In this unit we concentrate on the interactions between photons and electrons. We start with
some properties of each particle. Finally we look at how the two behave together. This period of
history starts around the 1880’s and will take us to about 1927.
I.
Properties of each
A. Electrons
1. J.J.Thomson
2. Robert Millikan
B. Photons
1. Max Planck
2. Albert Einstein
II.
Photoelectric Effect
A. Experiment and Data
B. Results and Conclusions
III.
Energy Level Diagrams
A. Absorption/Emission spectra
B. Bohr Model
IV.
De Broglie Wavelength
V.
X-Ray Production
VI.
Compton Scattering
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
Electrons & Photons
We start this unit with a look at the two key particles that are affected by the ideas and concepts
of Modern Physics. A particle model of light will be developed that describes each particle as a
photon. In 1897 the electron was discovered. In this lesson we review the properties of both
particles.
The Electron
In 1897 J.J. Thomson used a crude particle accelerator/ mass spectrometer to measure the
properties of a negatively charged particle. The best that he could do was to arrive at a value of
the charge to mass ratio of the particle. About a decade later, Robert Millikan developed the
now famous “Millikan Oil Drop Experiment (pp 483-485). The result of this experiment
determined the charge of an electron to be q = -1.6E-19C. Now coupled with the mass to charge
ratio of Thomson’s work it was determined that electrons have a mass of +9.11E-31Kg with an
immeasurably small size.
The Photon
In the last quarter of the 1800’s scientists had been working on a way to take what they knew
about mechanics, thermodynamics, electricity & magnetism and optics to describe how hot
objects cool by giving off light in various parts of the electromagnetic spectrum. The study of
this effect is known as blackbody radiation.
If an object is hotter than its surroundings it will cool by giving off light. In order to
study this effect scientist had to eliminate the other modes of cooling. Blocks of graphite were
hollowed and a small hole was drilled into the carbon. Although the outside of the carbon block
would cool by convection as well as radiation, the inside would cool by mostly radiation alone.
The intensity of each wavelength of light emitted by the inside of the hot, black boxes was
studied, hence the name- blackbody radiation. It was noticed that at each temperature there was
a specific wavelength that was most intense. Wavelengths slightly higher or lower would be less
intense so that radiation curves could be plotted for different temperatures of the carbon. See
figure 27.2. Observe that as the temperature gets higher the curve becomes more peaked with a
shorter width. Also notice that the maximum value of the peak moves to a shorter wavelength as
the temperature rises. A scientist named Wien determined that you could determine the
temperature of an object by merely measuring the brightest wavelength emitted and then using
formula [27.1] to find the temperature of the object. Chemists, metallurgists and astronomers all
used Wien’s formula but no classical theory would completely explain the other parts of the
curve.
In 1900 Max Planck developed a mathematical model that fit the data without using any
known theory. The idea was to find what function worked and let it tell him what the theory
should be. Much to the surprise of Planck, the mathematical model that worked called for light
to be jumping off the hot objects in
E = h f where
bits and pieces like particles instead of waves. Each bit of
h = 6.63E-34 Js
energy, called “quanta then but photons now” had a specific
h
= 4.14E-15 eVs
value of energy that is determined using the formula in the
box to the right. This is the equation for the energy of a
photon. Store the value of Planck’s constant, h=6.63E-34Jsec, in your calculator under [alpha],
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
[H]. Planck publicly refused to believe his own conclusion and makes mention of using the
equation “until a better idea is found”.
In 1905 Albert Einstein tells of a few more properties of the particle of light called
“photon”. Photons travel at v=c in a vacuum and there is nothing you can do to make photons go
faster or slower. Photons have no rest mass, which is unique to the photon so far as we know.
