Download Quantum Mechanics

Notes : _____
Quantum Mechanics
AP Physics B
Quantum?
Quantum mechanics is the study of processes
which occur at the atomic scale.
The word "quantum" is derived
From Latin to mean BUNDLE.
Therefore, we are studying the motion of
objects that come in small bundles called
quanta. These tiny bundles that we are
referring to are electrons traveling around
the nucleus.
“Newton, forgive me..”, Albert Einstein
At the atomic scale Newtonian Mechanics
cannot seem to describe the motion of
particles. An electron trajectory between
two points for example IS NOT a perfect
parabolic trajectory as Newton's Laws
predicts. Where Newton's Laws end
Quantum Mechanics takes over.....IN A
BIG WAY!
One of the most popular concepts
concerning Quantum Mechanics is called
, “The Photoelectric Effect”. In 1905,
Albert Einstein published this theory for
which he won the Nobel Prize in 1921.
What is the Photoelectric Effect?
In very basic terms, it is when electrons are
released from a certain type of metal upon
receiving enough energy from incident light.
So basically, light comes down and strikes the
metal. If the energy of the light wave is
sufficient, the electron will then shoot out of the
metal with some velocity and kinetic energy.
The Electron-Volt = ENERGY
Before we begin to discuss the photoelectric
effect, we must introduce a new type of unit.
Recall:
This is a very useful unit as it shortens our calculations and allows us
to stray away from using exponents.
The Photoelectric Effect
"When light strikes a material, electrons are
emitted. The radiant energy supplies the work
necessary to free the electrons from the
surface."
Photoelectric Fact #1
The LIGHT ENERGY (E) is in the form of quanta
called PHOTONS. Since light is an
electromagnetic wave it has an oscillating
electric field. The more intense the light the
more the field oscillates. In other words, its
frequency is greater.
Light Review
c  f
c  speed of light  constant(v acuum)
c  3 x108 m / s
c
 f , inverse relationsh ip between  & f

if  , f  and vice versa...
More on Fact #1
Make sure you USE the correct constant!
h
6.63x10-34 Js
4.14x10-15 eVs
E
E  f  E  hf   h
f
c
hc
 f E


hc
1.99x10-25 Jm
1.24x103 eVnm
Planck’s Constant is the SLOPE of an
Energy vs. Frequency graph!
Photoelectric Fact #2
The frequency of radiation must be above a certain value
before the energy is enough. This minimum frequency
required by the source of electromagnetic radiation to just
liberate electrons from the metal is known as threshold
frequency, f0.
The threshold frequency
is the X-intercept of the
Energy vs. Frequency
graph!
Photoelectric Fact #3
Work function, f, is defined as the least energy
that must be supplied to remove a free electron
from the surface of the metal, against the
attractive forces of surrounding positive ions.
Shown here is a PHOTOCELL. When
incident light of appropriate frequency
strikes the metal (cathode), the light
supplies energy to the electron. The
energy need to remove the electron
from the surface is the WORK!
Not ALL of the energy goes into work!
As you can see the electron then
MOVES across the GAP to the anode
with a certain speed and kinetic
energy.
Photoelectric Fact #4
The MAXIMUM KINETIC ENERGY is the energy difference between
the MINIMUM AMOUNT of energy needed (ie. the work function)
and the LIGHT ENERGY of the incident photon.
Light Energy, E
The energy NOT used
to do work goes into
KINETIC ENERGY as
the electron LEAVES
the surface.
WORK done to
remove the electron
THE BOTTOM LINE: Energy Conservation must still hold true!
Putting it all together
E  hf
K  W  hf
K  hf  W  K  hf  f
y  mx  b
KINETIC ENERGY can be plotted on the y axis and FREQUENCY on the xaxis. The WORK FUNCTION is the y – intercept as the THRESHOLD
FREQUNECY is the x intercept. PLANCK‘S CONSTANT is the slope of the
graph.



KE max = hf – hfo
KE max = hf ( energy of the original photon )
– hfo (Work function- property of the metal)
Work function is the minimum energy needed
to free an electron.
PROBLEM




Light with frequency of 2 X 10 15 hertz is
incident on a piece of copper.
A. what is the energy of light in joules and in
electron volt?
If the work function for copper is 4.5 eV ,
what is the maximum kinetic energy , in
electron volts of the emitted electrons ?
Note : 1 eV = 1.6 X 10-19 J
A. the energy in joules is given by E= hf
E = hf = (6.63X 10 -34 ) ( 2 X 1015) =
E= 1.326X10-16J

