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8/12/2019 Ch 35 Bohr Theory of Hydrogen
1/14
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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35-4 Bohr Theory of Hydrogen
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.
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35-6 Bohr Theory of Hydrogen
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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35-8 Bohr Theory of Hydrogen
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.
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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.
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35-10 Bohr Theory of Hydrogen
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.
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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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35-12 Bohr Theory of Hydrogen
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.
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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
Classical hydrogen atom 35-2
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
Classical hydrogen atom 35-2
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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