the photoelectric effect
what is light?
photons and gravity
Recall—photoelectric effect, classical theory and observations:
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?
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.
2.4 What is Light
Remember relativity, where we found that Newtonian
mechanics was an approximation to the generally correct
theory of relativity?
Now we have the wave theory of light and the particle theory
of light. Who can tell me which one is correct?
All these fifty years of conscious brooding have brought me no nearer to
the answer to the question, “What are light quanta?”—A. Einstein
Light has physical reality. Different experiments see different
aspects of that reality. Some experiments see wave-like
aspects of light. Others see particle-like aspects of light.
That doesn't mean there is anything wrong or unreal about
light. Our senses just aren't equipped to fully appreciate all
aspects of the physical reality of light.
http://www.colorado.edu/physics/2000/schroedinger/two-slit3.html
We can do experiments involving the wave nature of light
(reflection, refraction, interference, diffraction)
…or we can do
diffraction)…
experiments involving the particle nature of light (photoelectric
effect).
Recently we’ve even developed
experiments involving both at once.*
But the experiments can’t tell us
whether light is a wave or a particle.
Is there something “wrong” with us?
Are most of the experiments wrong?
Is something “wrong” with light?
No, no, and no. Our senses and intuition are the problem.
Don’t try to make light into something black and white!
*http://www.nobel.se/physics/articles/ekspong/
We have been talking about light, but light is just E&M waves
from a relatively narrow band of the spectrum. All of our
conclusions about light apply to E&M radiation.
If you use an equation with
Planck's constant in it, you are
doing quantum mechanics. An
equation without Planck's
constant can be derived using
only classical physics.
The equations for photon momentum and energy (which I will
summarize on the next slide) contain both wave and particle
aspects together.
The mathematics of light.
photon energy
E = hf
photon rest mass
m=0
photon velocity
v =c
E & M wave speed
c=f λ
*photon "mass"
photon momentum
p=h / λ
photon charge = 0
"particle intensity" I = n h f
wave speed
v=f λ
E hf
h
m= 2 = 2 =
c
c
cλ
*Of course, a photon doesn’t really have any mass.
Don’t even think about walking outside these halls and telling
everybody photons have mass. No. No. No.
Where did pphoton=h/ come from?
E = p c + mc
2
2 2