Since 1eV = 1.6 X 10-19 J
E= 8.28 eV
B. To find the maximum kinetic energy , we simply
subtract the work function from the photon energy .
KEmax = 8.28 V – 4.5 V = 3.79eV
Can we use this idea in a circuit?
We can then use this photoelectric effect idea to
create a circuit using incident light. Of course,
we now realize that the frequency of light must
be of a minimum frequency for this work.
Notice the + and – on the photocell itself. We
recognize this as being a POTENTIAL
DIFFERENCE or Voltage. This difference in
voltage is represented as a GAP that the
electron has to jump so that the circuit works
What is the GAP or POTENTIAL DIFFERENCE is too large?
Photoelectric Fact #5 - Stopping Potential
If the voltage is TOO LARGE the electrons WILL NOT have
enough energy to jump the gap. We call this VOLTAGE point
the STOPPING POTENTIAL.
If the voltage exceeds this value, no photons will be emitted no
matter how intense. Therefore it appears that the voltage has
all the control over whether the photon will be emitted and thus
has kinetic energy.
Wave-Particle Duality
The results of the photoelectric effect allowed
us to look at light completely different.
First we have Thomas Young’s
Diffraction experiment proving that
light behaved as a WAVE due to
constructive and destructive
interference.
Then we have Max Planck who allowed Einstein to build his
photoelectric effect idea around the concept that light is composed of
PARTICLES called quanta.
This led to new questions….
If light is a WAVE and is ALSO a particle, does
that mean ALL MATTER behave as waves?
That was the question that Louis de Broglie
pondered. He used Einstein's famous equation to
answer this question.
YOU are a matter WAVE!
Basically all matter could be said to
have a momentum as it moves.
The momentum however is
inversely proportional to the
wavelength. So since your
momentum would be large
normally, your wavelength would
be too small to measure for any
practical purposes.
An electron, however, due to it’s
mass, would have a very small
momentum relative to a person
and thus a large enough
wavelength to measure thus
producing measurable results.
This led us to start using the Electron
Microscopes rather than traditional
Light microscopes.
Problem
Find the de Broglie wavelength for each of the
following :
a. A 10g stone moving with a velocity of 20m/s
b. An electron (9.1 X 10-31) kg moving with a
velocity of 1 X 107 m/s
since m =10g = 0.01 kg and λ = h/mv
λ = 6.63 X 10 -34
----------------= 3.315 X 10 -33 m
(0.01)( 20)
b. In this part , we have
λ = h / mv = 6.63 X 10 -34
----------------- =
7.3 X 10-11 m
(9.1 X 10-31)(1X 10-7)
a.
The electron microscope
After the specimen is prepped. It
is blasted by a bean of
electrons. As the incident
electrons strike the surface,
electrons are released from
the surface of the specimen.
The deBroglie wavelength of
these released electrons vary
in wavelength which can then
be converted to a signal by
which a 3D picture can then
be created based on the
signals captured by the
detector.
CW: Problems
1. What is the Broglie wavelength for a proton (
m = 1.67X 10-27 kg ) with a velocity of
6 X 10 7 m/s
2. What is the momentum associated with
yellow light that has wavelength of 5,500 Å
1Å = Å X 10-10 m

1.

2.
6.6 X 10 -15 m
1.2 X 10 -27 kg m/s
Atomic & Nuclear Physics
AP Physics B
Objectives: After completing this
module, you should be able to:
• Define and apply the concepts of mass
number, atomic number, and isotopes.
• Calculate the mass defect and the binding
energy per nucleon for a particular isotope.
• Define and apply concepts of radioactive decay
and nuclear reactions.
• State the various conservation laws, and
discuss their application for nuclear reactions.
Life and Atoms
Every time you breathe you are taking
in atoms. Oxygen atoms to be
exact. These atoms react with the
blood and are carried to every cell
in your body for various reactions
you need to survive. Likewise,
every time you breathe out carbon
dioxide atoms are released.
The cycle here is interesting.
TAKING SOMETHING IN.
ALLOWING SOMETHING OUT!
The Atom
As you probably already know an
atom is the building block of all
matter. It has a nucleus with
protons and neutrons and an
electron cloud outside of the
nucleus where electrons are
orbiting and MOVING.
Depending on the ELEMENT, the
amount of electrons differs as
well as the amounts of orbits
surrounding the atom.
Composition of Matter
All of matter is composed of at least three fundamental particles
(approximations):
Particle
Fig.Sym
Mass
Charge
Size
9.11 x 10-31 kg -1.6 x 10-19 C

