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Making Light
We can also make light by exciting atoms.
From experiment, we see that different atoms
emit different light. But each type of atom
emits a very specific set of wavelengths
called a spectrum.
The hydrogen atoms emit three visible
wavelengths: one in the red, one in the
blue-green, and one in the violet.
Making Light
We need a model of the atom that will explain
why atoms emit only certain wavelengths.
First of all, what is the size of a typical atom?
Let’s take water (although that is a
molecule, we know a lot about water: its
mass density: 1 gm / 1 cc, it is H2O so it has
18 grams/mole, and we know Avagadro’s
number = 6.02 x 1023 molecules/mole.
Making Light from Atoms
(1 cc / 1 gm) * (18 gms / mole) * (1 mole /
6.02 x 1023 molecules)
= 18 x 10-6 m3 / 6 x 1023 molecules
= 3 x 10-29 m3 = 30 x 10-30 m3 .
Therefore, the size is about (30 x 10-30 m3)1/3
= 3 x 10-10 m. Thus the size of an atom
should roughly be about 0.1 nm .
Making Light from Atoms
Now that we know the size of an atom, how
much mass does the atom have?
From the mass spectrograph, we know that the
mass of an atom comes in integer values of
1 amu = 1.66 x 10-27 kg. (In fact, this is
important in getting Avagadro’s number!)
Making Light from Atoms
Now that we know the size and mass, what
parts does an atom consist of?
We know that the atom has electrons of very
small mass (me = 9.1 x 10-31 kg), about 2000
times smaller than one amu and a negative
charge of -1.6 x 10-19 Coul.
Making Light from Atoms
We also know that the atom is neutral, so the
part of the atom that is not the electrons
must have essentially all the mass and a
positive charge to cancel that of the
electrons.
But what is the structure of these electrons
and this other part of the atom?
Making Light from Atoms
Two possibilities come to mind:
• The planetary model, where the very light
electron orbits the heavy central nucleus.
• The plum pudding model, where the very
light and small electrons are embedded (like
plums) in the much more massive pudding
of the rest of the atom.
Making Light from Atoms
The Planetary Model:
If the light electron does go around the
central, heavy nucleus, then the electron is
accelerating (changing the direction of its
velocity). But an accelerating electron
should emit electromagnetic radiation (its
electric field is wobbling).
Making Light from Atoms
• If the electron is emitting E&M radiation, it
is emitting energy.
• If the electron is emitting energy, it should
then fall closer to the nucleus.
• The process should continue until the
electron falls into the nucleus and we have
the plum pudding model
Making Light from Atoms
• In addition, the frequency of the E&M
radiation (light) emitted by the accelerating
(orbiting) electron should continuously vary
in frequency as the frequency of the
electron continuously varies as it spirals
into the nucleus. This does not agree with
the experimental results: the spectrum of
hydrogen.
Making Light from Atoms
The plum pudding model has no such problem
with accelerating electrons, since the
electrons are just sitting like plums in the
pudding.
Rutherford Scattering
To test the plum pudding model, Rutherford
decided to shoot alpha particles
(mass = 4 amu’s; charge = +2e; moving very fast)
at a thin gold foil and see what happens to the
alpha particles.
(gold can be made very thin - only several atoms
thick; thus there should be very few multiple
scatterings)
Rutherford Scattering
If the plum pudding model was correct, then
the alphas should pretty much go straight
through - like shooting a cannon ball at a
piece of tissue paper. The positive charge of
the atom is supposed to be spread out, so by
symmetry it should have little effect. The
electrons are so light that they should
deflect the massive alpha very little.
Rutherford Scattering
Results:
• Most of the alphas did indeed go straight
through the foil.
• However, a few were deflected at
significant angles.
• A very few even bounced back!
(Once in a while a cannot ball bounced back
off the tissue paper!
Rutherford Scattering
The results of the scattering were consistent
with the alphas scattering off a tiny positive
massive nucleus rather than the diffuse
positive pudding.
The results indicated that the positive charge
and heavy mass were located in a nucleus
on the order of 10-14 m (recall the atom size is
on the order of 10-10 m).
