Transcript PHYS 342: Modern Physics

PHYS 141:
Principles of Mechanics
I. Blackbody Radiation A. Properties
1.
Definition
a) A blackbody is a perfect absorber of light at all
wavelengths.
b)
Wien’s Law: The peak wavelength of emission for a bb
decreases as temperature increases :
lp = 2.898 x 10-3 (m-K)/T,
c)
of
(I.A.1)
The Stefan-Boltzmann Law: The power = (E/s) = P/m2
a blackbody is given by:
Pbb = sAT4,
where s = 5.67 x 10-8 W/(m2K4).
(I.A.2)
I. Blackbody Radiation A. Properties
c) The Stefan-Boltzmann Law
*
Further define an energy Flux, F = P/A:
F = sT4
(I.A.3)
 Hotter blackbodies emit more total power per area.
 Properties of blackbody described principally by T.
d) Radiance: radiated power as a function of wavelength and
temperature (Max Planck).
2 hc2
Rbb (l , T )  5 ( hc /lkT )
.
l (e
 1)
(I.A.5-6)
(I.A.4)
h  6.63  1034 J  s.
k  1.38  1023 J / K .
I. Blackbody Radiation A. Properties
Radiance
Light bulb off: No light at any l.
UV
Visible
IR
Radio
I. Blackbody Radiation A. Properties
Radiance
Light bulb on: T = 3000 K.
=
The combined
brightness at each
l (color)
determines what
color a BB appears.
Dim in Violet
And Blue
Fairly bright in
Red & Orange
UV
Visible
IR
lmax= 1000 nm
Radio
I. Blackbody Radiation A. Properties
Radiance
Light bulb on: T = 6000 K.
=
Bright in Y&G
Fairly bright in
Red & Violet
UV
Visible
lmax= 500 nm
IR
Radio
I. Blackbody Radiation A. Properties
Radiance
Light bulb on: T = 30,000 K.
=
Very Bright in
V,I,B
Bright in
G,Y,O,R
UV
Visible
lmax= 100 nm
IR
Radio
I. Blackbody Radiation A. Properties
e) The Ultraviolet Catastrophe!
*
*
*
Classical thermal physics predicted infinite flux and
infinite intensity at small wavelengths
The solution: Planck’s hypothesis
“Electric resonators” responsible for bb radiation
have discrete (quantized) energies:
En = nhf.
where h is Planck’s constant,
(I.A.7)
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e
-e
-e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e
-e
-e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e-e-e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e -e -e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e
-e
-e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e
-e
-e
I. Blackbody Radiation A. Properties
B.
The Photoelectric Effect
1.
Einstein proposed the particle model of light: photons.
Ephoton = hf = nhc/l.
2.
(I.B.1)
Extension of Planck’s work on molecular oscillators
-e
-e
-e
I. Blackbody Radiation A. Properties
3.
Particle-Theory of Light predictions: RIGHT!
a)
As the frequency of light increases, the maximum K of
electrons also increases:
Kmax = hf - f,
(B.2)
Where f is the “Work Function” that holds the electrons
to the plate.
b)
If f < f0 = “cutoff frequency” = f/h, no electrons ejected.
c)
Maximum electron energy (as measured by the “stopping
potential ,” VS) is independent of intensity.
I. Blackbody Radiation A. Properties
I. Blackbody Radiation A. Properties
4.
Example: Light shines on a substance with f = 1 eV.
If l = 500 nm, are electrons ejected from the surface?
Step 1: Find the energy of the photons. For l = 500 nm, we
have
E = hc/l
= (6.6 x 10-34 J-s)(3.0 x 108 m/s)/(5 x 10-7 m)
= 4.0 x 10-19 J;
E = (4.0 x 10-19 J)(1 eV/1.6 x 10-19 J) = 2.5 eV.
Step 2: Find Kmax = 2.5eV - 1.0 eV = 1.5 eV > 0.
So, yes, electrons are ejected.
II. Light, More Light
A.
Compton Effect (1923)
1.
Scattered light has lower frequency than incident light.
l’
l
e-

