einstein were led to the

picture that light consists of photons rather than

Maxwell's waves of electric and magnetic force.

To construct a theory of hydrogen, Bohr knew the

following fact. Hydrogen gas at room temperature

emits no light. To get radiation, it has to be heated to

rather high temperatures. Then you get distinct spectral

lines rather than the continuous radiation spectrum

expected classically. The visible spectral lines are the

H , H and Hlines we saw in the hydrogen spec-

trum experiment. These and many infra red lines we

saw in the spectrum of the hydrogen star, Figure (33-28) reproduced below, make up the Balmer series of

lines. Something must be going on inside the hydrogen

atom to produce these sharp spectral lines.

Viewing the light radiated by hydrogen in terms of

Einstein's photon picture, we see that the hydrogen

atom emits photons with certain precise energies. As

an exercise in the last chapter you were asked to

calculate, in eV, the energies of the photons in the H ,

H and Hspectral lines. The answers are

EH = 1.89 eV

EH = 2.55 eV

EH = 2.86 eV (9)

The question is, why does the electron in hydrogen emit

only certain energy photons? The answer is Bohr's

main contribution to physics. Bohr assumed that the

electron had, for some reason, only certain allowed

energies in the hydrogen atom. He called theseallowed

energy levels. When an electron jumped from one

energy level to another, it emitted a photon whose

energy was equal to the difference in the energy of the

two levels. The red 1.89 eV photon, for example, was

radiated when the electron fell from one energy level to

another level 1.89 eV lower. There was a bottom,

lowest energy level below which the electron could not

fall. In cold hydrogen, all the electrons were in the

bottom energy level and therefore emitted no light.

When the hydrogen atom is viewed in terms of Bohrs

energy levels, the whole picture becomes extremely

simple. The lowest energy level is at -13.6 eV. This isthe total energy of the electron in any cold hydrogen

atom. It requires 13.6 eV to ionize hydrogen to rip an

electron out.

Figure 33-28

Spectrum of a hydrogen star

13.6

3.40

1.51

.850.544

0

n = 1

n = 2

n = 3

n = 4n = 5

H H H

3.65 10 3.70 10 3.75 10 3.80 10

H9H10H11H12H13H14H15H20H30H40

wavelength 3.85 105 5 5 5 5

Figure 3

Energy level

diagram for thehydrogen atom.

All the energylevels are given bythe simple formula

En = 13.6 / n2 eV.

All Balmer series

lines result fromjumps down to the

n = 2 level. The 3jumps shown giverise to the three

visible hydrogenlines.


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35-5

The first energy level above the bottom is at 3.40 eV

which 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 energy

levels needed to explain every spectral line emitted by

hydrogen are given by the formula

En = 13.6 eVn2(10)

where n takes on the integer values 1, 2, 3, .... These

energy levels are shown in Figure (3).

Exercise 1

Use Equation 10 to calculate the lowest 5 energy levels

and compare your answer with Figure 3.

Let us see explicitly how Bohr's energy level diagram

explains the spectrum of light emitted by hydrogen. If,for example, an electron fell from the n=3 to the n=2

level, the amount of energy E32 it would lose and

therefore the energy it would radiate would be

E32 = E3 E2

= 1.51 eV ( 3.40 eV)

= 1.89 eV

=energy lost in falling

from n = 3 to n = 2 level

(11)

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

Exercise 2

Show that the Hand Hlines correspond to jumps to

the n = 2 level from the n = 4 and the n = 5 levels

respectively.

From Exercise 2 we see that the first three lines in the

Balmer series result from the electron falling from the

third, 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 down

to the second energy level. For historical interest, let us

see how Balmer's formula for the wavelengths in this

series follows from Bohr's formula for the energy

levels. For Balmer's formula, the lines we have been

callingH , H andHareH3 , H4 , H5 . An arbitrary

line in the series is denoted by Hn , where n takes on the

values starting from 3 on up. The Balmer formula for

the wavelength of the Hn line is from Equation 33-6

n = 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 has

the value

En E 2 =13.6 eV

n2 13.6 eV

22

energy lost byelectrongoing

from nth to

second level

This must be the energy E Hn carried out by the

photon in the Hn spectral line. Thus

E Hn = 13.6 eV

1

4

1

n2

= 13.6 eVn2 4

4n2

(12)

We now use the formula

= 12.4 10

5cm eV

Ephoton in eV(34-8)

relating the photon's energy to its wavelength. Using

Equation 12 for the photon energy gives

n =

12.4 10 5cm eV13.6 eV

4n2

n2 4

n = 3.65 10

5cmn2

n2 4

which is Balmer's formula.