2 2
E = pc
E =pf
but Einstein says :
E =hf
hf =pf
pf =hf
p = h
h
p=

0
Sections we are skipping…
2.5 X-rays
X-rays are E&M radiation of a higher energy than visible light.
Of course, they are also photons.
A common way to produce x-rays: bombard a metal surface
with high energy electrons. Some of the resulting x-rays form
a more or less continuous spectrum. Others have very specific
energies. We will study the latter x-rays in Chapter 7.
2.6 X-ray Diffraction
Because x-rays are waves, they can be diffracted. X-ray
diffraction is an extremely powerful technique for investigating
condensed matter.
2.7 The Compton Effect
The increase in x-ray wavelength upon scattering by matter
can be explained only by the quantum theory of light.
Compton’s theory was initially met by some doubt, but upon
experimental verification, it convinced remaining doubters of
the particle aspect of light, and won Compton the Nobel Prize
in 1927.
2.8 Pair Production
In the presence of a nucleus,* a photon (pure energy) can
materialize into an electron and a positron (particles).
Also, an electron and a positron can annihilate and become
pure energy (a photon).
*Required for conservation of energy and momentum.
2.9 Photons and Gravity; Black Holes
What is mass?
The “thing” that goes into F = dp/dt?
The “thing” that goes into F = Gm1m2/r2?
Are these two “things” (inertial mass and gravitational mass)
the same?
OR! Is there theoretical or experimental basis for demanding
they are the same?
Interesting digression…
Hungarian physicist, Lorand Eötvös, carried
out precise experiments between 1906 and
1909 to compare gravitational and inertial
mass.
He concluded the two were the same to
within 1 part in 200,000,000 (give or take a
factor or 2).
A 1964 experiment1 showed the two were the same to within
1 part in 100,000,000,000
1. P. G. Roll, R. Krotkov, R. H. Dicke, Annals of Physics, New York, 26, 442,
1964.
This experimental result is the basis for one of the postulates
of general relativity.
The Principle of Equivalence: an observer in a closed
laboratory cannot distinguish between the effects produced by
a gravitational field and those produced by an acceleration of
the laboratory.
force
a=g
uniform gravitational
field, downwards
w = mg
uniform upward
acceleration = g
We’ll return to consequences of this, after I finish my digression.
“The original Eötvös experiment was designed to measure the
ratio of the gravitational mass to the inertial mass of different
substances.”
“Eötvös found the ratio to be one, to within approximately one
part in a million.”
“Fischbach and his collaborators reanalyzed1 Eötvös' data and
found a composition dependent effect, which they interpreted
as evidence for a Fifth Force.”
What do you think of this?
Remember: right now we believe there are only four
fundamental forces: strong, weak, E&M, and gravitational.
1. E. Fischbach et al., Phys. Rev. Letters, 56, 2424, 2426, 1986; E. Fischbach,
D. Sudarsky, A. Szafer, C. Talmadge, S. H. Aronson, 57, 1959, 1986.
Quotes from http://plato.stanford.edu/entries/physics-experiment/notes.html#A4-1.
“Fischbach and his collaborators reanalyzed1 Eötvös' data and
found a composition dependent effect, which they interpreted
as evidence for a Fifth Force.”
If this is true, two 1-kilogram masses of copper attract each
other with a different force than two 1-kilogram masses of
aluminum!
Revolutionary! Instant Nobel Prize (on confirmation)! Your
name goes down in the annals of physics next to Newton,
Maxwell, and Einstein!
Two experiments carried out in 1987 gave conflicting results:
one for the fifth force, one against it.
What’s a responsible physicist to do?
Suspend judgment until experiment (supported by theory)
offers conclusive evidence.
Ultimately, many experiments were carried out which
demonstrated conclusively that there is no fifth force.
In the reanalysis of the Eötvös
experiment, a trend in the data
which suggested the fifth force
turned out to be a result of
statistical fluctuations. The data
had a trend, but it was accidental
and statistically insignificant.
Physics works!
Interesting discussion: http://plato.stanford.edu/entries/physicsexperiment/app4.html.
End of digression. Back to Physics 201. The upshot of all this
is that mass is mass, and the Principle of Equivalence still
works.
So we investigate some consequences of the Principle of
Equivalence.
Inertial mass and gravitational mass of an object are the same,
as far as we can tell.
Light, like other good “particles,” should be affected by gravity,
because
hf
2
E = hf and E = mc so m = 2 .
c
In other words, a photon doesn't have a rest mass, but it has an
"inertial mass" as suggested above, so it ought to be affected by
gravity.
Experimental evidence for effect of gravity on photons:
Light from distant star is
deflected by the sun's
gravitational field; first
observed during solar
eclipse, not long after
Einstein proposed the
theory.
The deflection observed
was exactly that predicted
by Einstein.
Graphic from http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/grel.html.
If I drop a book, what does it do?
Duh, falls down.
If I drop a photon, what does it do?
Has to fall down!
Where do you know of photons
falling?
In this room!
What happens when you drop the
book?
Accelerates.
What happens when you drop the
photon?
Accelerates.
What else happens when you drop
the book?
Gains KE.
What else happens when you drop
the photon?
Gains Energy.
In 1960, Pound and Rebka measured the increase in frequency
of photons from the time they were emitted at the top of a
22.5 m tower to the time they were detected at the ground.*
Unfortunately, if you do a search for Pound and Rebka, one of the first sites
you hit is by the Absolute Motion Institute. They use impressive-sounding
arguments to explain why this experiment, and virtually everything we
have studied so far, is nonsense.
If you read their arguments carefully, and with understanding, you find flaw
after flaw.
It is good to constantly question the validity of accepted science theories
and experiments.
It is not good to use falsehoods to prove ideas you dislike are incorrect.
Use Internet sources of information with extreme care!
A photon of initial energy hf ought to gain an energy equal to
“m”gH when it falls through a height H. It’s not really mass,
hence “m.”
Since the photon's "mass" is “m”=hf/c2, the energy gain is
(hf/c2)gH. This gives
hf  = hf ( 1 +
Ef
Ei
gH
).
2
c
energy gained (when
multiplied by hf)
Page 86 gives a sample calculation for a photon falling 22.5
meters. Caution: if you plug the numbers into the above
equation, your calculator precision will probably result in no
difference between hf and hf '. You have to do the algebra
and solve for f' – f first, as on page 86.
Gravitational red shift and black holes.
Consider a photon with gravitational
mass equal to its inertial mass.
Suppose it was emitted at a distance R
from an object of mass M, and it had an
initial frequency f0.
When the photon is at a distant point infinitely removed from
the object of mass M, its kinetic energy is equal to its initial
kinetic energy minus the gravitational potential energy
(GM1M2/R) at R:
M  hf 

