PHL424: Nuclear Shell Model. Indian Institute of Technology Ropar

Size: px

Start display at page:

Download "PHL424: Nuclear Shell Model. Indian Institute of Technology Ropar"

Ariel Weaver
5 years ago
Views:

1 PHL424: Nuclear Shell Model

Themes and challenges in modern science Complexity

all its apparent complexity and diversity can be

and their interactions individual excitations of

How the world of complex systems can display such

2 Themes and challenges in modern science Complexity out of simplicity Microscopic How the world, with all its apparent complexity and diversity can be constructed out of a few elementary building blocks and their interactions individual excitations of nucleons Simplicity out of complexity Macroscopic How the world of complex systems can display such remarkable regularity and simplicity vibration rotation fission

3 The nuclear force The nuclear force is short-range, but does not allow for compression of nuclear matter. Yukawa potential: VV 0 rr = gg ss 1 rr mm ππcc ee ħ rr ω,ρ π σ m(π) 140 MeV/c 2 m(σ) MeV/c 2 m(ω) 784 MeV/c 2

The deuteron mass (MeV/c 2 ) 1875.61 charge (e) 1 I π 1 + binding energy (MeV) 2.

8798 μ N = μ S ( HH 1 2 ) the deuteron can not be a pure s state! ~ 96% s and 4% d.

4 The deuteron mass (MeV/c 2 ) charge (e) 1 I π 1 + binding energy (MeV) n p magnetic moment (μ N ) μ N = μ S ( HH 1 2 ) the deuteron can not be a pure s state! ~ 96% s and 4% d. quadrupole moment (b) not spherical consistent with s/d-ratio = 96/4 The deuteron is an ideal candidate for tests of our basic understanding of nuclear physics

5 Structure of the nuclear force Structure of the nuclear force is more complex than e.g. Coulomb force. It results from its structure as residual interaction of the colorless nucleons. central force V 0 (r) results from deuteron properties (96% 3 S 1 state) 2SS+1 LL JJ spin dependent central force results from neutron-proton scattering (spin-spin interaction) not central tensor force results from deuteron properties (4% 3 D 1 state) 2SS+1 LL JJ spin-orbit (l s) term results from scattering of polarized protons (left/right asymmetry) VV rr = VV 0 rr +VV ssss rr ss 1 ss 2 1 ħ 2 +VV TT rr 3 ħ 2 ss 1 xx ss 2 xx rr 2 ss 1 ss 2 +VV lss rr ss 1 + ss 2 l 1 ħ 2 central potential spin-spin interaction tensor force spin-orbit interaction

Structure of the nuclear force spin-spin force: 1 2 s = 0, l = 1 ~VV ssss rr ss 1 ss 2 /ħ 2 different eigenvalues for triplet and singlet states 1 2 + s = 1, l = 0 tensor force: xx ~VV TT rr 3 ħ 2 ss

6 Structure of the nuclear force spin-spin force: 1 2 s = 0, l = 1 ~VV ssss rr ss 1 ss 2 /ħ 2 different eigenvalues for triplet and singlet states s = 1, l = 0 tensor force: xx ~VV TT rr 3 ħ 2 ss 1 xx ss 2 xx rr 2 ss 1 ss 2 small deformation of deuterium maximum magnetic dipole moments xx attractive repulsive l s coupling: ~VV lss rr l ss scattering of protons on polarized protons asymmetry of counting rates - left scattering: l ss > 0 - right scattering: l ss < 0 l s coupling: - no net contribution in the center of nucleus - radial dependence at the surface of the nucleus VV lss rr 1 rr dddd dddd

7 Many-body forces internal forces governing a 3 He nucleus The force on one nucleon does not only depend on the position of the other nucleons, but also on the distance between the other nucleons! These are called many-body forces. tidal effects lead to 3-body forces in earth-sun-moon system Remember: Nucleons are finite-mass composite particles, can be excited to resonances. Dominant contribution Δ(1232 MeV)

8 The Fermi gas model proton potential neutrons protons neutron potential The Fermi gas model assumes that protons and neutrons are moving freely within the nuclear volume. They are distinguishable fermions (s = ½) filling two separate potential wells obeying the Pauli principle ( -pair). The model assumes that all fermions occupy the lowest energy states available to them to the highest occupied state (Fermi energy), and that there is no excitation across the Fermi energy (i.e. zero temperature). The Fermi energy is common for protons and neutrons in stable nuclei. If the Fermi energy for protons and neutrons are different then the β-decay transforms one type of nucleons into the other until the common Fermi energy (stability) is reached.

