1.  Nature of nuclear forces, cont.  Nuclear Models lecture 3&4

2. Yukawa noticed that the range of nuclear forces r 0 =1.4 fm, corresponds to the exchange of intermediate particle of mass =120MeV. The discovery of that particle in cosmic rays was a decisive step forward in the understanding of nuclear forces. 4.1 The nature of nuclear forces: H. Yukawa proposed corresponding potential: where m is mass of intermediate particle and ћ /mc is its Compton wave length. We put Compton length equal to range R of nuclear forces and we determine mass of intermediate particle:

3. Intermediate particles with similar masses were discovered and named as π mesons. In quantum field theory, forces between particles are described by the exchange of virtual particles: n → p + π - → n, p → n + π + → p, The nature of nuclear forces: p → p + π 0 → p, n → n+ π 0 → n Protons and neutrons emit and absorb mesons(pions).

4. 1. A deuteron ( 2 H nucleus) consists of a neutron and a proton. (A neutral atom of 2 H is called deuterium.) 2. It is the simplest bound state of nucleons and therefore gives us an ideal system for studying the nucleon-nucleon interaction. 3. An interesting feature of the deuteron is that it does not have excited states because it is a weakly bound system. The Deuteron

5. Constituents 1 proton 1 neutron Mass 2.014732 u Binding energy 2.224589 ± 0.000002 MeV Angular momentum 1 Magnetic moment 0.85741 ± 0.00002 μ N Electric quadrupole moment +2.88 x 10 –3 bar RMS separation 4.2 fm The Deuteron The deuteron, composed of a proton and a neutron, is a stable particle. abundance of 1.5 x 10-4 compared to 0.99985 for ordinary hydrogen.

6. The Deuteron - Angular momentum 1. In analogy with the ground state of the hydrogen atom, it is reasonable to assume that the ground state of the deuteron also has zero orbital angular momentum L = 0 2. However the total angular momentum is measured to be I = 1 (one unit of h/2π) thus it follows that the proton and neutron spins are parallel. s n +s p = 1/2 + 1/2 = 1 3. The implication is that two nucleons are not bound together if their spins are anti- parallel 4. The parallel spin state is forbidden by the Pauli exclusion principle in the case of identical particles 5. The nuclear force is thus seen to be spin dependent. I = s n + s p + l where s n and s p are individual spins of the neutron and proton and its parity is even.

7. There are four ways to couple s n, s p, and l to get a total I of 1. (a) s n and s p parallel with l = 0 (b) s n and s p antiparallel with l = 1 (c) s n and s p parallel with l = 1 (d) s n and s p parallel with l = 2 parallel antiparallel ● Since we know that the parity of the deuteron is even and the parity associated with orbital motion is determined by (-1) l we are able to rule out some options. ● Orbital angular momentum l = 0 and l = 2 give the correct parity determined from experimental observations. ● The observed even parity allows us to eliminate the combinations of spins that include l =1, leaving l = 0 and l = 2 as possibilities.

8. In the context of the present discussion we can ascribe the tiny discrepancy to the small mixture of d state ( l = 2) in the deuteron wave function: (24) (25) Calculating the magnetic moment from this wave function gives The observed value is consistent with (26) This means that the deuteron is 96% l = 0 ( s orbit) and only 4% l = 2 (d orbit).

9. The bare neutron and proton have no electric quadrupole moment, and so any measured nonzero value for the quadrupole moment must be due to the orbital motion. ― The pure l = 0 wave function would have a vanishing quadrupole moment. The electric quadrupole moment of the deuteron The observed quadrupole moment for the deuteron is When the mixed wave function [equation (24)] is used to calculate the quadrupole moment of the deuteron (Q) the calculation gives two contribution terms. One is proportional to (a d ) 2 and another proportional to the cross-term (a s a d ). (28) (27) where To calculate Q we must know the deuteron d-state wave function and it is obtainable from the realistic phenomenological potentials. The d-state admixture is of several percent in this calculation and is consistent with the 4% value deduced from the magnetic moment.

