iii. nuclear physics that determines the properties of the universe

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1 ar Physics that determines the properties of the Un Part I: Nuclear Masses . Why are masses important ? 1. Energy generation nuclear reaction A + B C if m A +m B > m C then energy Q=(m A +m B -m C )c 2 is generated by reaction 2. Stability there is a reaction A B + C with Q>0 ( or m A > m B + m C ) hen decay of nucleus A is energetically possible. eus A might then not exist (at least not for a very long time) 3. Equilibria a nuclear reaction in equilibrium abundances scale with e -Q (Saha equation) es become the dominant factor in determining the outcome of nucleosy “Q-value” Q = Energy generated (>0) or consumed (<0) by reaction

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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 m A +m B > m C then energy Q=(m A +m B -m C )c 2 is generated by reaction. - PowerPoint PPT Presentation

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Page 1: III. Nuclear Physics that determines the properties of the Universe

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III. Nuclear Physics that determines the properties of the UniversePart 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

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

“Q-value” Q = Energy generated (>0) or consumed (<0) by reaction

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2. Nucleons Mass Spin Charge

Proton 938.272 MeV/c2 1/2 +1e

Neutron 939.565 MeV/c2 1/2 0

size: ~1 fm

3. Nucleia bunch of nucleons bound together create a potential for an additional :nucleons attract each other via the strong force ( range ~ 1 fm)

neutron proton(or any other charged particle)

V

r

R

V

rR

Coulomb Barrier Vc

R

eZZVc

221

Pote

ntia

l

Pote

ntia

l

Nucleons in a Box:Discrete energy levels in nucleus

R ~ 1.3 x A1/3 fm

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4. 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.• B is roughly ~A

2),( BcNmZmNZm np

Most tables give atomic mass excessin MeV:

Masses are usually tabulated as atomic masses

2/ cAmm u (so for 12C: =0)

Nuclear Mass~ 1 GeV/A

Electron Mass511 keV/Z

Electron Binding Energy13.6 eV (H)to 116 keV (K-shell U) / Z

m = mnuc + Z me + Be

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Can be understood in liquid drop mass model: (Weizaecker Formula) (assumes incompressible fluid (volume ~ A) and sharp surface)

AaAZB V),(

3/2Aas

3/1

2

A

ZaC

A

AZaA

2)2/(

2/1 Aapx 1 eex 0 oe/eox (-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 neutronsprotons

lower totalenergy =more bound

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:

something is missing !

(Bertulani & Schechter)

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Shell model:(single nucleon energy levels)

Magic numbers

are not evenly spaced shell gaps

more boundthan average

less boundthan average

need to addshell correction termS(Z,N)

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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

3/4

2

3/1

1/

A

Za

AaaAB CSV

~surface/volume ratiofavours large nuclei

Coulomb repulsion has long range- the more protons the more repulsionfavours small (low Z) nuclei

constas strong force hasshort range

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) : decay rate (decays per second and parent nucleus)

NdtdN therefore ttNtN e)0()(

lifetime

half-life T1/2 = ln2 = ln2/ 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 nucleusthere is a Coulomb barrier

V

rR

Coulomb Barrier Vc

R

eZZVc

221

Pote

ntia

l

unboundparticle

Example: for 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• n decay• p decay• decay• fission

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6.1. decay (rates later …)

p n conversion within a nucleus via weak interaction

Modes (for a proton/neutron in a nucleus):

+ decay

electron capture

- decay

p n + e+ + e

e- + p n + e

n p + e- + e

Electron capture (or EC) of atomic electrons or, in astrophysics, of electrons in the surrounding plasma

Q-values for decay of nucleus (Z,N)

Q+ / 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)

Q- / 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 > Q by 1.022 MeV

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Note:2/ cAmm u

Q-values for reactions that conserve the number of nucleons can also be calculated directly using the tabulated values instead of the masses