Number of nucleon states Heisenberg Uncertainty Principle: xx pp 1 2 ħ The volume of one particle in phase space: 2ππ ħ The number of nucleon states in a volume V: pp mmmmmm 0 nn = dd3 rr dd 3 pp 2ππ

9 Number of nucleon states Heisenberg Uncertainty Principle: xx pp 1 2 ħ The volume of one particle in phase space: 2ππ ħ The number of nucleon states in a volume V: pp mmmmmm 0 nn = dd3 rr dd 3 pp 2ππ ħ 3 = VV 4ππ pp2 dddd 2ππ ħ 3 states in phase space At temperature T = 0, i.e. for the nucleus in its ground state, the lowest states will be filled up to the maximum momentum, called the Fermi momentum p F. The number of these states follows from integration from 0 to p max = p F. nn = pp FF VV 4ππ pp2 dddd 0 2ππ ħ 3 = VV 4ππ pp 3 FF 2ππ ħ 3 3 nn = VV pp FF 3 6ππ 2 ħ 3 Since an energy state can contain two fermions of the same species, we can have nnnnnnnnnnnnnnnn: NN = VV pp FF nn 3 pp 3 3ππ 2 ħ 3 pppppppppppppp: ZZ = VV pp FF 3ππ 2 ħ 3 pp FF nn is the Fermi momentum for neutrons, pp FF pp for protons

Fermi momentum Use RR = rr 0 AA 1 3 ffff VV = 4ππ 3 RR3 = 4ππ rr 3 0 3 AA The density of nucleons in a nucleus = number of nucleons in a volume V: nn = 2 VV pp FF 3 6ππ 2 ħ 3 =

0 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 p F.

10 Fermi momentum Use RR = rr 0 AA 1 3 ffff VV = 4ππ 3 RR3 = 4ππ rr AA The density of nucleons in a nucleus = number of nucleons in a volume V: nn = 2 VV pp FF 3 6ππ 2 ħ 3 = 2 4ππ 3 3 rr 0 3 pp FF AA 6ππ 2 ħ 3 = 4AA rr pp FF 9ππ ħ 3 two spin states Fermi momentum p F : pp FF = 6ππ2 ħ 3 nn 2VV 1/3 = 9ππħ3 4AA nn rr 0 3 1/3 = 9ππ nn 4AA 1/3 ħ rr 0 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 p F. 1/3 pp FF = pp FF nn = pp FF pp = 9ππ 8 ħ rr MMMMMM/cc The nucleons move freely inside the nucleus with large momenta Fermi energy: EE FF = pp FF 2 2mm NN 33 MMMMMM m N = 938 MeV/c 2 the nucleon mass

11 Nucleon potential The difference B between the top of the well and the Fermi level is the average binding energy per nucleon B/A = 7 8 MeV. The depth of the potential V 0 and the Fermi energy are independent of the mass number A: VV 0 = EE FF + BBB 40 MMMMMM Heavy nuclei have a surplus of neutrons. Since the Fermi level of the protons and neutrons in a stable nucleus have to be equal (otherwise the nucleus would enter a more energetically favorable state through β-decay) this implies that the depth of the potential well as it is experienced by the neutron gas has to be larger than of the proton gas. Protons are therefore on average less strongly bound in nuclei than neutrons. This may be understood as a consequence of the Coulomb repulsion of the charged protons and leads to an extra term in the potential: VV CC = ZZ 1 αα ħcc RR Protonen: 33MeV + 7MeV, Neutronen: 43MeV + 7 MeV

The Fermi gas model and the neutron star Assumption: neutron star as cold neutron gas with constant density - 1.5 sun masses: M = 3 10 30 kg (m N = 1.67 10-27 kg), number of neutrons: n = 1.