f
e-
2.
Recall that relativistic energy is E = pc for a massless
particle. Thus, the relativistic momentum for a photon is
p = E/c = hf/c = h/l.
(A.1)
II. Light, More Light
l’
l
e-

f
e-
Conserve momentum & energy for electron initially at rest:
l’= l + (h/mc)(1 - cos).
Dl  (h/mc)(1 - cos).
and (h/mc) = lC = Compton Wavelength.
(A.2)
(A.3)
(A.4)
II. Light, More Light
3.
Example: How much energy is lost by a 1 MeV photon that
scatters off an electron with an angle  = 60o?
Ei= hf = hc/l = 1.6 x 10-13 J*; l = 0.0012 nm.
l’
= l + (h/mec)(1 - cos).
= 1.2 x 10-12 m + (2.4 x 10-12 m)(1 - .5)
= 2.4 x 10-12 m = 0.0024 nm.
Since the wavelength doubled, the Ef = hc/l’ = 1/2Ei = 0.5 MeV.
*1
MeV = 1.602 x10-13 J.
II. Light, More Light B. Light as a Wave
1.
And yet: light also behaves as a wave
Low
Energy
a)
b)
b)
Young Double Slit experiment: interference
Maxwell’s Equations
Connection between frequency and wavelength
c)
lf = c.
“Electromagnetic Spectrum”
(B.1)
High
Energy
II. Light, More Light C. Waves of Matter?
1.
de Broglie (1923): If light is observed to have wavelike
properties some times (interference) and particle-like
properties other times (photoelectric effect), then what
about matter?
l = h/mv
2.
“de Broglie Wavelength”
(I.C.1)
Example: what is l for an electron moving at .01c?
l = (6.6 x 10-34 J-s)/(9.11 x 10-31 kg*3.0 x 106 m/s);
l = 2.4 x 10-10 m = 0.24 nm.
II. Light, More Light D. Atomic Spectra
1. Spectral Analysis
a) a) SA: The identification of a chemical substance
by its unique spectral lines.
b) b)
c) c)
Joseph Fraunhofer (1815) : Hundreds of dark
lines in the Solar Spectrum
The Value of Spectral Analysis
* Composition, abundance
* Temperature
* Motion
II. Light, More Light
2. The visible Hydrogen series (Balmer Series)
1/l = R(1/22 -1/n2), n = 3, 4, 5…
(I.D.1)
Where R = 1.097 x 107 m-1 is the Rydberg constant.
Balmer lines
are transitions
to/from n = 2
656 nm (B, n = 3).
486 nm (B, n = 4).
434 nm (B, n = 5).
II. Light, More Light E. The Bohr Model
1. General aspects of the model
a) Hydrogen Atom as simple example
b) Electron orbits are quantized; not all orbits
stable.
c) To move from one orbit to the next, an electron
needs to absorb or emit an exact DE.
d) Larger orbital differences mean more energy
required to move
e) Packet of energy: a photon!
f) Model later modified from strict orbits about
the nucleus to “energy levels”
II. Light, More Light
2.
Eu
hf
eEl
2. The mathematics of quantization
a) a) Frequency of radiation absorption/emission:
Eu - El = hf.
(I.E.1)
b) b) Quantum condition: only specific values of
electron orbital angular momentum allowed:
L = mevrn = n(h/2),
n = 1, 2, 3, … (I.E.2)
II. Light, More Light
c)
q2
r
Electrostatic Force:
q1
F = kq1q2/r2
PE = -kq1q2/r.
with
(I.E.2a)
(I.E.2b)
k = 8.99 x 109 Nm2/C2 is the Coulomb constant