Electron
e-
Proton
p
1.673 x 10-27 kg +1.6 x 10-19 C 3 fm
Neutron
n
1.675 x 10-31 kg
0
3 fm
The mass of the proton and neutron are close, but they are about 1840 times the
mass of an electron.
The Atomic Nucleus
Compacted nucleus:
4 protons
5 neutrons
Since atom is electri-cally
neutral, there must be 4
electrons.
4 electrons
Beryllium Atom
Modern Atomic Theory
The Bohr atom, which is
sometimes shown with
electrons as planetary
particles, is no longer a valid
representation of an atom, but
it is used here to simplify our
discussion of energy levels.
The uncertain position of an
electron is now described as a
probability distribution—loosely
referred to as an electron cloud.
Definitions
A nucleon is a general term to denote a nuclear particle - that is, either a
proton or a neutron.
The atomic number Z of an element is equal to the number of protons in the
nucleus of that element.
The mass number A of an element is equal to the total number of nucleons
(protons + neutrons).
The mass number A of any element is equal to the sum of the atomic
number Z and the number of neutrons N :
A=N+Z
Symbol Notation
A convenient way of describing an element is by giving its mass number and
its atomic number, along with the chemical symbol for that element.
A
Z
X
Mass number
Atomic number
For example, consider beryllium (Be):
 Symbol 
9
4
Be
Example 1: Describe the nucleus of a lithium atom
which has a mass number of 7 and an atomic
number of 3.
A = 7; Z = 3; N = ?
N=A–Z= 7-3
neutrons: N = 4
Protons:
Z=3
Electrons: Same as Z
7
3
Li
Lithium Atom
When the atom gets excited or NOT
To help visualize the atom think of it like a ladder. The bottom of
the ladder is called GROUND STATE where all electrons would
like to exist. If energy is ABSORBED it moves to a new rung on
the ladder or ENERGY LEVEL called an EXCITED STATE.
This state is AWAY from the nucleus.
As energy is RELEASED the electron can relax by moving to a
new energy level or rung down the ladder.
Energy Levels
Yet something interesting happens as
the electron travels from energy
level to energy level.
If an electron is EXCITED, that means
energy is ABSORBED and
therefore a PHOTON is absorbed.
If an electron is DE-EXCITED, that
means energy is RELEASED and
therefore a photon is released.
We call these leaps from energy level
to energy level QUANTUM LEAPS.
Since a PHOTON is emitted that
means that it MUST have a certain
wavelength.
Energy of the Photon
We can calculate the ENERGY
of the released or absorbed
photon provided we know the
initial and final state of the
electron that jumps energy
levels.
Energy Level Diagrams
To represent these
transitions we can
construct an ENERGY
LEVEL DIAGRAM
Note: It is very important to understanding that these transitions DO NOT
have to occur as a single jump! It might make TWO JUMPS to get back to
ground state. If that is the case, TWO photons will be emitted, each with a
different wavelength and energy.
Example
An electron releases energy
as it moves back to its
ground state position. As a
result, photons are
emitted. Calculate the
POSSIBLE wavelengths of
the emitted photons.
Notice that they give us the
energy of each energy
level. This will allow us to
calculate the CHANGE in
ENERGY that goes to the
emitted photon.
This particular sample will release three
different wavelengths, with TWO being
the visible range ( RED, VIOLET) and
ONE being OUTSIDE the visible range
(INFRARED)
Energy levels Application: Spectroscopy
Spectroscopy is an optical technique by which we can
IDENTIFY a material based on its emission
spectrum. It is heavily used in Astronomy and
Remote Sensing. There are too many subcategories
to mention here but the one you are probably the
most familiar with are flame tests.
When an electron gets excited inside
a SPECIFIC ELEMENT, the electron
releases a photon. This photon’s
wavelength corresponds to the
energy level jump and can be used
to indentify the element.
Different Elements = Different Emission
Lines
Emission Line Spectra
So basically you could look at light
from any element of which the
electrons emit photons. If you
look at the light with a diffraction
grating the lines will appear as
sharp spectral lines occurring at
specific energies and specific
wavelengths. This phenomenon
allows us to analyze the
atmosphere of planets or
galaxies simply by looking at the
light being emitted from them.
Nuclear Physics - Radioactivity
Before we begin to discuss the specifics of radioactive
decay we need to be certain you understand the
proper NOTATION that is used.
To the left is your typical radioactive
isotope.
Top number = mass number = #protons
+ neutrons. It is represented by the
letter "A“
Bottom number = atomic number = # of
protons in the nucleus. It is represented
by the letter "Z"
Nuclear Physics – Notation & Isotopes
An isotope is when you have
the SAME ELEMENT, yet
it has a different MASS.
This is a result of have
extra neutrons. Since
Carbon is always going to
be element #6, we can
write Carbon in terms of its
mass instead.
Carbon - 12
Carbon - 14
Isotopes of Elements
Isotopes are atoms that have the same number of protons (Z1= Z2), but a
different number of neutrons (N). (A1  A2)
3
2
He
Helium - 3
Isotopes of
helium
4
2
He
Helium - 4
Nuclides
Because of the existence of so many isotopes, the term
element is sometimes confusing. The term nuclide is better.
A nuclide is an atom that has a definite mass number A and
Z-number. A list of nuclides will include isotopes.
The following are best described as nuclides:
3
2
He
4
2
He
12
6
C
13
6
C
Atomic Mass Unit, u
One atomic mass unit (1 u) is equal to one-twelfth of the mass of the
most abundant form of the carbon atom--carbon-12.
Atomic mass unit: 1 u = 1.6606 x 10-27 kg
Common atomic masses:
Proton: 1.007276 u
Neutron: 1.008665 u
Electron: 0.00055 u
Hydrogen: 1.007825 u
Einstein – Energy/Mass Equivalence
In 1905, Albert Einstein publishes a 2nd major
theory called the Energy-Mass
Equivalence in a paper called, “Does the
inertia of a body depend on its energy
content?”
Einstein – Energy/Mass Equivalence
Looking closely at Einstein’s equation we see that he
postulated that mass held an enormous amount of
energy within itself. We call this energy BINDING
ENERGY or Rest mass energy as it is the energy
that holds the atom together when it is at rest. The
large amount of energy comes from the fact that the
speed of light is squared.
Energy Unit Check
2
m
EB  mc  Joule  kg  2
s
W  Fx  Joule  Nm
m
Fnet  ma  N  kg  2
s
2
m
m
E  W  kg  2  m  kg  2
s
s
2
Mass Defect
The nucleus of the atom is held together by a STRONG NUCLEAR
FORCE.
The more stable the nucleus, the more energy needed to break it apart.
Energy need to break to break the nucleus into protons and neutrons is
called the Binding Energy
Einstein discovered that the mass of the separated particles is greater
than the mass of the intact stable nucleus to begin with.
This difference in mass (m) is called the mass defect.

Particle Charge
Mass (g) Mass (amu)

Proton +1
1.6727 x 10-24 g
1.007316

Neutron 0
1.6750 x 10-24 g
1.008701

Electron -1 9.110 x 10-28 g
0.000549
Mass Defect - Explained
The extra mass turns into energy
holding the atom together.
Binding Energy is proportional to the mass
defect .
The energy equivalent of 1 atomic mass unit υ
Is equal to 931.5 million electron volts ( MeV)
Mass Defect = Predicted – Isotope Mass or
experimental value
Binding Energy = Mass Defect X 931.5 MeV
Avg BE = BE / A
A = mass number
Exampe 2: The average atomic mass of Boron-11 is
11.009305 u. What is the mass of the nucleus of one
boron atom in kg?
11
5
B = 11.009305
Electron: 0.00055 u
The mass of the nucleus is the atomic mass less the mass of Z = 5
electrons:
Mass = 11.009305 u – 5(0.00055 u)
1 boron nucleus = 11.00656 u
 1.6606 x 10-27 kg 
m  11.00656 u 

