The mass of a nucleus
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The mass of a nucleus. Energy generation in stars which nuclei are stable which nuclei exist in principle. SOHO, 171A Fe emission line. …. …. Nucleons in a Box: Discrete energy levels in nucleus. Nucleons. size: ~1 fm. Nuclei.

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The mass of a nucleus

  • Energy generation in stars

  • which nuclei are stable

  • which nuclei exist in principle

SOHO, 171A Fe emission line

Nucleons in a Box:Discrete energy levels in nucleus


size: ~1 fm


nucleons attract each other via the strong force ( range ~ 1 fm)

a bunch of nucleons bound together create a potential for an additional :


proton(or any other charged particle)



Coulomb Barrier Vc

R ~ 1.3 x A1/3 fm







 Nucleons are bound by attractive force. Therefore, mass of nucleus is smaller than the total mass of the nucleons by the binding energy dm=B/c2

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 very roughly ~A

Masses are usually tabulated as atomic masses

m = mnuc + Z me + Be

Nuclear Mass

~ 1 GeV/A

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

Electron Mass

511 keV/Z

Most tables give atomic mass excess D in MeV:

(so for 12C: D=0) (see nuclear wallet cards for a table)


Energy released in a nuclear reaction (>0 if energy is released, <0 if energy is used)

Example: The sun is powered by the fusion of hydrogen into helium:


4p  4He + 2 e+ + 2ne

Mass difference dMreleased as energydE = dM c2

(using nuclear masses !)

In practice one often uses mass excess D and atomic masses.

Q-value with mass excess D

As A is always conserved in nuclear reactions the mass excess D can alwaysbe used instead of the masses (the Amu term cancels)

(as nucleon masses cancel on both sides, its really the binding energies thatentirely determine the Q-values !)

Q-value with atomic masses:

If Z is conserved (no weak interaction) atomic masses can be used insteadof nuclear masses (Zme and most of the electron binding energy cancels)

Otherwise: For each positron emitted subtract 2me /c2= 1022 MeV from the Q-value

4p  4He + 2 e+ + 2ne


Z changes an 2 positrons are emitted

With atomic masses

With atomic mass excess

The liquid drop mass model for the binding energy: (Weizaecker Formula)

(assumes incompressible fluid (volume ~ A) and sharp surface)

(each nucleon gets bound by about same energy)

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)

lower totalenergy =more bound





and in addition: p-n more bound than p-p or n-n (S=1,T=0 morebound than S=0,T=1)

x 1 ee

x 0 oe/eo

x (-1) oo

Pairing term: even number of like nucleons favoured

(e=even, o=odd referring to Z, N respectively)

Binding energy per nucleon along the “valley of stability”

Fission generatesenergy

Fusion generatesenergy

Best fit values stability” (from A.H. Wapstra, Handbuch der Physik 38 (1958) 1)

in MeV/c2

Deviation (in MeV) to experimental masses:

(Bertulani & Schechter)

something is missing !

Shell model: stability”(single nucleon energy levels)

are not evenly spaced

shell gaps

less boundthan average

more boundthan average

need to addshell correction termS(Z,N)

Magic numbers

Const A cut stability”

The valley of stability

Magic numbers


Z=82 (Lead)

Valley of stability

(location of stable nuclei)

Z – number or protons

Z=50 (Tin)

Z=28 (Nickel)

Z=20 (Calcium)

Z=8 (Oxygen)

Z=4 (Helium)

N-number of neutrons

valley of stability stability”

Const. A cut:

Binding energy per nucleon along const A due to asymmetry term in mass formula





(Bertulani & Schechter)

What happens when a nucleus outside the valley of stability is created ?(for example in a nuclear reaction inside a star ?)

Decay - energetics and decay law

Decay of A in B and C is possible if reaction A B+C has positive Q-value

(again – masses are critcical !)

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)

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


lifetime t=1/l

half-life T1/2 = t ln2 = ln2/l is time for half of the nuclei present to decay

Decay modes is created ?

for anything other than a neutron (or a neutrino) emitted from the nucleusthere is a Coulomb barrier


Coulomb Barrier Vc





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

p n + e is created ?+ + ne

n p + e- + ne

e- + p n + ne

b decay



conversion within a nucleus via weak interaction

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

b+ decay

Favourable for n-deficient nuclei

electron capture

Favourable for n-rich nuclei

b- decay

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

= m(Z,N) - m(Z-1,N+1) - 2me

Qb+ / c2 = mnuc(Z,N) - mnuc(Z-1,N+1) - me

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

odd A is created ?isobaric chain

even Aisobaric chain

Typical part of the chart of nuclides

red: proton excessundergo b+ decay

blue: neutron excessundergo b- decay



Typical is created ?b decay half-lives:

very near “stability” : occasionally Mio’s of years or longer

more common within a few nuclei of stability: minutes - days

~ milliseconds

most exotic nuclei that can be formed:

Proton or neutron decay: is created ?

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 becomeproton 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

6.4. is created ?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:

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

large A therefore favored

lightest is created ?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 !

yelloware a emitter

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

6.5. Fission is created ?

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) nucleuscan be deformed sufficiently so that coulomb repulsion wins over nucleon-nucleon

attraction and nucleus fissions.


(from Meyer-Kuckuk, Kernphysik)

Real fission barriers: is created ?

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”

Example for parametrization in

Moller et al. Nature 409 (2001) 485

Fission fragments: is created ?

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

Example from

Moller et al. Nature 409 (2001) 485

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 occur as a decay mode

Valley of stability

(location of stable nuclei)


Fission ?


Z=82 (Lead)

b+ & EC decay


drip line

Z – number or protons

Z=50 (Tin)

b- decay

Neutron drip line

Z=28 (Nickel)

Z=20 (Calcium)

Z=8 (Oxygen)

Z=4 (Helium)

N-number of neutrons

Solar abundances and nuclear physics occur as a decay mode

Z=82 (Lead)

Sharp peaks at n-shells

Very small amounts of nuclei beyond Fe

Z=50 (Tin)


Broad peaks “below” n-shells


Z=28 (Nickel)


Nuclear physics also determinesset of nuclei that can be foundin nature (stable nuclei)

Note that EVERY stable nucleusseems to have been producedsomewhere in the universe

Peak at 56Fe





Z=4 (Helium)

Peaks at multiplesof 4He (though not at 2x4He=8Be)


99% H,He