Rutherford Scattering
If the electric repulsion of the gold nucleus is
the only force acting on the alpha
(remember both alpha and the nucleus are positively
charged)
then the deflection of the alpha can be
predicted.
Rutherford Scattering
The faster we fire the alpha, the closer the
alpha should come to the gold nucleus.
1/2 m v2 = q(kqgold/r)
We will know that we have “hit” the nucleus
(and hence know its size) when the
scattering differs from that due to the purely
electric repulsion. This also means that
there must be a “nuclear force”!
Rutherford Scattering
Note how small the nucleus is in relation to
the atom: the nuclear radius is 10-14 m
versus the atomic radius of 10-10 m - a
difference in size of 10,000 and a difference
in volume of 1012 (a trillion!).
The electron is even smaller. It is so small
that we can’t yet say how small, but it is
less than 10-17 meters in radius.
Rutherford Scattering
If the mass takes up only 1 trillionth of the
space, why can’t I walk right through the
wall?
Rutherford Scattering
The electric repulsion between the orbiting
electrons of the wall and the orbiting
electrons of me - and the electric repulsion
between the nuclei of the atoms in the wall
and the nuclei of my atoms, these repulsions
keep me and the wall separate.
The nuclear force does not come into play.
We’ll say more about the nuclear force in
part V of this course.
Making Light from Atoms
We now know that the atom seems to have a
very tiny nucleus with the electrons
somehow filling out the size of the atom just what the planetary model of the atom
would suggest.
However, we still have the problem of how
the electrons stay in those orbits, and how
the atom emits its characteristic spectrum of
light.
The Bohr Theory
Let’s start to consider the planetary model for
the simplest atom: the hydrogen atom.
Use Newton’s Second Law:
Fel = macircular , or ke2/r2 = mv2/r
(one equation, but two unknowns: v,r)
[Note that the theory should predict both v
and r.]
The Bohr Theory
We need more information, so try the law of
Conservation of Energy:
E = KE + PE = (1/2)mv2 + -ke2/r = E
(a second equation, but introduce a third
unknown, E; total unknowns: v, r, E)
The Bohr Theory
Need more information, so consider
Conservation of Angular Momentum:
L = mvr
(a third equation, but introduce a fourth
unknown, L; unknowns: v, r, E and L.)
The Bohr Theory
We have three equations and four unknowns.
Need some other piece of information or some
other relation.
Bohr noted that Planck’s constant, h, had the
units of angular momentum: L = mvr
(kg*m2/sec = Joule*sec)
so he tried this: L = nh
(quantize angular momentum).
The Bohr Theory
Actually, what he needed was this:
L = n*(h/2n*
where  = h/2 (called h-bar) 
This gave him four equations for four
unknowns (treating the integer, n, as a
known). From these he could get
expressions for v, r, E and L.
The Bohr Theory
In particular, he got:
r = n22/(meke2) = (5.3 x 10-11 m) * n2
(for n=1, this is just the right size radius for the
atom)
and
E = [-mek2e4/22]*(1/n2) = -13.6 eV / n2
(where 1 eV = 1.6 x 10-19 Joules).
This says the electron energy is QUANTIZED
The Bohr Theory
In particular, when the electron changes its
energy state (value of n), it can do so only
from one allowed state (value of ninitial) to
another allowed state (value of nfinal).
E = hf = [-13.6 eV]*[(1/ni2) - (1/nf2)] .
The Bohr Theory
E = hf = [-13.6 eV]*[(1/nf2) - (1/ni2)]
In the case of ni = 3, and nf = 2,
E = (-13.6 eV)*(1/4 - 1/9) = 1.89 eV
E = hf = hc/ , so in this case,
emitted = hc/E =
(6.63x10-34 J-sec)*(3x108 m/s)/(1.89 x 1.6x10-19 J)
= 658 nm (red light).
The Bohr Theory
Similarly, when ni = 4 and nf = 2, we get
E = 2.55 eV, andemitted = 488 nm
(blue-green); and
when ni = 5 and nf = 2, we get
E = 3.01 eV, andemitted = 413 nm
(violet).