10. Some comments concerning the d-state admixture obtained from the studies of magnetic moment μ and the quadrupole moment Q: 1. In the case of the magnetic dipole moment, there is no reason to expect that it is correct to use the free-nucleon magnetic moments in nuclei. 2. Spin-orbit interactions, relativistic effects, and meson exchanges may have greater effects on μ than the d-state admixture (but may cancel one another’s effect). 3. For the quadrupole moment, the poor knowledge of the d-state wave function makes the deduced d-state admixture uncertain. 4. Other experiments, particularly scattering experiments using deuterons as targets, also give d-state admixtures in the range of 4%. Thus our conclusions from the magnetic dipole and electric quadrupole moments may be valid after all. 5. It is important that we have an accurate knowledge of the d-state wave function because the mixing of l values in the deuteron is the best evidence for the noncentral (tensor) character of the nuclear force.

11. 4.3 Nuclear models and stability The aim of this chapter is to understand how certain combinations of N neutrons and Z protons form bound states and to understand the masses, spins and parities of those states. 4.3.1 The Liquid-Drop Model One of the first nuclear models, proposed in 1935 by Bohr, is based on the short range of nuclear forces, together with the additivity of volumes and of binding energies. It is called the liquid-drop model. Nucleons interact strongly with their nearest neighbors, just as molecules do in a drop of water. Therefore, one can attempt to describe their properties by the corresponding quantities, i.e. the radius, the density, the surface tension and the volume energy. The Bethe–Weizs¨acker mass formula An excellent parameterization of the binding energies of nuclei in their ground state was proposed in 1935 by Bethe and Weizs¨acker. This formula relies on the liquid-drop analogy but also incorporates two quantum ingredients we mentioned in the previous section.

12. One is an asymmetry energy which tends to favor equal numbers of protons and neutrons. The other is a pairing energy which favors configurations where two identical fermions are paired. The mass formula of Bethe and Weizs¨acker is 4.3.1 The Liquid-Drop Model, cont.

13. The numerical values of the parameters must be determined empirically (other than a c ), but the A and Z dependence of each term reflects simple physical properties. 4.3.1 The Liquid-Drop Model, cont. The first term is a volume term which reflects the nearest-neighbor interactions, and which by itself would lead to a constant binding energy per nucleon B/A ∼ 16 MeV. The term a s, which lowers the binding energy, is a surface term. Internal nucleons feel isotropic interactions whereas nucleons near the surface of the nucleus feel forces coming only from the inside. Therefore this is a surface `tension term, proportional to the area 4πR 2 ∼ A 2/3.

14. The term a c is the Coulomb repulsion term of protons, proportional to Q 2 /R, i.e. ∼ Z 2 /A 1/3. This term is calculable. It is smaller than the nuclear terms for small values of Z. It favors a neutron excess over protons. Conversely, the asymmetry term aa favors symmetry between protons and neutrons (isospin). In the absence of electric forces, Z = N is energetically favorable. Finally, the term δ(A) is a quantum pairing term. 4.3.1 The Liquid-Drop Model, cont. The existence of the Coulomb term and the asymmetry term means that for each A there is a nucleus of maximum binding energy found by setting ∂B/∂Z = 0. As we will see below, the maximally bound nucleus has Z = N = A/2 for low A where the asymmetry term dominates but the Coulomb term favors N >Z for large A.

15. For Light nuclei can undergo exothermic fusion reactions until they reach the most strongly bound nuclei in the vicinity of A ∼ 56. These reactions correspond to the various stages of nuclear burning in stars. For large A’s, the increasing comparative contribution of the Coulomb term lowers the binding energy. This explains why heavy nuclei can release energy in fission reactions or in α-decay. In practice, this is observed mainly for very heavy nuclei A > 212 because lifetimes are in general too large for smaller nuclei. 4.3.1 The Liquid-Drop Model, cont.