Download Physics 200 Class #1 Outline

Physics 200 Class #16 Notes
October 31, 2005
o Completion of Class 15 notes
o Review and questions for Class #15
o The Bohr Model of the Hydrogen Atom
Early in the 20th century, there were two models:
Plum Pudding: Electrons embedded in positive charge oscillate about equilibrium to radiate.
One of the major problems with this model was that it could not predict the spectral series
observed for hydrogen and other atoms.
Planetary Model: Initially, the nuclear model was also unpromising. In the first place,
classical physics showed that electrons bound to a positively charged nucleus must revolve in
orbits. But why would they have only those particular orbits that represented the special set
of frequencies observed in the spectrum of hydrogen? Even more of a problem was how they
could stay in orbit. Any orbiting charged particle must radiate electrical energy due to the
acceleration it undergoes as it is bent into the circular, or perhaps elliptical, orbital path. Thus
orbiting electrons would constantly lose energy and spiral in to the nucleus, just as Earthorbiting satellites that are low enough to encounter some atmosphere gradually lose energy
and spiral in to Earth. In the case of atomic electrons, though, the time to decay would be
microseconds or less, not months or years! Classical physics had no way to explain why we
have atoms at all or why they have the sizes they do.
Neils Bohr’s Model for the hydrogen atom (~1912)
Bohr decided to ignore classical physics' inability to account for stable atoms. He gave up
trying to relate the hydrogen spectra directly to internal motions, and he took the nuclear
model at its face value. If there had to be stable orbits, he reasoned, one should make stability
a basic property of the model rather than worrying about how orbits could not be stable.
The approach described here is not one that he originally used but it is one that he
proposed after his initial successes using what is now called the correspondence principle.
We will use this approach because it is easy to follow. The approach can be summarized by
three assumptions.
1. (Condition on orbits)
The electron moves about the proton in discrete circular orbits (energy levels) in which the
electron’s angular momentum mvr is an integral multiple of Planck’s constant divided by 2π.
[We will investigate angular momentum a bit more later]
Electric Force = ma produces circular orbits
nh
mvr =
condition on the orbits
2
2. (Break from classical physics)
Maxwell’s theory does not apply to the electron’s motion in these special states so it will not
radiate light while orbiting.
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3. (Radiation condition)
The electron moves from one orbit to another by an instantaneous process called a quantum
jump. When this jump occurs, a photon is emitted. The frequency of the light is given by the
change in energy divided by Planck’s constant.
f=
Einitial  E final
h
or
c


Einitial  E final
h
If a photon hits an atom with the right energy, it can make the electron jump to a different
state. This is "photon absorption" and is just a reverse of the photon emission process. Both
processes are shown in the above drawing.
Now, the problem is to find an expression for the energy of a given orbit (state). The following is a
schematic outline of what we do. We will avoid the algebra.
I The total energy
ke2
2
E = KE + PE
E = mv + ( 
)
r
kQq
NOTE: Remember Potential Energy PE =
(joules, J) , in this case the charges are
r
the charge on an electron and the charge on a proton, which we can write as +e and -e.
(actually, different people interchange the + and -. Just remember it's attractive, so the
potential energy is negative.
But we don’t know the velocity (v) and radius (r), but we have two more conditions.
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II Condition on orbits
angular momentum = n
h
2
mvr = n
h
2
III The electric force
ke2
v2
=
ma
=
m
r2
r
Electric Force = mass x acceleration
Using these three simultaneous equations, Bohr found expressions for the radius and total energy:
h2
2
rn = n
= n2 ao = n2 (0.0528) nm
2
2
4 mke
E
2 2 mk 2 e 4
13.6eV
En =
  21  
2 2
n h
n
n2
These special values of E that can occur are called "energy states." The lowest energy E1, i.e., n =
1, is often referred to as the "ground state" of the atom. Because of this association of the energy
states of the hydrogen atom with the integers n, it is customary to label the energy states with the
index n. Because the restriction of n to integer values forces there to be a finite difference, a quantum
of difference, between energies, the index is called a "quantum number." For atoms more complicated
than hydrogen, n continues to be very important, but other such indices are needed as well, and they
are all called quantum numbers. To distinguish the n quantum number from the others, it is called the
"principal quantum number," or, sometimes, the "radial quantum number."
Connection to The Ritz-Rydberg Formula
The difference in energy between a lower-energy state of quantum number nf and a higher-energy
state of quantum number ni , is just:
 1
1
Ei - Ef = 13.6 eV  2  2 
n

