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FYSN440 Nuclear astrophysics (5 ECTS)
Ari Jokinen Office: FL 242
Email: [email protected] http://users.jyu.fi/~ajokinen/FYSN440/
1
Practical information • Lectures (22+ h): 17.3.2014-12.5.2014
– Mon 10:15-12:00 (FYS 2) and Wed 10:15-12:00 (FYS 2)
• Exercises (12 h) – Wed 08:15-10:00 (FL 140), starting 26.3.2014 – Problem sets to be solved. Usually based on the lectures from the previous week – Mark which problems you have solved and be ready to show your results on the
blackboard
• Team work projects ???: – About 10 slides (10-15 min presentation) about one interesting topic related to nuclear
astrophysics – You can suggest your own topic of interest or select from a list – Can be done individually or in teams of two or three students
• Literature: C. Iliadis, Nuclear Physics of Stars, Wiley-VCH 2007. C. E. Rolfs and W. S. Rodney, Cauldrons in the Cosmos.
• A course material is based on the lectues given by Dr. A. Kankainen and
many of the used slides are based on Prof. Hendrik Schatz’s Nuclear astrophysics course at the NSCL, MSU, USA.
2
How to pass the course?
• Final exam 16.5.2014: 48 points • Exercise points: 12 points
– Scaled based on how many exercise problems you have done ( 90% to get full 12 points)
• Team work projects, if organized, corresponds to one exercise.
• Note! You get points similarly as in other exercises but the points will be given by the lecturer
• Total maximum: 60 points • 30 points needed to pass the course
3
Main goals for the course
• To understand the role of nuclear physics in astrophysics
• To deepen your knowledge in nuclear physics
• To learn about recent studies in nuclear
astrophysics
4
5
the 3rd minute cataclysmic binaries
stellar evolution
Supernovae AGB stars
Origin and fate of the elements in our universe Origin of radiation and energy in our universe
Nuclear Astrophysics
Goals and methods in nuclear astrophysics
Nuclear properties
Properties of the Universe
HOW?
Nuclear theory
Astronomical observations
Astrophysical models
Nuclear physics
experiments
WHAT?
JYFL Accelerator Laboratory
6
The composition of the Universe
72%
23%
4.6 %
Only 0.4% visible as stars Interstellar gas (about 4%)
TOPIC OF THIS COURSE!
7
…but important things are made of that 4%!
What kind of nuclei are there in the Universe? What are abundances of different elements or different nuclei?
My goal is simple. It is a complete understanding of the universe, why it is as it is and why it exists at all. - Stephen Hawking
8
The origin of elements?
9
Back to basics…
Mass Spin Charge
Proton 938.272 MeV/c2 1/2 +e
Neutron 939.565 MeV/c2 1/2 0
Nucleons attract each other via the strong force ( range ~ 1 fm)
Nuclei consist of neutrons (N) and protons (Z) AX A=Z+N Z
The diameter of an atom is about 104 times the diameter of a nucleus
The atomic mass is almost entirely due to the nucleus (Compare: me=0.510 998 910(13) MeV/c2)
7Li 3
10
The nuclear potential
A bunch of nucleons bound together create a potential for an additional :
neutron proton (or any other charged particle)
V
r
R
V
r R
Coulomb Barrier Vc
ReZZVc
221=
…
…
Nucleons in a Box: Discrete energy levels in nucleus
R ~ 1.3 x A1/3 fm
nucleons are bound by an attractive force mass of a nucleus is smaller than the total mass of the nucleons
by its binding energy 11
Nuclear masses and binding energy Energy that is released when a nucleus is assembled from neutrons and protons
mp = proton mass, mn = neutron mass, m(Z,N) = mass of nucleus with Z,N
• B>0 • with B the mass of the nucleus is determined
2/),( cBNmZmNZm np −+=
Most tables give atomic mass excess ∆ in MeV:
Masses are usually tabulated as atomic masses
2/ cAmm u ∆+=(so for 12C: ∆=0) (see nuclear wallet cards for a table)
Nuclear Mass ~ 1 GeV/A
Electron Mass 511 keV/Z
Electron Binding Energy 13.6 eV (H) to 116 keV (K-shell U) / Z
m = mnuc + Z me - Be
12
Nuclear masses
• Energy generation in stars • Which nuclei are stable • Which nuclei exist in principle
JYFLTRAP Penning trap at IGISOL • High-precision atomic mass measurements (10 ppb) • Strong static magnetic field (7 T) and RF electric fields • Frequency ratios between a reference with a well-
known mass and the ion of interest
13
Why are masses important? 1. Energy generation
