Never play leapfrog with a unicorn.Anonymous

1
Blackbody Radiation and the Ultraviolet
Catastrophe The Photoelectric Effect
Never play leapfrog with a unicorn.Anonymous
2
Comments on quiz grading
Trivial errors will make your 8-point problem
wrong. This is the way I want it!
Full problem -1 for math error -2 for minor
non-physics mistake -3 for physics error You
lose more if I cant tell what your mistake was
than if I can tell.
If your code name is your student number, SSN,
initials, UMR e-mail address, or anything else
that might identify you personally, please e-mail
me with a new code name!
3
2.2 Blackbody Radiation
Lets take another look at the electromagnetic
spectrum.
4
Blackbody radiation is electromagnetic radiation
emitted by objects in thermal equilibrium at a
non-zero (absolute) temperature.
If we claim to understand electromagnetic
radiation, we should be able to explain blackbody
radiation.
We will discuss blackbody radiation in more depth
in chapter 9, when we have the mathematical tools
to handle it. Well skim the mathematics
presented in this chapter.
The title of this section, but not really what
it is about.
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All objects emit radiation. For many real-world
objects, the radiation can be approximated by
blackbody radiation.
The characteristics of the radiation depend on
the properties of the object and conditions such
as temperature. Not surprisingly, the ability of
an object to absorb radiation is closely tied to
its ability to absorb radiation.
Physicists like to study idealized scenarios. It
would be nice to find a system where the emitted
radiation was independent of the details of the
radiating object.
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A blackbody is an example of an idealized
scenario. It is both a perfect absorber and
perfect emitter of radiation. As its name
suggests, it might be black.
But wait! If the object is black, how can it be
a good emitter of radiation? Good question.
Who knows the answer?
Just because an object is black does not mean it
emits no radiation.
Another objectionwhat do you mean perfect?
How can you make anything perfect?
You can make an extremely good approximation to a
blackbody by simply making a small hole in an
opaque box.
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http//www.egglescliffe.org.uk/physics/astronomy/b
lackbody/bbody.html
Who should understand blackbody radiation?
Who should understand blackbody
radiation? Physicists.
Who should understand blackbody
radiation? Physicists. Astronomers.
Who should understand blackbody
radiation? Physicists. Astronomers. Meterologists.
Who should understand blackbody
radiation? Physicists. Astronomers.
Meterologists. Politicians.
The principles of blackbody radiation apply
quite well to the Sun, stars, and planets. Each
can be treated as a radiating blackbody, giving
off energy at a rate commensurate with its
temperature. Therefore, we can apply this model
to finding the rate of energy output (power) from
the sun, and other stars. Once we know the power
coming out of a star at a given temperature we
can use some simple geometry and algebra to
determine the radii of distant stars.
http//www.cosmicshell.com/miranda/lab/BBradiatio
n/BBradiation.html dead link January 2005?
The principles of blackbody radiation are at the
heart of both the most simple and the most
complex climate models, including the general
circulation models (GCMs) which are run on
supercomputers and used for predicting global
climate change. The principles of blackbody
radiation also form the foundation for the
science of remote sensing. Passive and active
sensing of the earth's surface and atmosphere
usually take advantage of the radiative spectrum,
measuring intensity of emitted shortwave,
visible, longwave, and radiowave radiation. To
understand global warming or the workings of
meteorological satellites which monitor climate
change, we need to understand the principles of
blackbody radiation. http//www.imsa.edu/edu/geop
hysics/atmosphere/energy/energy1.html
The principles of blackbody radiation are at the
heart of both the most simple and the most
complex climate models, including the general
circulation models (GCMs) which are run on
supercomputers and used for predicting global
climate change. The principles of blackbody
radiation also form the foundation for the
science of remote sensing. Passive and active
sensing of the earth's surface and atmosphere
usually take advantage of the radiative spectrum,
measuring intensity of emitted shortwave,
visible, longwave, and radiowave radiation. To
understand global warming or the workings of
meteorological satellites which monitor climate
change, we need to understand the principles of
blackbody radiation. http//www.imsa.edu/edu/geop
hysics/atmosphere/energy/energy1.html
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EM radiation enters cavity.