and the unit of charge is the Coulomb:
e = 1.602 x 10-19 C.
II. Light, More Light
d)
Total energy of hydrogen atom:
+e rn -e
PE = -ke2/rn. (Electric Potential energy)
KE = (1/2)mev2.
Thus, E = PE + KE = -ke2/r +(1/2)mev2.
Now assume that the electron orbit is circular:
F = ke2/rn2 = mev2/rn = 2rn(KE), so
KE = ke2/(2rn), and
E = -ke2/(2rn).
(I.E.5)
(I.E.6)
(I.E.7)
(I.E.3)
(I.E.4)
II. Light, More Light
e)
Orbital sizes: use the quantum condition (E.1):
rn = n(h/2)/mev, and
rn2 = n2(h/2)2/me2v2.
From (E.5),
ke2/rn2 = mev2/rn, so v2 = (ke2/me)/rn. Thus,
rn = n2(h/2)2/(ke2me).
(I.E.8)
Define “the Bohr radius” = r1 = r(n=1) = 0.0529 nm, so
r n = n2 r 1 .
(I.E.9)
II. Light, More Light
e) Energy levels of orbits:
En = -{mek2e4/(2(h/2)2}(1/n2), or
(I.E.9)
En = (-13.6 eV)/n2 = E1/n2.
(I.E.10)
Energy required for ionization:
DEi = E(n = infinity) + (13.6 eV)/n2 = (13.6 eV)/n2.
(I.E.11)
II. Light, More Light
F. Quantization and atomic spectra (H)
a) Ground state: n = 1, E1 = -13.6 eV.
b) First excited state: n = 2, E2 = -3.4 eV.
c) Energy required to move from E1 to E2:
DE = -3.4 eV - -13.6 eV = 10.2 eV.
3.
Hydrogenic atoms: Z = number of protons in nucleus.
En = (Z/n)2E1.
rn = (n2/Z)r1.
(F.1)
(F.2)
II. Light, More Light
3.
Photon frequency/wavelength:
or
f = DE/h = [{meke2e4}/{4(h/2)3}](1/nf2 - 1/ni2),
f = cR(1/nf2 - 1/ni2), and
(I.F.3)
1/l = R(1/nf2 - 1/ni2).
(I.F.4)
II. Light, More Light
3.
Quantization and atomic spectra
f) Spectral types explained!
Unique electron orbits =>
unique energy differences =>
unique patterns
Absorption/Emission spectra
understood as upward/
downward transitions
of electrons.
II. Light, More Light
G. The Wave Nature of Matter (revisited)
1. de Broglie’s wave description of electrons can now
be understood in terms of standing waves
surrounding a nucleus: only waves that close back
on themselves can constructively interfere and
“survive”
III. Quantum Mechanics
The Failure of the Bohr Model
A.
1.
2.
3.
4.
Couldn’t be applied to more complicated atoms
Couldn’t explain “fine structure”
Couldn’t explain solids, liquids, molecules
Quasi-classical mechanics
The Bohr Model, while substantial, was incomplete.
III. Quantum Mechanics B. Wave Mechanics
Define a Wave Function, , that describes a particle.
1.
a)
b)
c)
d)
Wave function contains information about a particle’s state:
position, speed, momentum & energy, spin, etc.
“Matter Wave” or “Matter Field”
Depends on position and time:   (x,y,z,t).
Probabilities:
  “Probability density”
2 ~ Probability of finding an electron at
position (x,y,z) and time t.
III. Quantum Mechanics B. Wave Mechanics
Calculations: Solution to the “Schroedinger Wave Equation”
2.
a)
One dimensional case, Cartesian coordinates. Probability of
finding a particle between x and x + dx
P( x)   2dx.
b)
Normalization: The particle must be SOMEWHERE.
Therefore

