“I've been giving this lecture to first-year classes for over twenty-five years. You'd think they would begin to understand it by now.”—English mathematician John Littlewood

the photoelectric effect

what is light?

photons and gravity

Recall—photoelectric effect, classical theory and observations:

Predict Observe

PA2, PI 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:

remember, we’ll use f, not

straight line; y = mx + b

The plots of Kmax vs. f obey the relationship

max 0K = hf - hf ,

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 Js = 4.14x10-15 eVs.

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.

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.

photonE = hf .

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*

where Kmax is the maximum photoelectric energy and hf0 is the work function energy.

max 0hf = K + hf ,

*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 won the 1921 Nobel Prize for explaining the photoelectric effect. He never won a Nobel Prize for his work in relativity!

The equation is just an expression of conservation of energy; the “big deal” is the idea of the photon.

energy you start with energy you use to escapeenergy you leave with

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:

00

hcφ = hf = .

λ

Remember, for an E&M wave, c = f . c = fλ , E&M wave.

Now you can use our OSE:

max 0hf = K + hf

max0

hchf = K +

λ

max0

hc hc = K +

λ λ

max0

hc = 1

hcλ K +

λ

This is not the only way to solve it. I do suggest you do the algebra first, then plug in numbers at the end.

max0

hcλ

hcK +

λ

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

All these fifty years of conscious brooding have brought me no nearer to the answer to the question, “What are light quanta?”—A. Einstein

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?

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)…

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!

Recently we’ve even developed experiments involving both at once.*

*http://www.nobel.se/physics/articles/ekspong/

We can do experiments involving the wave nature of light (reflection, refraction, interference, diffraction) …or we can do experiments involving the particle nature of light (photoelectric effect).

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.

The equations for photon momentum and energy (which I will summarize on the next slide) contain both wave and particle aspects together.

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 mathematics of light.

photon energy E = hf photon momentum p = h / λ

photon rest mass m = 0

2 2

E hf hphoton "mass" m = = =

c c*

cλ

photon charge = 0

photon velocity v = c "particle intensity" I = n h f

E&M wave speed c = f λ wave speed v = f λ

*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. Don’t even think about walking outside these halls and telling everybody photons have mass. No. No. No.

Where did pphoton=h/ come from?

22 2 2 2E =p c + mc

0

E=pc

E=pf

but Einstein says: E=hf

hf =pf

pf =hf

p =h

hp=

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 DiffractionBecause x-rays are waves, they can be diffracted. X-ray diffraction is an extremely powerful technique for investigating condensed matter.

2.7 The Compton EffectThe 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 ProductionIn the presence of a nucleus,* a photon (pure energy) can materialize into an electron and a positron (particles).

*Required for conservation of energy and momentum.

Also, an electron and a positron can annihilate and become pure energy (a photon).

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.

Winter 2005, skip digression.

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.

w = mg

uniform gravitationalfield, downwards

a=g

uniform upwardacceleration = g

force

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.”

Quotes from http://plato.stanford.edu/entries/physics-experiment/notes.html#A4-1.

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.

What do you think of this?Remember: right now we believe there are only four fundamental forces: strong, weak, E&M, and gravitational.

“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/physics-experiment/app4.html.

End of digression. Back to Physics 107. 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

22

hfE = hf and E = mc so m = .

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 photon?

Gains Energy.

What else happens when you drop the book?

Gains KE.

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.**Sorry, but I have to get on my soapbox for a minute.

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

2

gHhf = hf ( 1 + ) .

c

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.

EiEf

energy gained (when multiplied by hf)

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:

2

M hfhf = hf - G ,

R c

2

GMhf = hf 1 - .

c R

2

GMhf = hf 1 -

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.

2

GMhf = hf 1 -

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.

s 2

2GMR =

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.

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.

http://imgsrc.hubblesite.org/hu/db/1995/47/images/b/formats/large_web.jpg

Electrons will be ejected from the silver ball.

max 0K = hf - hf = hf - φ

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.

Homework Problem 2-14A 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 stop escaping when

maxK = eV = hf - φ

hf - φV =

e

8

-15-9

3×10 m/s4.14×10 eV×s - 4.7eV

200×10 mV =

e

ch - φ

λV = e

8

-15-9

3×10 m/s4.14×10 eV×s - 4.7eV

200×10 mV =

e

V = 1.51 V

My mathcad solution is rather sloppy.

symbol 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.

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.

The Big Winners, Chapter 2:

Chapter 2 OSE’s

photonE = hf max 0hf = K + hf

c = f λphotonp = hf / c = h / λ

s 2

2GMR =

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.