1. No category

Do Now (2/21/14):
What does the word “quantized” mean?
 Where have we seen quantization in
Physics?
 What is the structure of an atom?

Objectives

Define photoelectric effect and evidence of
particle properties of light.
 Define work function.
 Calculate energy of a photon and an
electron.
 Determine Planck’s constant.
Particles and Waves
Quantum Theory

Max Planck (1900) recognized
electromagnetic radiation is quantized as
E=hf.
 1905, Einstein proposed photon theory of
light. Supported by work in photoelectric
effect.
Photoelectric Effect

E = KE + W
 Energy of impinging light equals KE of
electron plus the work function.
 Intensity increases will increase current.
 Frequency changes affect KE.
2/28/12
Newton

Thought of light as particles
Maxwell’s Theory
 Light
is composed of crossed electric and
magnetic fields which make up a wave.
Experiments show that when
light shines on a metal surface,
the surface emits electrons.
Planck’s Work
In 1900, Max Planck came up with a formula to
explain radiation from objects, but the formula
only made sense if the energy of a vibrating
molecule was quantized.
What are some other
examples of
“quantization”?
Planck’s Constant
Einstein’s Theory
 Based
on Planck's work,
Einstein proposed that
light also delivers its
energy in chunks
 light consists of particles
(quanta) called photons,
each with an energy of
Planck's constant times its
frequency
Photon
a light quantum that is massless, has energy
and momentum, and travels at the speed
of light
The Photoelectric Effect
the emission of electrons
produced when
electromagnetic radiation
falls on certain materials
Threshold Frequency f0
the minimum frequency
of incident light which
can cause photo
electric emission
Energy of a photon
E=hf
h=Planck’s constant
f=frequency
Electron Volts
1 eV=
E
-19
1.6x10
hc


J
1240eV  nm
λ=wavelength

Example:
Calculate the wavelength and the energy of a photon of
light with frequency equal to 1.984 x 1014 Hz.
• Calculating the wavelength, from :
c=fλ  3x108=λ (1.984 x 1014 )
= 1.51 x 10-6 m
• Calculating the energy of the photon:
E = hf
E = 6.628 x 10-34 x 1.984 x 1014
= 1.31 x 10-19 J
KE of photon
hc hc
KE = E photon -  = hf - hf 0 =
-

hf0=min. energy to release electron
0
Stopping Potential (Vo)
The negative potential at which the photo electric
current becomes zero
Example:
The stopping potential of a certain photocell is
4 V. What is the KE given to the electrons
by the incident light?
KE=-W
KE=-qV0
KE=-(1.6x10-19)(4)=+6.4x10-19J
Work Function ϕ0
Minimum amount of energy which is necessary
to start photo electric emission.
It is a property of material. Different materials
have different values of work function.
Einstein’s Theory
hf =  + ½
hf : energy of
each photon
2
mv
Source: http://www.westga.edu/~chem/courses/chem410/410_08/sld017.htm
Kinetic energy of emitted electron
vs. Light frequency



Higher-frequency photons have more
energy, so they make electrons come out
faster; same intensity but a higher
frequency increases the max KE of the
emitted electrons.
If frequency is the same but intensity
higher , more electrons come out
(because there are more photons to hit
them), but they won't come out faster,
because each photon still has the same
energy.
if the frequency is low enough, then none
of the photons will have enough energy to
knock an electron out. If you use really
low-frequency light, you shouldn't get any
Source: http://online.cctt.org/physicslab/
content/PhyAPB/lessonnotes/dualnature/
electrons, no matter how high the
photoelectric.asp
intensity is. if you use a high frequency,
you should still knock out some electrons
even if the intensity is very low.
Simple Photoelectric Experiment
Source: http://sol.sci.uop.edu/~jfalward/particlesandwaves/phototube.jpg
Photoelectric Effect
Applications
Applications