2

 dx  1.

III. Quantum Mechanics B. Wave Mechanics
Calculations: Solution to the “Schroedinger Wave Equation”
2.
c)
Radial Probability distribution. Probability of finding a
particle between r and r + dr
P(r)  4 r 2 2dr.
d)
(B.1)
Normalization: The particle must be SOMEWHERE.
Therefore

 4 r 
2
0
2
dr  1.

1
0 r  dr  4 .
2
2
(B.2)
III. Quantum Mechanics B. Wave Mechanics
3.
Example: Ground State of Hydrogen
a)
Wave function for ground state
 (r)  Ae r/a .
(B.3)
0
b)
Step 1: Find the constant by normalization

 4 r
0
2
2  2 r / a0
Ae
dr  1.

r e
0
A  1 /  a0 3 .
2 2 r / a0
1
dr 
. (B.4a,b)
2
4 A
(B.5)
III. Quantum Mechanics B. Wave Mechanics
3.
Example: Ground State of Hydrogen
c)
Now ask: What is the probability of finding an electron
inside the Bohr Radius, a0? (see pg. 649 and Appendix A the
for integral)
0
4


2
r
/
a
2 2 r / a0
P( r  a0 )   4 r 2 A2e 0 dr 
r
e
dr  0.323
3 
 a0 0
0
a0
a
III. Quantum Mechanics C. HUP
Heisenberg Uncertainty (Indeterminancy) Principle
C.
1.
2.
There is an inherent uncertainty in pairs of correlated variables
Momentum-Position Uncertainty Principle
DxDp ≥ (h/2).
(III.C.1)
The original argument comes from measurement of an
electron. The uncertainty in the electron’s position is roughly
the wavelength of observation (l), and the photon will give
some unknown portion of its momentum to the electron (h/l)
when it strikes the electron. Trying to narrow the position
means using smaller l, but that imparts more (unknown)
momentum to the electron.
III. Quantum Mechanics C. HUP
3.
Energy-Time Uncertainty Principle
DEDt ≥ (h/2).
(III.C.2)
In this UP, the energy of a particle is uncertain (DE), over a
time scale Dt ~ (h/2)/DE.
Warning: Part of the problem is that we tend to view
electrons, etc. exclusively as particles. Indeterminancy is a
direct result of electrons, protons, etc. having both wave AND
particle properties.
III. Quantum Mechanics C. HUP
4.
Example: What is the maximum precision for measuring the
position of an electron moving with a speed equal to
v = (3.00 ± 0.01) x 106 m/s?
DxDp ≥ (h/2) =>
Dxmin = (6.6 x 10-34 J-s/2)/[(9.1 x 10-31 kg)(104 m/s)],
= 1.2 x 10-8 m,
= 12 nm, or about 100x the size of an atom.
III. Quantum Mechanics C. HUP
5.
Example: What is the lifetime of a particle with a spread in
energy equal to 1kev?
DEDt ≥ (h/2) =>
Dt
= (6.6 x 10-34 J-s/2)/[(1000eV)(1.6 x 10-19 J/eV)],
= 6.6 x10-19 s.
III. Quantum Mechanics D. Quantum Numbers
1.
Principal quantum number: n (e.g., En = E1/n2).
a) Related to energy of electron
b) n = 1, 2, 3, … 
2.
Orbital Quantum number: l
a) related to orbital angular momentum vector of electron
L = {l(l +1)}1/2 h , where h = h/(2), and
(D.1)
l = 0, 1, 2, …, n-1.
(D.2)
III. Quantum Mechanics D. Quantum Numbers
3.
Magnetic quantum number: ml.
a) Related to direction of electron’s angular momentum
- l, - l + 1, - l + 2, …, 0, … l - 2, l - 1, l.
(D.3)
b)
Angular momentum along the “z-axis:”
Lz
h
0
-h
ml = 1.
ml = 0.
ml = -1.
Lz = ml h
(D.4)
Total number of states for a given value of
l depends on projection Lz.
III. Quantum Mechanics D. Quantum Numbers
3.
Magnetic quantum number: ml.
c) Energy levels are split in the presence of magnetic field.
n=2
l=1
ml = 1.
ml = 0.
ml = -1.
n=1
l=0
ml = 0.
III. Quantum Mechanics D. Quantum Numbers