ALL THREE MATCH THE ACTUAL
SPECTRUM OF HYDROGEN!
The Bohr Theory
This matching of theory with experiment is
the reason Bohr made his assumption that
L = n (instead of L = nh).
The Bohr Theory
• Note that we have quantized energy states
for the orbiting electron.
• Note that for all nfinal = 1, we only get UV
photons.
• Note that for all nfinal > 2, we only get IR
photons.
The Bohr Theory
Problems with the Bohr Theory:
• WHY is angular momentum quantized
(WHY does L=n need to be true.)
• What do we do with atoms that have more
than one electron? (The Bohr theory does
work for singly ionzed Helium, but what
about normal Helium with 2 electrons?)
DeBroglie Hypothesis
Problem with Bohr Theory: WHY L = n ?
• have integers with standing waves:
n(/2) = L
• consider circular path for standing wave:
n = 2r and so from Bohr theory:
L = mvr = n = nh/2get 2r = nh/mv = n
which means  = h/mv = h/p .
DeBroglie Hypothesis
DeBroglie = h/mv = h/p
In this case, we are considering the electron
to be a WAVE, and the electron wave will
“fit” around the orbit if the momentum is
just right (as in the above relation). But this
will happen only for specific cases - and
those are the specific allowed orbits and
energies that are allowed in the Bohr
Theory!
DeBroglie Hypothesis
The Introduction to Computer Homework on
the Hydrogen Atom (Vol. 5, number 5)
shows this electron wave fitting around the
orbit for n=1 and n=2.
What we now have is a wave/particle duality
for light (E&M vs photon), AND a
wave/particle duality for electrons!
DeBroglie Hypothesis
If the electron behaves as a wave, with
 = h/mv, then we should be able to test this
wave behavior via interference and
diffraction.
In fact, experiments show that electrons DO
EXHIBIT INTERFERENCE when they
go through multiple slits, just as the
DeBroglie Hypothesis indicates.
DeBroglie Hypothesis
Even neutrons have shown interference
phenomena when they are diffracted from a
crystal structure according to the DeBroglie
Hypothesis:  = h/p .
Note that h is very small, so that normally 
will also be very small (unless the mv is
also very small). A small  means very
little diffraction effects [1.22  = D sin()].
Quantum Theory
What we are now dealing with is the Quantum
Theory:
• atoms are quantized (you can have 2 or 3,
but not 2.5 atoms)
• light is quantized (you can have 2 or 3
photons, but not 2.5)
• in addition, we have quantum numbers
(L = n , where n is an integer)
Heisenberg Uncertainty Principle
There is a major problem with the
wave/particle duality:
a) a wave with a definite frequency and
wavelength (a nice sine wave) does not
have a definite location. [At a definite
location at a specific time the wave would
have a definite phase, but the wave would
not be said to be located there.]
[ a nice travelling sine wave = A sin(kx-t) ]
Heisenberg Uncertainty Principle
b) A particle does have a definite location at a
specific time, but it does not have a
frequency or wavelength.
c) Inbetween case: a group of sine waves can
add together (via Fourier analysis) to give a
semi-definite location: a result of Fourier
analysis is this: the more the group shows
up as a spike, the more waves it takes to
make the group.
Heisenberg Uncertainty Principle
A rough drawing of a sample inbetween case,
where the wave is somewhat localized, and
made up of several frequencies.
Heisenberg Uncertainty Principle
A formal statement of this (from Fourier
analysis) is: x * k
(where k = 2/, and  indicates the
uncertainty in the value)
But from the DeBroglie Hypothesis,  = h/p,
this uncertainty relation becomes:
x * (2/) = x * (2p/h) = 1/2 , or
x * p = /2.
Heisenberg Uncertainty Principle
x * p = /2
The above is the BEST we can do, since there
is always some experimental uncertainty.
Thus the Heisenberg Uncertainty Principle
says: x * p > /2 .
