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 functions 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. Didn’t quite realize
photons were a particle
• 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  photon is a
particle
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)/me c2
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 momentum
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
• Brehmstrahlung becomes more important
with higher energy or higher Z
• from Rev. of Particle Properties
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Photon Cross Section vs E
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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
• already went over kinematics
• 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).
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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 quantizes energy, and 2D angular
momentum (one component and absolute
magnitude)
• for transitions
1 1
E  E0 ( 2 - 2 )
ni n j
P460 - Early Modern
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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