hf  = hf - G   2  ,
 R  c 
  GM  

hf = hf 1 -  2   .
  c R 
  GM  

hf = hf 1 -  2  
  c R 
Note that the photon has been shifted to a lower frequency.
Lower frequency means more red, so this is a "gravitational
red shift."
This is not to be confused with the "red shift" due to the
Doppler effect, which you have probably heard of before. The
gravitational red shift is a very weak effect, barely measurable.
  GM  

hf = hf 1 -  2  
  c R 
Note that if GM/c2R=1, the photon has its frequency shifted to
zero.
If GM/c2R is greater than 1, a photon can never escape from
the object, because it didn't have enough energy to
overcome the gravitational field
You can solve for R, which is the radius from which light
cannot escape from an object of mass M. We won’t,
because…
The above argument was a classical one. To get the right
answer, you must use general relativity.
The result from general relativity has an extra 2 in the
numerator.
2GM
Rs = 2
c
This is called the Schwarzshild radius. It is
the radius inside which light can never escape
-- hence "black hole." The sphere of radius
RS is called the "event horizon." Nothing, not
even light, can pass the event horizon (from
inside to out, of course).
Graphic from http://heasarc.gsfc.nasa.gov/docs/blackhole.html.
http://imgsrc.hubblesite.org/hu
/db/1995/47/images/b/formats
/large_web.jpg
Light falling into a very
deep gravitational potential
has a blue shift, and light
escaping from a very deep
gravitational potential has a
red shift.
Keep in mind that photons have momentum but not mass.
When we talk about photon "mass" we are really referring to
the mass of an "ordinary" particle having the same momentum
as the photon.
Homework Problem 2-14
A silver ball is suspended by a string in a vacuum chamber and
ultraviolet light of wavelength 200 nm is directed at it. What
electrical potential will the ball acquire as a result.
Electrons will be ejected from the silver ball.
If the ball is in a vacuum, suspended by a nonconducting
string. As electrons are ejected, the ball acquires a positive
charge and positive potential.
The ball will build up a positive charge until no electrons have
enough KE to escape, at which point a steady-state situation
will exist and the potential will not change.
K max = hf - hf0 = hf - φ
Electrons stop escaping when
K max = eV = hf - φ
hf - φ
V=
e
c
h -φ
V= λ
e
V=

4.14×10-15
 3×108 m/s 
eV×s 
 - 4.7eV
-9
 200×10 m 
e

V=

4.14×10-15
 3×108 m/s 
eV×s 
 - 4.7eV
-9
 200×10 m 
e

V = 1.51 V
symbol
My mathcad solution is rather sloppy.
units
Suppose that a 60 W light bulb radiates primarily at a
wavelength of 1000 nm, a number just above the optical
range. Find the number of photons emitted per second.
power = total energy/time
power = (energy/photon) (N of photons/time)
(N of photons/time) = power / (energy/photon)
n = power / (energy of a photon)
n = P / hf = P / (hc/)
n = 60 / [(6.63x10-34)(3x108)]/(1000x10-9)
If you stick with SI units, n will be in units of s-1; i.e. photons/s.
The Big Winners, Chapter 2:
Planck, Nobel Prize, energy quanta, 1918.
Einstein, Nobel Prize, law of the photoelectric effect, 1921.
Millikan, Nobel Prize, charge of electron and photoelectric effect, 1923.
Chapter 2 OSE’s
Ephoton = hf
hf = K max + hf0
pphoton = hf / c = h / λ
c=f λ
2GM
Rs = 2
c
All our other equations are derived from these, the equations
of relativity, and the equations of classical physics.
I will give you more than these few equations on your quiz.