Transcript Monday, March 8, 2010

Physics 3313 – Review 1
Monday March 8 , 2010
Dr. Andrew Brandt
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Time Dilation Example
• Muons are essentially heavy electrons (~200 times heavier)
• Muons are typically generated in collisions of cosmic rays in upper
atmosphere and, unlike electrons, decay ( t  2.2 sec)
0
• For a muon incident on Earth with v=0.998c, an observer on Earth would
see what lifetime of the muon?
• 2.2 sec?
 
1
1
2
 16
v
c2
• t=35 sec
• Moving clocks run slow so when an outside observer measures, they see a
longer time than the muon itself sees. (It’s who is the observer that
matters not who is actually moving)
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More about Muons
•
They are typically produced in atmosphere about 6 km above surface of Earth and
frequently have velocities that are a substantial fraction of speed of light, v=.998 c
for example
8 m
6
vt0  2.994 x10
•
•
•
•
•
sec
x 2.2 x10 sec  0.66km
How do they reach the Earth?
Standing on the Earth, we see the muon as moving so it has a longer life (moving
clocks run slow) and we see it living t=35 sec not 2.2 sec , so it can travel 16
times further than 0.66 km that it “thinks” it can travel, or about 10 km, so it can
easily cover the 6km necessary to reach the ground.
But riding on a muon, the trip takes only 2.2 sec, so how do they reach the
ground???
Muon-rider sees the ground moving, so the length contracts and is only
L0
 6 /16  0.38km so muon can go .66km, but reaches the ground in only .38km

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Velocity Addition Example
• Lance is riding his bike at 0.8c relative to observer. He throws a ball at
0.7c in the direction of his motion. What speed does the observer see?
vx'  v
vx 
vv '
1  2x
c
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vx 
.7c  .8c
 0.962c
.7c  .8c
1
c2
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Relativistic Momentum Example
• A meteor with mass of 1 kg travels 0.4c
• Find its momentum, what if it were going twice as fast? compare with
classical case
a)
v
 0.4
c
  1.09
v
b)
 0.8
c
  1.67
c ) p  mv  1.2  108
d )  2.4  108
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p   mv  1.09 1kg  0.4  3  108
m
kg  m
 1.31 108
sec
sec
p  1.67  1 0.8  3 108  4.01108
kg  m
sec
kg  m
sec
kg  m
sec
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Relativistic Energy
E   mc 2  KE  mc 2
KE  (  1)mc 2
E  (mc 2 ) 2  p 2 c 2
• Ex. 1.6: A stationary bomb (at rest) explodes into two fragments each with
1.0 kg mass that move apart at speeds of 0.6 c relative to original bomb.
Find the original mass M.
Ei  Mc2  KEi  Mc2  0  E f   1m1c 2  1m2c2
Mc 
2
2m1c 2
1
2
 M  2kg / 0.8  2.5kg
v
c2
• What if the bomb were not stationary?
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Properties of Photoelectric Effect
• Existence of photoelectric effect was not a surprise, but the details were
surprising
1) Very little time (nanoseconds) between arrival of light pulse and emission
of electron
2) Electron energy independent of intensity of light
3) At higher frequency  get higher energy electrons
Minimum frequency (0) required for photoelectric effect depends on material:
slope=E/ =h
(planck’s constant)
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Einstein Explains P.E. Effect
• Einstein explained P.E. effect: energy of light not distributed evenly over
classical wave but in discrete regions called quanta and later photons
1) EM wave concentrated in photon so no time delay between incident
photon and p.e. emission
2) All photons of same frequency have same energy E=h, so changing
intensity changes number (I=Nh, where N is rate/area) but not energy
3) Higher frequency gives higher energy
h  6.626 1034 J sec
• Electrons have maximum KE when all energy of photon given to electron.
•  is work function or minimum energy required to liberate electron from
material ( =h 0 )
KE max  h    h  h 0
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Example of PE for Iron
•
a) Find  given
 0  1.11015 Hz
  h 0
15
15 1
  4.14 10 eV sec1.110 s   4.5eV
•
b) If p.e.’s are produced by light with a wavelength of 250 nm, what is stopping
potential?
KEmax  eV0
•
•
•
6
 h    1.24 10 eV m  4.5eV
9
250 10 m
=4.96-4.5 =0.46 eV (is this your final answer?)
NO! it is V0=0.46V (not eV)
x-ray is inverse photo-electric effect, can neglect binding energy, since x-ray is very
energetic
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Photon Energy Loss
   
h
(1  cos  )
mc
 2me c 2  .511MeV  2
e e  
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De Broglie
E  pc  h
hv h
h
 
 p

p
c 
• Tiger example
• He hits a 46 g golf ball with a velocity of 30 m/s (swoosh) .
• Find the
; what do you expect?
dB
•
6.6 x1034 J sec
h
v c   1


 4.8 1034 m
mv .046kg 30m / sec
• The wavelength is so small relevant to the dimensions of the golf ball that it
has no wave like properties
• What about an electron with v  107 m / sec ?
• This large velocity is still not relativistic so
h
6.63 1034 J s
11