1
u


m = 1.83 x 10-26 kg
Mass and Energy
Recall Einstein’s equivalency formula for m and E:
E  mc ; c  3 x 10 m/s
2
8
The energy of a mass of 1 u can be found:
E = (1 u)c2 = (1.66 x 10-27 kg)(3 x 108 m/s)2
E = 1.49 x 10-10 J
When converting amu to
energy:
Or
E = 931.5 MeV
c  931.5
2
MeV
u
Example 3: What is the rest mass energy of a proton
(1.007276 u)?
E = mc2 = (1.00726 u)(931.5 MeV/u)
Proton: E = 938.3 MeV
Similar conversions show other rest mass energies:
Neutron: E = 939.6 MeV
Electron: E = 0.511 MeV
The Mass Defect
The mass defect is the difference between the rest mass of a nucleus
and the sum of the rest masses of its constituent nucleons.
The whole is less than the sum of the parts! Consider the carbon-12
atom (12.00000 u):
Nuclear mass = Mass of atom – Electron masses
= 12.00000 u – 6(0.00055 u)
= 11.996706 u
The nucleus of the carbon-12 atom has this mass.
(Continued . . .)
Mass Defect (Continued)
Mass of carbon-12 nucleus: 11.996706
Proton: 1.007276 u
Neutron: 1.008665 u
The nucleus contains 6 protons and 6 neutrons:
6 p = 6(1.007276 u) = 6.043656 u
6 n = 6(1.008665 u) = 6.051990 u
Total mass of parts: = 12.095646 u
Mass defect mD = 12.095646 u – 11.996706 u
mD = 0.098940 u
The Binding Energy
The binding energy EB of a nucleus is the energy required to
separate a nucleus into its constituent parts.
EB = mDc2 where c2 = 931.5 MeV/u
The binding energy for the carbon-12 example is:
EB = (0.098940 u)(931.5 MeV/u)
Binding EB for C-12:
EB = 92.2 MeV
Binding Energy per Nucleon
An important way of comparing the nuclei of atoms is finding their binding
energy per nucleon:
Binding energy per
nucleon
EB
 MeV 
= 

A
 nucleon 
For our C-12 example A = 12 and:
EB 92.2 MeV
MeV

 7.68 nucleon
A
12
Formula for Mass Defect
The following formula is useful for mass defect:
Mass defect mD
mH = 1.007825 u;
mD   ZmH  Nmn   M 
mn = 1.008665 u
Z is atomic number; N is neutron number; M is mass of atom
(including electrons).
By using the mass of the hydrogen atom, you avoid the necessity of subtracting
electron masses.
Example 4: Find the mass defect for the nucleus of
4
helium-4. (M = 4.002603 u)
2 He
Mass defect mD
mD   ZmH  Nmn   M 
ZmH = (2)(1.007825 u) = 2.015650 u
Nmn = (2)(1.008665 u) = 2.017330 u
M = 4.002603 u (From nuclide tables)
mD = (2.015650 u + 2.017330 u) - 4.002603 u
mD = 0.030377 u
Example 4 (Cont.) Find the binding energy per
nucleon for helium-4. (mD = 0.030377 u)
EB = mDc2 where c2 = 931.5 MeV/u
EB = (0.030377 u)(931.5 MeV/u) = 28.3 MeV
A total of 28.3 MeV is required To tear apart the nucleons from the He-4
atom.
Since there are four nucleons, we find that
EB 28.3 MeV

 7.07
A
4
MeV
nucleon
Mass Defect –
Example
Problem :
Helium -4 nucleus contains 2 protons( each
with a mass of 1.0087u ) and 2 neutrons ( each
has a mass of 1.0087 u)
The total predicted mass would be 4.0330 u .
Experimentally , the Helium -4 is found to be
4.0026u .
What is the Mass defect ?
What is the Binding energy in MeV?
Radioactivity
When an unstable nucleus releases energy and/or
particles.
Radioactive Decay
There are 4 basic types of
radioactive decay
 Alpha – Ejected Helium
 Beta – Ejected Electron
 Positron – Ejected AntiBeta particle
 Gamma – Ejected Energy
You may encounter protons
and neutrons being emitted
as well
4
2
He
0
1
0
1
e
e
0
0