Although photons seem to have no
mass they do carry momentum with them. In order
p=h/
to calculate a photon momentum use the equation in
the box to the right. In 1905 the above equation worked for photons only. Particles that had
mass still used p = mv. Do not attempt to use p =mv on a photon because you have no value for
“m”. By combinations of E=hf
E = hc/  or E = pc
and c=f one can arrive at finding photon energy
in terms of wavelength or momentum as well as
frequency. The product of Planck’s constant and the speed of light show up so often that the AP
exam will have a value listed in the constants table. The above equations work only for the
mass-less photon. Do not attempt to use these on particles that have mass like electrons and
protons. Given any one of frequency, wavelength, momentum or energy of a photon you should
be able to calculate the other three values.
Suppose that you are given information concerning the power of a monochromatic
source. Could you find all of the above as well as the number of photons/sec being radiated?
We finish this lesson with an example problem.
Example Problem #1
A 3-milliwatt pen laser radiates at 633 nm. Find values for the following:
a) Frequency of light emitted,
b) energy of a single photon in joules,
c) energy of a photon in electron volts,
d) momentum of a single photon,
e) number of photons emitted in a quarter second burst of laser light and
f) length of the beam.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
a) c = f or
f = c/ = 3E+8m/s / 633E-9m
f = 4.74E+14 Hz
d) p = h/ = 6.63E-34Jsec/ 6.33E-9m
or p = E/c = 3.14E-19J / 3E8m/s
p = 1.05E-27 Nsec
b) E = hf = 6.63E-34Jsec(4.74E+14s-1)
E = 3.14E-19 J
e) Power = Nhf where N is photons/sec.
N= Power/(hf)= 3E-3J/sec/ 3.14E-19J
N = 9.55E+15 photons/sec
Total = 9.55E+15 photons/sec0.25s
= 2.39E+15 photons
c) E = hf = 4.14E-15eVs(4.74E+14s-1)
E = 1.96 eV
The length of the beam, (f) can be found using d=ct = 3E8m/s(0.25sec) = 75E+6m. Once you
divide the length by the number of photons you find that there are about 20 photons/meter of
beam.
Photoelectric Effect
While verifying that Maxwell’s electromagnetic waves are the same form as light waves
Heinrich Hertz noticed that under the right conditions UV light could cause sparks to fly from
metal surfaces. This phenomenon was labeled photoelectric effect. The irony of Hertz’s
discovery is that while doing the very last step in the history of light as a wave he had also taken
the first step towards for light as a particle nature!
Experimental Setup
Shown below is a slight variation on figure 27.4. The idea is that you have a light source that
you can vary in both intensity and frequency. In the circuit you have a variable voltage source
that is used to vary the amount of voltage needed to suppress electrons from jumping from the
emitter to the collector.
Variable Light
Source
emitter
V
Collector
A

With the battery turned off and the light source turned on the light can cause electrons to be
knocked off the surface of the emitter and onto the surface of the collector.
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Serway’s College Physics.



The ammeter will read the photocurrent. If 20 million electrons/sec are ejected from the
emitter to the collector then the ammeter will read a photocurrent of 3.2 pAmps.
Increasing battery voltage creates a more negative collector reducing the photocurrent.
Finding the voltage that reduces the photocurrent to zero yields the stopping voltage or VS.
VS can be used to find the maximum kinetic energy of the ejected electrons according to
e(VS) = KEMAX. If the stopping voltage is 3.4 volts for example then the maximum kinetic
energy is 3.4 eV or 5.44E-19J. See equation [27.4].
The two independent variables are the light intensity
Variables VS IPC
and light frequency or wavelength of the source.
