Blackbody Radiation
and the
Ultraviolet Catastrophe
The Photoelectric Effect
Dr. Mahmoud Ahmad
2.2 Blackbody Radiation
Let’s take another look at the electromagnetic spectrum.
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. We’ll
skim the mathematics presented in this chapter.
*The title of this section, but not really what it is about.
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.
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 objection—what 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.
http://www.egglescliffe.org.uk/physics/astronomy/blackbody
/bbody.html
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/BBradiation/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/geophysics/atmosphere/energy/energy1.html
E&M 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.
The Ultraviolet Catastrophe
Using the conceptual design on the previous slide, one can
make a blackbody and investigate its radiation.
The intensity also depends
on the temperature of the
blackbody.
Why would you guess the
blue curve corresponds to
the hotter blackbody?
A range of colors is emitted.
Intensity depends on frequency.
Solar Spectrum
Intensity (W/m2/m m)
2.50E+03
ASTM E490 Air Mass Zero solar spectral
irradiance is based on data from
satellites (1999)
2.00E+03
1.50E+03
1.00E+03
5.00E+02
0.00E+00
0
1
2
3
4
WaveLength (micro-meters)
5
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.” Don’t worry
about the mathematics for now. We will return to the math in
That’s a “pi” in the
Chapter 9. Here’s the result:
8πkT 2
U(f) df =
f df .
3
c
font I’m 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
8πkT 2
U(f) df =
f df .
3
c
U(f) is the energy per unit volume inside the blackbody. T is
the temperature of the blackbody, k is Boltzmann’s 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. Let’s always use f for frequency in this class.
theory
experiment
well… The Ultraviolet Catastrophe!
What’s 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
E&M radiation. E&M 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.
Rayleigh and Jeans assumed that the oscillators could later
A valid assumption.*
emit their energy at any frequency.
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
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.
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.
Planck’s 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.
Here is Planck’s formula for blackbody radiation:
8πh
f3
U(f) df = 3 hf
df .
c e kT  1
We’ll 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.
Show that at low frequencies Plank formula becomes
as Rayleight Jeans formula? Answer Page61
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)
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.
2.3 The Photoelectric Effect
Heinrich Hertz in 1887-8 studied the photoelectric effect and
generated E&M 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.
An example of the kind of apparatus used to study the
photoelectric effect (but not Hertz’s apparatus) is described in
your text.
Start with a glass tube.
Insert a couple of metal plates, with vacuum feedthroughs
carrying wires to the outside.
Pump out “all”* the air and other gases.
*You never get all the gases out, but you can get
most out.
-
+
x
V
A
Seal off the tube (or keep pumping to maintain a vacuum).
Connect the metal plates to a variable high voltage power
supply. + is the anode and - is the cathode.*
Connect an ammeter to measure the current.
*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.
-
+
x
V
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 minute—aren’t electrons attracted to + and repelled from -?”
They are, but if their initial KE is large enough, they can reach the cathode
anyway.
-
+
x
V
A
You can increase the retarding potential (make the cathode
voltage more negative relative to the anode) and see how that
affects the current.
-
+
x
V
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.”
Hertz’s experiments with the photoelectric effect confirmed
that light consists of electromagnetic waves…
…until you looked at the details.
Remember,
ρ 2 2
Power 
ω A vp.
2
What does this predict?
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.
Let’s put these in a table, with shorthand notation:
Predict
PA2, PI  Eelectron  Ilight
Eelectron  flight2
Vextinction  flight2
tescape = “very long”
no limit to electron KEmax,
no Imin to produce e-
Observe
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.
Let’s add the experimental results to our table:
Predict
Observe
PA2, PI  Eelectron  Ilight
N(e-)  Ilight, Ee- independent of Ilight
Eelectron  (flight)2
Eelectron  (flight)
Vextinction  (flight)2
Vextinction  (flight)
tescape = “very long”
tescape = “instantaneous”
no limit to electron KEmax,
no Imin to produce e-
there is a maximum electron KE,
there is an Imin needed to produce e-
Theorists: your comments?
Experimentalists: your comments?
We have a problem here, don't we? What to do?
The experiment (correctly done, of course) always provides
guidance for the theorist. Let’s 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?)
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?)
Here’s a plot of the maximum photoelectron energy versus
frequency of incident light:
straight line; y = mx + b
remember, we’ll use f, not 
The plots of Kmax vs. f obey the relationship
K max = hf - hf0 ,
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. Planck’s constant = h = 6.63x10-34 Js
= 4.14x10-15 eVs.
It sounds like we’re on to something, doesn’t 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?”
So we have a theory full of holes (Rayleigh/Jeans)…
…and an empirical equation that works only because we’ve
thrown in a fudge factor.
Who you gonna call for help?
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 E=hf. An electron can absorb
all of a photon's energy or none of it, but nothing in between.
Ephoton = hf .
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.
Thus, according to Einstein, the empirical equation for the
photoelectric effect really says*
hf = K max + hf0 ,
energy you start with
energy you leave with
energy you use to escape
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!
Einstein brilliantly explained all of the features of the
photoelectric effect, but his ideas were so revolutionary in 1905
that they weren’t really accepted until 1916 when Millikan
provided conclusive experimental verification.
Actually, Millikan viewed Einstein’s 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 it’s any consolation, Millikan won the 1923 Nobel Prize for proving
Einstein right.
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:
hc
φ = hf0 =
.
λ0
Remember, for an E&M wave, c = f .
c = fλ , E&M wave.
Now you can use our OSE:
hf = K max + hf0
hf = K max
hc
+
λ0
hc
hc
= K max +
λ
λ0
hc

hc 
 λ   K max + λ 
0 

=1
This is not the only way
to solve it. I do suggest
you do the algebra first,
then plug in numbers at
the end.
hc
λ
K max
hc
+
λ0
Plug in the numbers, get the answer!
Now we have light (and E&M waves) all figured out.
It (light) has all the properties of a wave…
…except sometimes it has the properties of a particle.