6/14


It does not take great intuition to suspect that there are

other series of spectral lines beyond the Balmer series.

The photons emitted when the electron falls down to

the lowest level, down to -13.6 eV as indicated in

Figure (4), form what is called theLyman series. In this

series 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 the

other 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 and

see a Balmer series photon like H , H or H,

eventually a 10.2 eV Lyman series photon must be

emitted before the hydrogen can get back down to its

ground state. With telescopes on earth we see many

hydrogen stars radiating Balmer series lines. We do not

see the Lyman series lines because these ultravioletphotons do not make it down through the earth's

atmosphere. But the Lyman series lines are all visible

using orbiting telescopes like the Ultraviolet Explorer

and the Hubble telescope.

Another series, all of whose lines lie in the infra red, is

the Paschen series, representing jumps down to the

n = 3 energy level at -1.55 eV, as indicated in Figure (5).

There are other infra red series, representing jumps

down to the n = 4 level, n = 5 level, etc. There are many

series, each containing many spectral lines. And all

these lines are explained by Bohr's conjecture that the

hydrogen atom has certain allowed energy levels, all

given by the simple formula En = (13.6/n2) eV .

This one simple formula explains a huge amount of

experimental data on the spectrum of hydrogen.

Exercise 3

Calculate the energies (in eV) and wavelengths of the 5

longest wavelength lines in

(a) the Lyman series

(b) the Paschen series

On a Bohr energy level diagram show the electron

jumps corresponding to each line.

Exercise 4

In Figure (33-28), repeated 2 pages back, we showed

the spectrum of light emitted by a hydrogen star. The

lines get closer and closer together as we get to H40 and

just beyond. Explain why the lines get closer together

and calculate the limiting wavelength.

13.6

3.40

1.51

.850.544

0

n = 1

n = 2

n = 3

n = 4n = 5

Figure 4

The Lyman series

consists of all jumpsdown to the 13.6eV

level. (Since this is asfar down as theelectron can go, this

level is called the

ground state.)

Figure 5

The Paschen series

consists of all jumpsdown to the n = 3level. These are all in

the infra red. 1.51

.850

.544

0

n = 3

n = 4

n = 5.378 n = 6.278 n = 7


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35-7

Lymanse

ries

Ba

lme

r

se

ries

Pasc

hen

serie

s

r2

r1

r3

THE BOHR MODEL

Where do Bohr's energy levels come from? Certainly

not from Newtonian mechanics. There is no excuse in

Newtonian mechanics for a set of allowed energy

levels. But did Newtonian mechanics have to be

rejected altogether? Planck was able to explain theblackbody radiation formula by patching up classical

physics, by assuming that, for some reason, light was

emitted and absorbed in quanta whose energy was

proportional to the light's frequency. The reason why

Planck's trick worked was understood later, with

Einstein's proposal that light actually consisted of

particles whose energy was proportional to frequency.

Blackbody radiation had to be emitted and absorbed in

quanta because light itself was made up of these quanta.

By 1913 it had become respectable, frustrating per-

haps, but respectable to modify classical physics in

order to explain atomic phenomena. The hope was that

a deeper theory would come along and naturally ex-

plain the modifications.

What kind of a theory do we construct to explain the

allowed energy levels in hydrogen? In the classical

picture we have a miniature solar system with the

proton at the center and the electron in orbit. This can

be simplified by restricting the discussion to circular

orbits. From our earlier work with the classical model

of hydrogen, we saw that an electron in an orbit of

radius r had a total energy E(r) given by

E(r) = e2

2r

total energy of

an 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 of

radius rn given by

En = e2

2rn(13)

The rn are the radii of the famous Bohr orbits. This

leads to the rather peculiar picture that the electron can

exist in only certain allowed orbits, and when the

electron jumps from one allowed orbit to another, it

emits a photon whose energy is equal to the differencein energy between the two orbits. This model is

indicated schematically in Figure (6).