The Photoelectric effect has numerous applications, for
example night vision devices take advantage of the effect.
Photons entering the device strike a plate which causes
electrons to be emitted, these pass through a disk consisting of
millions of channels, the current through these are amplified
and directed towards a fluorescent screen which glows when
electrons hit it. Image converters, image intensifiers, television
camera tubes, and image storage tubes also take advantage of
the point-by-point emission of the photocathode. In these
devices an optical image incident on a semitransparent
photocathode is used to transform the light image into an
“electron image.” The electrons released by each element of
the photoemitter are focused by an electron-optical device
onto a fluorescent screen, reconverting it in the process again
into an optical image
Applications: Night Vision
Device
http://www.lancs.ac.uk/ug/jacksom2/
Photoelectric Effect Applications

Source: http://chemistry.about.com/cs/howthingswork/a/aa071401a.htm
Photoelectric Detectors In one type of
photoelectric device, smoke can block a light
beam. In this case, the reduction in light
reaching a photocell sets off the alarm. In the
most common type of photoelectric unit,
however, light is scattered by smoke particles
onto a photocell, initiating an alarm. In this type
of detector there is a T-shaped chamber with a
light-emitting diode (LED) that shoots a beam of
light across the horizontal bar of the T. A
photocell, positioned at the bottom of the vertical
base of the T, generates a current when it is
exposed to light. Under smoke-free conditions,
the light beam crosses the top of the T in an
uninterrupted straight line, not striking the
photocell positioned at a right angle below the
beam. When smoke is present, the light is
scattered by smoke particles, and some of the
light is directed down the vertical part of the T to
strike the photocell. When sufficient light hits the
cell, the current triggers the alarm.
Photoelectric Smoke Detector
Source: http://www.bassburglaralarms.com/images_products/d350rpl_addressable_duct_smoke_detector_b10685.jpg
Applications

Solar panels are nothing more
than a series of metallic plates
that face the Sun and exploit the
photoelectric effect. The light
from the Sun will liberate
electrons, which can be used to
heat your home, run your lights,
or, in sufficient enough
quantities, power everything in
your home.
Source: www.futureenergy.org/ picsolarpannelsmatt.jpg
Work Cited
Amar, Francois G. The Photoelectric Effect. 25 Sep 2003. Section of Chemistry 121 for fall
03. 11 May 2006
<http://chemistry.umeche.maine.edu/~amar/fall2003/photoelectric.html>
Blawn, Jeramy R. and Colwell, Catharine H. Physics Lab: Photoelectric Effect. 10 Jun 2003.
Mainland High School: Online Physics Labs. 11 May 20006
<http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/dualnature/photoelectric.
asp>
Helmenstine, Anne Marie. Photoelectric & Ionization Smoke Detector. 25 Feb 2006.
About.com. 11 May 2006
<http://chemistry.about.com/cs/howthingswork/a/aa071401a.htm>
Einstein, Albert. “Concerning an Heuristic Point of View Toward the Emission and
Transformation of Light.” American Journal Of Physics 5 May 1965: 137.
Nave, Rod. HyperPhysics. 19 Aug. 2000. Georgia State University. 06 May 2006
<http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html> .
Thornton T., Stephen, and Rex, Andrew. Modern Physics for Scientists and Engineers.
Canada : Thomson Brooks/Core, 2006
Photoelectric Effect. 24 Apr. 2006. Wikipedia Free Encyclopedia. 05 May 2006.
<http://en.wikipedia.org/wiki/Photoelectric_effect>.
Do Now (2/25/14):
In your own words, describe the
photoelectric effect. Use the
words “work function,”
“threshold frequency,”
“electron,” and “photon,” at
least once in your paragraph.
Agenda:

Finish competition
 Complete Quantum lecture
 Complete Chapter 27 Multiple Choice
Questions
 Introduce tomorrow’s lab
White Board Competition!



