Transcript Early Modern Physics

Pre-quantum mechanics
Modern Physics
• Historical “problems” were resolved by modern
treatments which lead to the development of
quantum mechanics
• need special relativity
• EM radiation is transmitted by massless photons
which have energy and momentum.
Mathematically use wave functions (wavelength,
frequency, amplitude, phases) to describe
• “particles” with non-zero mass have E and P and
use wave functioms to describe
SAME
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Blackbody Radiation
• Late 19th Century: try to derive Wien and StefanBoltzman Laws and shape of observed light spectra
• used Statistical Mechanics (we’ll do later in 461) to
determine relative probability for any wavelength l
• need::number of states (“nodes”) for any l
- energy of any state
probability versus energy
• the number of states = number of standing waves
= N(l)dl = 8pV/l4 dl with V = volume
• Classical (that is wrong) assigned each node the
same energy E = kt and same relative probability
this gives energy density u(l) = 8p/l4*kT
u -> infinity as
u
wavelength ->0
wavelength
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Blackbody Radiation II
• Modern, Planck, correct: E = hn = hc/ l Energy
and frequency are the same
• From stat. Mech -- higher energy nodes/states
should have smaller probability
try 1: Prob = exp(-hn/kt) -wrong
try 2: Prob(E) = 1/(exp(hn/kt) - 1) did work
• will do this later. Planck’s reasoning was obscure
but did get correct answer…..Bose had more
complete understanding of statistics
• Gives u(l) = 8p/ l4 * hc/ l * 1/(exp(hc/lkt) - 1)
Agrees with experimental
observations
u
Higher Temp
wavelength
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•
Photoelectric
effect
Photon absorbed by electron in a solid
(usually metal or semiconductor as “easier”
to free the electron) . Momentum conserved
by lattice
• if Eg > f electron emitted with Ee = Eg - f
(f is work function)
• Example = 4.5 eV. What is largest
wavelength (that is smallest energy) which
will produce a photoelectron? Eg = f or
l = hc/f = 1240 eV*nm/4.5 eV = 270 nm
0
Ee in
Ee
f
Eg
f
Conduction band
solid
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4
Compton Effect
 g + e -> g’+ e’ electron is quasifree. What
are outgoing energies and angles?
• Conservation of E and p
Einitial = Efinal or E + me = E’ + Ee’
x: py = p(e’)cosf + p( g’) cosq
y: 0 = p(e’)sinf - p (g’)sinq
• 4 unknowns (2 angles, 2 energies) and 3
eqns. Can relate any 2 quantities
• 1/Eg’ - 1/Eg = (1-cosq)/mec2
g
g
Feymann diag
q
f
e
e
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g
e
e
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Compton Effect
· if .66 MeV gamma rays are Comptoned
scattered by 60 degrees, what are the
outgoing energies of the photon and
electron?
• 1/Eg’ - 1/Eg = (1-cosq)/mec2
1/Egamma’ = 1/.66 MeV + (1-0.5)/.511
or Egamma’ = 0.4 MeV and
Te = kinetic energy = .66-.40 = .26 MeV
g’
g Z
q
f
e
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Brehmstrahlung + X-ray Production
· e+Z -> g’+ e’+Z
electron is accelerated in
atomic electric field and emits a photon.
• Conservation of E and p. atom has
momentum but Eatom =p2/2/Matom. And so
can ignore E of atom.
Einitial = Efinal or Egamma = E(e)- E(e’)
Ee’ will depend on angle -> spectrum
E+Z->e+Z+ g Brem
e +g -> e + g Compton
e+Z+g -> e+Z photoelectric
Z+g -> Z+e+e pair prod
energy
e
e z
e
e
g
g
Z
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brehm g
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Pair Production
· A photon can convert its energy to a particle