7.3

10
m
mv 9.11031 kg 107 m/s
11
• Now, the radius of a hydrogen atom (proton + electron) is 5 10 m
• Thus, the wave character of the electron is the key to understanding atomic
structure and behavior
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De Broglie Example 3.2
•
•
What is the kinetic energy of a proton with a 1fm wavelength
Rule of thumb, need relativistic calculation unless
pc
pc 
hc

mp c2  .938GeV
4.136 1015 ev s 3 108 m / s

 1.24GeV
11015 m
•
Is pc = energy? Units are right, but pc .ne. energy
•
Since
•
This implies a relativistic calculation is necessary so can’t use KE=p2/2m
pc  mp c2 need relativistic calculation
E  mc 2  p 2 c 2
 (.938) 2  (1.24) 2
 1.55GeV
KE  E  mc 2
 1.55  0.938  0.617GeV  617MeV
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Wave Velocities
•
•
•
•

Definition of phase velocity: v p    2 / 2  k
 d 

Definition of group velocity: vg 
k dk
v

v

c
For light waves : g p
d
For de Broglie waves
vg 
dk
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Particle in a Box Still
2
2
1
p
h
2
• General expression for non-rel Kinetic Energy: KE  mv 

2
2
2
m
2
m

• With no potential energy in this model, and applying
constraint on wavelength gives:
h2
2 2
2
En 

n
h
/
8
mL
2m(2 L / n)2
• Each permitted E is an energy level, and n is the quantum number
• General Conclusions:
1) Trapped particle cannot have arbitrary energy like a free particle—only
specific energies allowed depending on mass and size of box
2) Zero energy not allowed! v=0 implies infinite wavelength, which means
particle is not trapped
3) h is very small so quantization only noticeable when m and L are also
very small
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Example 3.5
• 10 g marble in 10 cm box, find energy levels
En  n 2 h 2 / 8mL2
•
•
•
•
n2 (6.63 1034 J s)2
64 2


5.5

10
n J
2
1
2
8 10 kg  (10 m)
For n=1 Emin  5.5 1064 J and v  3.3  1031 m / s
Looks suspiciously like a stationary marble!!
At reasonable speeds n=1030
Quantum effects not noticeable for classical phenomena
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Uncertainty Principle
a)
For a narrow wave group the position is
accurately measured, but wavelength and thus
momentum cannot be precisely determined
b) Conversely for extended wave group it is easy to
measure wavelength, but position uncertainty is
large
• Werner Heisenberg 1927, it is impossible to
know exact momentum and position of an object
at the same time
xp
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E t
2
16
Rutherford Scattering
K
N ( ) 
sin 4

2
KE 2
• The actual result was very different—
although most events had small angle
scattering, many wide angle scatters were
observed
• “It was almost as incredible as if you fired a 15 inch
shell at a piece of tissue paper and it came back at
you”
• Implied the existence of the nucleus.
• We perform similar experiments at
Fermilab and CERN to look for fundamental
structure
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Spectral Lines
• For Hydrogen Atom (experimental observation):
1
1
1
 R( 2  2 )

nf
ni
• where nf and ni are final and initial quantum states
• R=Rydberg Constant 1.097 107 m1  0.01097 nm 1
• Balmer Series nf = 2 and ni=3,4,5 visible wavelengths
in Hydrogen spectrum 656.3, 486.3,…364.6 (limit as
n)
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Bohr Energy Levels
E
e 2
En 
 21
8 0 r m n
How much energy is required to raise an electron from the ground state
of a Hydrogen atom to the n=3 state?
E  E f  Ei 
E1 E1
1 1
8
 2  E1 ( 2  )  13.6( )  12.1eV
2
nf
ni
3 1
9
Rydberg atom has r=0.01 mm what is n? E?
rn  n r1
2
n
rn
1105

 435
a0
5.3 1011
En 
E1
 7.19 105 eV
2
n
Bohr atom explains energy levels
E  E f  Ei 
E1 E1

 h  hc / 
n f 2 ni 2
1


 E1 1
1
( 2  2)
ch n f
ni
Correspondence principle: For large n Quantum Mechanics  Classical Mechanics
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Expectation Values
• Complicated QM way of saying average
• After solving Schrodinger Eq. for particle under particular condition 
contains all info on a particle permitted by uncertainty principle in the form
of probabilities.
• This is simply the value of x weighted by its probabilities and summed over
all possible values of x, we generally write this as

x 
  *( x, t ) x( x, t )dx

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Eigenvalue Examples
d2
• If operator is G  2 and wave function is
dx
d 2 2x
d  d 2x  d
2x
G


e

e



2
e
•
2

dx
dx  dx  dx

•
•
•
•
•
So eigenvalue is 4
How about sin(kx)?
eigenvalue is –k2
How about x4?
Not an eigenvector since
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
  e2 x find eigenvalue
4e 2 x
 G  4
d2 4
G  2  x   12 x 2  cx 4
dx
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