Binding 1 1 m(N,Z) = E(N,Z) = NM + ZM − B(N, Z) c2 n H c 2 The binding energy contributes significantly (~1%) to the mass of a nucleus. This implies that the constituents of two (or more) nuclei can be rearranged to yield a different and perhaps greater binding energy and thus points towards the existence of nuclear reactions in close analogy with chemical reactions amongst atoms.
B is positive for a bound system!
€ Nuclear liquid drop The semi-empirical mass formula, based on the liquid drop model, considers five contributions to the binding energy (Bethe-Weizacker 1935/36) (N − Z) 2 Z 2 B = a A − a A2/3 − a − a −δ(A) vol surf sym A C A1/3 15.68 18.56 28.1 0.717
pairing term
$ −34 A−3 / 4 for even - even & δ(A) = % 0 for even - odd & −3/4 '& 34A for odd - odd
Leptodermous expansion B. Grammaticos, Ann. Phys. (NY) 139, 1 (1982)
€ € Odd-even effect Separation energies
A = 21 isobaric chain one-nucleon separation energies
two-nucleon separation energies
Using http://www.nndc.bnl.gov/chart/ find one- and two-nucleon separation energies of 8He, 11Li, 45Fe, 141Ho, and 208Pb. Discuss the result. Odd-even effect
J. Phys. G 39 093101 (2012) chemical potential Pairing energy
N-odd B(N +1, Z)+ B(N −1, Z) Δ = − B(N, Z) n 2 B(N, Z +1)+ B(N, Z −1) Δ = − B(N, Z) p 2 Z-odd
A common phenomenon in mesoscopic systems! Example: Cooper electron pairs in superconductors…
Find the relation between the neutron pairing gap and one-neutron separation energies. Using http://www.nndc.bnl.gov/chart/ plot neutron pairing gaps for Er (Z=68) isotopes as a function of N. The semi-empirical mass formula, based on the liquid drop model, compared to the data American Journal of Physics -- June 1989 -- Volume 57, Issue 6, p. 552 # ∂B & stability valley proton % ( = 0 $ ∂N ' A= const drip line isotones # ∂B& % ( = 0 isobars $ ∂Z' N = const
neutron drip line
# ∂B & % ( = 0 isotopes $ ∂N ' Z = const
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€ Neutron star, a bold explanation
A lone neutron star, as seen by NASA's Hubble Space Telescope
2 2 2/3 (N − Z) Z 3 G 2 B = avolA − asurf A − asym − aC 1/3 −δ(A) + 1/3 M A A 5 r0 A Let us consider a giant neutron-rich nucleus. We neglect Coulomb, surface, and pairing energies. Can such an object exist?
3 G 2 B = avolA − asym A + 1/3 (mn A) = 0 limiting condition 5 r0 A 3 G m2 A2/3 = 7.5MeV ⇒ A ≅ 5×1055, R ≅ 4.3 km, M ≅ 0.045M n ⊙ 5 r0
More precise calculations give M(min) of about 0.1 solar mass (M⊙). Must neutron stars have R ≅10 km, M ≅1.4 M⊙
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€ Binding (summary) • For most nuclei, the binding energy per nucleon is about 8MeV. • Binding is less for light nuclei (these are mostly surface) but there are peaks for A in multiples of 4. (But note that the peak for 8Be is slightly lower than that for 4He. • The most stable nuclei are in the A~60 mass region • Light nuclei can gain binding energy per nucleon by fusing; heavy nuclei by fission. • The decrease in binding energy per nucleon for A>60 can be ascribed to the repulsion between the (charged) protons in the nucleus: the Coulomb energy grows in proportion to the number of possible pairs of protons in the nucleus Z(Z-1)/2 • The binding energy for massive nuclei (A>60) grows roughly as A; if the nuclear force were long range, one would expect a variation in proportion to the number of possible pairs of nucleons, i.e. as A(A-1)/2. The variation as A suggests that the force is saturated; the effect of the interaction is only felt in a neighborhood of the nucleon.
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