Heisenberg Uncertainty Principle
A similar relation from Fourier analysis for
time and frequency: t *  = 1/2 leads to
another part of the Uncertainty Principle
(using E = hf):
t * E > /2 .
There is a third part: * L > /2 (where L
is the angular momentum value).
All of this is a direct result of the
wave/particle duality of light and matter.
Heisenberg Uncertainty Principle
Let’s look at how this works in practice.
Consider trying to locate an electron
somewhere in space. You might try to “see”
the electron by hitting it with a photon. The
next slide will show an idealized diagram,
that is, it will show a diagram assuming a
definite position for the electron.
Heisenberg Uncertainty Principle
We fire an incoming photon at the electron,
have the photon hit and bounce, then trace
the path of the outgoing photon back to see
where the electron was.
incoming
photon
electron
Heisenberg Uncertainty Principle
screen
slit so we can
determine direction
of the outgoing
outgoing
photon
photon
electron
Heisenberg Uncertainty Principle
• Here the wave-particle duality creates a
problem in determining where the electron
photon hits here
was.
slit so we can
determine direction
of the outgoing
photon
electron
Heisenberg Uncertainty Principle
• If we make the slit narrower to better
determine the direction of the photon (and
hence the location of the electron, the wave
nature of light will cause the light to be
diffracted. This diffraction pattern will
cause some uncertainty in where the photon
actually came from, and hence some
uncertainty in where the electron was .
Heisenberg Uncertainty Principle
We can reduce the diffraction angle if we
reduce the wavelength (and hence increase
the frequency and the energy of the photon).
But if we do increase the energy of the
photon, the photon will hit the electron
harder and make it move more from its
location, which will increase the uncertainty
in the momentum of the electron.
Heisenberg Uncertainty Principle
Thus, we can decrease the x of the electron
only at the expense of increasing the
uncertainty in p of the electron.
Heisenberg Uncertainty Principle
Let’s consider a second example: trying to
locate an electron’s y position by making it
go through a narrow slit: only electrons that
make it through the narrow slit will have the
y value determined within the uncertainty of
the slit width.
Heisenberg Uncertainty Principle
But the more we narrow the slit (decrease y),
the more the diffraction effects (wave
aspect), and the more we are uncertain of
the y motion (increase py) of the electron.
Heisenberg Uncertainty Principle
Let’s take a look at how much uncertainty
there is: x * p > /2 .
Note that /2 is a very small number
(5.3 x 10-35 J-sec).
Heisenberg Uncertainty Principle
If we were to apply this to a steel ball of mass
.002 kg +/- .00002 kg, rolling at a speed of
2 m/s +/- .02 m/s, the uncertainty in
momentum would be 4 x 10-7 kg*m/s .
From the H.U.P, then, the best we could be
sure of the position of the steel ball would
be: x = 5.3 x 10-35 J*s / 4 x 10-7 kg*m/s
= 1.3 x 10-28 m !
Heisenberg Uncertainty Principle
As we have just demonstrated, the H.U.P.
comes into play only when we are dealing
with very small particles (like individual
electrons or photons), not when we are
dealing with normal size objects!
Heisenberg Uncertainty Principle
If we apply this principle to the electron going
around the atom, then we know the electron
is somewhere near the atom,
(x = 2r = 1 x 10-10 m)
then there should be at least some uncertainty
in the momentum of the atom:
px > 5 x 10-35 J*s / 1 x 10-10 m = 5 x 10-25 m/s
Heisenberg Uncertainty Principle
Solving for p = mv from the Bohr theory
[KE + PE = Etotal, (1/2)mv2 - ke2/r = -13.6 eV
gives v = 2.2 x 106 m/s ] gives
p = (9.1 x 10-31 kg) * (2.2 x 106 m/s)
= 2 x 10-24 kg*m/s;
this means px is between -2 x 10-24 kg*m/s and
2 x 10-24 kg*m/s, with the minimum px
being 5 x 10-25 kg*m/s, or 25% of p.
Heisenberg Uncertainty Principle
Thus the H.U.P. says that we cannot really
know exactly where and how fast the
electron is going around the atom at any
particular time.