1
1
p
1
0
n
Alpha Decay
240
94
Pu U  He
236
92
4
2
Alpha Decay Applications
241
95
Am He  ?
4
2
A
Z
Americium-241, an alpha-emitter, is used in smoke detectors. The alpha particles
ionize air between a small gap. A small current is passed through that ionized air.
Smoke particles from fire that enter the air gap reduce the current flow, sounding
the alarm.
Beta Decay
There aren’t really any
applications of beta decay
other than Betavoltaics which
makes batteries from beta
emitters. Beta decay, did
however, lead us to discover
the neutrino.
228
88
Ra  e Ac
0
1
228
89
Beta Plus Decay - Positron
Isotopes which undergo this decay
and thereby emit positrons include
carbon-11, potassium-40, nitrogen13, oxygen-15, fluorine-18, and
iodine-121.
230
91
Pa e Th
0
1
230
90
Beta Plus Decay Application - Positron
emission tomography (PET)
Positron emission tomography
(PET) is a nuclear medicine
imaging technique which
produces a three-dimensional
image or picture of functional
processes in the body. The
system detects pairs of gamma
rays emitted indirectly by a
positron-emitting radionuclide
(tracer), which is introduced
into the body on a biologically
active molecule. Images of
tracer concentration in 3dimensional space within the
body are then reconstructed by
computer analysis.
Gamma Decay
240
94
Pu Pu  
240
94
0
0
Gamma Decay Applications
Gamma rays are the most dangerous type of radiation
as they are very penetrating. They can be used to
kill living organisms and sterilize medical equipment
before use. They can be used in CT Scans and
radiation therapy.
Gamma Rays are used to view stowaways inside of a truck. This
technology is used by the Department of Homeland Security at many ports
of entry to the US.
Significant Nuclear Reactions - Fusion
2
1
H  H  He  n
3
1
4
2
1
0
nuclear fusion is the process by which multiple like-charged atomic nuclei
join together to form a heavier nucleus. It is accompanied by the release or
absorption of energy.
Fusion Applications - IFE
In an IFE (Inertial Fusion Energy) power plant, many (typically
5-10) pulses of fusion energy per second would heat a lowactivation coolant, such as lithium-bearing liquid metals or molten
salts, surrounding the fusion targets. The coolant in turn would
transfer the fusion heat to a power conversion system to produce
electricity.
CW: PROBLEMS
COMPLETE THE EQUATIONS:
1.
238
4
92U
2He
+ ?
What decay is this ?
2.
234
234
90 Th
91
Pa + ?
What kind of decay is this ?
3.
64
29Cu
o
+
-1e
?
What process is the above equation ?
4.
64
29
o
Cu
1
What decay is this ?
e + ?
1.
234
91 Th
Alpha particle decay
2. o
-1 e
Beta decay
3.
64
Ni
electron capture
28
3. Transmutation
4.
64
Ni
positron emission
28
Significant Nuclear Reactions - Fission
n U  Ba  Kr 3 n  energy
1
235
141
92
1
0
92
56
36
0
Nuclear fission differs from other forms of radioactive decay in that it can be
harnessed and controlled via a chain reaction: free neutrons released by
each fission event can trigger yet more events, which in turn release more
neutrons and cause more fissions. The most common nuclear fuels are 235U
(the isotope of uranium with an atomic mass of 235 and of use in nuclear
reactors) and 239Pu (the isotope of plutonium with an atomic mass of 239).
These fuels break apart into a bimodal range of chemical elements with
atomic masses centering near 95 and 135 u (fission products).
Fission Bomb
One class of nuclear weapon, a fission
bomb (not to be confused with the
fusion bomb), otherwise known as
an atomic bomb or atom bomb, is a
fission reactor designed to liberate
as much energy as possible as
rapidly as possible, before the
released energy causes the reactor
to explode (and the chain reaction to
stop).
A nuclear reactor is a device in
which nuclear chain fission reactions
are initiated, controlled, and
sustained at a steady rate, as
opposed to a nuclear bomb, in which
the chain reaction occurs in a
fraction of a second and is
uncontrolled causing an explosion.
Atomic Physics &
Quantum Effects
AP Physics B
Blackbody radiation

A blackbody absorbs all incident light rays.

All bodies, no matter how hot or cold, emit electromagnetic
waves. We can see the waves emitted by very hot objects
because they are within the visible spectrum (light bulb filament;
red-hot metal). At lower temps we can’t see the waves but they
are still there. For example, the human body emits waves in the
infrared range. This is why we can use infrared detecting
devices to “see” in the dark.

The distribution of energy in blackbody radiation is
independent of the material from which the blackbody is
constructed – it depends only on the temperature.
The diagrams below show the intensity of various
frequencies of EM radiation emitted by blackbodies of
various temperatures. Note that as the temperature
increases, the energy emitted (area under curve) increases
and the peak in the radiation shifts to higher frequencies.
This is important because classical physics predicts a
completely different curve that increases to infinite
intensity in the ultraviolet region. (thus called the
Ultraviolet Catastrophe). The only way to make sense of
this finding is by saying energy is quantized (Planck’s
quantum hypothesis)
Quantum?
Quantum mechanics is the study of processes
which occur at the atomic scale.
The word "quantum" is derived
From Latin to mean BUNDLE.
Therefore, we are studying the motion of
objects that come in small bundles called
quanta. These tiny bundles that we are
referring to are electrons traveling around
the nucleus.
“Newton, forgive me..”, Albert Einstein
At the atomic scale Newtonian Mechanics
cannot seem to describe the motion of
particles. An electron trajectory between
two points for example IS NOT a perfect
parabolic trajectory as Newton's Laws
predicts. Where Newton's Laws end
Quantum Mechanics takes over.....IN A
BIG WAY!
One of the most popular concepts
concerning Quantum Mechanics is called
, “The Photoelectric Effect”. In 1905,
Albert Einstein published this theory for
which he won the Nobel Prize in 1921.
What is the Photoelectric Effect?
In very basic terms, it is when electrons are
released from a certain type of metal upon
receiving enough energy from incident light.
So basically, light comes down and strikes the
metal. If the energy of the light wave is
sufficient, the electron will then shoot out of the
metal with some velocity and kinetic energy.
The Electron-Volt = ENERGY
Before we begin to discuss the photoelectric
effect, we must introduce a new type of unit.
Recall:
This is a very useful unit as it shortens our calculations and allows us
to stray away from using exponents.
The Photoelectric Effect
"When light strikes a material, electrons are
emitted. The radiant energy supplies the work
necessary to free the electrons from the
surface."
Photoelectric Fact #1
The LIGHT ENERGY (E) is in the form of quanta
called PHOTONS. Since light is an
electromagnetic wave it has an oscillating
electric field. The more intense the light the
more the field oscillates. In other words, its
frequency is greater.
Light Review
c  f
c  speed of light  constant(v acuum)
c  3 x108 m / s
c
 f , inverse relationsh ip between  & f