4.
Spin quantum number: ms.
a) originally thought of as intrinsic spin of electron
b) ms = ±1/2, but s = ½ always (for an electron)
c) “fine structure”
Electron spin vector is given by S  s(s  1) .
n=2
l=0
ml = 0.
n=1
l=0
ml = 0.
(D.5).
III. Quantum Mechanics D. Quantum Numbers
5.
Total Angular Momentum: Vector sum of J = L + S:
For l ≠ 0
For l = 0
j=l±½
j=½
J
(parallel vs. antiparallel)
j( j  1) .
(D.6)
(D.7)
(D.8)
Total number of angular momentum states depends on projection Jz:
-j, -j+1, …, j-1, j => 2j + 1 states.
Nomenclature: n(l)j,
(D.9)
Where l = 0, 1, 2, 3, 4 => S, P, D, F, G
III. Quantum Mechanics D. Quantum Numbers
6.
Selection Rules and Probability Distributions
a) “Allowed” transitions: Dl = ±1.
l
n
l and ml change the
shape of the probablitiy
distribution function.
III. Quantum Mechanics E. Other Atoms
1.
The Pauli Exclusion Principle
a) Electrons in the same atom cannot have the same quantum
state--each electron has a unique set of quantum numbers.
b)
Usually electrons are found in the lowest possible state.
c)
Similar exclusion principle for any particle that has half
integer values of spin (e.g., protons & neutrons): Fermions.
d)
Particles with integer values of spin do not obey EP
(e.g., photons): Bosons.
III. Quantum Mechanics E. Other Atoms
e)
Examples
He (Z = 2).
2e:
(n, l, ml, ms ) = (1, 0, 0, 1/2).
(n, l, ml, ms ) = (1, 0, 0, -1/2).
Li (Z = 3).
3e:
(n, l, ml, ms ) = (1, 0, 0, 1/2).
(n, l, ml, ms ) = (1, 0, 0, -1/2).
(n, l, ml, ms ) = (2, 0, 0, 1/2).
Be (Z = 4).
4e:
(n, l, ml, ms ) = (1, 0, 0, 1/2).
(n, l, ml, ms ) = (1, 0, 0, -1/2).
(n, l, ml, ms ) = (2, 0, 0, 1/2).
(n, l, ml, ms ) = (2, 0, 0, -1/2).
IV. Nuclear Physics A. Radioactive Decay
1. Radioactive Decay Equation
N/N0 = e-lt,
(A.1)
where l is the decay rate.
2. Half-life: The time for half the sample to decay from
parent isotope to daughter isotope = “t1/2.”
N / N0  2t /t1/2.
(A.2)
t1/2  0.693 / l.
(A.3)
IV. Nuclear Physics A. Radioactive Decay
Rewriting the RDE, we have
N/N0 = exp(-0.693t/t1/2).
3.
(A.4)
Define activity as the number of decays/second,
DN/Dt = -lN = (0.693/t1/2)N = (DN/Dt)0e-lt.
4.
Decay channels:
-particle emission: He nucleus (2p, 2n)
-particle emission: e- + 
-particle emission: high energy photon
=> Z - 2, A - 4
=> Z+1, A
=> Z, A
IV. Nuclear Physics A. Radioactive Decay
Example: 238U as a half-life of 4.5 billion years. If an initial
sample contains 1018 atoms, what is the initial activity?
l = 0.693/t1/2 ~ 5 x 10-18 s-1.
The initial activity for the sample is
(DN/Dt)0 = lN = (2 x 10-18 s-1)(1018 nuclei) = 5 decays/second.
What is the activity after 13.5 billion years?
(DN/Dt) = (DN/Dt)0 e-lt = 0.6 decays/second ~ 1/8(DN/Dt)0
IV. Nuclear Physics A. Radioactive Decay
Example: 238U as a half-life of 4.5 billion years. If an initial
sample contains 1018 atoms, what is the initial activity?
After 13.5 billion years, how much 238U is left?
N/N0 = exp(-0.693t/t1/2) = exp(-0.693(3)) = 1/8.
3 half-lives => (1/2)(1/2)(1/2).
IV. Nuclear Physics B. Nuclear Structure
1.
Nomenclature
Z ≡ proton (element) number
N ≡ neutron number
A ≡ atomic mass number, e.g., 147N or 136C
2.
Spin
a)
Each nucleon (p,n) is a fermion
b)
Nuclear spin is quanitized (I, Iz): resonances & NMR
c)