Exercise 5

From Equation 13 and the fact that E1= 13.6eV,

calculate the radius of the first Bohr orbit r1. [Hint: first

convert eV to ergs.] This is known as the Bohr radius

and is in fact a good measure of the actual radius of a

cold hydrogen atom. [The answer is

r1 =.529 10 8cm=.529A .] Then show that rn=n

2 r1 .

Figure 6

The Bohr orbits are determined by

equating the allowed energy

En = 13.6n2 13.6n2 to the energy En = e

2 2rn e2 2rn

for an electron in an orbit of radius rn.The Lyman series represents all jumps

down to the smallest orbit, the Balmerseries to the second orbit, the Paschen

series 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 E

n.)


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Angular Momentum in the Bohr Model

Nothing in Newtonian mechanics gives the slightest

hint as to why the electron in hydrogen should have

only certain allowed orbits. In the classical picture

there is nothing special about these particular radii.

But ever since the time of Max Planck, there was aspecial unit of angular momentum, the amount given

by Planck's constant h. Since Planck's constant keeps

appearing whenever Newtonian mechanics fails, and

since Planck's constant has the dimensions of angular

momentum, perhaps there was something special about

the electron's angular momentum when it was in one of

the allowed orbits.

We can check this idea by re expressing the electron's

total energy not in terms of the orbital radius r, but in

terms of its angular momentum L.

We first need the formula for the electron's angular

momentum when in a circular orbit of radius r. Back

in Equation 4, we found that the speed v of the electron

was given by

v = emr

(4 repeated)

Multiplying this through by m gives us the electron's

linear momentum mv

mv = memr = e mr (14)

The electron's angular momentum about the center of

the circle is its linear momentum mv times the lever

arm r, as indicated in the sketch of Figure (7). The result

is

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 angular

momentum L. Squaring Equation 13 gives

L2 = e2mr

or

r =

L2

e2m (16)

Finally we can eliminate the variable r in favor of the

angular momentum L in our formula for the electron's

total energy. We get

total energy

of the electronE =

e2

2r

= e2

2 L2 e2mL2 e2m

= e2

2e2mL2

= e4m

2L2(17)

In the formula e4m/2L2 for the electron's energy,

only the angular momentum L changes from one orbit

to another. If the energy of the nth orbit is En , then

there must be a corresponding value Ln for the angular

momentum of the orbit. Thus we should write

En = e4m2Ln

2(18)

v

L = mvr

r

m

Figure 7

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


9/14

35-9

At this point, Bohr had the clue as to how to modify

Newtonian mechanics in order to get his allowed

energy levels. Suppose that angular momentum is

quantized in units of some quantity we will call L0. In

the 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 orbit

it would have n units

Ln = nL0

quantization

of 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's

energy level formula

En = 13.6 eV1n2

(10 repeated)

we see that we can explain the energy levels by

assuming that the electron in the nth energy level has n

units of quantized angular momentum L0. We can also

evaluate the size of L0by equating the constant factors

in Equations 10 and 20. We get

e4m2L0

2= 13.6 eV (21)

Converting 13.6 eV to ergs, and solving for L0gives

e4m

2L02= 13.6 eV 1.6 10 12

ergs

eV

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 2 .

L0 =

h2

=6.63 10 27gm cm/sec

2

= 1.05 10 27gm cmsec

This quantity, Planck's constant divided by 2 , ap-pears so often in physics and chemistry that it is given

the special name h barand is written h

h h2

"h bar" (23)

Using h for L0in the formula for En , we get Bohr's

formula

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 in

esu, m in grams, and h is erg-sec.]

Exercise 6

Use Equation 21 to evaluate L0 .

Exercise 7

What is the formula for the first Bohr radius in terms of the

electron 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 8

Starting from Newtonian mechanics and the Coulomb

force law F =e2/r2, write out a clear and concise deriva-

tion of the formula

En = e4m

2h2

1

n2

Explain the crucial steps of the derivation.

A day or so later, on an empty piece of paper and a clean

desk, see if you can repeat the derivation without

looking at notes. When you can, you have a secure

knowledge of the Bohr theory.