Intensity
1
2
These values are set and then the experiment is run in
Frequency
3
4
order to measure dependent variables such as stopping
voltage and photocurrent. Some experiments with
time are also critical. The relationships among the four variables in the above table are briefly
described. (1) The intensity of the light has no affect on the stopping voltage. This contradicts
the wave model of light because a greater intensity is equivalent to waves of higher amplitude
and therefore more kinetic energy of the ejected electrons. (2) The intensity of the light source is
linear with the photocurrent. (3) The frequency of the light source is linear with the stopping
voltage as long as the source is above a certain frequency. See figure 27.6. Below a certain
frequency no electrons are ejected no matter how long the experiment runs and no matter how
intense the light source is. The minimum frequency to cause ejected electrons is called the cutoff frequency. The cut-off frequency is different for different metals. (4) Frequency has no
bearing on the photocurrent. (5) The photocurrent develops within nanoseconds of the light
source being turned on or not at all. The “start-up” time for photocurrent should be frequency or
intensity dependent according to wave models since a lower frequency requires more delivery
time for waves in order to get enough energy to free electrons from a metal.
Albert Einstein’s Explanation
In 1905 Albert Einstein gave a very simple explanation of the photoelectric effect. He was
eventually awarded the Nobel Prize in Physics for “His explanation of the photo-electric effect
and other contributions in physics.” According to Einstein, light is acting like particles that
eventually became known as photons. Each electron can absorb a single photon. When you
increase the intensity of light more photons are created and liberate more electrons. This
explains the liner relationship between the intensity of the light source and the photocurrent.
A single photon brings to the metal surface an energy of E =hf using Planck’s equation
for photon energy. Recall that only five years earlier Planck and the rest of the world refused to
publicly accept that this equation was correct. Some of the photon energy is used to free the
electron from the metal surface. The amount of energy needed to free the electron from the
surface is called the “work
hf = KEe + 
function” and is designated . See Table 27.1. The value
of  is different for different metals. Once the photon
liberates the electron from the surface, remaining energy shows up as kinetic energy of the
ejected electron. Suppose that a 6 eV photon strikes the surface of a metal where the work
function is 1.5 eV. What is the maximum kinetic energy of the ejected electron? The answer is
4.5 eV. Suppose that a 1.3eV photon strikes the same surface. What happens? Since the photon
energy is below the value of the work function no electron is liberated. This is the way in which
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
the cut-off frequency is explained. Replace the equation for the kinetic energy of the electron
with e(VS).
Use algebra to isolate the stopping voltage. You get the
equation shown in the text box to the right. If Einstein is
correct then every metal should give a graph of stopping
voltage as a function of frequency that has the same slope. The slope should be Planck’s
constant divided by the charge of the electron. This is exactly the slope of the graph for all
metals so Einstein was correct. But wait, there’s more! Different metals should have a different
y-intercept. According to Einstein the negative y-intercept of any metal is its work function
divided by the charge of an electron. He was correct again. The cut-off frequency is the xintercept of the graph. If Einstein is correct the cut-off frequency for any metal is the work
function divided by Planck’s constant. Again this is true! Albert Einstein had the courage to
use Planck’s equation in his explanation. A simple but bold explanation was presented to the
world. Review example 27.4. Note that if you are in eV’s then the charge of the electron is 1 in
the above equations and you must use the correct value of “h”.
X-Rays and Wave Particle Duality
After 1905 the world admits that electrons can absorb photons. But can electrons emit or radiate
photons in a manner opposite to the photoelectric effect? A bigger question should have arisen
immediately after 1905. Light can sometimes act like a wave and sometimes act like a particle;
does any other particle also have a wave-particle duality? These two questions are the focus of
today’s lesson.
X-Ray Production
In 1895 Wilhelm Roentgen generated the first man-made x-rays. X-rays are the highest energy
photons that are associated with electron emission. The diagram of an x-ray machine is very
simple. Refer to figure 27.8a. A high voltage is used to accelerate electrons from rest. The final
kinetic energy of the electrons depends on the accelerating voltage according to W=qV or
eV. Upon gaining all of the kinetic energy the electron can slow down by emitting one or more
energetic photons. Suppose for example that an electron has 1000 eV of kinetic energy. The
electron could emit a single 1000eV photon or two 500eV photons or three 333eV photons or so
forth. In order to have the electron give up its energy it must decelerate or stop. In figure 27.10
electrons slam into a Tungsten target giving up most of their kinetic energy in a short amount of
time by x-ray production. There is no way to predict if individual electrons will emit a single
photon or two or three or so forth. The way to get around the problem is to add words like
“maximum frequency” or “minimum wavelength”.