Radiation is absorbed by cavity walls. The
energy is then emitted by the walls.
Only radiation which fits in the box can live
inside, and eventually exit the cavity.
9
The Ultraviolet Catastrophe
Using the conceptual design on the previous
slide, one can make a blackbody and investigate
its radiation.
Why would you guess the blue curve corresponds to
the hotter blackbody?
A range of colors is emitted.
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Solar Spectrum
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We think we have a basic experimental
understanding of the measured properties of
blackbody radiation. How about coming up with a
theoretical understanding?
Rayleigh and Jeans in the late 19th century
developed a mathematical description of blackbody
radiation by modeling it with standing waves set
up inside a cavity.
Beiser gives the result and a brief derivation.
Dont worry about the mathematics for now. We
will return to the math in Chapter 9. Heres the
result
Thats a pi in the font Im using!
The essential fact is that all the pictures
which science now draws of nature, and which
alone seem capable of according with
observational facts, are mathematical
pictures.James Jeans
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U(f) is the energy per unit volume inside the
blackbody. T is the temperature of the
blackbody, k is Boltzmanns constant, and f is
the frequency of the radiation emitted.
We should be able to compare this theoretical
formula with the experimental data shown
previously.
Beiser uses the Greek nu symbol ? for
frequency. It is too easy to confuse with v for
velocity. Lets always use f for frequency in
this class.
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The Ultraviolet Catastrophe!
well
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Whats wrong with the theory?
Rayleigh and Jeans assumed the radiation was
absorbed and emitted by oscillators in the
blackbody walls.
A valid assumption.
This is not as odd an idea as it may sound at
first. Light is EM radiation. EM radiation can
accelerate electrons. Electrons in the atoms of
the blackbody wall will act like little balls on
springs (harmonic oscillators) when you pull on
them with light.
Electrons absorb light energy. Now they are
excited. After a very brief time, they will
get rid of their excess energy.
The excess energy may come out at any valid
oscillator frequency, i.e., at any frequency of
light corresponding to a valid oscillator energy.
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Rayleigh and Jeans assumed that the oscillators
could later emit their energy at any frequency.
A valid assumption.
Rayleigh and Jeans assumed that only wavelengths
which could fit inside the cavity could exist
there.
A valid assumption.
Rayleigh and Jeans assumed that the radiation
exiting the cavity was the same as the radiation
inside.
A valid assumption.
according to classical physics
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Planck spent many years investigating blackbody
radiation, and discovered that he could explain
the blackbody radiation distribution by assuming
the blackbody to be made up of an enormous number
of oscillators, with each oscillator vibrating at
a fixed frequency, but with a wide range (from 0
to infinity) of possible frequencies.
However, the oscillators could not take on any
arbitrary frequency. Instead, they could
oscillate only in integral multiples of a
frequency f which depended on the blackbody
temperature.
Same as Rayleigh and Jeans, so far.
A revolutionary idea.
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These oscillators emit energy in units of hf,
which Planck called "quanta" of energy. A
quantum of energy is E hf, and h is called
Planck's constant This h is the same constant
which will appear in the next section in the
equation for the photoelectric effect.
The fact that the oscillators in the cavity walls
can interchange energy with standing waves only
in units of hf is a dramatic departure from
classical physics. Plancks theory explained
blackbody radiation, but even Planck believed
that later on somebody would reconcile blackbody
radiation with classical physics.
...the whole procedure was an act of despair
because a theoretical interpretation had to be
found at any price, no matter how high that might
be.Max Planck
Nevertheless, Planck won the 1918 Nobel prize
for his discovery of quanta.
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Here is Plancks formula for blackbody radiation
Well derive it in Chapter 9. Planck was right
to be suspicious of it, because when he found it,
there was no theoretical basis for it. However,
it does describe blackbody radiation accurately.
