PHYS 141: Principles of Mechanics

1.

I. Blackbody Radiation A. Properties

Definition a) A blackbody wavelengths.

is a perfect absorber of light at all b) Wien’s Law : The peak wavelength of emission for a bb decreases as temperature increases : l p = 2.898 x 10 -3 (m-K)/T, (I.A.1) c) of The Stefan-Boltzmann Law: The power = (E/s) = P/m 2 a blackbody is given by: P bb = s AT 4 , where s = 5.67 x 10 -8 W/(m 2 K 4 ).

(I.A.2)

I. Blackbody Radiation A. Properties

c) The Stefan-Boltzmann Law * Further define an energy Flux, F = P/A: F = s T 4 (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).

R bb

l  2 

hc

2 l 5 (

e

(

hc

/ l

kT

)  1) .

(I.A.4) (I.A.5-6)

h

   34

k

 1.38 10  23  .

.

I. Blackbody Radiation A. Properties

Light bulb off: No light at any l .

UV Visible IR Radio

I. Blackbody Radiation A. Properties

Light bulb on: T = 3000 K.

= The combined brightness at each l (color) determines what color a BB appears.

UV Visible Dim in Violet And Blue Fairly bright in Red & Orange IR l max = 1000 nm Radio

I. Blackbody Radiation A. Properties

UV Light bulb on: T = 6000 K.

= Bright in Y&G Fairly bright in Red & Violet Visible l max = 500 nm IR Radio

I. Blackbody Radiation A. Properties

Light bulb on: T = 30,000 K.

= Very Bright in V,I,B Bright in G,Y,O,R UV Visible l max = 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: E n = nhf.

where h is Planck’s constant, (I.A.7)

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light:

photons

.

2.

E photon = hf = nhc/ l. (I.B.1) Extension of Planck’s work on molecular oscillators

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e -e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e-e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e -e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e -e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e -e

I. Blackbody Radiation A. Properties

B.

1.

The Photoelectric Effect Einstein proposed the particle model of light: photons.

2.

E photon = hf = nhc/ l .

(I.B.1) Extension of Planck’s work on molecular oscillators -e -e -e

3.

I. Blackbody Radiation A. Properties

Particle-Theory

of Light predictions: RIGHT!

a) As the frequency of light increases, the maximum K of electrons also increases: K max = hf f , (B.2) Where f is the “Work Function” that holds the electrons to the plate.

b) c) If f < f 0 = “cutoff frequency” = f /h, no electrons ejected.

Maximum electron energy (as measured by the “stopping potential ,” V S ) is independent of intensity.

I. Blackbody Radiation A. Properties

4.

I. Blackbody Radiation A. Properties

Example: Light shines on a substance with f If l = 1 eV. = 500 nm, are electrons ejected from the surface?

Step 1: Find the energy of the photons. For l have = 500 nm, we E = hc/ l E = (4.0 x 10 -19 = (6.6 x 10 -34 = 4.0 x 10 -19 J; J-s)(3.0 x 10 8 J)(1 eV/1.6 x 10 -19 m/s)/(5 x 10 J) = 2.5 eV.

-7 m) Step 2: Find K max = 2.5eV - 1.0 eV = 1.5 eV > 0. So, yes, electrons are ejected.

II. Light, More Light

A.

1.

Compton Effect (1923) Scattered light has lower frequency than incident light.

2.

l ’ l e  f e 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) = l C =

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  = 60 o ?

E i = hf = hc/ l = 1.6 x 10 -13 J * ; l = 0.0012 nm.

l ’ = l + (h/m e c)(1 - cos  ).

= 1.2 x 10 -12 = 2.4 x 10 -12 m + (2.4 x 10 -12 m = 0.0024 nm.

m)(1 - .5) Since the wavelength doubled, the E f = hc/ l ’ = 1/2E i = 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 a) b) b) c) Young Double Slit experiment: interference Maxwell’s Equations Connection between frequency and wavelength l f = c.