e(V) = hfMAX = hc/ MIN
The use of superlatives for extremes implies that all
of the electron kinetic energy is emitted in a single
photon in order to produce the highest energy and smallest wavelength. Be sure to use the
correct form of Planck’s constant in the above equations. Review example 27.5 and the
following example problem.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
Example Problem
A dental x-ray machine has been set to 120kV at 12 mA. Assume that a minimum number of
photons are emitted in a 4sec burst from the x-ray machine.
A) What is the wavelength for the x-rays produced?
B) How many x-ray photons were generated?
A) From above equation,  = hc/(eV) = (6.63E-34Js3E+8m/s)/(1.6E-19C120,000J/C) =
10.4pm. Alternatively,  = hc/(eV) = (4.14E-15eVs3E+8m/s)/(120,000eV)=10.4pm.
B) (12E-3C/sec)(1electron/1.6E-19C)(1photon/1electron)4sec = 300E+9 photons or
about 300 billion x-ray photons that are quite revealing!
Wave-particle Duality for All
After the study of blackbody radiation, photoelectric effect and Compton scattering (tomorrow’s
lesson) there is no doubt that light sometimes behaves as a wave and sometimes behaves as a
particle. Up until the early 1900’s it was believed that if something acted like one then it could
not behave in the other sense. This is because a particle is confined to a very small, finite space
and can be described as having a single position at any single value of time. Particles are
concentrated in a small space. Waves on the other hand distribute their energy and their position
over many points in space at the same instant. How can something have a single position for a
single instant of time like a particle and for some other circumstance be everywhere at once?
But light does indeed act sometimes like the particle and sometimes like a wave.
I would like to interject the following bit of advice. Return to figure 21.22 on page 675.
You recall having to know the entire spread of the electromagnetic spectrum? At the longer
wavelengths like radio and microwaves the wavelength is much too large to easily observe the
particle nature. Physicists who work at this end of the spectrum think almost entirely as “light
acts like a wave”. They are very much concerned with diffraction and interference and so forth.
Consider the opposite end of the spectrum. An x-ray is on the order of the size of an atom and
gamma rays are on the order of the size of the nucleus. Physicists at this end of the spectrum
think almost exclusively that light is made of particles called photons. And particle views work
very well here with the photons having such a short wavelength. Indeed the energy is focused in
a very small region of space much like a particle. And guess where we are most observant as
humans? In the middle where both wave and particle characteristics are both easily seen.
In the early 1920’s Louis de Broglie (Fresnel was also from Broglie, France) proposed
that if wave-particle duality was good for light then why not for matter particles as well? The
idea that momentum can be used to
p = mv = h/
determine the wavelength of any particle, even
those with mass such as electrons was being proposed.
If you study the two example problems from the text, examples 27.8 & 9, you will see that the
authors stress the following idea. Objects with large, everyday masses will have very small
wavelengths and appear almost exclusively to behave like particles. But as masses approach
smaller values such as those for electrons the wavelengths can become quite noticeable. This is
true because Planck’s constant is so small, (E-34Jsec). Had the value been larger we would have
realized the ideas of quantum mechanics much sooner.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
In 1927, Davisson and Germer performed experiments where low energy electron beams were
fired at nickel crystals. The scattered electrons from the beam were distributed in a pattern
similar to that of x-rays with the corresponding wavelength. Davisson and Germer had
experimentally verified de Broglie’s hypothesis. Now all objects have wave-particle duality.
