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Nuclear Models: the Liquid Drop Model Fermi-Gas Model

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Nuclear Models: the Liquid Drop Model Fermi-Gas Model LectureLecture 22 NuclearNuclear models:models: TheThe liquidliquid dropdrop modelmodel FermiFermi --GasGas ModelModel WS2012/13 : ‚Introduction to Nuclear and Particle Physics ‘, Part I 1 NuclearNuclear modelsmodels Nuclear models Models with strong interaction between Models of non-interacting the nucleons nucleons Liquid drop model Fermi gas model ααα-particle model Optical model Shell model … … Nucleons interact with the nearest Nucleons move freely inside the nucleus: neighbors and practically don‘t move: mean free path λ ~ R A nuclear radius mean free path λ << R A nuclear radius 2 I.I. TheThe liquidliquid dropdrop modelmodel 3 TheThe liquidliquid dropdrop modelmodel The liquid drop model is a model in nuclear physics which treats the nucleus as a drop of incompressible nuclear fluid first proposed by George Gamow and developed by Niels Bohr and John Archibald Wheeler The fluid is made of nucleons (protons and neutrons), which are held together by the strong nuclear force . This is a crude model that does not explain all the properties of the nucleus, but (!) • does explain the spherical shape of most nuclei. • It also helps to predict the binding energy of the nucleus . George Gamow Niels Henrik David Bohr John Archibald Wheeler (1930-1968) (1885-1962) (1911-2008) 4 TheThe liquidliquid dropdrop modelmodel The parametrisation of nuclear masses as a function of A and Z, which is known as the Weizsäcker formula or the semi-empirical mass formula , was first introduced in 1935 by German physicist Carl Friedrich von Weizsäcker: M ( A, Z ) === Zm p +++ Nm n −−− E B EB is the binding energy of the nucleus : Volum term Surface term Coulomb term Assymetry term Pairing term Empirical parameters: 5 Binding energy of the nucleus Volume energy (dominant term): E V === aV A A - mass number Coefficient aV ≈≈≈ 16 MeV •The basis for this term is the strong nuclear force . •The strong force affects both protons and neutrons this term is independent of Z. •Because the number of pairs that can be taken from A particles is A(A-1)/2 , one might expect a term proportional to A2 . However, the strong force has a very limited range , and a given nucleon may only interact strongly with its nearest neighbors and next nearest neighbors. Therefore, the number of pairs of particles that actually interact is roughly proportional to A. 6 Binding energy of the nucleus 2 / 3 Surface energy: E S === −−− a S A Coefficient a S ≈≈≈ 20 MeV •This term, also based on the strong force , is a correction to the volume term. •A nucleon at the surface of a nucleus interacts with less number of nucleons than one in the interior of the nucleus, so its binding energy is less. This can also be thought of as a surface tension term , and indeed a similar mechanism creates surface tension in liquids. •The surface energy term is therefore negative and is proportional to the surface area : if the volume of the nucleus is proportional to A (V=4/3 πππR3), then the radius should be proportional to A1 / 3 (R~A1 / 3) and the surface area to A2 / 3 (S= π R222=π=π=πΑ=π ΑΑΑ222 / 3 ). 7 Binding energy of the nucleus Z 2 Coulomb (or electric) energy: E === −−− a C C A /1 3 Coefficient a ≈≈≈ 75.0 MeV C p •The basis for this term is the electrostatic repulsion between p protons. n • The electric repulsion between each pair of protons in a nucleus contributes toward decreasing its binding energy : from QED - interaction energy for the charges q1, q 2 inside the ball q q E ~ 1 2 int R Here R - empirical nuclear radius: R~A1 / 3 Z 2 Thus, E ~ C A 1 / 3 8 Binding energy of the nucleus Asymmetry energy (also called Pauli Energy): (((N −−− Z )))2 E === −−−a asym Sym A Coefficient a sym ≈≈≈ 21 MeV An energy associated with the Pauli exclusion principle : two fermions can not occupy exactly the same quantum state . At a given energy level, there is only a finite number of quantum states available for particles. Thus, as more nucleons are added to the nuclei, these particles must occupy higher energy levels, increasing the total energy of the nucleus (and decreasing the binding energy). Protons and neutrons, being distinct types of particles, occupy different quantum states. One can think of two different ‚pools‘ of states, one for protons and one for neutrons. For example, if there are significantly more neutrons than protons in a nucleus, some of the neutrons will be higher in energy than the available states in the proton pool. The imbalance between the number of protons and neutrons causes the energy to be higher than it needs to be, for a given number of nucleons. This is the basis for the asymmetry term. 9 Binding energy of the nucleus δδδ Pairing energy : E === −−− pair A /1 3 •An energy which is a correction term that arises from the effect of spin-coupling . •Due to the Pauli exclusion principle the nucleus would have a lower energy if the number of protons with spin up will be equal to the number of protons with spin down. This is also true for neutrons. •Only if both Z and N are even, both protons and neutrons can have equal numbers of spin up and spin down particles. •An even number of particles is more stable (δ<δ<δ< 0 for even-even nuclei) than an odd number (δδδ>0). 