10/14


Exercise 9

An ionized helium atom consists of a single electron

orbiting a nucleus containing two protons as shown in

Figure (8). Thus the Coulomb force on the electron has

a magnitude

Fe = e 2er2

= 2e2r2

e

2e

a) Using Newtonian mechanics, calculate the total

energy 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 angular

momentum L. (First calculate L in terms of r, solve for

r, and substitute as we did in going from Equations 16

to 17.)

c) Find the formula for the energy levels of the electron

in ionized helium, assuming that the electron's angular

momentum is quantized in units of h.

d) Figure out whether ionized helium emits any visible

spectral lines (lines with photon energies between 1.8eV and 3.1 eV.) How many visible lines are there and

what are their wavelengths?)

Exercise 10

You can handle all single electron atoms in one calcu-

lation by assuming that there are z protons in the

nucleus. (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 z

protons in the nucleus. (There is no simple formula for

multi electron atoms because of the repulsive forcebetween the electrons.)

DE BROGLIE'S HYPOTHESIS

Despite its spectacular success describing the spectra

of hydrogen and other one-electron atoms, Bohr's

theory represented more of a problem than a solution.

It worked only for one electron atoms, and it pointed to

an explicit failure of Newtonian mechanics. The ideaof correcting Newtonian mechanics by requiring the

angular momentum of the electron be quantized in

units of h , while successful, represented a bandaid

treatment. It simply covered a deeper wound in the

theory. For two centuries Newtonian mechanics had

represented a complete, consistent scheme, applicable

without exception. Special relativity did not harm the

integrity of Newtonian mechanicsrelativistic New-

tonian mechanics is a consistent theory compatible

with the principle of relativity. Even general relativity,

with its concepts of curved space, left Newtonianmechanics intact, and consistent, in a slightly altered

form.

The framework of Newtonian mechanics could not be

altered to include the concept of quantized angular

momentum. Bohr, Sommerfield, and others tried

during the decade following the introduction of Bohr's

model, 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, seen

in the 2-slit experiment and Maxwell's theory, and aparticle nature seen in Einstein's explanation of the

photoelectric effect. Physicists could not explain how

light 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 less

ignored for nearly 20 years.

De Broglie's idea was that, if light can have both a

particle and a wave nature, perhaps electrons can too!

Perhaps the quantization of the angular momentum of

an electron in the hydrogen atom was due to the wavenature of the electron.

The main question de Broglie had to answer was how

do you determine the wavelength of an electron wave?

Figure 8

Ionized helium has anucleus with two protons,

surrounded by one electron.


11/14

35-11

An analogy with photons might help. There is, how-

ever, a significant difference between electrons and

photons. Electrons have a rest mass energy and pho-

tons do not, thus there can be no direct analogy between

the total energies of the two particles. But both particles

have mass and carry linear momentum, and the amount

of momentum can vary from zero on up for both

particles. Thus photons and electrons could have

similar formulas for linear momentum.

Back in Equation 34-13 we saw that the linear momen-

tum p of a photon was related to its wavelengthby thesimple equation

= hpde Broglie

wavelength (34-13)

De Broglie assumed that this same relationship alsoapplied to electrons. An electron with a linear momen-

tum p would have a wavelength = h/p . This is nowcalled thede Broglie wavelength. This relationship

applies not only to photons and electrons, but as far as

we know, to all particles!

With a formula for the electron wavelength, de Broglie

was able to construct a simple model explaining the

quantization of angular momentum in the hydrogen

atom. In de Broglie's model, one pictures an electron

wave chasing itself around a circle in the hydrogenatom. If the circumference of the circle, 2r did nothave an exact integral number of wavelengths, then the

wave, 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 exact

integral number of wavelengths as illustrated in Figure

(10), there would be no cancellation. This would

therefore be one of Bohr's allowed orbits shown in

Figure (6).

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

n = 2rn (25)

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

n hp = 2rn (26)

Multiplying both sides by p and dividing through by

2 gives

n h2 = prn (27)

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

Equation 27 gives

nh = prn = L n (28)

Equation 28 tells us that for the allowed orbits, the

orbits in which the electron wave does not cancel, the

angular momentum comes in integer amounts of the

angular momentum h . The quantization of angular

momentum is thus due to the wave nature of theelectron, a concept completely foreign to Newtonian

mechanics.

r

Figure 9

De Broglie picture of anelectron wave cancelling

itself out.

Figure 10

If the circumference of the orbit isan integer number of wavelengths,

the electron wave will go aroundwithout any cancellation.

