Transcript III. Nuclear Physics that determines the properties of the Universe Part I: Nuclear Masses 1.

III. Nuclear Physics that determines the properties of the Universe
Part I: Nuclear Masses
1. 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 reaction
“Q-value” Q = Energy generated (>0) or consumed (<0) by reaction
2. Stability
if there is a reaction
A
B+C
with Q>0 ( or mA> mB + mC)
then decay of nucleus A is energetically possible.
nucleus A might then not exist (at least not for a very long time)
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
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2. Nucleons
3. Nuclei
Mass
Spin
Charge
Proton
938.272 MeV/c2
1/2
+1e
Neutron
939.565 MeV/c2
1/2
0
nucleons attract each other via the strong force ( range ~ 1 fm)
a bunch of nucleons bound together create a potential for an additional :
neutron
proton
(or any other charged particle)
V
V
Coulomb Barrier Vc
R ~ 1.3 x A1/3 fm
Z1 Z 2 e 2
Vc 
R
R
…
size: ~1 fm
…
r
R
r
Nucleons in a Box:
Discrete energy levels in nucleus
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4. Nuclear Masses and Binding Energy
Energy that is released when a nucleus is assembled from neutrons and protons
m(Z , N )  Zmp  Nmn Bc2
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.
• B is roughly ~A
Masses are usually tabulated as atomic masses
m = mnuc + Z me + Be
Nuclear Mass
~ 1 GeV/A
Electron Mass
511 keV/Z
Most tables give atomic mass excess D in MeV:
Electron Binding Energy
13.6 eV (H)
to 116 keV (K-shell U) / Z
m  Amu  D / c2
(so for 12C: D=0)
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Can be understood in liquid drop mass model: (Weizaecker Formula)
(assumes incompressible fluid (volume ~ A) and sharp surface)
B(Z , A)  aV A
 as A2 / 3
Volume Term
Surface Term ~ surface area (Surface nucleons less bound)
Z2
 aC 1/ 3
A
Coulomb term. Coulomb repulsion leads to reduction
uniformly charged sphere has E=3/5 Q2/R
( Z  A / 2) 2
 aA
A
Asymmetry term: Pauli principle to protons: symmetric filling
of p,n potential boxes has lowest energy (ignore Coulomb)
lower total
energy =
more bound
protons neutrons
1/ 2
 ap A
x 1 ee
x 0 oe/eo
x (-1) oo
protons neutrons
Pairing term: even number of like nucleons favoured
(e=even, o=odd referring to Z, N respectively)
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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:
(Bertulani & Schechter)
something is missing !
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Shell model:
(single nucleon
energy levels)
are not evenly spaced
shell gaps
less bound
than average
more bound
than average
need to add
shell correction term
S(Z,N)
Magic numbers
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Understanding the B/A curve:
neglect asymmetry term (assume reasonable asymmetry)
neglect pairing and shell correction - want to understand average behaviour
then
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1
Z
B / A  aV  aS 1/ 3  aC 4 / 3
A
A
const
as strong force has
short range
~surface/volume ratio
favours large nuclei
Coulomb repulsion has long range
- the more protons the more repulsion
favours small (low Z) nuclei
maximum around ~Fe
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5. Decay - energetics and decay law
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 to happen in a given time interval.
N: Number of nuclei A (Parent)
l : decay rate (decays per second and parent nucleus)
dN  l Ndt
therefore
N (t )  N (t  0) el t
lifetime t1/l
half-life T1/2 = t ln2 = ln2/l is time for half of the nuclei present to decay
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6. Decay modes
for anything other than a neutron (or a neutrino) emitted from the nucleus
there is a Coulomb barrier
V
Coulomb Barrier Vc
Z1 Z 2 e 2
Vc 
R
unbound
particle
R
r
If that barrier delays the decay beyond the lifetime of the universe (~ 14 Gyr)
we consider the nucleus as being stable.
Example: for 197Au -> 58Fe + 139I has Q ~ 100 MeV !
yet, gold is stable.
not all decays that are energetically possible happen
most common:
• b decay
• n decay
• p decay
• a decay
• fission
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6.1. b decay
p
(rates later …)
n conversion within a nucleus via weak interaction
Modes (for a proton/neutron in a nucleus):
b+ decay
electron capture
b- decay
p
e- + p
n
n + e+ + ne
n + ne
p + e- + ne
Electron capture (or EC) of atomic electrons or, in astrophysics, of electrons in
the surrounding plasma
Q-values for decay of nucleus (Z,N)
with nuclear masses
with atomic masses
Qb+ / 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)
Qb- / c2 = mnuc(Z,N) - mnuc(Z+1,N-1) - me = m(Z,N) - m(Z+1,N-1)
Note: QEC > Qb by 1.022 MeV
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Note: Q-values with D values
m  Amu  D / c2
Q-values for reactions that conserve the number of nucleons can also be
calculated directly using the tabulated D values instead of the masses
Example: 14C -> 15N + e +ne
Q  m14C  m14N
 14mu  D14C  14mu  D14 N
 D14C  D14 N
(for atomic D’s)
Q-values with binding energies B
m(Z , N )  Zmp  Nmn Bc2
Q-values for reactions that conserve proton number and neutron number
can be calculated using -B instead of the masses
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b decay basically no barrier -> if energetically possible it usually happens
(except if another decay mode dominates)
therefore: any nucleus with a given mass number A will be converted into the most
stable proton/neutron combination with mass number A by b decays
(Bertulani & Schechter)
valley of stability
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Typical part of the chart of nuclides
red: proton excess
undergo b decay
odd A
isobaric
chain
even A
isobaric
chain
blue: neutron excess
undergo b decay
Z
N
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Typical b decay half-lives:
very near “stability” : occasionally Mio’s of years or longer
more common within a few nuclei of stability: minutes - days
most exotic nuclei that can be formed: ~ milliseconds
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6.2. Neutron decay
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
then no more neutrons can be bound, the neutron drip line is reached
beyond the neutron drip line, neutron decay occurs:
(Z,N)
(Z,N-1) + n
Q-value: Qn = m(Z,N) - m(Z,N-1) - mn
(same for atomic and nuclear masses !)
Neutron Separation Energy Sn
Sn(Z,N) = m(Z,N1) + mn  m(Z,N) = Qn for n-decay
Neutron drip line:
Sn= 0
beyond the drip line Sn<0
the nuclei are neutron unbound
As there is no Coulombbarrier, and n-decay is governed by the strong force,
for our purposes the decay is immediate and dominates all other possible decay modes
Neutron drip line very closely resembles the border of nuclear existence !
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Example: Neutron Separation Energies for Z=40 (Zirconium)
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add 37 neutrons
Sn (MeV)
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valley of stability
10
5
0
-5
30
40
50
60
70
neutron number N
80
90
100
neutron drip
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6.3. Proton decay
same for protons …
Proton Separation Energy Sp
Sp(Z,N) = m(Z1,N) + mp  m(Z,N)
Proton drip line:
Sp= 0
beyond the drip line Sn<0
the nuclei are neutron unbound
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N=40 isotonic chain:
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add 7protons
Sp (MeV)
15
10
5
0
-5
10
20
30
proton number Z
40
50
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Main difference to neutron drip line:
• 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)
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).
Note: have to go beyond Z=50 to prolong half-lives of proton emitters so that they
can be studied in the laboratory (microseconds)
for Z<50 nuclei beyond the drip line are very fast proton emitters (if they
exist at all)
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Example: Proton separation energies of Lu (Z=71) isotopes
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Sp (MeV)
p-emitter
EC/b decay
a-emitter
0
35ms 80ms
stabilizing effects of barriers slow down p-emission
-5
75
85
80
90
neutron number Z
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Pb (82)
proton dripline
Mass known
Half-life known
nothing known
Sn (50)
Fe (26)
neutron dripline
protons
H(1)
neutrons
note odd-even effect in drip line !
(p-drip: even Z more bound - can take away more n’s)
(n-drip: even N more bound - can take away more p’s)
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6.4. a decay
emission of an a 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
Q-value for a decay:
Qa  m( Z , A)  m( Z  2, A  4)  ma
  B( Z , A)  B( Z  2, A  4)  ma
<0, but closer to 0 with larger A,Z
large A therefore favored
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lightest a emitter: 144Nd (Z=60)
(Qa1.9 MeV but still T1/2=2.3 x 1015 yr)
beyond Bi a emission ends the valley of stability !
yellow
are a emitter
the higher the Q-value the easier the
Coulomb barrier can be overcome
(Penetrability ~ exp( const E 1/ 2 ) )
and the shorter the a-decay half-lives
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6.5. Fission
Very heavy nuclei can fission into two parts (Q>0 if heavier than ~iron already)
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.
Separation
(from Meyer-Kuckuk, Kernphysik)
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Real fission barriers:
Fission barrier depends on how shape is changed (obviously, for
example. it is favourable to form a neck).
Five Essential Fission Shape Coordinates
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”
Example for parametrization in
Moller et al. Nature 409 (2001) 485
d
ef1
M1
ef2
M2
Q2
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
20