This should cause a brief bit of panic and anxiety on the reader’s life. You do know what
happens when a wave passes through vertical slit? According to diffraction the wave is spread
horizontally across space. So what happens to a student walking through a vertical door if he/she
exhibits their wave nature at the precise moment of entering the doorframe? Do they get
diffracted all over the hallway? And who cleans up the mess? Do not worry. Your wavelength
is much smaller than the width of the door so that diffraction problems are not a concern. But if
you were to pass through a door at 1 m/s that had a width comparable to your de Broglie
wavelength then you may actually diffract. To observe this however you must get through a
door smaller than the width of an atom and that will never happen at your school!
Compton Scattering
In the early 1920’s Arthur Compton demonstrated that x-rays scattered by electrons or other
charged particles will have a longer wavelength due to transfer of momentum and energy from
the incident photon to the charged particle. Refer to figure 27.16. All of the initial momentum
and energy is in the incident photon. The following statements can be made about the system.
The initial momentum in the x direction is px = h/ o. The initial momentum in the y-direction is
py = 0. The initial energy is E = hc/ o. The following three equations can be established from
the Laws of Conservation of Momentum and Energy:
Conservation of Momentum for x-direction: h/ o = h/  (cos ) + meve (cos )
Conservation of Momentum for y-direction: 0 = h/  (sin ) + meve (sin )
For the previous equation use absolute value of angles since negative signs are already in
place.
Conservation of Energy: hc/ o = hc/  + ½ mev2
The previous steps leave you with three equations for finding up to three unknowns. In tpast the
AP exams equations are decoupled the by giving scattering angles and so forth.
There is an alternate approach to setting up
po
equations using a momentum triangle for the
vectors of initial and final momentum. After


drawing the triangle you can use Law of
Sines, Law of Cosines and Conservation of
p
pe
Energy to determine unknowns. If this
approach is used write the electron
momentum as pe and electron kinetic energy
Recall that p = hc/ for initial and final
as pe2/2me.
photons but p=mv for electron.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
Using either the components method for momentum or the Law of Sines and Law of Cosines
method for momentum leads to the following result as stated in your textbook without proof. If a
wavelength is not the initial then it
 = h/(mec) (1 - cos)
must be the final wavelength. The equation shows the
where h/(mc) = 2.426pm
shift in wavelength of the scattered photon with
for electrons
reference to the photon scattering angle. The constants
in front of the equation combine to have a unit of length. The value is 2.426 picometers and is
known as the Compton wavelength. Beware of scattering high-energy photons off of protons.
That would require use of a different mass in the above box and would lead to a different
coefficient for h/mc. Review example 27.7. Warning about doing the homework, the aboveboxed equation is nice to use but it is not given on most exams where the previous equations may
be more efficient. Try to use the basics on some of the homework problems (blue ones) in order
to prepare for your exam.
Energy Level Diagrams
Atomic Structure
When the early work on the electron was competed by Thomason and Millikan a model of a
neutral atom had to be constructed. The model had to explain a neutral charge state for the atom
while acknowledging the existence of electrons. Thomson’s model, a.k.a. the plum pudding
model, is shown in figure 28.1.
A few years later, two students of Ernest Rutherford scattered alpha particles off gold foil
(figure 28.2) and determined that the atom had a very small, very dense, positively charged
nucleus. A new model had to be constructed that accounted for the positive nucleus and the
electrons. The Rutherford model proposed that the electrons orbited the positive nucleus of a
neutral atom much like planets orbit the sun. Rutherford’s planetary model had one big flaw.
According to the laws of mechanics and Maxwell’s equations accelerating electrons should emit
light and lose energy. Any electrons orbiting the nucleus should continuously emit light
spiraling into the nucleus. A new model had to be constructed that accommodated the positive
nucleus, negative electrons and have stable orbits that did not spiral into the nucleus. The new
model that replaced the planetary model is known as the Bohr Model after Niels Bohr. In this
lesson we focus upon the structure of the atom and the orbiting electrons. In the next unit we
will turn our attention to the nucleus of the atom.