– Nuclear reaction A+B C – If mA+mB > mC then energy Q=(mA+mB-mC )c2 is generated by the reaction – Q-value= energy generated (Q> 0) or consumed
(Q<0) in the reaction 2. Stability
– If Q>0 (mA > mB+mC ) for A B+C, then the decay of the nucleus A is possible
3. Equilibria – For a nuclear reaction in equilibrium abundances scale
with e-Q (Saha equation) – Masses become the dominant factor in determining
the outcome of nucleosynthesis
A B
C
∆E=∆Mc2
14
Average binding energy per nucleon
FUSION generates
energy
FISSION generates
energy
15
The liquid drop mass model for the binding energy (Weizsäcker Formula)
AaAZB V=),(3/2Aas−
3/1
2
AZaC−
AAZaA
2)2/( −−
2/1−+ Aap
x 1 ee x 0 oe/eo x (-1) oo
Volume Term
Surface Term ~ surface area (Surface nucleons less bound)
Coulomb term. Coulomb repulsion leads to reduction uniformly charged sphere has E=3/5 Q2/R
Asymmetry term: Pauli principle to protons: symmetric filling of p,n potential boxes has lowest energy (ignore Coulomb)
protons neutrons neutrons protons
lower total energy = more bound
Pairing term: even number of like nucleons favoured
(e=even, o=odd referring to Z, N respectively)
(each nucleon gets bound by about same energy)
and in addition: p-n more bound than p-p or n-n (S=1,T=0 more bound than S=0,T=1)
Assumes incompressible fluid (volume ~ A) and sharp surface
16
Liquid drop model
17
Best fit values (from A.H. Wapstra, Handbuch der Physik 38 (1958) 1)
(in MeV/c2) aV aS aC aA aP 15.85 18.34 0.71 92.86 11.46
Deviation (in MeV) to experimental masses:
something is missing !
(Bertulani & Schechter)
18
Shell model: (single nucleon energy levels)
Magic numbers
are not evenly spaced shell gaps
more bound than average
less bound than average
need to add shell correction term S(Z,N)
Understanding the B/A curve
19
• Neglect asymmetry term (assume reasonable asymmetry) • Neglect pairing and shell correction (want to understand average behaviour) • Then:
3/4
2
3/1
1/AZa
AaaAB CSV −−=
Constant as the strong force has
a short range
Surface/volume ratio favours large nuclei
Coulomb repulsion has a long range. The more protons, the more repulsion favours small (low Z) nuclei
Maximum around Fe
Modern mass models
20
1) Microscopic – Macroscopic mass models • Macroscopic part: liquid drop, droplet, or their refinements • Microscopic part: shell correction, pairing correction, refinement of surface
term accounting for finite range of nuclear force …
2) Microscopic mass models • based on some (parametrized) nucleon-nucleon interaction • Problem: not very accurate due to limitations of current microscopic theories • Solution: Fit parameters of interaction specifically to masses to obtain a mass model
3) Local mass “models”
• extrapolations based on neighboring masses (Atomic Mass Evaluation) • mirror symmetry: Coulomb shifts, IMME • Garvey-Kelson …
Mass measurements have sufficiently progressed so that global mass models are mainly needed only for very neutron rich nuclei (r-process, neutron star crusts)
GLOBAL
LOCAL
Modern mass models – rms deviations
21
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2149 nuc70 new
18 new n-rich
rms
devi
atio
n (M
eV)
HFB-7FRDM
Example: mic model: HFB series (Goriely, Pearson) currently at HFB-15 (2008) mic-mac : Finite Range Droplet Model FRDM (Moller et al.) unchanged since 1993
Compare rms deviations )= root-mean-square deviations to experiment:
Important is not how well the model fits known masses, but how well it predicts unknown masses !
Predicted masses for Zr isotopes
22
What about mass differences?
23
Neutron capture Q-values for Zr isotopes (neutron separation energy Sn)
The valley of stability
24 N-number of neutrons
Z=82 (Lead)
Z=50 (Tin)
Z=28 (Nickel)
Z=20 (Calcium)
Z=8 (Oxygen)
Z=4 (Helium)
Magic numbers
Valley of stability (location of stable nuclei)
N=Z
Const A cut
Constant A cut
25
valley of stability (Bertulani & Schechter)
Binding energy per nucleon along const A due to asymmetry term in mass formula
decay decay decay decay
Decay – energetics and decay law
26
Decay of A in B and C is possible if reaction A B+C has positive Q-value
BUT: there might be a barrier that prolongs the lifetime
Decay is described by quantum mechanics and is a pure random process, with a constant probability for the decay of a single nucleus to happen in a given time interval.
