Nuclear Stability A sufficient condition for nuclear stability is that, for a collection of A nucleons, there exists no more Lecture 4 tightly bound aggregate. • E. g., a single 8Be nucleus. though it has finite binding energy, Basic Nuclear Physics – 2 has less binding energy than two 4He nuclei, hence 8Be quickly splits into two heliums. • An equivalent statement is that the nucleus AZ is stable if there is Nuclear Stability, no collection of A nucleons that weighs less. The Shell Model • However, one must take care in applying this criterion, because while unstable, some nuclei live a very long time. An operational definition of unstable is that the isotope has a measurable abundance and no decay has ever been observed (ultimately all nuclei heavier than the iron group are unstable, but it takes almost forever for them to decay). One must also include any lepton masses emitted or absorbed in a weak decay. Most collections of nucleons have finite positive binding energy, but a nucleus is still considered “unbound” if it can gain binding by ejecting a neutron or proton. If energetically feasible, this ejection occurs on a very short time scale The neutron and proton “drip lines” are defined by Pb Even a nucleus that is bound is commonly unstable to weak decay or alpha-decay. Examples: 2He - diproton - BE < 0 unbound 3 -decay: (preserves isospin) He BE = 7.718 MeV bound and stable BE(n) = BE(p) = 0 Classification of Decays • emission of Helium nucleus 4He 28.296 bound and stable • Z ! Z-2 5 −22 He 27.56 unbound 7.6 ×10 s • N ! N-2 6 6 • A ! A-4 He 29.27 bound but decays to Li in 807 ms 7 −21 ! EC - He 28.86 unbound 3 × 10 s + e -decay (or -decay) Protons Protons e • emission of e- and " • Z ! Z+1 5Li 26.33 unbound →4 He + p in 3 ×10−22 s • N ! N-1 6Li 31.99 bound and stable • A = const 7Li 39.24 bound and stable + e -decay 8Li 41.27 bound (but decays to 8Be in 840 ms) • emission of e+ and " 8Be 56.50 (barely) unbound - decays to 2 4He " • Z ! Z-1 17 • N ! N+1 in 6.7 ×10− sec • A = const etc Neutrons Electron Capture (EC) - • absorbtion of e and emiss " The difference in binding energies for reactions other • Z ! Z-1 than weak interactions is also the "Q-value for the reaction" • N ! N+1 5 3 4 • A = const e.g. He(n,γ ) He Q= 20.56 MeV The mass of the neutral atom, aka the “atomic mass” is Energy can be released by adding nucleons or other nuclei to produce a more tightly bound product: - nuclear part (but mH contains e ) /c2 - Both 56Fe and 57Fe 2 are stable /c electronic binding energy For Fe the neutron drip line is found at A = 73; the proton drip is at A = 45. Nuclei from 46Fe to 72Fe are stable against strong decay. glected. More commonly used is the Atomic Mass Excess A M( Z) 115 pages i.e., m + m p e the amomic mass neutral atoms 15.994915 amu or M(A Z) = A + Δ amu's A is an integer see also This automatically includes the electron masses Wilhelm Ostwald suggested O in 1912 (before isotopes were known) In 1961 the carbon-12 standard was adopted. O was not really pure 16O The binding energy (MeV)is given in terms of the mass excess by the previous definition of mass excess n → p+ e− +ν (neglecting electronic binding energy) e unstable if BE A Add Z-1 electron 2 = Z mH+ N mn - M( Z) Nuclear masses c masses Atomic masses Δ Mass excesses M(AZ) = A + amu's (1 amu) c2 = 931.49... MeV 931.49... Δ = M − A p → n + e+ +ν BE(MeV) Δ(AZ) e = Z (1.007825 amu) + N (1.008649 amu) - Z - N - 931.49.. 931.49... Δ(AZ) = Z (0.007825 BEam u) + N(0.008649 amu) - 931.49... Add Z+1 electron masses A BE = Z ΔH + N Δn - Δ( Z) where ΔH = 7.288969 MeV = mass excess of H in amu × 931.49... MeV Δn = 8.071323 MeV = mass excess of n in amu × 931.49... MeV eg. 4He Δ = +2.425 Audi and Wapstra, Nuc. Phys A., 595, 409 (1995) BE = 2(8.07132)+2(7.2889) - 2.425 http://t2.lanl.gov/nis/data/astro/molnix96/massd.html = 29.296 MeV p + e− →n +ν e xxxx + Add Z electrons Also possible at high T + e +n → p +νe positron capture 2 −2mec 2 2mec =1.02 MeV Δ Z N 13B 16.562 5 8 13 C 3.125 6 7 13N 5.345 7 7 13 O 23.114 8 5 Frequently