if  , f  and vice versa...
More on Fact #1
Make sure you USE the correct constant!
h
6.63x10-34 Js
4.14x10-15 eVs
E
E  f  E  hf   h
f
c
hc
 f E


hc
1.99x10-25 Jm
1.24x103 eVnm
Planck’s Constant is the SLOPE of an
Energy vs. Frequency graph!
Photoelectric Fact #2
The frequency of radiation must be above a certain value
before the energy is enough. This minimum frequency
required by the source of electromagnetic radiation to just
liberate electrons from the metal is known as threshold
frequency, f0.
The threshold frequency
is the X-intercept of the
Energy vs. Frequency
graph!
Photoelectric Fact #3
Work function, f, is defined as the least energy
that must be supplied to remove a free electron
from the surface of the metal, against the
attractive forces of surrounding positive ions.
Shown here is a PHOTOCELL. When
incident light of appropriate frequency
strikes the metal (cathode), the light
supplies energy to the electron. The
energy need to remove the electron
from the surface is the WORK!
Not ALL of the energy goes into work!
As you can see the electron then
MOVES across the GAP to the anode
with a certain speed and kinetic
energy.
Photoelectric Fact #4
The MAXIMUM KINETIC ENERGY is the energy difference between
the MINIMUM AMOUNT of energy needed (ie. the work function)
and the LIGHT ENERGY of the incident photon.
Light Energy, E
The energy NOT used
to do work goes into
KINETIC ENERGY as
the electron LEAVES
the surface.
WORK done to
remove the electron
THE BOTTOM LINE: Energy Conservation must still hold true!
Putting it all together
E  hf
K  W  hf
K  hf  W  K  hf  f
y  mx  b
KINETIC ENERGY can be plotted on the y axis and FREQUENCY on the xaxis. The WORK FUNCTION is the y – intercept as the THRESHOLD
FREQUNECY is the x intercept. PLANCK‘S CONSTANT is the slope of the
graph.
Can we use this idea in a circuit?
We can then use this photoelectric effect idea to
create a circuit using incident light. Of course,
we now realize that the frequency of light must
be of a minimum frequency for this work.
Notice the + and – on the photocell itself. We
recognize this as being a POTENTIAL
DIFFERENCE or Voltage. This difference in
voltage is represented as a GAP that the
electron has to jump so that the circuit works
What is the GAP or POTENTIAL DIFFERENCE is too large?
Photoelectric Fact #5 - Stopping Potential
If the voltage is TOO LARGE the electrons WILL NOT have
enough energy to jump the gap. We call this VOLTAGE point
the STOPPING POTENTIAL.
If the voltage exceeds this value, no photons will be emitted no
matter how intense. Therefore it appears that the voltage has
all the control over whether the photon will be emitted and thus
has kinetic energy.
Importance of photoelectric effect:

Classical physics predicts that any frequency of light can eject
electrons as long as the intensity is high enough.

Experimental data shows there is a minimum (cutoff frequency)
that the light must have.

Classical physics predicts that the kinetic energy of the ejected
electrons should increase with the intensity of the light.

Again, experimental data shows this is not the case; increasing
the intensity of the light only increases the number of electrons
emitted, not their kinetic energy.

THUS the photoelectric effect is strong evidence for the photon
model of light.
Wave-Particle Duality
The results of the photoelectric effect allowed
us to look at light completely different.
First we have Thomas Young’s
Diffraction experiment proving that
light behaved as a WAVE due to
constructive and destructive
interference.
Then we have Max Planck who allowed Einstein to build his
photoelectric effect idea around the concept that light is composed of
PARTICLES called quanta.
The momentum of the photon


Combining E=mc2 and p=mv, you get:
p / E = v / c2
The photon travels at the speed of light, so
v = c and p / E = 1 / c


Therefore the momentum, p, of the photon is
p=E/c
But we also know that E = hf and λ=c/f, so
hf
h
p

c

This led to new questions….
If light is a WAVE and is ALSO a particle, does
that mean ALL MATTER behave as waves?
That was the question that Louis de Broglie
pondered. He used Einstein's famous equation to
answer this question.
YOU are a matter WAVE!
Basically all matter could be said to
have a momentum as it moves.
The momentum however is
inversely proportional to the
wavelength. So since your
momentum would be large
normally, your wavelength would
be too small to measure for any
practical purposes.
An electron, however, due to it’s
mass, would have a very small
momentum relative to a person
and thus a large enough
wavelength to measure thus
producing measurable results.
This led us to start using the Electron
Microscopes rather than traditional
Light microscopes.
The electron microscope
After the specimen is prepped. It
is blasted by a bean of
electrons. As the incident
electrons strike the surface,
electrons are released from
the surface of the specimen.
The deBroglie wavelength of
these released electrons vary
in wavelength which can then
be converted to a signal by
which a 3D picture can then
be created based on the
signals captured by the
detector.
The Compton Effect:



when an X-ray photon strikes an electron in a piece of graphite,
the X-ray scatters in one direction, and the electron recoils in
another direction after the collision (like two billiard balls colliding
on a pool table)
the scattered photon has a frequency f ’ that is smaller than the
frequency f of the incident photon, thus the photon loses energy
in the collision
the difference between the two frequencies depends on the
angle at which the scattered photon leaves the collision
Similar to the analysis for the kinetic energy and work for
photoelectric effect, we find:
 The electron is assumed to be initially at rest and essentially free
(not bound to the atoms of the material)
 According to principle of conservation of energy:
hf = hf ’ + K
or
energy of incident photon = energy of scattered photon + KE of e

For an initially stationary electron, conservation of total linear
momentum requires that:
 Momentum of incident photon = momentum of scattered photon +
momentum of electron

From this point, these equations are combined with the
relativistic equations for energy and momentum to derive the
equation for Compton Scattering.