10 TheThe liquidliquid dropdrop modelmodel The different contributions to the binding energy per nucleon versus mass number A: The horizontal line at ≈ 16 MeV represents the contribution of the volume energy . This is reduced by the surface energy , the asymmetry energy and the Coulomb energy to the effective binding energy of ≈ 8 MeV(lower line ). The contributions of the asymmetry and Coulomb terms increase rapidly with A, while the contribution of the surface term decreases. The Weizsäcker formula (= > liquid drop model) is based on some properties known from liquid drops: constant density, short-range forces, saturation, deformability and surface tension. An essential difference, however, is found in the mean free path of the particles: for molecules in liquid drops, this is far smaller than the size of the drop; but for nucleons in the nucleus, it is λ ~ R A . 11 Binding energy of the nucleus A graphical representation of the semi-empirical binding energy formula: (as a contour plot): the binding energy per nucleon in MeV (highest numbers in dark red, in excess of 8.5 MeV per nucleon) is plotted for various nuclides as a function of Z, the atomic number (on the Y-axis), vs. N, the atomic mass number (on the X-axis). The highest numbers are seen for Z = 26 (iron). Z binding energy per nucleon in MeV N 12 II.II. FermiFermi --GasGas ModelModel 13 TheThe basicbasic conceptconcept ofof thethe FermiFermi --gasgas modelmodel The theoretical concept of a Fermi-gas may be applied for systems of weakly interacting fermions , i.e. particles obeying Fermi-Dirac statistics leading to the Pauli exclusion principle • Simple picture of the nucleus: — Protons and neutrons are considered as moving freely within the nuclear volume. The binding potential is generated by all nucleons — In a first approximation, these nuclear potential wells are considered as rectangular : it is constant inside the nucleus and stops sharply at its end — Neutrons and protons are distinguishable fermions and are therefore situated in two separate potential wells — Each energy state can be ocupied by two nucleons with different spin projections — All available energy states are filled by the pairs of nucleons no free states , no transitions between the states — The energy of the highest occupied state is the Fermi energy EFFF — The difference B‘ between the top of the well and the Fermi level is constant for most nuclei and is just the average binding energy per nucleon B‘/A = 7–8 MeV. 14 NumberNumber ofof nucleonnucleon statesstates Heisenberg Uncertainty Principle: The volume of one particle in phase space: 2πππ h pF The number of nucleon states in a volume V: pmax V ⋅⋅⋅ 4πππ p2dp d 3r d 3 p ∫∫∫ ~ ∫∫∫∫∫∫ 0 (1) 3 === === (V === ∫∫∫ d )r n 3 3 ((()(2πππh))) ((()(2πππh))) At temperature T = 0, i.e. for the nucleus in its ground state, the lowest states will be filled up to a maximum momentum , called the Fermi momentum pF. The number of these states follows from integrating eq.(1) from 0 to pmax =pF: pF V ⋅⋅⋅ 4πππ p2dp ∫∫∫ 3 3 V ⋅⋅⋅ 4πππ p V ⋅⋅⋅ p (2) n~ === 0 === F ⇒⇒⇒ n~ === F ((()(2πππh)))3 ((()(2πππh)))3 ⋅⋅⋅ 3 6πππ 2h 3 Since an energy state can contain two fermions of the same species, we can have n 3 p 3 V ⋅⋅⋅ (((pF ))) V ⋅⋅⋅ (((pF ))) Neutrons: N === Protons: Z === 3πππ 2h3 3πππ 2h3 n p 15 pF is the fermi momentum for neutrons, pF – for protons FermiFermi momentummomentum 4πππ 4πππ Let‘s estimate Fermi momentum p Use R = R .A1/3 fm , V === R3 === R 3 A F: 0 3 3 0 The density of nucleons in a nucleus = number of nucleons in a volume V (cf. Eq.(2)): 3 3 3 3 V ⋅⋅⋅ p 4πππ 3 p 4 A R p n === 2 ⋅⋅⋅ n~ === 2 ⋅⋅⋅ F === 2 ⋅⋅⋅ R A⋅⋅⋅ F === 0 F (3) 6πππ 2h 3 3 0 6πππ 2h 3 9πππ h 3 two spin states ! Fermi momentum pF: 3/1 3/1 6πππ 2h3n 9πππh3 n 9πππ ⋅⋅⋅ n 3/1 h p === === === ⋅⋅⋅ (4) F 3 2V 4A R0 4A R0 After assuming that the proton and neutron potential wells have the same radius, we find for a nucleus with n=Z=N =A/ 2 the Fermi momentum pF: 3/1 h n p 9πππ The nucleons move freely pF === pF === pF === ⋅⋅⋅ ≈≈≈ 250 MeV c 8 R0 inside the nucleus with large momentum! p2 Fermi energy: E === F ≈≈≈ 33 MeV F 2M M =938 MeV- the mass of nucleon 16 NucleonNucleon potentialpotential The difference B‘ between the top of the well and the Fermi level is constant for most nuclei and is just the average binding energy per nucleon B/A = 7–8 MeV .
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