“Electromagnetic Spectrum” (B.1) Low Energy High Energy

II. Light, More Light C. Waves of Matter?

1.

2.

de Broglie (1923): about matter?

If light is observed to have wavelike properties some times (interference) and particle-like properties other times (photoelectric effect), then what l = h/mv “de Broglie Wavelength” (I.C.1) Example: what is l for an electron moving at .01c?

l l = (6.6 x 10 -34 = 2.4 x 10 -10 J-s)/(9.11 x 10 m = 0.24 nm.

-31 kg*3.0 x 10 6 m/s);

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) Joseph Fraunhofer (1815) : Hundreds of dark lines in the Solar Spectrum c) c) The Value of Spectral Analysis * Composition, abundance * Temperature * Motion

II. Light, More Light

2. The visible Hydrogen series (Balmer Series) 1/ l = R(1/2 2 -1/n 2 ), n = 3, 4, 5… (I.D.1) Where R = 1.097 x 10 7 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 required to move e) Packet of energy: a photon!

the nucleus to “ energy levels ” D E.

d) Larger orbital differences mean more energy f) Model later modified from strict orbits about

II. Light, More Light

E u e E l 2.

2. The mathematics of quantization a) a) Frequency of radiation absorption/emission: hf E u - E l = hf.

(I.E.1) b) b) Quantum condition: only specific values of electron orbital angular momentum allowed: L = m e vr n = n(h/2  ), n = 1, 2, 3, … (I.E.2)

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c) Electrostatic Force: q 2 r q 1 F = kq 1 q 2 /r 2 PE = -kq 1 q 2 /r. with (I.E.2a) (I.E.2b) k = 8.99 x 109 Nm 2 /C 2 is the Coulomb constant and the unit of charge is the Coulomb: e = 1.602 x 10 -19 C.

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d) Total energy of hydrogen atom: +e r n -e PE = -ke 2 /r n . (Electric Potential energy) KE = (1/2)m e v 2 .

Thus, E = PE + KE = -ke 2 /r +(1/2)m e v 2 . Now assume that the electron orbit is circular: (I.E.3) (I.E.4) F = ke 2 /r n 2 = m e v 2 /r n KE = ke 2 /(2r n ), and E = -ke 2 /(2r n ).

= 2r n (KE), so (I.E.5) (I.E.6) (I.E.7)

II. Light, More Light

e) Orbital sizes: use the quantum condition (E.1): r n = n(h/2  )/m e v, and r n 2 = n 2 (h/2  ) 2 /m e 2 v 2 .

From (E.5), ke 2 /r n 2 = m e v 2 /r n , so v 2 = (ke 2 /m e )/r n . Thus, r n = n 2 (h/2  ) 2 /(ke 2 m e ).

(I.E.8) Define “the Bohr radius” = r 1 = r(n=1) = 0.0529 nm, so r n = n 2 r 1 .

(I.E.9)

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e) Energy levels of orbits: E n = -{m e k 2 e 4 /(2(h/2  ) 2 }(1/n 2 ), or E n = (-13.6 eV)/n 2 = E 1 /n 2 .

(I.E.9) (I.E.10) Energy required for ionization: D E i = E(n = infinity) + (13.6 eV)/n 2 = (13.6 eV)/n 2 .

(I.E.11)

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F. Quantization and atomic spectra (H)

a) Ground state: n = 1, E 1 = -13.6 eV.

b) c) First excited state: n = 2, E 2 = -3.4 eV.

Energy required to move from E 1 to E 2 : D E = -3.4 eV - -13.6 eV = 10.2 eV.

3.

Hydrogenic atoms: Z = number of protons in nucleus.

E n r n = (Z/n) 2 E 1 .

= (n 2 /Z)r 1 .

(F.1) (F.2)

II. Light, More Light

3.

Photon frequency/wavelength: or f = D E/h = [{m e k e 2 e 4 }/{4  (h/2  ) 3 }](1/n f 2 - 1/n i 2 ), f = cR(1/n f 2 - 1/n i 2 ), and (I.F.3) 1/ l = R(1/n f 2 - 1/n i 2 ).