12 The Fermi gas model and the neutron star Assumption: neutron star as cold neutron gas with constant density sun masses: M = kg (m N = kg), number of neutrons: n = Fermi momentum p F for cold neutron gas: pp FF = 9ππ nn 4 1/3 ħ RR R is the radius of the neutron star Average kinetic energy per neutron: EE kkkkkk NN = pp FF = 2mm NN 9ππ nn 4 2/3 3ħ mm NN RR 2 = CC RR 2 Gravitational energy of a star with constant density has an average potential energy per neutron: EE pppppp NN = 3 5 GG nn mm nn 2 = DD RR RR Minimum total energy per neutron: dd dddd dd dddd EE/NN = dd dddd EE kkkkkk/nn + EE pppppp /NN = 0 CC RR 2 DD RR = 2CC RR 3 + DD RR 2 = 0 RR = 2CC DD RR = ħ2 9ππ/4 2/3 GG mm NN 3 nn 1/3 GG = mm 3 kkkk ss 2 radius of a neutron star ~ 10.7 km

13 Shell structure in nuclei Deviations from the Bethe-Weizsäcker mass formula: Neutron Proton B/A (MeV per nucleon) mass number A especially stable: 4 2 He O Ca Ca Pb

14 Shell structure in nuclei deviations from the Bethe-Weizsäcker mass formula: large binding energies 208 Pb 132 Sn

15 2-neutron binding energies = 2-neutron separation energies 25 SS 2nn = BBBB NN, ZZ BBBB NN 2, ZZ N = S(2n) MeV N = 84 Sm Hf Ba Pb Sn Neutron Number N = 126

16 Shell structure in nuclei EE 21 + Nuclei with magic numbers of neutrons/protons high energies of the first excited 2 + state small nuclear deformations transition probabilities measured in single particle units (spu) BB EEE;

17 Shell structure in nuclei EE 21 + BB EEE; Maria Goeppert-Mayer J. Hans D. Jensen

18 Nuclear potential AA HH = pp ii 2 ii=1 AA 2mm ii + VV rr ii, rr jj ii<jj AA HH = pp ii 2 ii=1 2mm ii + VV rr ii AA + VV rr ii, rr jj ii<jj AA VV rr ii ii=1 ħ2 2mm 2 + VV rr εε Ψ rr = 0 VV rr = VV ee rr RR 0 aa Ψ rr = uu l rr rr YY lmm θθ, φφ Χ mmss In the average nuclear potential V(r): a) harmonic oscillator b) square well potential c) Woods-Saxon potential the nucleons move freely

19 Nuclear shell model AA HH = pp ii 2 ii=1 2mm ii + VV rr ii harmonic oscillator square-well potential realistic potential + spin-orbit coupling V 0 V 0 r 6 ω 5 ω 4 ω 3 ω 2 ω 1 ω 0 ω s 3d 2g 3p 1i 2f 1h/3s 2d 1g 2p 1d 2s 1d 1p 1s s1/2 3d3/2 2g7/2 3d5/2 1i11/2 2g9/2 2f5/2 3p1/2 1i13/2 2p3/2 2f7/2 1h9/2 3s1/2 1h11/2 2d1/2 2d5/2 1g7/2 1g9/2 2p1/2 3p3/2 1f5/2 1f7/2 2g1/2 1d3/2 1d5/2 1p1/2 1p3/2 1s1/2

Woods-Saxon potential Woods-Saxon does not reproduce the correct magic numbers (2, 8, 20, 40, 70, 112, 168) WS (2, 8, 20, 28, 50, 82, 126) exp Meyer und Jensen (1949): strong spin-orbit interaction

20 Woods-Saxon potential Woods-Saxon does not reproduce the correct magic numbers (2, 8, 20, 40, 70, 112, 168) WS (2, 8, 20, 28, 50, 82, 126) exp Meyer und Jensen (1949): strong spin-orbit interaction ħ2 2mm 2 + VV rr + VV lss rr l ss εε Ψ rr = 0 VV lss rr ~ λλ 1 rr dddd dddd mmmmmm λλ > 0 JJ dv ( r) dr ss l V ( r) r The spin-orbit term has its origin in the relativistic description of the single particle motion inside the nucleus

