Met

al

e-

e-

e-

e-

e-

e+

12x p

Consider a nearly enclosed container at uniform temperature:

•Light gets produced in hot interior•Bounces around randomly inside before escaping•Should be completely random by the time it comes out•Pringheim measures spectrum, 1899

u() = energy/ volume /nm

Black Body Radiation: Light in a Box

•Quantum effects almost always involve individual particles•These particles typically have charges like +e or –e •The only practical way to push on them is with electric and magnetic fields

The Electron Volt

U q V

•Charge is e and voltage is measured in volts•Define:

1 eV Ve 191.602 10 C V 191.602 10 J .

•This unit is commonly used in quantum mechanics

•Statistical Mechanics was a relatively new branch of physics• It explained some things, like the kinetic theory of gasses

•Physicists tried to explain black body radiation in terms of this theory

Attempting to Explain the Result

•It said that the energy in black body could have any possible wavelength•There are many, many ways to fit short wavelengths inside the box•The amount of energy in any given wavelength could be any number from 0 to infinity•Therefore, there should be a lot more energy at shortwavelengths then at long•Prediction #1: There will be much more energyat short wavelengths than long•Prediction #2: The total amount of energywill be infinity

• The “ultraviolet catastrophe”

4

8 Bk Tu

Comparison Theory vs. Experiment:Theory

Experiment

What went wrong?•Not truly in thermal equilibrium?•Possible state counting done wrong?•The amount of energy in any given wavelength could be any number from 0 to infinity

Max Planck’s strategy (1900):•Assume energy E must always be an integer multiple of frequency f times a constant h

•E = nhf, where n = 0, 1, 2, …•Perform all calculations with h finite•Take limit h 0 at the end

4

8 Bk Tu

Why this might help:•Assume energy E must always be an integer multiple of frequency f times a constant h

Theory

Experiment

•Notice the problem is at short wavelength = high frequency•Without this hypothesis, energy can be small without being zero

Ener

gy

0 Average Energy

Average Energy

•Now add levels•It can no longer have a little bit of energy•For high frequency, it has almost no energy

Planck’s Black Body Law

5

8 1exp 1B

hcuhc k T

Max Planck’s strategy (1900):•Take limit h 0 at the end•Except, it fit the curve with finite h!

34

15

6.626 10 J s

4.136 10 eV s

h

h

E hf

Planck Constant

•Looks like light comes in chunks with energy E = hf - PHOTONS

Photoelectric Effect: Hertz, 1887•Metal is hit by light•Electrons pop off•Must exceed minimum frequency

•Depends on the metal•Brighter light, more electrons•They start coming off immediately

•Even in low intensity

Met

al

e-

e-

e-

e-

Einstein, 1905•It takes a minimum amount of energy to free an electron•Light really comes in chunks of energy hf•If hf < , the light cannot release any electrons from the metal•If hf > , the light can liberate electrons

•The energy of each electron released will be K = hf –

Photoelectric Effect•Will the electron pass through a charged plate that repels electrons?•Must have enough energy

•Makes it if:

Met

al

e-

+––+V

K e V

hf e V

K hf

f

Vmax

slope =

h/e

Nobel Prize, 1921

Sample Problem

K hf

When ultraviolet light of wavelength 227 nm strikes calcium metal, electrons are observed to come off with a kinetic energy of 2.57 eV.

1. What is the work function for calcium?2. What is the longest wavelength that can free electrons from calcium?3. If light of wavelength 312 nm were used instead, what would be the

energy of the emitted electrons?

We need the frequency:f c

cf

8

9

3.00 10 m/s227 10 m

15 11.32 10 s

hf K 15 15 14.136 10 eV s 1.32 10 s 2.57 eV

5.46 eV 2.57 eV 2.89 eV= Continued . . .

Sample Problem continued

K hf

2. What is the longest wavelength that can free electrons from calcium?3. If light of wavelength 312 nm were used instead, what would be the

energy of the emitted electrons?f c 2.89 eV

•The lowest frequency comes when K = 0 min0 hf

minfh

15

2.89 eV4.136 10 eV s

14 16.99 10 s

•Now we get the wavelength:cf

8

14 1

3.00 10 m/s6.99 10 s

74.29 10 m 429 nm

•Need frequency for last part:cf

8

9

3.00 10 m/s312 10 m

14 19.61 10 s

K hf 15 14 14.136 10 eV s 9.61 10 s 2.89 eV 1.08 eV

The de Broglie Relation

•We have a formula for the energy of a photon:

•Now, steal a formula from special relativity:

•Combine it with a formula from electromagnetic waves:

•And we get the de Broglie relation:

•Photons should have momentum too

E hf

E cp

c f

hf cp

hf f p

p h

Atom

The Compton Effect•By 1920’s X-rays were clearly light waves•1922 Arthur Compton showed they carried momentum

e-

e-

e-

Photon in

Photon out

•Conservation of momentum and energy implies a change in wavelength

1 coshmc

Photons carry energy and momentum, just

like any other particle

Light is . . .•Initially thought to be waves

•They do things waves do, like diffraction and interference•Wavelength – frequency relationship

•Planck, Einstein, Compton showed us they behave like particles (photons)•Energy and momentum comes in chunks•Wave-particle duality: somehow, they behave like both

Electrons are . . .•They act like particles

•Energy, momentum, etc., come in chunks•They also behave quantum mechanically•Is it possible they have wave properties as well?

E hf

Waves and Electrons

p h

The de Broglie Hypothesis

•Two equations that relate the particle-like and wave-like properties of light

E hf

p h 1924 – Louis de Broglie postulated that theserelationships apply to electrons as well•Implied that it applies to other particles as well•de Broglie was able to explain the spectrum of hydrogen using this hypothesis

The Davisson-Germer ExperimentSame experiment as scattering X-rays, except•Reflection probability from each layer greater

•Interference effects are weaker•Momentum/wavelength is shifted inside the material•Equation for good scattering identical

d

2 cosd m

e-

Quantum effects are weird•Electron must scatter off of all layers

What Objects are Waves?•1928: Electrons have both wave and particle properties•1900: Photons have both wave and particle properties•1930: Atoms have both wave and particle properties•1930: Molecules have both wave and particle properties•Neutrons have both wave and particle properties•Protons have both wave and particle properties•Everything has both wave and particle properties

Dr. Carlson has a mass of 82 kg and leaves this room at a velocity of about 1.3 m/s. What is his

wavelength?

hp

h

mv

346.626 10 J s

82 kg 1.3 m/s

366.22 10 m

It’s a Particle, It’s a Wave, No It’s a . . .

•Consider the two slit experiment•We can do it with photons or electrons, it doesn’t matter

•We can build a detector that counts individual photons or electrons•We can put through particles one at a time•We can count the number of photons on a screen•Over time, we build up an interference pattern

•Interference only works with both slits open•Every photon is going through both slits•Sometimes, we say we have wave-particle duality

•It acts sometimes like a particle, sometimes like a wave

•It is a quantum object – something completely new

Diffraction And Uncertainty

•A plane wave approaches a small slit, width a•Initially it is very spread out in space•But it has a very definite direction and wavelength•It therefore has a definite momentum

min a

x

•The uncertainty in a quantity is how spread out the possible value is

1T score 79.2 8.6 1 8.6T

0p

•After it passes through the slit, it has a more definite position•It now has a spread in angle

•This creates an uncertainty in its momentum in this direction

14x a

xp p 2

pa

ha

18x p h

The Uncertainty Relation

4hx p

•Waves, in general, are not concentrated at a point•They have some uncertainty x•Unless they are infinitely spread out, they also typically contain more than one wavelength•The have some uncertainty in wavelength•The have some uncertainty in momentum•Hard mathematical theorem:

• Make a precise definition of the uncertainty in position• Make a precise definition of the uncertainty in momentum• There is a theoretical limit on the product of these:

Sample Problem

An experimenter determines the position of a proton to an accuracy of 10.0 nm. 1. What is the corresponding minimum uncertainty in the momentum?2. As a consequence how far will the proton move (minimum) 1.00 ms later?

4hp

x

4hx p

34 2

8

6.626 10 m kg/s4 1.00 10 m

275.27 10 kg m/s

•This corresponds to an uncertainty in the velocity ofpv

m

27

27

5.27 10 kg m/s1.673 10 kg

3.15 m/s

•This means the proton will move a minimum distance: d vt 33.15 m/s 1.00 10 s 3.15 mm

New Equations for Test 4

Images

Quantum:

Light in Materials

cf vk n

1 1 2 2sin sinn n

Diffraction Grating

1 1 1p q f

E hfp h

83.00 10 m/sc

sin md

i r

Reflection/Refraction

2

1

sin cnn

Diffraction

Limit

a D

0, 1, 2,m 4hx p

K hf

End of material for Test 4