Each nucleus may be a fermion or a boson depending on
the number of nucleons. For even and equal numbers of
p & n, nucleons are paired=> spin 0
=> spin 0 (boson)
3 He => spin ½ (fermion)
2
4
2He
IV. Nuclear Physics B. Nuclear Structure
3.
Nuclear stability
a) Nuclei have “shell structure” analogous to atomic shells
b) Z,N = 2, 8, 20, 28, 50, 82, 126: stable
c) PEP applies to n,p =>forces nucleons into energy levels:
Ground State configurations
4
3
2He
2He
12
p
p
n
n
6C
p
s=0
n
s=½
E3
E2
E1
Even Z, N minimize
total energy and increase
stability of nucleus.
IV. Nuclear Physics C. Energy
1.
2.
Reactions
a + X => Y + b.
X(a,b)Y.
(IV.C.1)
(IV.C.2)
Reaction energy
Q
= (Ma + MX - Mb -MY)c2.
= DKE = KEb + KEY - KEa - KEX.
(IV.C.3)
(IV.C.4)
Q > 0: Exothermic (energy released).
Q < 0: Endothermic (energy required).
IV. Nuclear Physics C. Energy
1.
2.
Reactions
a + X => Y + b.
X(a,b)Y.
(IV.C.1)
(IV.C.2)
Reaction energy
Q
= (Ma + MX - Mb -MY)c2.
= DKE = KEb + KEY - KEa - KEX.
(IV.C.3)
(IV.C.4)
Q > 0: Exothermic (energy released).
Q < 0: Endothermic (energy required).
IV. Nuclear Physics C. Energy
3.
Binding Energy
a) Unified mass
1u = 931.5 MeV/c2.
b) Definition: The energy required to hold a nucleus together.
M(proton) = 1.00728 u
M(neutron) = 1.00866 u
M(42He) = 4.00153 u
BE = (2Mp + 2Mn)c2 – MHec2
= (4.03188 uc2 - 4.00153 uc2 )(931.5 MeV/uc2)
= 28.3 MeV.
IV. Nuclear Physics C. Energy
4.
Reaction Example:
n + 147N → 21H + X. What is X?
In this example, a neutron strikes a 147N nucleus producing
deuterium (21H) and another element. To find X, conserve
nucleon number and charge:
14 (nitrogen) + 1 (neutron) = A + 2 (deuterium).
+7e (in nitrogen nucleus) = qX + 1e (deuterium).
Thus, A = 13 and qX = 6, and X = 136C.
IV. Nuclear Physics C. Energy
5.
Example: What is the minimum proton kinetic energy required
for the reaction (11H + 136C → 137N + n) to take place?
Step one: compute rest energy on both sides:
M(136C) = 13.003355 u
M(11H) = 1.007825 u
M(137N) = 13.005738 u
M(n) = 1.008665 u
(Mafter - Mbefore)c2 = (0.003223 uc2)(931.5 MeV/uc2) = 3.00 MeV.
Thus, the proton would have to have
kinetic energy of at least ~ 3.00 MeV.
IV. Nuclear Physics D. Fission
1.
2.
Definition: the splitting of one nucleus into multiple
nuclei after particle bombardment.
Example: 23592U
a) n + 23592U → 23692U* → 14156Ba + 9236Kr + 3n.
b) n + 23592U → 8838Sr + 13654Xe + 12n.
3. Energy released in single fission event:
Efission = DMc2
(D.1)
IV. Nuclear Physics D. Fission
4.
Example: What is the energy released by the reaction
n + 23592U → 8838Sr + 13654Xe + 12n?
Before: Mi = MU + Mn = 235.0439 u + 1.008665 u = 236.053 u.
After:
Mf = MSr + MXe + M12n = 235.917 u.
Efission = DMc2 = (0.135 uc2)(931.5 MeV/uc2) = 126 MeV.
IV. Nuclear Physics D. Fission
Chain Reactions
5.
a)
b)
c)
d)
Sustained nuclear
reaction requires chain
reaction
Moderator to slow
neutrons
Enrichment
Critical mass ~
kilograms
IV. Nuclear Physics E. Fusion
1.
2.
Definition: the fusing of two or more nuclei into new
nucleus.
Example: Proton-Proton Chain in the Sun.
First Step in “Proton-Proton Chain”
1 H
1
+ 11H → 21H + e+ + e + 0.42 MeV.
Likelihood of interaction: cross-section of
interaction between two protons
(t ~ 109 years.)
Second Step in “Proton-Proton Chain”
2