Work in groups
For each correct question, make a tally in
the upper right hand corner of your board.
BE HONEST!!!
The teams with the most points at the end
will receive extra credit!
#1
According to Einstein, the energy of a
photon depends on the _________ of the
electromagnetic radiation.
A.momentum
B. speed
C. frequency
D. intensity
#2
The work function of iron is 4.7 eV.
What is the threshold wavelength
of iron?
A.2.60 nm
B. 260 nm
C. 470 nm
D. 2600 nm
#3
The stopping potential, V0, that prevents
electrons from flowing across a certain
photocell is 6.0 V. What is the kinetic
energy in J given to the electrons by the
incident light?
A.9.6 x 10-19 J
B.1.60 x 10-19 J
C.6.9 x 10-19 J
D. 6.4 x 10-19 J
#4
When light is directed on a metal surface, the
kinetic energies of the electrons
A. vary with the intensity of light
B. vary with the speed of light
C. vary with the frequency of the light
D. are random
#5
The threshold frequency for photoelectric
emission in copper is 1.1 x 1015 Hz. What is the
maximum kinetic energy in eV of the
photoelectrons when light of frequency 1.5 x
1015 Hz is directed on a copper surface?
A. 2.65 eV
B. 2.12 eV
C. 1.66 eV
D. 1.03 eV
#6
What will likely happen if a light whose
frequency is below the threshold frequency
hits a clean metal surface?
A. no electron will be ejected from the metal
B. fewer electrons will be ejected from the
metal
C. more electrons will be ejected from the
metal
D. ejected electrons will have higher kinetic
energy
#7
What is the work function of a
metal whose threshold frequency
is 3.5 x 1015 Hz?
A.2.32 x 10-18 J
B. 3.11 x 10-18 J
C. 3.65 x 10-18 J
D. 4.01 x 10-18 J
#8
What is the maximum wavelength of
light that will cause photoelectrons to
be emitted from sodium if the work
function of sodium is 2.3 eV?
A.1.75 x 10-7 m
B. 3.44 x 10-7 m
C. 5.40 x 10-7 m
D. 5.88 x 10-7 m
#9
What will the maximum kinetic energy
of the photoelectrons be if 200-nm
light falls on a sodium surface (work
function is 2.3 eV)?
A.2.96 x 10-19 J
B. 4.73 x 10-19 J
C. 5. 21 x 10-19 J
D. 6.26 x 10-19 J
#10
When 230-nm light falls on a metal, the current
through the photoelectric circuit is brought to
zero at a reverse voltage of 1.64 V. What is the
work function of the metal?
A. 4. 39 x 10-19 J
B. 5.38 x 10-19 J
C. 6.01 x 10-19 J
D. 7.11 x 10-19 J
#11
The current in a photoelectric effect experiment
decreases to zero when the retarding voltage is
raised to 1.25 V. What is the maximum speed of
the electrons?
A. 6.63 x 105 m/s
B. 5.53 x 105 m/s
C. 4.78 x 105 m/s
D. 4.19 x 105 m/s
#12
What is the maximum speed of an electron ejected
from a sodium surface whose work function is
2.28 eV when illuminated by light of wavelength
450 nm?
A. 3.25 x 105 m/s
B. 4.10 x 105 m/s
C. 4.85 x 105 m/s
D. 5.25 x 105 m/s
#13
Light is incident on the surface of metallic sodium,
whose work function is 2.3 eV. The maximum
speed of the photoelectrons emitted by the
surface is 1.2 x 106 m/s. What is the wavelength
of the light?
A. 1.95 x 10-7 m
B. 2.42 x 10-7 m
C. 2.86 x 10-7 m
D. 3.01 x 10-7 m
#14
Ultraviolet radiation (wavelength 250 nm) falls on a
metal target and electrons are liberated. If the
maximum kinetic energy of these electrons is 1.00
x 10-19 J, what is the lowest frequency of
electromagnetic radiation that will initiate a
photocurrent on this target?
A. 1.05 x 1015 Hz
B. 1.35 x 1015 Hz
C. 1.65 x 1015 Hz
D. 1.78 x 1015 Hz
#15
Photons of wavelength 220 nm on a metal target
and liberate electrons with kinetic energies
ranging from 0 to 61 x 10-20 J. Determine the
threshold wavelength of the metal.
A. 1.68 x 10-7 m
B. 1.95 x 10-7 m
C. 2.06 x 10-7 m
D. 6.77 x 10-7 m
#1

http://lrt.ednet.ns.ca/PD/ict_projects/photoelectric
/index.htm
Photoelectric Effect






When light shines on a surface (metal), electrons are
emitted from the surface.
E = KEe + W0
Energy of impinging light equals KE of electron plus the
work function.
Light Intensity increases will increase current (# of
electrons).
Frequency changes affect KEe.
Contributes to the theory of light as a particle. The
photons absorbed are “packets” of light energy.
Work Function