antiparticle pair. To conserve E and p
another particle (atom, electron) has to be
involved.
• Pair is “usually” electron+positron and
Ephoton = Ee+Epos > 2me > 1 MeV and
atom conserves momentum
g + Z -> e+ + e- + Z
can then annihilate electron-positron pair.
Need 2 photons to conserve energy
ALSO: Mu-mu pairs
e+ + e- -> g + g
p+cosmic MWB
Particle antiparticle
Z
Usually electron
g
positron
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Electron Cross section
• Brehmstrahlunf becomes more important
with higher energy or higher Z
• from Rev. of Particle Properties
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Photon Cross Section vs E
From Review of Particle properties
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Rutherford Scattering off Nuclei
· First modern. Gave charge distribution in
atoms. Needed l << atomic size. For 1 MeV
alpha, p=87 MeV, l=h/p = 10-12cm
•
kinematics: if Mtarget >> Malpha then little energy
transfer but large possible angle change. Ex: what
is the maximum kinetic energy of Au A=Z+N=197
after collision with T=8Mev alpha?
• Ptarget = Pin+Precoil ~ 2Pin (at 180 degrees)
• Ktarget = (2Pin)2/2/Mtarget = 4*2*Ma*Ka/2/Mau =
4*4/197*8MeV=.7MeV
recoil
a in
A
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target
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Rutherford Scattering II
· Assume nucleus has infinite mass. Conserve
Ea = Ta +2eZe/(4per)
•
conserve angular momentum
La = mvr = mv(at infinity)b
• E+R does arithmetic gives cot(q/2) = 2b/D where
D= zZe*e/(4pe Ka) is the classical distance of
closest approach for b=0
• don’t “pick” b but have all ranges 0<b<atom size
all alphas need to go somewhere and the cross
section is related to the area ds = 2pbdb (plus
some trigonometry) gives ds/dW =D2/16/sin4q/2
b=impact parameter
a
b
Z
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q
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Rutherford Scattering III
· Rutherford scattering can either be off a
heavier object (nuclei) ---> change in angle
but little energy loss --> “multiple
scattering”
• or off light target (electrons) where can transfer
energy but little angular change (energy loss due to
ionization, also produces “delta rays” which are
just more energetic electrons). Fall with increasing
velocity until a minimum then a relativistic rise
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Particles as Waves
• EM waves (Maxwell Eqs) are composed of
individual (massless) particles - photons - with
E=hf and p = h/l and E = pc
• observed that electrons scattered off of crystals had
a diffraction pattern. Readily understood if “matter”
particles (with mass) have the same relation
between wavelength and momentum as photons
• Bragg condition gives constructive interference
• 1924 DeBroglie hypothesis: “particles” (those with
mass as photon also a particle…) have wavelength
l = h/p
What is wavelength of K = 5 MeV proton ?
Non-rel p=sqrt(2mK) = sqrt(2*938*5)=97 MeV/c
l=hc/pc = 1240 ev*nm/97 MeV = 13 Fermi
p=50 GeV/c (electron or proton) gives .025 fm
(size of proton: 1 F)
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Bohr Model I
• From discrete atomic spectrum, realized something
was quantized. And the bound electron was not
continuously radiating (as classical physics)
• Bohr model is wrong but gives about right energy
levels and approximate atomic radii. easier than
trying to solve Schrodinger Equation….
• Quantized angular momentum (sort or right, sort of
wrong) L= mvr = n*hbar n=1,2,3... (no n=0)
• kinetic and potential Energy related by K = |V|/2
(virial theorem) gives
2
2
n