21 Woods-Saxon potential (jj-coupling) ȷȷ = l + ss l ss = 1 2 jj2 l 2 ss 2 ħ 2 = 1 2 jj jj + 1 l l + 1 ss ss + 1 ħ2 The nuclear potential with spin-orbit term: VV rr + l 2 VV lss ffffff jj = l + 1/2 VV rr l VV lss ffffff jj = l 1/2 spin-orbit interaction leads to a large splitting for large l. jj = l 1/2 l + 1 /2 VV lss jj = l ± 1/2 jj = l + 1/2 l/2 VV lss

Woods-Saxon potential l 1/2 l + 1/2 EE lss = 2l + 1 2 ħ 2 VV lss The spin-orbit term lowers the j = l+1/2 orbital from the

consequences: lowering orbitals from higher lying N+1 shell having different parity than orbitals from the N shell strong

22 Woods-Saxon potential l 1/2 l + 1/2 EE lss = 2l ħ 2 VV lss The spin-orbit term lowers the j = l+1/2 orbital from the higher oscillator shell (intruder states) reproduces the magic numbers large energy gaps very stable nuclei Important consequences: lowering orbitals from higher lying N+1 shell having different parity than orbitals from the N shell strong interaction preserves the parity. The lowered orbitals with different parity are rather pure states and do not mix within the shell

23 Shell model mass dependence of single-particle energies Mass dependence of the neutron energies: EE~RR 2 number of neutrons in each level: 2 2l + 1

24 Success of the extreme single-particle shell model Ground state spin and parity: Every orbital has 2j+1 magnetic sub-states, completely filled orbitals have spin J=0, they do not contribute to the nuclear spin. For a nucleus with one nucleon outside a completely occupied orbital the nuclear spin is given by the single nucleon. n l j J (-) l = π

25 Success of the extreme single-particle shell model Magnetic moments: The g-factor g j is given by: μμ jj = gg l l + gg ss ss = gg jj ȷȷ μμ jj = gg l l + gg ss ss ȷȷ jj ȷȷ jj with l 2 = ȷȷ ss 2 = ȷȷ 2 2 ȷȷ ss + ss 2 ss 2 = ȷȷ l 2 = ȷȷ 2 2 ȷȷ l + l 2 μμ jj = gg l jj jj l l + 1 3/4 + gg ss jj jj + 1 l l /4 2 jj jj + 1 ȷȷ gg jj = 1 2 gg l + gg ss l l + 1 ss ss + 1 2jj jj + 1 gg l gg ss Simple relation for the g-factor of single-particle states μ/μ N nucleus state J π model experiment μμ μμ NN = gg nnnnnnnnnnnnnn = gg l ± gg ss gg l 2l + 1 ffffff jj = l ± 1

26 Success of the extreme single-particle shell model magnetic moments: μμ zz = gg l jj gg ss μμ NN ffffff jj = l + 1/2 jj jj + 1 gg l jj gg ss μμ NN ffffff jj = l 1/2 g-factor of nukleons: proton: g l = 1; g s = neutron: g l = 0; g s = proton: μμ zz = jj μμ NN ffffff jj = l + 1/2 jj jj jj + 1 μμ NN ffffff jj = l 1/2 neutron: μμ zz = 1.91 μμ NN ffffff jj = l + 1/2 jj jj + 1 μμ NN ffffff jj = l 1/2

27 Magnetic moments: Schmidt lines magnetic moments: proton magnetic moments: neutron

28 The three structures of the shell model

30 Systematics of the Te isotopes (Z=52) Neutron number Val. Neutr. number

c E If photon Mass particle 8-1

c E If photon Mass particle 8-1 Nuclear Force, Structure and Models Readings: Nuclear and Radiochemistry: Chapter 10 (Nuclear Models) Modern Nuclear Chemistry: Chapter 5 (Nuclear Forces) and Chapter 6 (Nuclear Structure) Characterization