nuclei are unstable to both electron-capture and positron emission. Example: p(p,e+)2H" 1 Mass excess 2 H = 2 x 7.289 MeV = 14.578 MeV 2 Mass excess H = 13.136 M+ eV. This is a smaller number so the diproton is unstable to weak decay. The Q value is given by 14.578 -13.136 = 1.442 MeV 2 - 2mec = 0.420 MeV but the electron and positron annihilate and 2 so we get the 2mec back and the reaction yields 1.442 MeV But the neutrino carries away a variable amount Decays may proceed though excited states of energy that averages to 0.262 MeV so really only deposit 1.18 MeV of energy locally In terms of binding energy A A Qβ = BE( Z + 1)− BE( Z)+ 0.782MeV A A Qe+ = BE( Z − 1)− BE( Z)−1.804 MeV A A Qec = BE( Z − 1)− BE( Z)− 0.782 MeV ( The ones with the bigger (less negative) mass excesses are unstable. Proof 2 2 ⎛ a3 ⎞ 2 2 ⎛ a4 ⎞ ⎛ a ⎞ ΔBE = − Z − Z − ⎡A − 2Z ⎤ − ⎡A − 2Z ⎤ 3 2 ⎜ 1/3 ⎟ ( stab ) ⎜ ⎟ ⎣ ⎦ ⎣ stab ⎦ F = − 2ZZstab − 2Zstab ⎝ A ⎠ ⎝ A ⎠ ( ) ⎝⎜ A1/3 ⎠⎟ ( ) ⎛ a ⎞ 2 2 ⎛ 4a ⎞ = − 3 Z − Z 4 2 ⎜ 1/3 ⎟ ( stab ) − −2Zstab − AZ + AZstab + 2ZZstab ⎝ A ⎠ ⎝⎜ A ⎠⎟ ( ) ( ⎛ a ⎞ ⎛ a 4a ⎞ − 4 A2 − 4AZ + 4Z 2 − A2 + 4AZ − 4Z 2 2Z 3 4 Z Z ⎜ ⎟ ( stab stab ) = − stab 1/3 + − stab ⎝ A ⎠ ⎝⎜ A A ⎠⎟ ( ) ⎛ a ⎞ ⎛ 4a ⎞ − 4a Z − Z = − 3 Z 2 − Z 2 − 4 Z 2 − Z 2 − AZ + AZ ( 4 )( stab ) ⎜ 1/3 ⎟ ( stab ) ⎜ ⎟ ( stab stab ) ⎝ A ⎠ ⎝ A ⎠ ⎛ 2Z ⎞ stab a A2/3 4a Z Z 4a Z Z = ⎜ ⎟ ( 3 + 4 )( stab − )− ( 4 )( stab − ) ⎛ a ⎞ 2 2 2 ⎝ A ⎠ = − 3 Z − 2ZZ + Z + 2ZZ − 2Z ⎜ 1/3 ⎟ ( stab stab stab stab ) ⎝ A ⎠ ⎛ ⎞ ⎡ 2a4A ⎤ ⎜ 2⎢ 2/3 ⎥⎟ ⎛ 4a4 ⎞ 2 2 2 a A + 4a − Z − 2ZZ + Z − 2Z − AZ + AZ + 2ZZ = ⎜ ⎣ 3 4 ⎦⎟ a A2/3 + 4a Z − Z − 4a Z − Z ⎝⎜ A ⎠⎟ ( stab stab stab stab stab ) ⎜ A ⎟ ( 3 4 )( stab ) ( 4 )( stab ) ⎜ ⎟ ⎜ ⎟ 2 ⎛ a ⎞ 2 ⎝ ⎠ = K Z − Z − 3 2ZZ − 2Z ( stab ) ⎜ 1/3 ⎟ ( stab stab ) ⎝ A ⎠ = 0 ⎛ 4a ⎞ 2 − 4 −2Z 2 − AZ + AZ + 2ZZ =K Z − Z + F ⎜ A ⎟ ( stab stab stab ) ( stab ) ⎝ ⎠ Odd A. A=135 At constant A Single parabola even-odd and odd-even Only 135Ba is stable. β- β- β+ β- EC β- 52 54 56 58 Te I Xe Cs Ba La Ce Pr • Even A: • two parabolas Even A. A=102 Two parabolas separated by 2δ, • one for o-o & one for e-e odd-odd and even-even • lowest o-o nucleus often has two decay modes odd-odd • most e-e nuclei have two β+ stable isotopes • there are nearly no stable o-o β- nuclei in nature because these β+ β- can nearly all -decay to an e- β- e nucleus β+ even-even 42 44 46 48 Mo Tc Ru Rh Pd Ag Cd an even-even nucleus must decay to an odd-odd nucleus and vice versa. mass 64 96 Zr is stable mass 194 To summarize: odd A There exists one and only one stable isotope odd Z – odd N Very rarely stable. Exceptions 2H, 6Li, 10B, 14N. Large surface to volume ratio. Our liquid drop model is not really applicable. The Shell Model even Z – even N Frequently only one stable isotope (below sulfur). At higher A, frequently 2, and occasionally, 3. Z Neutron Magic • Doesn’t Predict Magic Numbers 126 Shortcomings of the Liquid Drop Model Numbers • Simple model does not apply for A < 20 82 (10,10) (N,Z) (6,6) Proton (2,2) Magic Numbers (8,8) 50 (4,4) • Magic Proton Numbers 28 (stable isotopes) 20 • Magic Neutron Numbers (stable isotones) 8 2 N Abundance patterns reflect magic numbers Ba Neutron separation energy in MeV • Neutron separation energies – saw tooth from pairing term N=126 N=126 Z=82 – step down when N N=82 goes across magic number at 82 Z=50 Z=50 N=50 N=50 iron mountain Z = N = 28 no A=5 or 8 Illiadis ∞ A highly idealized nuclear potential Shell Model – Mayer and Jensen 1963 Nobel Prize looks something like this infinite V Coulomb square well. -R r 0 R Our earlier discussions treated the nucleus as sets of As is common in such problems identical nucleons and protons comprising two degenerate one applies boundry conditions to Fermi gases. That is OK so far as it goes, but now we shall Schroedinger s equation. consider the fact that the nucleons have spin and angular V = −V r < R momentum and that, in analogy to electrons in an atom, are nuc r R in ordered discrete energy levels characterized by conserved = ∞ ≥ quantized variables – energy, angular momentum and spin.