N Number of nuclei A (Parent) λ Decay rate (decays per second and parent nucleus) τ=1/λ Lifetime
NdtdN λ−=
ttNtN λ−== e)0()(
Half-life is the time for half of the nuclei present to decay
(again – masses are critical !)
T1/2 = τln2 = ln2/λ
Decay modes
27
Coulomb barrier for anything other than a neutron (or a neutrino) emitted from the nucleus
V
r R
Coulomb Barrier Vc
ReZZVc
221=
unbound particle
Example: 197Au -> 58Fe + 139I has Q ~ 100 MeV ! yet, gold is stable.
If that barrier delays the decay beyond the lifetime of the universe (~ 14 Gyr) we consider the nucleus as being stable.
not all decays that are energetically possible
happen
Most common decay modes: • β decay • n decay • p decay • α decay • fission
Beta decay
28
p n conversion within a nucleus via weak interaction
Modes (for a proton/neutron in a nucleus):
β+ decay
Electron capture
β- decay
p n + e+ + νe
e- + p n + νe
n p + e- + νe
Electron capture (or EC) of atomic electrons or, in astrophysics, of electrons in the surrounding plasma
Q-values for decay of a nucleus (Z,N)
Qβ+ / c2 = mnuc(Z,N) - mnuc(Z-1,N+1) - me = m(Z,N) - m(Z-1,N+1) - 2me
QEC / c2 = mnuc(Z,N) - mnuc(Z-1,N+1) + me = m(Z,N) - m(Z-1,N+1)
Qβ- / c2 = mnuc(Z,N) - mnuc(Z+1,N-1) - me = m(Z,N) - m(Z+1,N-1)
with nuclear masses with atomic masses
Note: QEC > Qβ+ by 1.022 MeV!
Favourable for n-deficient nuclei
Favourable for n-rich nuclei
Calculating Q-values
• Usually atomic masses are used instead of nuclear masses
• Practical to use atomic mass excesses – Q values are directly obtained in keV – values well tabulated (Atomic Mass Evaluation = AME)
Nuclear reactions: A is always conserved the mass excess ∆ can always be used instead of the masses (the Amu term cancels)
Q-values with mass excess (∆)
29
∆ = (m-Au)c2
Example: 14C 14N+ e- + ne Q = [m(14C)-m(14N)]c2
= 14u+∆(14C)-14u-∆(14N) = ∆(14C)-∆(14N)
1 uc2= 931.4940090(71) MeV
30 30
Typical part of the chart of nuclides
Z
N
blue: neutron excess undergo β- decay
red: proton excess undergo β+ decay
even A isobaric chain
odd A isobaric chain
Typical beta-decay half-lives
31
• Very near “stability” : occasionally millions of years or longer - 40K β- decay to the stable 40Ca: 1.28 x 109 a • More common within a few nuclei of stability: minutes – days • Most exotic nuclei that can be formed: ~millisseconds
- superallowed beta decays, such as 62Ga EC decay ~116.121(21) ms - 69Kr ~32 ms
Proton or neutron decay
32
Usually, the protons and neutrons in a nucleus are bound Q-value for proton or neutron decay is negative
For extreme asymmetries in proton and neutron number nuclei become proton or neutron unbound Proton or neutron decay is then possible
A nucleus that is proton unbound (Q-value for p-decay > 0) is beyond the “proton drip line” A nucleus that is neutron unbound (Q-value for n-decay >0) is beyond the “neutron drip line”
NOTE: nuclei can exist beyond the proton and neutron drip line: • for very short time • for a “long” time beyond p-drip if Q-value for p-decay is small (Coulomb barrier !) • for a long time beyond n-drip at extreme densities inside neutron stars
Neutron decay
33
When adding neutrons to a nucleus eventually the gain in binding energy due to the volume term is exceeded by the loss due to the growing asymmetry term
No more neutrons can be bound the neutron drip line is reached
Neutron decay: (Z,N) (Z, N-1) + n
Q- value
Neutron separation energy = the energy needed to separate a neutron from the nucleus:
Sn = [m(Z,N-1) +m(n)-m(Z,N)]c2=-Qn
Qn = [m(Z,N)-m(Z,N-1)-m(n)]c2
Neutron drip line: Sn=0 Beyond the drip line: Sn < 0
No Coulomb barrier governed by the strong force The decay is immediate and dominates all other possible decay modes Neutron drip line very closely resembles the border of nuclear existence on the neutron-rich side!