1H
+ 11H → 32He +  + 5.49 MeV.
t ~ 1 second.
Third Step in “Proton-Proton Chain”
3
3 He → 4 He + 1 H + 12.86 MeV
He
+
2
2
2
1
t ~ 106 years.
Total products: 42He + 2e + 2 + 2e+
+26.7 MeV
“Bottlenecks”
Deuterium bottleneck: H to He Fusion
High Mass Bottleneck: no stable nuclei
with 5 or 8 nucleons (protons or
neutrons)
e.g, 42He + 42He → 84Be is endothermic
& the product has a half-life of 10-16 s.


Fusion reactions tend to favor steps built
on 4He, rather than “proton capture,”
(but process is SLOW.)
IV. Nuclear Physics E. Fusion
2.
Example: Proton-Proton Chain in the Sun.
a) “Deuterium Bottleneck”
b) Energy produced in one chain: ~ 27 MeV.
c) 2 gamma rays + positrons + 2 neutrinos
d) Requires high temperature (fast moving particles):
Tcore ~ 15 million K
Use of plasma with similar parameters as solar interior plasma for
energy production on Earth is completely impractical - as even modest
1 GW fusion power plant would require about 170 billion tons of
plasma occupying almost one cubic mile.
IV. Nuclear Physics E. Fusion
3.
Solar life-time estimate: How long can the Sun shine?
Lsun = 4 x 1026 W = 4 x 1026 J/s = 2.5 x 1039 MeV/s.
The mass of the Sun’s core is ~5 x 1029 kg, or ~ 2 x1056 protons.
In the p-p chain, 4p => He + 27 MeV. The number of reactions per
second is then ~ 2.5 x 1039 MeV/s/(27 MeV) ~ 1038.
Every second, 4 x 1038 H become He (~billion metric tons).
How long before there is no more hydrogen?
2 x 1056 p/(4 x 1038 p/s) = 5 x 1017 s ~ 10 billion years.
IV. Nuclear Physics E. Fusion
4.
Other Fusion reactions
a)
b)
c)
d)
C-N-O cycle
H-fusion in higher mass
stars and later evolution
of the Sun.
12C is a nuclear catalyst.
Yield is ~ 27 MeV.
IV. Nuclear Physics E. Fusion
e)
Example: How much energy is
released in the reaction
13 C(1 H,)14 N?
6
1
7
DM = M(13C) + M(1H) - M(14N),
= 0.008106 u
E = (0.008106uc2)(931.5 MeV/uc2),
= 7.55 MeV = 1.2 x 10-12 J.
l > hc/E = (6.6 x 10-34 Js)(3 x 108 m/s)/(1.2 x 10-12 J) = 0.000165 nm.
IV. Nuclear Physics E. Fusion
4.
Other Fusion reactions
b) “Triple-alpha”: He-Fusion