The minimum energy required is called the work
function, W0
 If hf < W0 then no electrons are emitted
 The lower the energy required to expel the electron,
the faster the electron will be moving away from the
surface.
 This makes it more likely be able to escape from the
material entirely.
practice

What is the work function when
monochromatic light of frequency
4.5x1015Hz releases the least tightly
held electrons from a metal with a
maximum KE of 13.10eV?
Do Now (2/25/14):

A sodium surface is illuminated with light
of wavelength 3 x 10-7 m. The work
function for sodium is 2.46 eV. Find (a) the
kinetic energy of the ejected photoelectron
and (b) the cutoff wavelength for sodium.
QUANTUM PHYSICS: DAY 2
Blackbody Radiation

An object an any temperature is known to
emit thermal radiation
 Stefan’s Law:
Star Temperatures
Stars approximate blackbody radiators and their visible
color depends upon the temperature of the radiator.
The curves show blue, white, and red stars. The white
star is adjusted to 5270K so that the peak of its
blackbody curve is at the peak wavelength of the sun,
550 nm.
Wien’s displacement law.

From the wavelength at the peak, the
temperature can be deduced from the Wien
displacement law.
Planck's Hypothesis

In 1900 Max Planck proposed a formula for the
intensity curve which did fit the experimental
data quite well. He set out to find a model that
would produce his formula.
 Instead of allowing energy to be continuously
distributed among all frequencies, Planck's model
required that the energy in the atomic vibrations
of frequency f was some integer times a small,
minimum, discrete energy, Emin = hf, where h is a
constant, now known as Planck's constant,
 h = 6.626176 x 10-34 J s
Planck’s Hypothesis

Planck's proposal requires that all the energy in the
atomic vibrations with frequency f can be written
as E = n h f, where n = 1, 2, 3, . . . No other values
of the energy were allowed. The atomic oscillators
could not have energy of (2.73) hf or (5/8) hf.
 This idea that something -- the energy in this case
-- can have only certain discrete values is
called quantization. We say that the energy
is quantized. This is referred to as
Planck's quantum hypothesis.
Planck’s Hypothesis

Planck did not realize how radical and farreaching his proposals were. He viewed his
strange assumptions as mathematical constructions
to provide a formula that fit the experimental data.
 It was not until later, when Einstein used very
similar ideas to explain the Photoelectric Effect in
1905, that it was realized that these assumptions
described "real Physics" and were much more than
mathematical constructions to provide the right
formula.
X-Rays

In 1895, Wilhelm Roentgen found a
mysterious radiation in his lab, which he
dubbed “x-rays.”
 They are produced when high speed
electrons are suddenly deccelerated
 Any accelerating voltage applied must be
higher than a certain threshold voltage
Bremsstrahlung
Electrons emit radiation
when they undergo a
deceleration in the target
The continuous
radiation is called
Bremsstrahlung
(Gernan for
“braking
radiation”).
X-Rays

Electron energy:
eV  hf max 
hc

X-Rays

In 1912, Max von Laue used a crystal lattice
to diffract X-Rays.
 This method become popular for analyzing
matter
Bragg’s Law

When x-rays are scattered from a crystal lattice,
peaks of scattered intensity are observed which
correspond to the following conditions:
– The angle of incidence = angle of scattering.
– The path length difference is equal to an integer
number of wavelengths.
2d sin   m
Bragg’s Law
The condition for maximum intensity contained in
Bragg's law above allow us to calculate details
about the crystal structure, or if the crystal
structure is known, to determine the wavelength
of the x-rays incident upon the crystal.
The Compton Effect

Compton deflected an x-ray of wavelength
λ0 toward a block of graphite.
 The reflected rays had a longer wavelength
than the incident rays.
 This change is called the Compton Shift.
The Compton Effect

Could only be explained using particles
(momentum)
h
1  cos  
    0 
me c
Example:

X-rays of wavelength 0.200000 nm are scattered
from a block of material. The scattered X-rays
are observed at an angle of 45° to the incident
beam. Calculate the wavelength of the x-rays
scattered at this angle.
λ=0.200710 nm
Wave Particle duality

Light exists as both photons and as
electromagnetic waves.
 We must accept both models to fully
describe it.
The Wave Properties of Particles

In 1932, Louis de Broglie postulated that
because photons have wave properties, all
matter could have wave properties.
The de Broglie Wavelength:

The wavelength of a particle, given by
h

p
where h is Planck's constant and p is
the momentum.
 In the nonrelativistic limit, this can be written
h

mv
where m is the particle mass and v is the velocity.
Momentum and Energy of a
photon:

No mass? How do we calculate?
Davisson-Germer

Measured the wavelength of electrons
 accidentally proved de Broglie’s hypothesis
 Low energy electrons were shot at a nickel
target, which became oxidized accidentally.
This made a diffraction grating for the
electron matter waves.
Example:

Calculate the de Broglie wavelength for an
electron moving at 10 7 m/s.
Example.

Calculate the de Broglie wavelength for a
50 g rock through with a speed of 40 m/s.
Schrodinger’s Cat

Schrödinger's cat is a thought
experiment (paradox) devised by
Austrian physicist Erwin
Schrödinger in 1935 that illustrates the
principle of quantum theory of
superposition.

Schrödinger's cat demonstrates the apparent conflict
between what quantum theory tells us is true about
the nature and behavior of matter on the
microscopic level and what we observe to be true
about the nature and behavior of matter on the
macroscopic level -- everything visible to the
unaided human eye.
Schrodinger’s Cat: (theoretical) experiment
We place a living cat into a steel chamber, along
with a device containing a vial of hydrocyanic
acid. There is, in the chamber, a very small
amount of hydrocyanic acid, a radioactive
substance. If a single atom of the substance
decays during the test period, a relay mechanism
will trip a hammer, which will, in turn, break the
vial and kill the cat.
Schrodinger’s Cat Experiment
The observer cannot know whether
or not an atom of the substance
has decayed, and consequently,
whether the vial has been broken,
the hydrocyanic acid released, and
the cat killed.
Since we cannot know, according
to quantum law, the cat is both
dead and alive, in what is called
a superposition of states.
Thought Experiment
It is only when we break open
the box and learn the
condition of the cat that the
superposition is lost, and the
cat becomes one or the other
(dead or alive). This is
sometimes called quantum
indeterminacy or the
observer's paradox: there is
no single outcome unless it
is observed.
Schodinger’s Cat
Superposition occurs at the
subatomic level, because there are
observable effects of interference,
in which a single particle
is demonstrated to be in multiple
locations simultaneously. What
that fact implies about the nature
of reality on the observable level
(cats, for example, as opposed
to electrons) is one of the stickiest
areas of quantum physics.
Schrödinger
himself is
rumored to
have said,
later in life,
that he
wished he
had never
met that cat.
The Wave Function
The Uncertainty Principle

Position and momentum of a particle cannot be
simultaneously measured with arbitrarily high
precision.
 There is a minimum for the product of the
uncertainties of these two measurements, as well
as for the product of the uncertainties of the
energy and time.
Uncertainty Principle

Not a statement about the inaccuracy of
measurement instruments, nor a reflection on the
quality of experimental methods
 Arises from the wave properties inherent in the
quantum mechanical description of nature.
 Even with perfect instruments and technique,
the uncertainty is inherent in the nature of
things.
The Heisenberg Uncertainty
Principle:

(h bar)
Example:

The speed of an electron is measured to be 5
x 103 m/s to an accuracy of 0.00300%. Find
the uncertainty in determining the position
of the electron.
Photoelectric Competition
Practice:

Complete the multiple choice problems in
Chapter 27
Do Now (2/26/14) (7 minutes):
1.
2.
3.
4.
What is the de Broglie wavelength of a 0.050
gram projectile fired at 180m/s?
What kind of wave properties could we see
from the wavelength in the above question?
In your own words, describe Schrodinger’s cat
and what it represents.
Which equation(s) of Einstein’s have you seen
before?
Pair Production and Annihilation

Pair production:

Pair annihilation:
Pair Production

The creation of an elementary particle and its
antiparticle, usually when a photon interacts
with a nucleus or another boson.
 For example, an electron and its
antiparticle, the positron, may be created.
Pair Production