Atomic Spectra
Since the mid 1800’s chemists and astronomers had known that each element of the periodic
table would emit a unique color pattern when heated at low pressure. By passing light from a
heated gas through a prism or diffraction grating the colors can be dispersed or resolved. The
color patterns of hydrogen, mercury and neon are shown on page 867, figure 28.3. The top three
patterns are known as emission spectra since they are the patterns of a gas emitting light. Since
stars tend to heat gas we can look at light from a star and determine their chemical composition
from their emission spectra. When this was done to our own star a pattern of light was noticed
that had never been seen. A new element was discovered, Helium.
Other astronomers were using absorption spectra instead of emission spectra. What they
had discovered is that if you pass white light through a cold gas all of the wavelengths will get
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through the gas except for those same wavelengths that were emitted for emission spectra.
Compare the pattern of emission spectra of hydrogen at the top of figure 28.3 with the white light
passed through cold hydrogen on the 4th row of figure 28.3. The absence of lines in a continuous
pattern is known as absorption spectra.
Hydrogen had the simplest pattern of lines and was therefore thought to have the easiest
atomic structure of all atoms. Hydrogen actually has three distinct groups of emission lines.
One group of lines is entirely in the UV part of the spectrum and is known as the Lyman Series.
A second group of lines is in the visible part of the spectrum and is known as the Balmer Series.
A third group of lines falls in the infrared and is known as the Paschen Series. Figure 28.4
shows the visible lines of the Balmer series and some of the UV lines of the Lyman Series.
Bohr Model of Hydrogen
In 1913 Niels Bohr developed a model for hydrogen that explained the existence of electrons, a
positive nucleus and the correct wavelengths of all three groups of the hydrogen emission
spectrum. Although the AP exam no longer requires you to know the assumptions of the model
they are still worth reviewing. The assumptions are listed on page 968. Most notable of these
assumptions to us are the first three. They can be summarized as follows.
 Electron orbits are only allowed at certain distances from the nucleus. While in these
allowed orbits the electron will not emit light.
 Electrons only emit light if they make a transition to a smaller orbit. The photon energy of
the emitted light will account for the
hf = Eo - Ef
electron energy difference between initial and
final orbits according to the formula to the right.
Energies to the right of the equals sign are for the orbital energies of the electron.
 Electrons can absorb photon energies but only if it is the exact amount to raise the electron to
an allowed orbit of greater radius.
Using these assumptions and starting with classical mechanics Bohr was able to calculate the
radius and the energy of each allowed orbit. The allowed orbits are numbered from n=1 to n=
with a smaller number representing a smaller orbit.
Allowed Radii
Allowed Electron Energies
The electron for hydrogen can orbit in any
When in any of the orbital radii listed from
of the following radii but nowhere in
the previous column the electron will have
between these values.
energies as listed below.
rn = n2 ao where ao = 0.0529nm
and n = 1, 2, 3, 4, …
The value of ao is known as the Bohr radius.
Bohr was able to calculate the smallest
radius of the hydrogen atom from the
constants as shown in equation [28.10].
This value had already been measured due
to corrections of the ideal gas law. Bohr
knew he was on the right track when the
correct value appeared.
En =  13.6 eV where n=1,2,3…
n2
Again, the energy required to ionize
hydrogen had already been known. This
was another sign that Bohr was moving in
the correct direction for describing atomic
structure. Using known constants the
ionization energy of 13.6eV was found from
equation [28.12].
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.
Energy Level Diagrams
Drawing the orbits of the electron to scale becomes a problem due to the “n2” nature of the
function. But something had to be done to describe the relationship of the electron orbits before
and after a transition. Both figures 28.7 and 28.8 demonstrate the idea of energy level diagrams
although the scales have been exaggerated or compressed in different places.