The difference between the wavelength λ’ of the
scattered photon and the wavelength λ of the
incident photon is related to the scattering angle θ
by:
h
1  cos  
   
mc

m is the mass of the electron. h/mc is the “Compton
wavelength of the electron” and is h/mc = 2.43x10-12 m.
It is interesting to note that according to Einstein’s
theory of relativity, the rest mass of a photon is zero.
However, it is never at rest, it is always moving (at
the speed of light!) so it does have a finite
momentum (even though p=mv doesn’t work)
The Davisson-Germer experiment demonstrated the wave nature of the
electron, confirming the earlier hypothesis of deBroglie.

Davisson and Germer measured the energies of electrons scattering from a metal
surface. Electrons from a heated filament were accelerated by a voltage and allowed
to strike the surface of nickel metal, which could be rotated to observe angular
dependence of the scattered electrons. They found that at certain angles there was a
peak in the intensity of the scattered electron beam. In fact, the electron beam was
scattered by the surface atoms on the nickel at the exact angles predicted for the
diffraction of x-rays according to Bragg's formula nλ=2dsinθ, with a wavelength given
by the de Broglie equation, λ=h/p. X-rays are accepted to be wavelike, thus this is
evidence for wavelike behavior of electron.
This is Davisson-Germer’s
data relating the
intensity of scattered
electrons as a function
of accelerating voltage
for a particular angle.
X-Rays


can be produced when electrons, accelerated through a large
potential difference, collide with a metal target (made from
molybdenum or platinum for example) contained within an
evacuated glass tube
a plot of X-ray intensity per unit wavelength versus wavelength
consists of sharp peaks or lines superimposed on a broad
continuous spectrum
X-ray spectrum
shown here is
produced when a
molybdenum target
is bombarded with
electrons that have
been accelerated
through a potential
difference







when the energetic electrons impact the target metal, they undergo a rapid
deceleration (braking). as the electrons suddenly come to rest they give off
high-energy radiation in the form of X-rays over a wide range of wavelengths.
This is referred to as “Bremsstrahlung continuum” (bremsstrahlung is German
for “braking radiation”) This is the base for the peaks seen in the graph
the sharp peaks are called characteristic lines or characteristic X-rays because
they are characteristic of the target material.
the characteristic lines are marked Kα and Kβ because they involve the n=1 or
K shell of a metal atom. (K shell is the innermost electron shell)
if an electron with enough energy strikes the target, one of the K-shell electrons
can be knocked entirely out of a target atom
an electron in one of the outer shells can then fall into the K shell, and an X-ray
photon is emitted in the process. Kα is a change from n=2 to n=1; Kβ is a
change from n=3 to n=1
there is a cutoff wavelength (as seen in the diagram). an impinging electron
cannot give up any more than all of its KE, thus an emitted X-ray photon can
have an energy no more than the KE of the impinging electron. the wavelength
that corresponds to this is the cutoff wavelength (max frequency-min
wavelength).
K=eV, E=hf thus eV=hf. f=c/λ, so…
hc
0 
eV
V is the potential difference applied across the X-ray tube; e is the charge of an
electron
Line Spectra



for a solid object the radiation emitted has a continuous range of
wavelengths, some of which are in the visible region of the
spectrum.
the continuous range of wavelengths is characteristic of the
entire collection of atoms that make up the solid.
in contrast, individual atoms, free of the strong interactions that
are present in a solid, emit only certain specific wavelengths,
rather than a continuous range. these wavelengths are
characteristic of the atom.
 a low-pressure gas in a sealed tube can be made to emit EM
waves by applying a sufficiently large potential difference
between two electrodes located within the tube
 with a grating spectroscope the individual wavelengths emitted by
the gas can be separated and identified as a series of bright
fringes or lines. these series of lines are called the LINE
SPECTRA.
Bohr Model of Hydrogen atom


Bohr tried to derive the formula that describes the line spectra that was
developed by Balmer using trial and error. he used Rutherford’s model
of the atom, the quantum ideas of Planck and Einstein, and the
traditional description of a particle in uniform circular motion.
assumptions of Bohr model:





electron in H moves in circular motion
only orbits where the angular momentum of the electron is equal to an
integer times Planck’s constant divided by 2π are allowed
electrons in allowed orbits do not radiate EM waves. Thus the orbits are
stable. (if it emitted radiation, it would lose energy and spiral into the
nucleus)
EM radiation is given off or absorbed only when an electron changes from
one allowed orbit to another. ΔE = hf
(ΔE is energy difference between
orbits and f is frequency of radiation emitted or absorbed)
Bohr theorized that a photon is emitted only when the electron changes
orbits from a larger one with a higher energy to a smaller one with a
lower energy.
Bohr Energy levels in electron volts

the centripetal force that provides the circular orbit is electrostatic force
(Coulomb’s Law). setting this equal to centripetal force we get:
mv 2
Ze 2
k 2
r
r
or
Ze 2
mv  k
r
2
note that electrostatic force between an electron and a nucleus with Z protons
is ke(Ze)/r2 = kZe2/r2. (Z = # of protons = atomic number)

angular momentum is L=rmv . the second assumption listed above states
the angular momentum of a given orbit, Ln, equals an integar, n, times h/2π;
or Ln=nh/2π. setting these equal and solving for v gives:
vn 


nh
2mrn
combining these two equations gives:
2
 nh 
Ze 2
  k
m
(note that n is the orbital number)
rn
 2mrn 
solving for rn gives us
Bohr orbital radius
2