Whats the BIG IDEA here?
Oscillators can oscillate only in integral
multiples of some fundamental frequency.
(Chapter 5)
These oscillators emit energy in units of hf,
called "quanta" of energy. A quantum of energy
is E hf. (Chapter 2)
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Here are a couple more visualizations
http//csep10.phys.utk.edu/guidry/java/planck/plan
ck.html
http//jersey.uoregon.edu/vlab/PlankRadiationFormu
la/PRF.html
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For your interest (not for test)
500K or less -- essentially all of the radiation
is at frequencies less than those of visible
light. (Infrared ROY G. BiV) The blackbody
is black. 2000K -- visible light of appreciable
intensity, but mostly at the red (low frequency)
end of the spectrum. 3000K -- red predominates,
but significantly more blue. This is about the
temperature of an incandescent lamp
filament. 6500K -- distribution is more nearly
uniform, therefore "white hot." Example -- the
surface of the sun. 10000K and above -- blue
light predominates. "Blue hot." Example -- some
of the hotter stars.
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2.3 The Photoelectric Effect
Heinrich Hertz in 1887-8 studied the
photoelectric effect and generated EM waves to
verify Maxwell's theory of the electromagnetic
nature of light.
Hertz observed that a spark would jump more
readily between two metal spheres when their
surfaces were illuminated by the light from
another spark. Any ideas why that should happen?
The actual experimental setup was more complex
than sparks and a couple of metal spheres, and
allowed for a number of different experiments
which verified the electromagnetic nature of
light.
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An example of the kind of apparatus used to study
the photoelectric effect (but not Hertzs
apparatus) is described in your text.
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Hint for remembering which is which. Cathode
ray tubes have electron guns. Electrons come
from the cathode. Electrons are negative.
Therefore the cathode is negative. The other
one must be positive. Chemists must have been
confused when they named cations and anions.
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x
A
Light striking the anode causes electrons to be
emitted from the anode.
The electrons are emitted with kinetic energy,
and some of them reach the cathode (in spite of
its negative voltage). You can measure the
current with the ammeter.
Wait a minutearent electrons attracted to
and repelled from -?
They are, but if their initial KE is large
enough, they can reach the cathode anyway.
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x
A
You can increase the retarding potential (make
the cathode voltage more negative relative to the
anode) and see how that affects the current.
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x
A
You can make the cathode voltage negative enough
so that no photoelectrons reach the cathode.
(When the cathode reaches the "extinction
voltage," no electrons are detected there.)
So, what's the big deal. Light carries energy.
Energy is transferred to electrons in the anode.
They escape the metal anode. If they get enough
energy, they can even reach the cathode.
Classical physics explains this perfectly.
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Hertzs experiments with the photoelectric effect
confirmed that light consists of electromagnetic
waves
until you looked at the details.
Remember,
What does this predict?
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Predictions, based on classical theory
(1) For a fixed light frequency, the
Ptransmitted?A2 i.e., the power should be
directly proportional to the intensity. The
energy of the electrons coming out therefore
ought to be directly proportional to the
intensity (brightness) of the light.
(2) Similarly, for a fixed light intensity,
KEelectron ?f2.
(3) The extinction voltage ought to depend on the
f2, or on the intensity of the light.
(4) As Beiser shows, it should take a "long" time
for the electrons to accumulate enough energy to
escape.
(5) I can't see anything here that says there
ought to be either a minimum light intensity
which causes emission of photoelectrons or a
maximum photoelectron energy.
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Lets put these in a table, with shorthand
notation
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Here's what we actually observe
(1) The number of electrons emitted, but not
their energy, depends on the brightness of the
light.
(2) The electron energy is proportional to the
first power of the frequency of the light (not
the square).
(3) The extinction voltage depends on the first
power of the frequency of the light.
(4) Electrons are emitted almost right away
(within ?10-9 s).
(5) For a given metal, there is a frequency of
light below which no photoelectrons are emitted.
Also, for each frequency f, there is a maximum
energy which photoelectrons can have.