15

15

15
Q2 ~ Elongation (fission direction)
ag ~ (M1-M2)/(M1+ M2) Mass asymmetry
ef1 ~ Left fragment deformation
ef2 ~ Right fragment deformation
d ~ Neck
 2 767 500 grid points  156 615 unphysical points
 2 610 885 physical grid points
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Fission fragments:
Naively splitting in half favourable (symmetric fission)
There is a 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
Mass Number A
100
120
140
80
160
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Nuclear Charge Yield in Fission of
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U
Yield Y( Z) (%)
15
10
5
Example from
Moller et al. Nature 409 (2001) 485
0
25
30
35
40 45 50 55
Proton Number Z
60
65
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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
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Understanding some solar abundance features from nuclear masses:
1. Nuclei found in nature are the stable ones (right balance of protons and neutron,
and not too heavy to undergo alpha decay or fission)
2. There are many more unstable nuclei that can exist for short times after their creation
(everything between the proton and neutron drip lines and below spont. Fission line)
3. Alpha particle is favored building block because
• comparably low Coulomb barrier when fused with other nuclei
• high binding energy per nucleon - reactions that use alphas have larger Q-values
explains peaks at nuclei composed of multiples of alpha particles in
solar system abundance distribution
4. Binding energy per nucleon has a maximum in the Iron Nickel region
in equilibrium these nuclei would be favored
iron peak in solar abundance distribution indicates that some fraction
of matter was brought into equilibrium (supernovae !)
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Surprises:
1. ALL stable nuclei (and even all that we know that have half-lives of the order of the
age of the universe, such as 238U or 232Th) are found in nature
Explanation ?
2. Nucleons have maximum binding in 62Ni.
But the universe is not just made out of 62Ni !
Explanation ?
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