Ignition T ~ 108 K
Occurs in the Sun when
H core is depleted.
4He
+ 4He => 8Be + 4He
=> 12C + γ + 7.367 MeV.
Side process:
12C
+ 4He → 16O+ γ
5.
Massive Stars & Nucleosynthesis: 25 MSUN Star
Saturn Orbit ~ 3 billion km
5.
Massive Stars & Nucleosynthesis: 25 MSUN Star
Earth Diameter ~ 12, 000 km
H fusion
He fusion
Fe
C fusion
O fusion
Ne fusion
Si fusion
5.
Massive Stars & Nucleosynthesis: 25 MSUN Star
Fe
5.
Massive Stars & Nucleosynthesis: 25 MSUN Star
10 km
n
Stellar fusion
Built from alpha capture
Neutron capture, decay products
BBN
spallation
Built from identical nucleus fusion + n, p release
V. Cosmology A. The Pillars of Modern Cosmology
1. General Relativity
a)
Spacetime is curved by the mass-energy in it. “Matter tells space
how to curve, Space tells Matter how to move.”
2. The Cosmological Principle
b) At the largest scales, the
Universe is both isotropic
and homogeneous.
V. Cosmology A. The Pillars of Modern Cosmology
The Shape of Space: Given GR and the CP, mass-energy density of
the Universe allows only three possible “geometries”
1.“Spherical” (positive curvature)
2.“Hyperbolic” (negative curvature)
3.“Flat”—Euclidean (zero curvature)
V. Cosmology A. The Pillars of Modern Cosmology
1. General Relativity
a)
Spacetime is curved by the mass-energy in it. “Matter tells space
how to curve, Space tells Matter how to move.”
2. The Cosmological Principle
b) At the largest scales, the
Universe is both isotropic
and homogeneous.
V. Cosmology B. Observations
1.
Observational Data and Cosmological Theory
a)
b)
c)
Galaxy Recession
Elemental Abundances
Cosmic Microwave Background
V. Cosmology B. Observations
2.
Galaxy Recession: Why are Galaxies Moving Away from Us?
a)
b)
c)
Vesto Slipher
(1910’s)
Hubble & Humason
(1920’s)
Doppler Shift
v
1
Dl
1 
c
z

1 
 1.
v
l
1


1
c
v
znonrelativistic    .
c
(B.1)
(B.2)
Similar shift in frequency to smaller
frequencies for recession and larger
frequencies for approach.
V. Cosmology B. Observations
3.
The Hubble Law
vCosmo = H0d, (B.4)
where v is in km/s and
d is in Mega-light years
or Mega-parsecs.
Current WMAP measurements give:
H 0  22  1 (km/s)/Mly.
V. Cosmology B. Observations
4.
Age of the Universe
We can estimate the age
of the Universe using the
Hubble constant:
[H0] = s-1, so
t(Universe) ~ 1/H0 = 13.7 ± 0.1 Gyr.
V. Cosmology C. The Big Bang
1.
Standard Model (prior to ~1980)
a)
b)
c)
d)
Combines CP, GR
Hubble Constant => Approximate Age
Temperature decreases with time
Expansion => mass-energy & Dark Energy
t < 10-43 s:
t = 10-43 s:
t = 10-35 s:
t = 10-12 s:
“after Big Bang”: General Relativity breaks down.
strong-electro-weak force + gravity, T = 1032 K.
electro-weak force + strong force + gravity, T = 1027 K.
four forces separate, T = 1015 K.
V. Cosmology C. The Big Bang
1.
Standard Model (prior to ~1980)
t = 10-6 s:
t = 1 s:
t = 3 min:
Quark confinement & production of p & n, T = 1013 K.
Universe becomes transparent to neutrinos, T = 1010 K.
Cosmic nucleosynthesis everywhere, T = 109 K.
DE ≥ ħ/Dt.
V. Cosmology C. The Big Bang
Side note on particles:
Leptons (F)
Hadrons: Baryons (F) & Mesons (B)
Field Particles (B)
Higgs (B)
+2/3
+2/3
-1/3
Also: particles and anti-particles
V. Cosmology C. The Big Bang
1.
Standard Model (prior to ~1980)
t = 10-6 s:
t = 1 s:
t = 3 min:
t = 4 x 105 yrs:
t = 4 x 108 yrs:
Quark confinement & production of p & n, T = 1013 K.
Universe becomes transparent to neutrinos, T = 1010 K.
Cosmic nucleosynthesis everywhere, T = 109 K.
Universe becomes transparent to photons, T = 3000 K
First galaxies, T = 30 K.
DE ≥ ħ/Dt.
V. Cosmology D. Universal Composition
VI. The END
A. The Ultimate Fate of the Universe
1. Before 1999
a) The Big Good Bye (?)
b) The Big Crunch (100 billion years)
c) The Big Freeze (+1014 years)
2. Post 1999
a) The Big Rip?
(20 Billion years)
VI. The END
A. The Ultimate Fate of the Universe
1. Before 1999
a) The Big Good Bye (?)
b) The Big Crunch (100 billion years)
c) The Big Freeze (+1014 years)
2. Post 1999
a) The Big Rip?
(20 Billion years)