 f ni 
 1
hc
1
hf =
= 13.6 eV  2  2 
n


 f ni 
energy of emitted photon
1

=
13.6 eV  1
1 

 n2 n2 
hc
i 
 f
1

 1
1
=R  2  2
n
 f
n i 
For nf = 2 we have the Balmer series.
(The wavelength order just happens to be reversed between these two pictures.)
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Energy-level Diagrams
It is possible to describe any atom in terms of its energy states. An important graphical aid for
representing the energies of an atom is the "energy-level diagram." As an example, consider the hydrogen
atom. You can calculate the energies En of its quantum states from the relation
En = 
13.6 eV
n2
and arrange them vertically on a scale, as shown in the figure below. The lines represent the "energy
levels" of the atom. In the diagram shown here, the zero of the scale is the energy for n equal to
infinity. This condition corresponds to an infinite separation between the electron and the nucleus,
i.e., the hydrogen atom would be ionized. This means that 13.6 e V of energy is required to free an
electron from the hydrogen atom in its ground state.
The energy-level diagram looks like a ladder with "rungs" getting closer together as they
approach the top. A photon is emitted by an atom when it drops from one energy level to a
lower one. The energy of the emitted photon is just equal to the difference in energy between
the two energy levels of the atom. We say that the atom has undergone a "transition" from
one energy state to another.
You can use this picture to understand the various spectral series that are emitted by gaseous
atomic hydrogen. A hydrogen atom can undergo a transition from any higher energy state to any
lower one. Different transitions have different probabilities of happening. If a hydrogen atom is
ionized, an unbound electron, i.e., an electron outside of the atom, can make a transition to any energy
level of the atom.
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In the laboratory when an electric current is passed through hydrogen gas in a thin glass tube, the
resulting electric discharge ionizes some hydrogen atoms and excites the electrons in others to higher
energy states. After excitation an atom loses its energy by a cascade of successive jumps. Which
jumps occur differ from one atom to the next, and there is no guarantee that any given atom will go
from any particular energy state to another. The process is quite random, and the atom can de-excite
by skipping steps. (Modern quantum theory can predict the probabilities of these transitions, but the
Bohr model cannot.) The result is that when a collection of atoms de-excites, photons of many
wavelengths will be emitted by different atoms. For any given atom, the whole sequence of photon
emissions down to the ground state usually takes a very short time, from microseconds to
nanoseconds. An observed spectrum is made up of photons emitted from a variety of different
transitions occurring in many different hydrogen atoms.
The transitions that produce the Balmer series of spectral lines are shown in the figure by the
vertical arrows connecting higher energy states with the n = 2 energy level. Note that the final state
of the Balmer line transitions is not the lowest possible energy state of the atom. After a
Balmer transition, the hydrogen atom will change into its lowest, most stable, energy state-its ground
state. In other words, it will make a transition from n = 2 to n = 1, and it will emit a photon that
has a wavelength too short (energy too high) to be visible to the eye.
The Balmer series is apparent in the laboratory because it is the only series that consists of
wavelengths of visible light. Even so, only three or four of the possible transitions to the n =
2 level are visible; the rest are in the ultraviolet and beyond the sensitivity of the eye.
The diagram suggests other possible series. For example, there should be a series that has
a final state n = 1. This is the Lyman series, consisting entirely of spectral lines with photon
energies of 10.2 eV or more.
Example: What photon energies would be produced from atoms that have been excited to
the n = 3 state?
The electron could go to the ground state by two routes. One would be to jump directly to n = 1,
emitting the second line of the Lyman series, or it could go from 3 to 2 and then to 1. These
transitions would be the first lines of the Balmer and Lyman series, respectively. Thus there are three
possible photon energies:
E3 - E1  eV
E3 - E2  eV
E2 - E1  eV
Exercises
1. Redraw the energy level diagram for hydrogen and show on it the possible Lyman series
transitions.
2. In the hydrogen spectrum the lines of the Paschen series are transitions to nf = 3. Using the
diagram, determine the energies of the first two lines of this series.
3. Two more hydrogen series in the infrared are given by the Brackett series with nf = 4 and
the Pfund series with nf = 5. Find the photon energies for the first line in each series.
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4. What are the wavelengths of the lines calculated in the previous problem?
5. Suppose a group of hydrogen atoms are all excited to n = 5. How many different photon energies
are emitted as the collection decays.
6. Find the energies of all the photons that could be emitted from hydrogen atoms excited to n = 4.
7. Use the Bohr Model to explain absorption spectra.
Limitations of the Bohr Model
1. It is a two-dimensional model, atoms are three dimensional.
2. It cannot predict the lifetimes of the excited states.
3. t cannot predict the relative intensities of the spectral lines.
4. It uses a mixture of classical physics with some assertions about quantized energy levels and nonradiating orbits.
5. It does not predict the “splitting” of spectral lines when a source is placed in a magnetic field.
A complete, consistent theory of the atom came with the development of quantum mechanics in the
1920s. This theory, which used several of Bohr’s fundamental ideas, completely superseded the
Bohr Model.
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