This is consistent with the idea that the
electron is actually a wave as it moves
around the electron.
Quantum Theory
But if an electron acts as a wave when it is
moving, WHAT IS WAVING?
When light acts as a wave when it is moving,
we have identified the
ELECTROMAGNETIC FIELD as waving.
But try to recall: what is the electric field?
Can we directly measure it?
Quantum Theory
Recall that by definition, E = F/q. We can
only determine that a field exists by
measuring an electric force! We have
become so used to working with the electric
and magnetic fields, that we tend to take
their existence for granted. They certainly
are a useful construct even if they don’t
exist.
Quantum Theory
We have four LAWS governing the electric
and magnetic fields: MAXWELL’S
EQUATIONS. By combining these laws
we can get a WAVE EQUATION for E&M
fields, and from this wave equation we can
get the speed of the E&M wave and even
more (such as reflection coefficients, etc.).
Quantum Theory
But what do we do for electron waves?
What laws or new law can we find that will
work to give us the wealth of predictive
power that MAXWELL’S EQUATIONS
have given us?
Quantum Theory
The way you get laws is try to explain
something you already know about, and
then see if you can generalize. A successful
law will explain what you already know
about, and predict things to look for that
you may not know about. This is where the
validity (or at least usefulness) of the law
can be confirmed.
Quantum Theory
Schrodinger started with the idea of
Conservation of Energy: KE + PE = Etotal .
He noted that
• KE = (1/2)mv2 = p2/2m, and that =h/p, so
that p = h/ = (h/2)*(2/) = k = p, so
KE = 2k2/2m
• Etotal = hf = (h/2)*(2f) = .
Quantum Theory
He then took a nice sine wave, and called
whatever was waving, :
(x,t) = A sin(kx-t) = Aei(kx-t) .
He noted that both k and  were in the
exponent, and could be gotten down by
differentiating. So he tried operators:
Quantum Theory
(x,t) = A sin(kx-t) = Aei(kx-t) .
pop = i[d/dx] = i[-ikAe-i(kx-t)] = k
= (h/2)*(2/)* = (h/ = p .
similary:
Eop= i[d/dt] = i[-iAei(kx-t)] = 
= ((h/2)*(2f)* = (hf = E .
Quantum Theory
Conservation of Energy: KE + PE = Etotal
becomes with the momentum and energy
operators:
-(2/2m)*(d2/dx2) + PE* = i(d/dt)
which is called SCHRODINGER’S
EQUATION. If it works for more than the
free electron, then we can call it a LAW.
Quantum Theory
What is waving here? 
What do we call ? the wavefunction
Schrodinger’s Equation allows us to solve for
the wavefunction. The operators then allow
us to find out information about the
electron, such as its energy and its
momentum.
Quantum Theory
To get a better handle on , let’s consider
light: how did the E&M wave relate to the
photon?
Quantum Theory
The photon was the basic unit of energy for
the light. The energy in the wave depended
on the field strength squared.
[Recall energy in capacitor, Energy = (1/2)CV2,
where for parallel plates, Efield = V/d and
C// = KoA/d, so that
Energy = (1/2)*(KoA/d)*(Efieldd)2
= KoEfield2 * Vol, or Energy Efield2.]
Quantum Theory
Since Energy is proportional to field strength
squared, AND energy is proportional to the
number of photons, THEN that implies that
the number of photons is proportional to the
square of the field strength.
This then can be interpreted to mean that the
square of the field strength is related to
the probability of finding a photon.
Quantum Theory
In the same way, the square of the
wavefunction is related to the probability of
find the electron!
Since the wavefunction is a function of both x
and t, the the probability of finding the
electron is also a function of x and t!
Prob(x,t) = (x,t)2
Quantum Theory
Different situations for the electron, like being
in the hydrogen atom, will show up in
Schrodinger’s Equation in the PE part.
Different PE functions (like PE = -ke2/r for
the hydrogen atom) will cause the solution
to Schrodinger’s equation to be different,
just like different PE functions in the
normal Conservation of Energy will cause
different speeds to result for the particles.