Anybody here remember electric eyes?
Oops, not that one, this one.
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Lets add the experimental results to our table
Theorists your comments?
Experimentalists your comments?
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We have a problem here, don't we? What to do?
The experiment (correctly done, of course) always
provides guidance for the theorist. Lets look
at the data.
These graphs show current vs. retarding voltage
for the same metal.
Graphs from http//www.chembio.uoguelph.ca/educmat
/ chm386/rudiment/tourexp/photelec.htm. (broken
link, January 2005?)
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If you take data for several metals and then plot
stopping potential vs. light frequency, you get
this
Electrons leave the metal with KE. The stopping
potential is proportional to the maximum KE.
Graph from http//www.chembio.uoguelph.ca/ educmat
/chm386/rudiment/tourexp/photelec.htm. (broken
link, January 2005?)
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Heres a plot of the maximum photoelectron energy
versus frequency of incident light
straight line y mx b
remember, well use f, not ?
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The plots of Kmax vs. f obey the relationship
where h is a constant, f is the frequency of the
incident light, and f0 is the threshold frequency
below which no photoelectrons are emitted.
The constant h has the same value for all metals,
but f0 depends on the metal. Plancks constant
h 6.63x10-34 J?s 4.14x10-15 eV?s.
It sounds like were on to something, doesnt it,
but keep in mind
this is an empirical equation i.e., it fits
the experiment, but we haven't explained anything.
Have you heard the term empirical parameter?
What does it mean?
So what is the real meaning of the term
empirical parameter?
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So we have a theory full of holes
(Rayleigh/Jeans)
and an empirical equation that works only
because weve thrown in a fudge factor.
Who you gonna call for help?
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Einstein's hypothesis and explanation for the
photoelectric effect.
Einstein postulated that a beam of light consists
of small bundles of energy, called "light quanta"
or "photons." The energy of a photon is given by
Ehf. An electron can absorb all of a photon's
energy or none of it, but nothing in between.
Some electrons may acquire enough energy to
escape from the illuminated metal surface (the
escape energy is called the work function of the
surface). Electrons escaping from the metal may
or may not use up additional energy in escaping.
The maximum energy electrons can leave the metal
with is equal to hf minus the work function.
Light of frequency f can't give an electron any
more energy than hf.
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Thus, according to Einstein, the empirical
equation for the photoelectric effect really says
where Kmax is the maximum photoelectric energy
and hf0 is the work function energy.
The equation is just an expression of
conservation of energy the big deal is the
idea of the photon.
Einstein won the 1921 Nobel Prize for explaining
the photoelectric effect. He never won a Nobel
Prize for his work in relativity!
This may look like just a rearrangement of our
previous equation. The difference is that this
equation is part of a testable theory. Huge
difference!
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Einstein brilliantly explained all of the
features of the photoelectric effect, but his
ideas were so revolutionary in 1905 that they
werent really accepted until 1916 when Millikan
provided conclusive experimental verification.
Actually, Millikan viewed Einsteins explanation
of the photoelectric effect as a direct attack on
the wave nature of electromagnetic waves, and
worked very hard for a decade to prove Einstein
wrong.
Instead, Millikan proved Einstein right.
If its any consolation, Millikan won the 1923
Nobel Prize for proving Einstein right.
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Example. Homework Problem 2.11. The maximum
wavelength for photoelectric emission in tungsten
is 230 nm. What wavelength of light must be used
in order for electrons with a maximum energy of
1.5 eV to be ejected?
The first step is to interpret the problem.
Photons with ?230 nm are lower in energy than
230 nm photons. The problem has given you the
minimum energy photon required to eject an
electron.
This minimum energy is equal to the work function
Remember, for an EM wave, c f ?.
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Now you can use our OSE
This is not the only way to solve it. I do
suggest you do the algebra first, then plug in
numbers at the end.
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Plug in the numbers, get the answer!
Now we have light (and EM waves) all figured out.
It (light) has all the properties of a wave
except sometimes it has the properties of a
particle.