(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

A.

The Failure of the Bohr Model 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

1.

Define a Wave Function,  , that describes a particle.

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

2.

a) Calculations: Solution to the “Schroedinger Wave Equation” One dimensional case, Cartesian coordinates. Probability of finding a particle between x and x + dx b)   2

dx

.

Normalization: The particle must be SOMEWHERE. Therefore     2

dx

 1.

III. Quantum Mechanics B. Wave Mechanics

2.

c) Calculations: Solution to the “Schroedinger Wave Equation” Radial Probability distribution. Probability of finding a particle between r and r + dr d)  4

r

2

dr

.

(B.1) Normalization: The particle must be SOMEWHERE. Therefore 0   4

r

2

dr

 1.

 0 

r

2  2

dr

 1 4  .

(B.2)

III. Quantum Mechanics B. Wave Mechanics

3.

a) Example: Ground State of Hydrogen Wave function for ground state  

Ae

 / 0 .

b) Step 1: Find the constant by normalization 0   4  2 2

r A e

 2 / 0

dr

 1.

 0  2

r e

 2 / 0

dr

 (B.3) 4  1

A

2 .

(B.4a,b)

A

 1 / 

a

0 3 .

(B.5)

III. Quantum Mechanics B. Wave Mechanics

3.

c) Example: Ground State of Hydrogen Now ask: What is the probability of finding an electron inside the Bohr Radius, a 0 ? (see pg. 649 and Appendix A the for integral) 

a

0 ) 

a

0 0  4  2 2

r A e

 2 / 0

dr

 4  

a

0 3

a

0 0  2

r e

 2 / 0

dr

 0.323

III. Quantum Mechanics C. HUP

C.

1.

2.

Heisenberg Uncertainty (Indeterminancy) Principle There is an inherent uncertainty in pairs of correlated variables Momentum-Position Uncertainty Principle D x D p ≥ (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 D E D t ≥ (h/2  ).

(III.C.2) In this UP, the energy of a particle is uncertain ( D E), over a time scale D t ~ (h/2  )/ D E.

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 10 6 m/s?

D x D p ≥ (h/2  ) => D x min = (6.6 x 10 -34 = 1.2 x 10 -8 m, J-s/2  )/[(9.1 x 10 -31 kg)(10 4 m/s)], = 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?

D E D t ≥ (h/2  ) => D t = (6.6 x 10 -34 = 6.6 x10 -19 s.

J-s/2  )/[(1000eV)(1.6 x 10 -19 J/eV)],

III. Quantum Mechanics D. Quantum Numbers 1.

a) b) Principal quantum number: n (e.g., E n Related to energy of electron n = 1, 2, 3, …  = E 1 /n 2 ).

2.

a) Orbital Quantum number: l related to orbital angular momentum vector of electron L = { l ( l +1)} 1/2 h , where = h/(2  ), and (D.1) l = 0, 1, 2, …, n-1.

(D.2)

III. Quantum Mechanics D. Quantum Numbers 3.

a) Magnetic quantum number: m .

l Related to direction of electron’s angular momentum l , l + 1, l + 2, …, 0, … l - 2, l - 1, l .

(D.3) h L z b) Angular momentum along the “z-axis:” L z =

m l

h

m l

= 1.

0 h

m l

= 0.

m l

= -1.

(D.4) l Total number of states for a given value of depends on

projection

L z .

III. Quantum Mechanics D. Quantum Numbers n = 2 l 3.

Magnetic quantum number: m .

l

c) Energy levels are split in the presence of magnetic field.

= 1

m l m l m l

= 1.

= 0.

= -1.

l n = 1 = 0

m l

= 0.

III. Quantum Mechanics D. Quantum Numbers 4.

Spin quantum number: m

s

.

a) originally thought of as intrinsic spin of electron b) c) m s = ±1/2, but s = ½ always (for an electron) “fine structure” Electron spin

vector

l n = 2 = 0 l n = 1 = 0

m m l l

= 0.