Picture the electron trapped in an orbit about the nucleus of hydrogen as being stuck at
the bottom of a well or hole in the ground. The depth of the hole represents how much more
energy the electron needs to get free of the nucleus. If the electron is in the smallest orbit, n=1,
then it needs the most energy, -13.6eV. If the electron is in orbit n=2 then it is farther from the
nucleus and needs less energy to get free, -3.4eV. In other words it is no longer at the bottom of
the hole but closer to the top, where E=0eV and the electron is free.
We can map out the energy level diagram by drawing horizontal lines at various depths to
represent the various energies. Tape two sheets of paper together on the 8.5 inch edges to make
one longer sheet about 21 inches long. Use a scale of 1eV = 3cm and construct the energy level
diagram of hydrogen for the first 6 orbits of hydrogen. Compare your energy level diagram
drawn to scale with figure 28.8.
Graphing Calculator Exercise
We are drawing horizontal lines so the x
Energy level diagrams can be
scale is really not important but leave room
mapped out on a graphing calculator by
for labeling n values and E values. I
taking the following steps. A) Go to
recommend the window below.
[STAT], [1] and enter the numbers 1,2,3,4,5
under list one, L1. B) Now move the cursor
of your calculator until it is exactly on top of
the heading of list 2, L2. C) Type –13.6/ L12
and [ENTER]. Note that L1 is [2nd], [1] on
your calculator. You now have in the
second list energy values for the first five
orbits. I have also listed the orbital radii for
the first five orbits in nanometers in list 3.
Graph the following in the [y=]:
Y1= L2/ (x  0) /(x  1)
Even with the graphing calculator, the pixel height becomes significant just like pencil width on
your exercise with paper and meter stick.
Transitions
With energy level diagrams constructed electron jumps are easier to describe and calculate. In
figure 28.8 for example the electron jumps for the Balmer series are shown. If an electron falls
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Serway’s College Physics.
down in an energy level diagram photons are emitted equivalent to the difference in energy
levels of before and after. If an electron moves up the energy level diagram then photons of the
proper energy must be absorbed. Some examples must be described. Refer to figure 28.8 for all
of these descriptions.
1) The electron in figure 28.8 jumps from n=3 to n=2; a 1.89 eV photon is emitted.
2) The electron going directly from n=4 to n=2 will emit a single, 2.55eV photon.
3) An electron from n=4 could do two jumps to get from n=4 to n=2. The electron could emit a
single, 0.66eV photon to go from n=4 to n=3 then emit a single 1.89eV photon to go from
n=3 to n=2.
4) To move the electron from n=2 to n=5 requires the hydrogen atoms to be exposed to 2.86eV
photons.
5) If electrons in the n=2 state are exposed to photons with an energy of 5.0eV the electrons will
be freed from the nucleus with a resulting kinetic energy of 1.6eV for the electrons leaving
the atom in an ionic state.
Bohr Model Failures
With Bohr’s model he was able to account for all of the lines of the Lyman series as transition to
n=1 states, hence the UV nature. The Balmer series are all transitions to n=2 states and the
Paschen series are all transitions to n=3 states. He also gave us a theoretical value for the both
hydrogen radius and ionization energy. But the model did not work for all atoms, merely
hydrogen like atoms or atoms with a single electron.
Bohr made some mistakes in his assumption. The biggest was that he treated the electron
as a particle instead of a wave. Almost one decade later de Broglie pointed out this flaw in the
Bohr Model. A few years after that Schrodinger came up with a three dimensional wave
equation that when applied to orbiting electrons gave us the entire periodic table and the fuzzy,
electron-cloud model of the atom as we know it today. There were other minor corrections
including relativistic effects. So why is this model so famous if it works so infrequently? We
recognize his great intuition because he led us into the idea of photon emission and absorption
only during electron transitions.
These notes are supplied courtesy of Mr. Wayne Mullins. Pages, problems, & figures reference
Serway’s College Physics.