m
e


2
K  mv / 2  


 m r  2 8pe 0 r
4pe 0  2 2
2
solve for radius rn 
n

a
n
0
2
me
· radius is quantized
· a0 is the Bohr radius = .053 nm = ~atomic size
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Bohr Model II
• En = K + V = E0/n2 where E0 = -13.6 eV for H
• E0 = -m/2*(e*e/4pehbar)2 = -m/2 a2
where
a is the fine structure constant (measure of the
strength of the EM force relative to hbar*c= 197 ev
nm)
e2 1
1
a

4pe 0 c 137
 0.0072973525
68  0.0000000000
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• Bohr model quantizes energy and radius and 1D
angular momentum.
Reality has energy, and 2D angular momentum
(one component and absolute magnitude)
• for transitions
1 1
E  E0 ( 2 - 2 )
ni n j
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levels i, j
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Bohr Model III
• E0 = -m/2*(e*e/4pehbar)2 = -m/2 a2 (H)
•
easily extend Bohr model. He+ atom, Z=2 and
En = 4*(-13.6 eV)/n2 (have (zZ)2 for 2 charges)
• reduced mass. 2 partlces (a and b)
m =ma*mb/(ma+mb) if other masses
En = m/(me )*E0(zZ/n)2
Atom
ep
m p
mass
.9995me
94 MeV
pm
60 MeV
bb quarks q=1/3
2.5 GeV
P460 - Early Modern
E(n=1)
-13.6 eV
2.6 keV
1.6 keV
.9 keV
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Developing Wave Equations
• Need wave equation and wave function for
particles. Schrodinger, Klein-Gordon, Dirac
• not derived. Instead forms were guessed at, then
solved, and found where applicable
• So Dirac equation applicable for spin 1/2
relativistic particles
• Start from 1924 DeBroglie hypothesis: “particles”
(those with mass as photon also a particle…)have
wavelength l = h/p
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Wave Functions
• Particle wave functions are similar to amplitudes
for EM waves…gives interference (which was used
to discover wave properties of electrons)
• probability to observe =|wave amplitude|2=|y(x,t)|2
• particles are now described by wave packets
• if y = A+B then |y|2 = |A|2 + |B|2 + AB* + A*B
giving interference. Also leads to
indistinguishibility of identical particles
t1
t2
merge
vel=<x(t2)>-<x(t1)>
(t2-t1)
Can’t tell apart
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Wave Functions
• Describe particles with wave functions
 y(x) = S ansin(knx) Fourier series (for example)
• Fourier transforms go from x-space to k-space
where k=wave number= 2p/l. Or p=hbar*k and
Fourier transforms go from x-space to p-space
• position space and momentum space are conjugate
• the spatial function implies “something” about the
function in momentum space
y ( x)  1 / 2p  f (k )e dk
ikx
f (k )  1 / 2p y ( x)e
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- ikx
dx
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Wave Functions (time)
• If a wave is moving in the x-direction (or -x) with
wave number k have
y k ( x, t )  C cos(kx - wt ) + D sin(kx - wt )
i ( kx-wt )
or Ae
-i ( kx-wt )
+ Be
• kx-wt = constant gives motion of wave packet
• the sin/cos often used for a bound state while the
exponential for a right or left traveling wave
k  2p / l w  2p / T  2p f  2pn
p  h / l  k E  hn  w
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Wave Functions (time)
• Can redo Transform from wave number space
(momentum space) to position space
y ( x, t )  1 / 2p  f (k )e
or y ( x, t )   A(k )e
i ( kx-wt )
i ( kx-wt )
dk
dk
• normalization factors 2p float around in Fourier
transforms
• the A(k) are the amplitudes and their squares give
the relative probability to have wavenumber k
• could be A(k,t) though mostly not in our book
• as different k have different velocities, such a wave
packet will disperse in time. See sect. 2-2. Not
really 460 concern…..
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Heisenberg Uncertainty Relationships
• Momentum and position are conjugate. The
uncertainty on one (a “measurement”) is related to
the uncertainty on the other. Can’t determine both
at once with 0 errors
xpx   / 2
• p = hbar k
• electrons confined to nucleus. What is maximum
kinetic energy? x = 10 fm
• px = hbarc/(2c x) = 197 MeV*fm/(2c*10 fm) =
10 MeV/c
px  ( px - p )2   px2 - ( px )2
• while <px> = 0
• Ee=sqrt(p*p+m*m) =sqrt(10*10+.5*.5) = 10 MeV
electron can’t be confined (levels~1 MeV)
proton Kp = .05 MeV….can be confined
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