= 0.

III. Quantum Mechanics D. Quantum Numbers 5.

Total Angular Momentum: Vector sum of

J = L + S:

For l For l ≠ 0 = 0 j = l ± ½ j = ½ (parallel vs. antiparallel) (D.6) (D.7)

J

  1) .

(D.8) Total number of angular momentum states depends on

projection

J z : -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.

a) Selection Rules and Probability Distributions “Allowed” transitions: D l = ±1.

l

n l and

m l

change the shape of the probablitiy distribution function.

III. Quantum Mechanics E. Other Atoms 1.

a) The Pauli Exclusion Principle 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).

Li (Z = 3).

Be (Z = 4).

2e: (n,

l, m l

,

m s

(n,

l, m l

,

m s

) = (1, 0, 0, 1/2).

) = (1, 0, 0, -1/2).

3e: 4e: (n,

l, m l

,

m s

(n,

l, m l

,

m s

(n,

l, m l

,

m s

) = (1, 0, 0, 1/2).

) = (1, 0, 0, -1/2).

) = (2, 0, 0, 1/2).

(n,

l, m l

,

m s

(n,

l, m l

,

m s

(n,

l, m l

,

m s

(n,

l, m l

,

m s

) = (1, 0, 0, 1/2).

) = (1, 0, 0, -1/2).

) = (2, 0, 0, 1/2).

) = (2, 0, 0, -1/2).

IV. Nuclear Physics A. Radioactive Decay 1. Radioactive Decay Equation N/N 0 = e l t , where l is the decay rate.

(A.1) 2. Half-life : The time for half the sample to decay from parent isotope to daughter isotope = “t 1/2 .” 0  2  / 1/2 .

(A.2) t 1/2  (A.3)

IV. Nuclear Physics A. Radioactive Decay Rewriting the RDE, we have 3.

4.

N/N 0 = exp(-0.693t/t 1/2 ).

(A.4) Define activity as the number of decays/second, D N/ D t = l N = (0.693/t 1/2 )N = ( D N/ D t) 0 e l t .

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: 238 U as a half-life of 4.5 billion years. If an initial sample contains 10 18 atoms, what is the initial activity?

l = 0.693/t 1/2 ~ 5 x 10 -18 s -1 .

The initial activity for the sample is ( D N/ D t) 0 = l N = (2 x 10 -18 s -1 )(10 18 nuclei) = 5 decays/second.

What is the activity after 13.5 billion years?

( D N/ D t) = ( D N/ D t) 0 e l t = 0.6 decays/second ~ 1/8( D N/ D t) 0

IV. Nuclear Physics A. Radioactive Decay Example: 238 U as a half-life of 4.5 billion years. If an initial sample contains 10 18 atoms, what is the initial activity?

After 13.5 billion years, how much 238 U is left?

N/N 0 = exp(-0.693t/t 1/2 ) = exp(-0.693(3)) = 1/8.

3 half-lives => (1/2)(1/2)(1/2).

1.

IV. Nuclear Physics B. Nuclear Structure Nomenclature Z N ≡ ≡ proton (element) number neutron number A ≡ atomic mass number, e.g., 14 7 N or 13 6 C 2.

a) b) c) Spin Each

nucleon

(p,n) is a fermion Nuclear spin is quanitized ( I, I z ): resonances & NMR 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 4 2 3 2 He => spin 0 (boson) He => spin ½ (fermion)

3.

IV. Nuclear Physics B. Nuclear Structure 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 2 He p n s = 0 3 2 He p n s = ½ Even Z, N minimize total energy and increase stability of nucleus.

12 6 C p n E 3 E 2 E 1

IV. Nuclear Physics C. Energy 1.

Reactions a + X => Y + b.

X(a,b)Y.

2. Reaction energy Q = (M a + M X = D KE = KE b - M b -M Y )c 2 .

+ KE Y - KE a - KE X .

Q > 0: Exothermic (energy released).

Q < 0: Endothermic (energy required).