 2
h

n
rn   2
2 
 4 mkZe 


energy of electron is sum of kinetic and potential energies:
E = K + U = ½ mv2 + U
the electrostatic potential energy is U = -kZe2/r and using the first equation
in this section we get:
1  Ze 2 
Ze 2
Ze 2
k
E   k
 k
2  r 
r
2r

substituting in the Bohr orbital radius, rn, from above:
 kZe2  4 2 mkZe2  1
kZe2

 2
En  
 
2
2rn
2
h


n

grouping all of the constants yields:
 2 2 mk 2 e 4  Z 2
En  


h
2
 2
n
plugging in values for the constants gives us:
Z2
En  (13.6eV ) 2
n
Z = atomic number; n = energy level (1,2,3,…)


when an electron in an initial orbit with a larger energy Ei, drops to a lower
orbit with energy Ef, the emitted photon has an energy of Ei – Ef , consistent
with the law of conservation of energy. combining this with Einstein’s E =
hf, we get
Ei – Ef = hf
To find the wavelengths in hydrogen’s line spectrum (Z=1), we apply that to
the equation above:
 2 2 mk 2 e 4  1
1 

E n  
 2
2
2

h
ni 

 n f

given that E = hc/λ :
 2 2 mk 2 e 4  1
1 

 
 2
3
2

hc
ni 

 n f
1

the value in the first parenthesis is only constants. calculating this value
gives the Rydberg Constant (1.097x107) – the exact value that Balmer
found
de Broglie Waves and Bohr Model



the angular momentum assumption in Bohr’s model is there because it
produces results in agreement with experiment.
however, de Broglie matter-wave relationship explains the significance:
think of matter (electron) waves as analogous to a wave on a string –
except that the string is a circle representing the electrons orbit. the
standing wave must fit an integral number of wavelengths into the
circumference of the orbit: nλ=2πr. combine this with p=h/λ:
p  mv 

h


nh
2r
rearranging and multiplying both sides of the equation by r gives us angular
momentum
nh
L  mvr 
2

Thus this condition of the Bohr model is a reflection of the wave nature of
matter
Life and Atoms
Every time you breathe you are taking
in atoms. Oxygen atoms to be
exact. These atoms react with the
blood and are carried to every cell
in your body for various reactions
you need to survive. Likewise,
every time you breathe out carbon
dioxide atoms are released.
The cycle here is interesting.
TAKING SOMETHING IN.
ALLOWING SOMETHING OUT!
The Atom
As you probably already know an
atom is the building block of all
matter. It has a nucleus with
protons and neutrons and an
electron cloud outside of the
nucleus where electrons are
orbiting and MOVING.
Depending on the ELEMENT, the
amount of electrons differs as
well as the amounts of orbits
surrounding the atom.
When the atom gets excited or NOT
To help visualize the atom think of it like a ladder. The bottom of
the ladder is called GROUND STATE where all electrons would
like to exist. If energy is ABSORBED it moves to a new rung on
the ladder or ENERGY LEVEL called an EXCITED STATE.
This state is AWAY from the nucleus.
As energy is RELEASED the electron can relax by moving to a
new energy level or rung down the ladder.
Energy Levels
Yet something interesting happens as
the electron travels from energy
level to energy level.
If an electron is EXCITED, that means
energy is ABSORBED and
therefore a PHOTON is absorbed.
If an electron is DE-EXCITED, that
means energy is RELEASED and
therefore a photon is released.
We call these leaps from energy level
to energy level QUANTUM LEAPS.
Since a PHOTON is emitted that
means that it MUST have a certain
wavelength.
Energy of the Photon
We can calculate the ENERGY
of the released or absorbed
photon provided we know the
initial and final state of the
electron that jumps energy
levels.
Energy Level Diagrams
To represent these
transitions we can
construct an ENERGY
LEVEL DIAGRAM
Note: It is very important to understanding that these transitions DO NOT
have to occur as a single jump! It might make TWO JUMPS to get back to
ground state. If that is the case, TWO photons will be emitted, each with a
different wavelength and energy.
Example
An electron releases energy
as it moves back to its
ground state position. As a
result, photons are
emitted. Calculate the
POSSIBLE wavelengths of
the emitted photons.
Notice that they give us the
energy of each energy
level. This will allow us to
calculate the CHANGE in
ENERGY that goes to the
emitted photon.
This particular sample will release three
different wavelengths, with TWO being
the visible range ( RED, VIOLET) and
ONE being OUTSIDE the visible range
(INFRARED)
Energy levels Application: Spectroscopy
Spectroscopy is an optical technique by which we can
IDENTIFY a material based on its emission
spectrum. It is heavily used in Astronomy and
Remote Sensing. There are too many subcategories
to mention here but the one you are probably the
most familiar with are flame tests.
When an electron gets excited inside
a SPECIFIC ELEMENT, the electron
releases a photon. This photon’s
wavelength corresponds to the
energy level jump and can be used
to indentify the element.
Different Elements = Different Emission
Lines
Emission Line Spectra
So basically you could look at light
from any element of which the
electrons emit photons. If you
look at the light with a diffraction
grating the lines will appear as
sharp spectral lines occurring at
specific energies and specific
wavelengths. This phenomenon
allows us to analyze the
atmosphere of planets or
galaxies simply by looking at the
light being emitted from them.
Line Spectra of Hydrogen Atom


Lyman series occurs when
electrons make transition s
from higher energy levels
with ni = 2, 3, 4, … to the
first energy level where nf =
1.
notice when an electron
transitions from n=2 to n=1,
the longest wavelength
photon in the Lyman series
is emitted, since the energy
change is the smallest
possible. when the
electron transitions from
highest to lowest, the
shortest wavelength is
emitted.