Download 投影片 1

lecture 3&4
 Nature of nuclear forces, cont.
 Nuclear Models
4.1 The nature of nuclear forces:
Yukawa noticed that the range of nuclear forces r0 =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.
H. Yukawa proposed corresponding potential:
e mcr
V(r) 
r

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:
mc 2 
c c 197MeVfm


 120MeV

R
1.7 fm
The nature of nuclear forces:
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:
p → p + π0 → p ,
n → n+ π0 → n
n → p + π - → n,
p → n + π + → p,
Protons and neutrons emit and absorb mesons(pions).
The Deuteron
1. A deuteron (2H nucleus) consists of a neutron and a proton. (A
neutral atom of 2H 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
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.
Constituents
Mass
Binding energy
Angular momentum
Magnetic moment
Electric quadrupole moment
RMS separation
1 proton 1 neutron
2.014732 u
2.224589 ± 0.000002 MeV
1
0.85741 ± 0.00002 μN
+2.88 x 10–3 bar
4.2 fm
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. sn+sp = 1/2 + 1/2 = 1
I = sn + sp + l
where sn and sp are individual spins of the neutron and proton and its
parity is even.
3. The implication is that two nucleons are not bound together if their spins are antiparallel
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.
There are four ways to couple sn, sp, and l to get a total I of 1.
(a) sn and sp parallel with l = 0
parallel
(b) sn and sp antiparallel with l = 1
(c) sn and sp parallel with l = 1
antiparallel
(d) sn and sp parallel with l = 2
●
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.
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:
  as (l  0)  ad (l  2)
(24)
Calculating the magnetic moment from this wave function gives
  as 2  (l  0)  ad 2  (l  2)
(25)
The observed value is consistent with
as  0.96 and ad  0.04
2
2
(26)
This means that the deuteron is 96% l = 0 ( s orbit)
and only 4% l = 2 (d orbit).
The electric quadrupole moment of the deuteron
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 observed quadrupole moment for the deuteron is
Q  0.00288  0.00002 b
(27)
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 (ad)2 and another proportional to the cross-term (asad).
Q
where
r2
sd
2
as ad r 2
10
sd
  r 2 Rs (r ) Rd (r )r 2dr

1 2 2
ad r
20
r2
dd
dd
(28)
  r 2 Rd (r ) Rd (r )r 2dr
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.
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.
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.
4.3.1 The Liquid-Drop Model, cont.
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.
The numerical values of the parameters must be determined empirically
(other than ac), but the A and Z dependence of each term reflects simple
physical properties.
• 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 as, 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
4πR2 ∼ A2/3.
4.3.1 The Liquid-Drop Model, cont.
• The term ac is the Coulomb repulsion term of protons, proportional
to Q2/R, i.e. ∼ Z2/A1/3. 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.
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.
4.3.1 The Liquid-Drop Model, cont.
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.