(IV.C.1) (IV.C.2) (IV.C.3) (IV.C.4)

IV. Nuclear Physics C. Energy 1.

Reactions a + X => Y + b.

X(a,b)Y.

2. Reaction energy Q = (M a + M X = D KE = KE b - M b -M Y )c 2 .

+ KE Y - KE a - KE X .

Q > 0: Exothermic (energy released).

Q < 0: Endothermic (energy required).

(IV.C.1) (IV.C.2) (IV.C.3) (IV.C.4)

IV. Nuclear Physics C. Energy 3.

Binding Energy a) Unified mass 1u = 931.5 MeV/c 2 .

b) Definition: The energy required to hold a nucleus together.

M(proton) = 1.00728 u M(neutron) = 1.00866 u M( 4 2 He) = 4.00153 u BE = (2M p + 2M n ) c 2 – M He c 2 = (4.03188 uc 2 - 4.00153 uc 2 )(931.5 MeV/uc 2 ) = 28.3 MeV.

IV. Nuclear Physics C. Energy 4.

Reaction Example: n + 14 7 N → 2 1 H + X. What is X?

In this example, a neutron strikes a 14 7 N nucleus producing deuterium ( 2 1 H) and another element. To find X, conserve nucleon number and charge: 14 (nitrogen) + 1 (neutron) = A + 2 (deuterium).

+7e (in nitrogen nucleus) = q X + 1e (deuterium).

Thus, A = 13 and q X = 6, and X = 13 6 C.

IV. Nuclear Physics C. Energy 5.

Example: What is the minimum proton kinetic energy required for the reaction ( 1 1 H + 13 6 C → 13 7 N + n) to take place?

Step one: compute rest energy on both sides: M( 13 6 C) = 13.003355 u M( 1 1 H) = 1.007825 u M( 13 7 N) = 13.005738 u M(n) = 1.008665 u (M after - M before )c 2 = (0.003223 uc 2 )(931.5 MeV/uc 2 ) = 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: 235 92 U a) n + 235 92 U → b) n + 235 92 U → 236 92 88 38 U* → 141 56 Ba + 92 36 Kr + 3n.

Sr + 136 54 Xe + 12n.

3. Energy released in single fission event: E fission = D Mc 2 (D.1)

IV. Nuclear Physics D. Fission 4.

Example: What is the energy released by the reaction n + 235 92 U → 88 38 Sr + 136 54 Xe + 12n?

Before: M i = M U + M n = 235.0439 u + 1.008665 u = 236.053 u.

After: M f = M Sr + M Xe + M 12n = 235.917 u.

E fission = D Mc 2 = (0.135 uc 2 )(931.5 MeV/uc 2 ) = 126 MeV.

IV. Nuclear Physics D. Fission 5.

a) b) c) d) Chain Reactions 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 1 H + 1 1 H → 2 1 H + e + +  e + 0.42 MeV.

Likelihood of interaction: cross-section of interaction between two protons (t ~ 10 9 years.)

 Second Step in “Proton-Proton Chain” 2 1 H + 1 1 H → 3 2 He +  + 5.49 MeV.

t ~ 1 second.

Third Step in “Proton-Proton Chain” 3 2 He + 3 2 He → 4 2 He + 1 1 H + 12.86 MeV t ~ 10 6 years.

Total products: 4 2 He + 2  e + 2  +26.7 MeV + 2e +

  “Bottlenecks” Deuterium bottleneck: H to He Fusion High Mass Bottleneck: no stable nuclei with 5 or 8 nucleons (protons or neutrons) e.g, 4 2 He + 4 2 He → 8 4 Be is endothermic & the product has a half-life of 10 -16 s.

Fusion reactions tend to favor steps built on 4 He, 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) d) 2 gamma rays + positrons + 2 neutrinos Requires high temperature (fast moving particles): T core ~ 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?

L sun = 4 x 10 26 W = 4 x 10 26 J/s = 2.5 x 10 The mass of the Sun’s core is ~5 x 10 29 39 MeV/s.

kg, or ~ 2 x10 56 protons.