Figure 10a--Movie

The standing waves on acircular metal band nicely

illustrate de Broglies waves


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When a graduate student does a thesis project, typically

the student does a lot of work under the supervision of

a thesis advisor, and comes up with some new, hope-

fully verifiable, results. What do you do with a student

that 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 the

summer of 1924 and was asked if de Broglie's thesis

should be accepted. Although doubtful himself about

a wave nature of the electron, Einstein recommended

that the thesis be accepted, for de Brogliejust might be

right.

In 1925, two physicists at Bell Telephone Laboratories,

C. J. Davisson and L. H. Germer were studying the

surface of nickel by scattering electrons from the

surface. The point of the research was to learn more

about metal surfaces in order to improve the quality ofswitches used in telephone communication. After

heating the metal target to remove an oxide layer that

accumulated following a break in the vacuum line, they

discovered that the electrons scattered differently. The

metal had crystallized during the heating, and the

peculiar scattering had occurred as a result of the

crystallization. Davisson and Germer then prepared a

target consisting of a single crystal, and studied the

peculiar scattering phenomena extensively. Their ap-

paratus is illustrated schematically in Figure (11), and

their experimental results are shown in Figure (12). For

their experiment, there was a marked peak in the

scattering when the detector was located at an angle of

50from the incident beam.

Davisson presented these results at a meeting in Lon-

don in the summer of 1927. At that time there was a

considerable discussion about de Broglie's hypothesis

that electrons have a wave nature. Hearing of this idea,

Davisson recognized the reason for the scattering peak.

The atoms of the crystal were diffracting electron

waves. The enhanced scattering at 50was a diffrac-tion peak, a maximum similar to the reflected maxima

we saw back in Figure (33-19) when light goes through

a diffraction grating. Davisson had the experimental

evidence that de Broglie's idea about electron waves

was correct after all.electron gun

detector

nickel crystal

electron

beam

=

50

Figure 11Scattering electrons from thesurface of a nickel crystal.

Figure 12

Plot of intensity vs. angle for electrons scattered by a

nickel crystal, as measured by Davisson and Germer.The peak in intensity at 50was a diffraction peak

like the ones produced by diffraction gratings. (Theintensity is proportional to the distance out from theorigin.)

Reflectedmaximum

transmittedmaximum

Figure 33-19

Laser beamimpinging on

a diffractiongrating.


13/14

35-13

Index

Symbols

13.6 eV, hydrogen spectrum 35-4

AAllowed orbits, Bohr theory 35-1Angular momentum

Bohr model 35-1, 35-8

Planck's constant 35-8Atoms

Classical hydrogen atom 35-2

BBalmer series

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

Hydrogen spectrum 35-4Bell Telephone Lab, electron waves 35-12Bohr model

Allowed orbits 35-1Angular momentum 35-1, 35-8

Chapter on 35-1De Broglie explanation 35-1

Derivation of 35-8Energy levels 35-4Planck's constant 35-1, 35-8

Quantum mechanics 35-1Rydberg constant 35-9

Bohr orbits, radii of 35-7

CCGS units

Classical hydrogen atom 35-2Circular orbit, classical hydrogen atom 35-2

Classical hydrogen atom 35-2Coulomb's law


DDavisson & Germer, electron waves 35-12

De Broglie

Electron waves 35-11Formula for momentum 35-11

Hypothesis 35-10Key to quantum mechanics 35-1

Wavelength, formula for 35-11Waves, movie of standing wave model 35-11

EElectromagnetic radiation

Energy radiated by classical H atom 35-3

Electron

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 35-1

Kinetic energyBohr model of hydrogen 35-3

Classical hydrogen atom 35-3Total energy

Classical H atom 35-3

Energy level diagramBalmer series 35-6

Bohr theory 35-4

Lyman series 35-6Paschen series 35-6

FForce

Electric force


Hh bar, Planck's constant 35-9

Hydrogen atomBohr theory 35-1

Classical 35-2Hydrogen atom, classical

Failure of Newtonian mechanics 35-3Hydrogen spectrum

Balmer series 35-4Lyman series 35-6Of star 35-4

Paschen 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 spectrum

Balmer formula 35-5Spectral lines, hydrogen

Bohr theory 35-4Lyman series, energy level diagram 35-6


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Ch 35 Bohr Theory of Hydrogen