Q-values with values

Q-values with binding energies B

2),( BcNmZmNZm np

Q-values for reactions that conserve proton number and neutron number can be calculated using -B instead of the masses

Example: 14C -> 15N + e +

N1414C

N1414C

14N14C

1414

uu mm

mmQ

(for atomic ’s)

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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 decays

valley of stability(Bertulani & Schechter)

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Typical part of the chart of nuclides

Z

N

blue: neutron excessundergo decay

red: proton excessundergo decay

even Aisobaric chain

odd Aisobaric chain

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Typical 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 dueto 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

Neutron drip line:

Sn= 0

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 !)

beyond the drip line Sn<0

Neutron Separation Energy Sn

Sn(Z,N) = m(Z,N1) + mn m(Z,N) = Qn for n-decay

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

the nuclei are neutron unbound

Neutron drip line very closely resembles the border of nuclear existence !

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30 40 50 60 70 80 90 100neutron num ber N

-5

0

5

10

15

20

Sn

(MeV

)

Example: Neutron Separation Energies for Z=40 (Zirconium)

neutron drip

valley of stability

add 37 neutrons

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6.3. Proton decay

same for protons …

Proton drip line:

Sp= 0

beyond the drip line Sn<0

Proton Separation Energy Sp

Sp(Z,N) = m(Z1,N) + mp m(Z,N)

the nuclei are neutron unbound

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10 20 30 40 50proton num ber Z

-5

0

5

10

15

20

Sp

(MeV

)N=40 isotonic chain:

add 7protons

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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

75 80 85 90neutron num ber Z

-5

0

5

Sp

(M

eV)

35ms 80ms

p-emitter

EC/ decay-emitter

stabilizing effects of barriers slow down p-emission

Page 22: III. Nuclear Physics that determines the properties of the Universe

22neutrons

protons

Mass knownHalf-life knownnothing known

H(1)

Fe (26)

Sn (50)

Pb (82)

neutron dripline

proton dripline

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. decay

emission of an 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

mAZBAZB

mAZmAZmQ

)4,2(),(

)4,2(),(

<0, but closer to 0 with larger A,Z

Q-value for decay:

large A therefore favored

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lightest emitter: 144Nd (Z=60) (QMeV but still T1/2=2.3 x 1015 yr)

beyond Bi emission ends the valley of stability !

yelloware emitter

the higher the Q-value the easier the Coulomb barrier can be overcome(Penetrability ~ )and the shorter the -decay half-lives

)constexp( 2/1 E

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6.5. Fission

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) nucleuscan 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)

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Fission barrier depends on how shape is changed (obviously, 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”

Real fission barriers:

Q 2

41 Q 2 ~ Elo ng a tio n (fissio n d ire c tio n)

20 g ~ (M 1-M 2)/(M 1+ M 2) M a ss a sym m e try

15 f1

~ Le ft fra g m e nt d e fo rm a tio n

f1

f2

15 f2

~ Rig ht fra g m e nt d e fo rm a tio n

15

d ~ Ne c k

d

Five Esse ntia l Fissio n Sha p e C o o rd ina te s

M 1 M 2

2 767 500 g rid p o ints 156 615 unp hysic a l p o ints

2 610 885 p hysic a l g rid p o ints

Example for parametrization inMoller et al. Nature 409 (2001) 485

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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 favouredif more bound due to magic neutron or proton number)

Both modes occur

Nuc le a r C ha rg e Yie ld in Fissio n o f 234U

25 30 35 40 45 50 55 60 65

80 100 120 140 160 M a ss Num b e r A

Pro to n Num b e r Z

0

5

10

15

20

Yie

ld Y

(Z

) (%

)

Example fromMoller et al. Nature 409 (2001) 485

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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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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

explains peaks at nuclei composed of multiples of alpha particles in solar system abundance distribution

Understanding some solar abundance features from nuclear masses:

• comparably low Coulomb barrier when fused with other nuclei• high binding energy per nucleon - reactions that use alphas have larger Q-values

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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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 ?

Surprises: