Transcript Objectives

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
1240 eV  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 - hf0 =
-
 0
hf0=min. energy to release electron
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.
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?
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









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



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.