Transcript Document

General Physics (PHY 2140)
Lecture 27
 Modern Physics
Quantum Physics
Blackbody radiation
Plank’s hypothesis
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 27
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If you want to know your progress so far, please
send me an email request at
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Lightning Review
Last lecture:
p
1. Modern physics
 Relativistic energy, momentum
 Relativistic addition of velocities
vab 
mv
1  v2 c2
vad  vdb
v v
1  ad 2 db
c
Review Problem: You are packing for a trip to another star, to which you will
be traveling at 0.99c. Should you buy smaller sizes of your clothing, because
you will be skinnier on the trip? Can you sleep in a smaller cabin than usual,
because you will be shorter when you lie down?
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Reminder (for those who don’t read syllabus)
Reading Quizzes (bonus 5%):
It is important for you to come to class prepared, i.e. be familiar with the
material to be presented. To test your preparedness, a simple five-minute
quiz, testing your qualitative familiarity with the material to be discussed in
class, will be given at the beginning of some of the classes. No make-up
reading quizzes will be given.
There could be one today…
… but then again…
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26.9 Relativistic Energy
The definition of kinetic energy requires modification in
relativistic mechanics
KE = mc2 – mc2


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The term mc2 is called the rest energy of the object and is
independent of its speed
The term mc2 is the total energy, E, of the object and depends
on its speed and its rest energy
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Relativistic Energy – Consequences
A particle has energy by virtue of its mass alone

A stationary particle with zero kinetic energy has an energy
proportional to its inertial mass
E = mc2
The mass of a particle may be completely convertible to
energy and pure energy may be converted to particles
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Energy and Relativistic Momentum
It is useful to have an expression relating total energy, E, to the
relativistic momentum, p

E2 = p2c2 + (mc2)2
When the particle is at rest, p = 0 and E = mc2
Massless particles (m = 0) have E = pc

This is also used to express masses in energy units
mass of an electron = 9.11 x 10-31 kg = 0.511 MeV
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Problem: relativistic proton
A proton in a high-energy accelerator is given a kinetic energy of
50.0 GeV. Determine
(a) the momentum and
(b) the speed of the proton.
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A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV.
Determine (a) the momentum and (b) the speed of the proton.
Given:
Recall that E2 = p2c2 + (mc2)2. This can be used to
solve for p:
2
2
p
E = 50.0 GeV
Thus,
p=?
v =?

c

 mc

2
 KE

 mc 2
c
 KE   2 mc 2
2
p
Find:

E 2  mc 2
  KE 
c
 50.9 GeV
c
Similarly with velocity:


E   mc 2   
1
1  v2 c2

E
mc 2
2
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 E 
v  1  1  2   0.9998c
 mc 
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QUICK QUIZ
A photon is reflected from a mirror. True or false:
(a) Because a photon has a zero mass, it does not exert a force on
the mirror.
(b) Although the photon has energy, it cannot transfer any energy to
the surface because it has zero mass.
(c) The photon carries momentum, and when it reflects off the mirror,
it undergoes a change in momentum and exerts a force on
the mirror.
(d) Although the photon carries momentum, its change in momentum
is zero when it reflects from the mirror, so it cannot exert a
force on the mirror.
(a)
(b)
(c)
(d)
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False
False
True
False
p  Ft
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Example 1: Pair Production
An electron and a positron
are produced and the photon
disappears

A positron is the antiparticle
of the electron, same mass
but opposite charge
Energy, momentum, and
charge must be conserved
during the process
The minimum energy
required is 2me = 1.04 MeV
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Example 2: Pair Annihilation
In pair annihilation, an
electron-positron pair
produces two photons

The inverse of pair
production
It is impossible to create a
single photon

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Momentum must be
conserved
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Quantum Physics
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Introduction: Need for Quantum Physics
Problems remained from classical mechanics that relativity
didn’t explain:
Blackbody Radiation

The electromagnetic radiation emitted by a heated object
Photoelectric Effect

Emission of electrons by an illuminated metal
Spectral Lines

Emission of sharp spectral lines by gas atoms in an electric discharge tube
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Development of Quantum Physics
1900 to 1930

Development of ideas of quantum mechanics
Also called wave mechanics
Highly successful in explaining the behavior of atoms, molecules, and nuclei
Quantum Mechanics reduces to classical mechanics when applied
to macroscopic systems
Involved a large number of physicists

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Planck introduced basic ideas
Mathematical developments and interpretations involved such people as
Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac
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26.1 Blackbody Radiation
An object at any temperature is known to emit
electromagnetic radiation


Sometimes called thermal radiation
Stefan’s Law describes the total power radiated
P   AeT 4
Stefan’s constant

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emissivity
The spectrum of the radiation depends on the temperature and
properties of the object
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Blackbody
Blackbody is an idealized system that absorbs incident
radiation of all wavelengths
If it is heated to a certain temperature, it starts radiate
electromagnetic waves of all wavelengths
Cavity is a good real-life approximation to a blackbody
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Blackbody Radiation Graph
Experimental data for
distribution of energy in
blackbody radiation
As the temperature increases,
the total amount of energy
increases

Shown by the area under
the curve
As the temperature increases,
the peak of the distribution
shifts to shorter wavelengths
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Wien’s Displacement Law
The wavelength of the peak of the blackbody distribution
was found to follow Wein’s Displacement Law
λmax T = 0.2898 x 10-2 m • K
λmax is the wavelength at the curve’s peak
T is the absolute temperature of the object emitting the radiation
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The Ultraviolet Catastrophe
Classical theory did not match
the experimental data
At long wavelengths, the
match is good

Rayleigh-Jeans law
P
2 ckT
4
At short wavelengths, classical
theory predicted infinite energy
At short wavelengths,
experiment showed no energy
This contradiction is called the
ultraviolet catastrophe
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Planck’s Resolution
Planck hypothesized that the blackbody radiation was produced by
resonators

Resonators were submicroscopic charged oscillators
The resonators could only have discrete energies
En = n h ƒ
n is called the quantum number
ƒ is the frequency of vibration
h is Planck’s constant, h=6.626 x 10-34 J s
Key point is quantized energy states
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