Neutron Separation Energies for Z=40 (Zr)
34
30 40 50 60 70 80 90 100neutron number N
-5
0
5
10
15
20
Sn
(MeV
)
neutron drip
valley of stability
add 37 neutrons
Proton decay
35
Proton decay: (Z,N) (Z-1, N) + p
Q- value
Proton separation energy = the energy needed to separate a proton from the nucleus:
Sp = [m(Z-1,N) +m(p)-m(Z,N)]c2=-Qp
Qp = [m(Z,N)-m(Z-1,N)-m(p)]c2
Proton drip line: Sp=0 Beyond the drip line: Sp < 0
Note! Also two-proton decays have been observed!
• When adding protons, asymmetry AND Coulomb term reduce the binding therefore steeper drop of proton separation energy - drip line reached much sooner • Coulomb barrier (and Angular momentum barrier) can stabilize decay, especially for higher Z nuclei (lets say > Z~50)
N=40 isotonic chain
36
10 20 30 40 50proton number Z
-5
0
5
10
15
20Sp
(MeV
)
add 7protons
Nuclei beyond (not too far beyond) can therefore have other decay modes than p-decay. One has to go several steps beyond the proton drip line before nuclei cease to exist (how far depends on absolute value of Z).
α decay
37
emission of an α particle (= 4He nucleus)
Coulomb barrier twice as high as for p emission, but exceptionally strong bound, so larger Q-value
emission of other nuclei does not play a role (but see fission !) because of • increased Coulomb barrier • reduced cluster probability
α
αα
mAZBAZBmAZmAZmQ+−−+−=
−−−−=
)4,2(),(
)4,2(),(
<0, but closer to 0 with larger A,Z
Q-value for a decay:
large A therefore favored
38 38
lightest α emitter: 144Nd (Z=60) (Qα=1.9 MeV but still T1/2=2.3 x 1015 yr)
beyond Bi α emission ends the valley of stability !
yellow are α emitter
the higher the Q-value the easier the Coulomb barrier can be overcome (Penetrability ~ ) and the shorter the α-decay half-lives
)constexp( 2/1−⋅− E
Fission
39
Very heavy nuclei can fission into two parts
For large nuclei surface energy less important - large deformations less prohibitive. Then, with a small amount of additional energy (Fission barrier) nucleus can be deformed sufficiently so that Coulomb repulsion wins over nucleon-nucleon attraction and nucleus fissions.
(Q>0 if heavier than ~iron already)
Separation
(from Meyer-Kuckuk, Kernphysik)
Real fission barriers
40
Fission barrier depends on how shape is changed (for example it is favourable to form a neck).
Real theories have many more shape parameters - the fission barrier is then a landscape with mountains and valleys in this parameter space. The minimum energy needed for fission along the optimum valley is “the fission barrier”
Q2
41 Q2 ~ Elongation (fission direction)
20 αg ~ (M1-M2)/(M1+ M2) Mass asymmetry
15 εf1 ~ Left fragment deformation
εf1 εf2
15 εf2 ~ Right fragment deformation
15⊗
⊗
⊗
⊗
d ~ Neck
d
Five Essential Fission Shape Coordinates
M1 M2
⇒ 2 767 500 grid points − 156 615 unphysical points⇒ 2 610 885 physical grid points
Example for parametrization in Moller et al. Nature 409 (2001) 485
Fission fragments
41
Naively splitting in half favourable (symmetric fission)
Asymmetric fission mode due to shell effects (somewhat larger or smaller fragment than exact half might be favoured if more bound due to magic neutron or proton number)
Both modes occur
Nuclear Charge Yield in Fission of 234U
25 30 35 40 45 50 55 60 65
80 100 120 140 160 Mass Number A
Proton Number Z
0
5
10
15
20
Yield
Y(
Z) (%
)
Example from Moller et al. Nature 409 (2001) 485
42
If fission barrier is low enough spontaneous fission can occur as a decay mode
green = spontaneous fission
spontaneous fission is the limit of existence for heavy nuclei
Summary
43 N-number of neutrons
Z=82 (Lead)
Z=50 (Tin)
Z=28 (Nickel)
Z=20 (Calcium)
Z=8 (Oxygen)
Z=4 (Helium)
Valley of stability (location of stable nuclei)
N=Z
Neutron drip line
Proton drip line
α-decay Fission ?
β- decay
β+ & EC decay
44
Solar abundances and nuclear physics
Z=82 (Lead)
Z=50 (Tin)
Z=28 (Nickel)
Z=20 (Calcium)
Z=8
Z=4 (Helium) N=8
N=20 N=28
N=50
N=82
Peak at 56Fe
Peaks at multiples of 4He (though not at 2x4He=8Be) 99% H,He
Very small amounts of nuclei beyond Fe
Sharp peaks at n-shells
Broad peaks “below” n-shells
Nuclear physics also determines set of nuclei that can be found in nature (stable nuclei) Note that EVERY stable nucleus seems to have been produced somewhere in the universe