The minimum energy that a photon must have to
produce a single electron-positron pair can be
found using conservation of energy by equating
the photon energy to the total rest energy of the
pair
Rest
hf min  2me c
2
energy
mass
Pair annihilation
Occurs when an electron (e−) and a positron (e+)
collide. The result is the annihilation of the electron
and positron, and the creation of gamma
ray photons or, at higher energies, other particles:
e− + e+ → γ + γ
It must satisfy a number of laws, including:
 Conservation of electric charge.
 Conservation of linear momentum and total energy.
 Conservation of angular momentum.
Electrons and positrons may also interact with each other
without annihilating.
Agenda

Finish photoelectric competition
 Photoelectric Notes Sheet
 Units Sheet
 Conceptual Question
 Multiple Choice (if time)
Photoelectric Notes Sheet

Fill in the blanks on the note sheet; you may use
your peers and the book to help you
 Ten minutes in, pink copies of the answer sheet
will be passed around. Check your work.
 Complete one problem from each section.
 When you finish, raise your hand so you may
receive a stamp
 Complete any additional problems for extra
credit
 We will discuss any questions afterwards.
Units Sheet

Complete one problem from each section.
 When you finish, raise your hand so you
may receive a stamp
 Complete any additional problems for extra
credit
Conceptual Questions

Work with your group to complete the
conceptual questions.
 Use a different color writing utensil for each
group member
Do Now (2/27/14): (6 Min)

Our next topic will be atomic physics. In that
topic, we will see that electrons in atoms can
be found in higher states of energy called
excited states for short periods of time. If the
uncertainty of the average time that an electron
exists in one of these states is 1.00 x10-8 s,
what is the minimum uncertainty in energy of
the excited state?
Agenda:

Complete Lecture Notes sheet
 WHEN FINISHED, continue working on
Units Practice Sheet
 Check answers for lecture notes
 Complete photoelectric mini lab
Do Now (2/28/14): (8 min)

Find the maximum kinetic energy of
photoelectrons from a certain material if the
work function is 2.3 eV and the frequency of
radiation is 3 x 1015 Hz.
KE  hf  
Agenda:
Quantum “test.”
 Conceptual questions

Quantum Challenge

Work with your group using only your AP
formula sheet and your calculators. Check
your work with Ms. Timson when complete
 The first group to get 100% gets bonus
points
Do Now (3/4/14):

In one sentence only, describe the
following:
– The Photoelectric Effect
– The Compton Effect
– The de Broglie Wavelength

What was the most productive thing you did
over the snow weekend?
Agenda:

Breifly review Quantum Challenge
 Group work – conceptual questions
 AP free response practice
Conceptual Questions

Work with your group to complete the
conceptual questions.
 Use a different color writing utensil for each
group member
 #13 – 18 are bonus, as well as #10 on the
back (the last question)
Do Now (3/5/14): (on your Do
Now sheet)

Complete parts a, b, & c from the AP free
response problem you received yesterday
(pink sheet)
 You may complete d for extra credit
Review:

Summary
Compton Effect

Short wavelength light (x-rays) scattered
from materials had a lower frequency than
the incident light.
 Wave nature of light would not have shown
this shift in wavelength. Explained only
through particle explanations.
Wave Particle Duality

Apparently conflicting observations of
wave nature and particle nature of light.
 Principle of Complementarity (Niels Bohr)

E=hf is a nice bridge since it incorporates
both particle and wave properties.
Wave Nature of Matter

Louis DeBroglie
 = h/(mv)
 Electrons vs. macroscopic matter

practice

What is the de Broglie wavelength of
a .050gram projectile fired at
180m/s?
Photons and Matter

4 possible interactions of photon with matter:
– Scattering (Compton effect) with lower frequency
but same speed (c).
– Photoelectric effect
– Excitation of electron (if energy too small to
ionize)
– Pair production-photon creates matter through
production of an electron and a positron
Do Now (3/3/14):
Atomic Structure

J.J. Thomson
 Ernest Rutherford
 Niels Bohr
 Energy level diagrams
 E = hf and c=f
 Lowest n has lowest energy. (Most
negative)
Big Ideas