In the p-p chain, 4p => He + 27 MeV. The number of reactions per second is then ~ 2.5 x 10 39 MeV/s/(27 MeV) ~ 10 38 .

Every second, 4 x 10 38 H become He (~billion metric tons). How long before there is no more hydrogen?

2 x 10 56 p/(4 x 10 38 p/s) = 5 x 10 17 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.

12 C is a nuclear catalyst.

Yield is ~ 27 MeV.

IV. Nuclear Physics E. Fusion e) Example: How much energy is released in the reaction 13 6 C( 1 1 H,  ) 14 7 N?

D M = M( 13 C) + M( 1 H) - M( 14 N), = 0.008106 u E = (0.008106uc

2 )(931.5 MeV/uc 2 ), = 7.55 MeV = 1.2 x 10 -12 J.

l > hc/E = (6.6 x 10 -34 Js)(3 x 10 8 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 ~ 10 8 K Occurs in the Sun when H core is depleted.

4 He + 4 He => 8 Be + 4 He => 12 C + γ + 7.367 MeV.

Side process: 12 C + 4 He → 16 O+ γ

5.

Massive Stars & Nucleosynthesis: 25 M SUN Star Saturn Orbit ~ 3 billion km

5.

Massive Stars & Nucleosynthesis: 25 M SUN Star Earth Diameter ~ 12, 000 km H fusion He fusion Fe C fusion O fusion Ne fusion Si fusion

5.

Massive Stars & Nucleosynthesis: 25 M SUN Star Fe

5.

Massive Stars & Nucleosynthesis: 25 M SUN 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) Galaxy Recession b) c) 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

z

 D l l  1 

v

1 

c v c

 1  1  1     1 .

z nonrelativ istic



v 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 v Cosmo = H 0 d , (B.4) where v is in km/s and d is in Mega-light years or Mega-parsecs.

Current WMAP measurements give:

H

0  (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/H 0 = 13.7 ± 0.1 Gyr.

V. Cosmology C. The Big Bang 1. Standard Model (prior to ~1980) a) Combines CP, GR b) c) Hubble Constant => Approximate Age Temperature decreases with time d) Expansion => mass-energy & Dark Energy t < 10 -43 t = 10 -43 t = 10 -35 s: s: s: t = 10 -12 s: “after Big Bang”: General Relativity breaks down.

strong-electro-weak force + gravity, T = 10 32 K.

electro-weak force + strong force + gravity, T = 10 27 four forces separate, T = 10 15 K.

K.

V. Cosmology C. The Big Bang 1. Standard Model (prior to ~1980) t = 10 -6 t = 1 s: s: t = 3 min: Quark confinement & production of p & n, T = 10 13 Universe becomes transparent to neutrinos, T = 10 10 Cosmic nucleosynthesis everywhere, T = 10 9 K.

K.

K.

D E ≥ ħ/ D t.

V. Cosmology C. The Big Bang Side note on particles: +2/3 +2/3 Leptons (F) Hadrons: Baryons (F) & Mesons (B) Field Particles (B) Higgs (B) -1/3 Also: particles and anti-particles

V. Cosmology C. The Big Bang 1. Standard Model (prior to ~1980) t = 10 -6 t = 1 s: s: t = 3 min: t = 4 x 10 5 t = 4 x 10 8 Quark confinement & production of p & n, T = 10 Universe becomes transparent to neutrinos, T = 10 Cosmic nucleosynthesis everywhere, T = 10 yrs: First galaxies, T = 30 K. 9 K.

13 10 K.

K.

yrs: Universe becomes transparent to photons, T = 3000 K D E ≥ ħ/ D t.

V. Cosmology D. Universal Composition

VI. The END A. The Ultimate Fate of the Universe 1. Before 1999 a) The Big Good Bye (?) b) c) The Big Crunch (100 billion years) The Big Freeze (+10 14 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) c) The Big Crunch (100 billion years) The Big Freeze (+10 14 years) 2. Post 1999 a) The Big Rip? (20 Billion years)