Electromagnetic radiation Electromagnetic radiation behaves as particlesbehaves as particles

1.1. Notes of the problem discussed Tuesday.Notes of the problem discussed Tuesday.2.2. Quiz 9.11 and a few comments on quiz Quiz 9.11 and a few comments on quiz

9.09.9.09.3.3. Topics in EM waves as particles:Topics in EM waves as particles:

Blackbody radiation and Planck’s constant, Blackbody radiation and Planck’s constant, Planck’s Nobel Prize in physics.Planck’s Nobel Prize in physics.

The Photoelectric Effect and Einstein's Nobel The Photoelectric Effect and Einstein's Nobel Prize in physics.Prize in physics.

The X-rays and Roentgen’s Nobel Prize in The X-rays and Roentgen’s Nobel Prize in physics.physics.

The Compton Effect and Compton’s Nobel Prize The Compton Effect and Compton’s Nobel Prize in physics.in physics.

Pair production, energy to mass conversion and Pair production, energy to mass conversion and Anderson’s Nobel Prize in physics.Anderson’s Nobel Prize in physics.

The wave-particle duality and the door to yet The wave-particle duality and the door to yet another new world.another new world.

today

Tuesday

One page review of Special One page review of Special RelativityRelativity

SS and and S’S’ system: system:

For a particle with velocity in S:For a particle with velocity in S:

The Doppler effect:The Doppler effect:

x

x'y

z

S

y'

z'

S '

S’ moves with velocity v in S along the x-axis.

vx' x vt

y' y

z' z

2v

vt' x t

c

1

21 xx x

u vu ' u vc

1

21 xy y v

u vu ' uc

1

21 xz z v

u vu ' uc

2

1

1v

v,

c

p uum

12 21u u c

2

uE mc 21uKE mc

propervt t

proper vL L

u

2 2 2 2 4E p c m c

1 cossource

obsv

ff

When When θ =0, the course is =0, the course is moving away from the moving away from the observer.observer.

v

problem 33, page 64problem 33, page 64

Step 1, choose reference systemsStep 1, choose reference systems

Step 2, interpreter proper length and time in their systemsStep 2, interpreter proper length and time in their systems

in S:in S: in S’:in S’:

Ground: S On the muon: S’ 10 9950 0 9954 0 9952 10 22

2 vv . . . c, .

0 186 89mvL L .

0 2 2μs.

Solve it in S:

0 22 48μsv .

0 1910mL

08

1910s 6 40μs

0 9952 3 10

Lt .

v .

The time it takes the muons to cover the distance L0:

0

3950 75

527

N.

N

0

t

N N e

Now verify:

6 40

22 48 0 75.

.e .

is verified.0

t

N N e

So

Solve it in S’:

The time it takes the muons to cover the distance L:

8

186 89s 0 63μs

0 9952 3 10

L .t' .

v .

0 63

2 2 0 75.

.e .Since

00

t'

N N e

and

is verified.

0

3950 75

527

N.

N

The Compton EffectThe Compton Effect When two waves meet, they superposition When two waves meet, they superposition

each other. When two particles meet, they each other. When two particles meet, they collide. collide.

When X-rays meet electrons When X-rays meet electrons Compton Compton scatter, behave exactly as two particles scatter, behave exactly as two particles collide:collide: The energy of a photon with frequency The energy of a photon with frequency ff is: is:

So its momentum So its momentum

In the x-ray electron system, the initial In the x-ray electron system, the initial momentum ismomentum is

The final moment is The final moment is xx-component-componenty-componenty-component

Based on moment conservation:Based on moment conservation:

And energy conservation:And energy conservation:

Solve the above three equations (all algebraic Solve the above three equations (all algebraic manipulation) We have:manipulation) We have:

E hfE hf h

pc c

p h

cos cosu eh ' m u

yy

xxsin sinu eh ' m u

cos cosu eh h ' m u

0 sin sinu eh ' m u

2 2e u ech m c ch ' m c

1 cose

h'

m c

2 2 2 2 4E p c m c hint

Example 3.3Example 3.3Compton scattering and incident photon’s Compton scattering and incident photon’s wavelength is known.wavelength is known.

Asked: momenta of incident photon, scattered Asked: momenta of incident photon, scattered photon and the electron.photon and the electron.

Relevant formulas and concept:Relevant formulas and concept:

Work on the blackboard.Work on the blackboard.

p h 1 cose

h'

m c Momentum conservation

Example 3.4Example 3.4

Employ energy and momentum conservation Employ energy and momentum conservation laws.laws.

Work on the blackboard.Work on the blackboard.

2 20 62He . c D

ch m c m c

0 62 0 6 cosphoton . c D

hp m . c

Pair production, energy to mass Pair production, energy to mass conversionconversion

When photon with energy above the When photon with energy above the rest mass of two electrons ( ) rest mass of two electrons ( ) interact with the electric field of a interact with the electric field of a nucleus, this photon may be turned nucleus, this photon may be turned into a pair of electron and positron. into a pair of electron and positron. This process is called pair production This process is called pair production through which energy gets turned through which energy gets turned into mass.into mass.

Positron is the anti-particle of Positron is the anti-particle of electron: it has the same mass as an electron: it has the same mass as an electron but the opposite charge. electron but the opposite charge.

When a particle and anti-particle When a particle and anti-particle meet, they annihilate into a photon, meet, they annihilate into a photon, the process that mass converts into the process that mass converts into energy.energy.

Example 3.5: energy conservation.Example 3.5: energy conservation.

22 em c

F E v Bq q

What is the B-field direction?

Pair production in a bubble Pair production in a bubble chamberchamber

What is the magnetic field direction?What is the magnetic field direction?

electron positron

The wave-particle dualityThe wave-particle duality Wave or particle? Depends on the Wave or particle? Depends on the

wavelength ( ) and the dimensions (wavelength ( ) and the dimensions ( ) in ) in your experiment.your experiment.

The double-slit experimentThe double-slit experiment

: particle

: wave

D

D

D

Wave

Particle as Wave

p h

E hf

Example 3.6Example 3.6

How do we solve this problem?How do we solve this problem?

For (c), from

We calculate the power intensity at this place, and then repeat the procedures in (a).

20

1cos

2I I

cE nh

For (a), we have wavelength and the power intensity, For (a), we have wavelength and the power intensity, and this formula: and this formula:

The number of photons is what in the question. For (b), the minimum at the interference pattern has no light no photons

Review questionsReview questions

List the arguments in this chapter List the arguments in this chapter that photon is a particle. that photon is a particle.

List the arguments in this chapter List the arguments in this chapter that photon is a wave.that photon is a wave.

When you treat photon as a wave? When you treat photon as a wave? When as a particle?When as a particle?

Preview for the next Preview for the next classclass

Text to be read:Text to be read: In chapter 4:In chapter 4:

Section 4.1Section 4.1 Section 4.2Section 4.2 Section 4.3Section 4.3

Questions:Questions: What is the Bragg Law?What is the Bragg Law? An electron is a particle. One cannot have half an An electron is a particle. One cannot have half an

electron. How do you interpret the amplitude of an electron. How do you interpret the amplitude of an electron wave?electron wave?

What is the form of the Shroedinger equation for What is the form of the Shroedinger equation for free particle? free particle?

Homework 5, due by 9/23Homework 5, due by 9/23

1.1. Problem 31 on page 94.Problem 31 on page 94.

2.2. Problem 35 on page 94.Problem 35 on page 94.

3.3. Problem 40 on page 94.Problem 40 on page 94.

4.4. Problem 42 on page 94. Problem 42 on page 94.