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Millikan
Planck
Rutherford
DeBroglie
Bohr
Compton
Atomic Spectra
Photo-electric Effect
Wave particle duality
Atomic Structure

J.J. Thomson

Millikan

Ernest Rutherford
Cathode Ray and the Electron

F=evB

Accurately determined the charge carried by an
electron using his oil-drop experiment (1.602x10-19
coulomb)
 Proved that this quantity is a constant
 Experimentally verified Einstein’s photoelectric
equation and made the first direct photoelectric
determination of Planck’s constant
 Explored the region of the spectrum between
ultraviolet and X-radiation, extending the ultraviolet
spectrum far beyond the known limit
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
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Two parallel metal plates
acquire charge when electric
current is applied.
Atomizer sprays mist of oil
droplets, which then fall
slowly through a small hole.
Space between plates ionized
by radiation and electrons
attach themselves to oil
droplets, giving them a
negative charge
Ernest Rutherford
Rutherford History

Ernest Rutherford, 1st Baron Rutherford of Nelson,
OM, FRS (30 August 1871 – 19 October 1937) was a New
Zealand chemist who became known as the father of
nuclear physics. He discovered that atoms have a small
charged nucleus, and thereby pioneered the Rutherford
model (or planetary model, which later evolved into the
Bohr model or orbital model) of the atom, through his
discovery of Rutherford scattering with his gold foil
experiment. He was awarded the Nobel Prize in Chemistry
in 1908.
The Experiment
Rutherford Scattering


This experiment showed that the positive matter in atoms was
concentrated in an incredibly small volume and gave birth to the idea of
the nuclear atom. In so doing, it represented one of the great turning
points in our understanding of nature.
It also put a rest to the Thompson model of the atom because of the
angle’s at which the particles were scattered away from the nucleus of the
atoms was greater than the Thompson model said it could be.
Quantum theory – Max Planck

In 1900 Planck postulated that energy is
radiated in small, discrete units, which he
called quanta.
 he discovered a universal constant of nature,
Planck's constant. Planck's law states that
the energy of each quantum is equal to the
frequency of the radiation multiplied by the
universal constant.
E=hf
Planck’s constant

E=hf
 E=nhf

E= energy
 n=integer (1,2,3…)
 h=constant= 6.626 *10-34 J*s
 f= frequency
practice

a.
b.
c.
d.
e.
According to Plank’s quantum hypothesis,
which of the following could be the
energy of molecular vibrations in a
radiating object with a wavelength of λ?
4λhc
hc/2λ
4hc/λ
2λc/h
λhc/2
Atomic Structure

Niels Bohr
 Bohr model of the atom
 Energy level diagrams
Bohr and Quantum
Hypothesis

Discharge spectra
 hf=Eu – Ei where Eu is energy of the upper
state.
 Orbit closest to the nucleus has lowest
energy (most negative). An electron at
infinite distance has energy of 0 eV.
Energy Level Diagrams

Minimum energy to remove an electron is
binding energy or ionization energy.
 13.6eV – energy required to remove an
electron from the lowest state E1= -13.6eV
up to E=0.
 Lyman series, Balmer series, Paschen series
for hydrogen atoms. – pg 848.
The diagram above shows the lowest four discrete energy levels of
an atom. An electron in the n = 4 state makes a transition to the n =
2 state, emitting a photon of wavelength 121.9 nm.
(a) Calculate the energy level of the n = 4 state.
The diagram above shows the lowest four discrete energy levels of
an atom. An electron in the n = 4 state makes a transition to the n =
2 state, emitting a photon of wavelength 121.9 nm.
(b) Calculate the momentum of the photon.
The diagram above shows the lowest four discrete energy levels of
an atom. An electron in the n = 4 state makes a transition to the n =
2 state, emitting a photon of wavelength 121.9 nm.
The photon is then incident on a silver surface in a photoelectric
experiment, and the surface emits an electron with maximum
possible kinetic energy. The work function of silver is 4.7 eV.
(c) Calculate the kinetic energy, in eV, of the emitted electron.
The diagram above shows the lowest four discrete energy levels of
an atom. An electron in the n = 4 state makes a transition to the n =
2 state, emitting a photon of wavelength 121.9 nm.
(d) Determine the stopping potential for the emitted electron.