Part II Nuclear Physics Handout 2 V.Gibson Lent Term 2004

Handout 2 r

V.Gibson Lent Term 2004

Lent Term 2004 Nuclear / V.Gibson 62 Section III The Nucleon Force

Nucleons are made of spin 1/2 point-like quarks. Quarks are held together by the strong interaction arising from the exchange of other quarks and spin 1 gluons (see Particles course). The force between nucleons (the strong nuclear force) is a many-body problem in which

 quarks do not behave as if they were completely independent inside the nuclear volume  nor do they behave as if they were completely bound to form protons and neutrons. e.g. p-p interaction pp p p u u u d u d

d u uu d u u u The nuclear force is therefore not calculable in detail at the quark level and can only be deduced empirically from nuclear data.

Lent Term 2004 Nuclear / V.Gibson 63 General Features

The fact that a nucleus exists implies that the nuclear force is:

 Strong: stronger than the electromagnetic, weak and gravitational forces.

 Short range: nuclei experience the strong interaction at short distances (~2 fm ) as they start to overlap.

 Attractive.

 Repulsive core: Volume~A, nucleus doesn’t collapse to ∞ density.

 Saturates: B/A~constant; in a nucleus the nucleons are only attracted by nearby nucleons.

 Charge independent: No distinction between protons and neutrons. Evidence seen from tendency for small nuclei to have N=Z and similarities of the low-lying energy levels of pairs of mirror nuclei.

Lent Term 2004 Nuclear / V.Gibson 64 Mirror Nuclei 23 23 Example ( 11Na, 12Mg)

JP + + ⎛ 3 5 ⎞ 3 ⎜ , ⎟ 2.982 2 2.908 ⎝ 2 2 ⎠ + − 2.704 9 2.771 1 2 2 + 2.640 1 − 2.715 9 2 2 + 1 + 1 2 2.359 2 2.391 + + 7 2.051 7 2.076 2 2

+ + 5 0.451 5 0.440 2 2 + 0 3 + 0 3 2 2 23 23 11Na 12Mg

Illustrates charge symmetry, p-p interaction ≡ n-n interaction Does not imply p-n = p-p or n-n because the number of p-n pairs is the same in both nuclei.

Lent Term 2004 Nuclear / V.Gibson 65 The Nucleon-Nucleon Potential v v V(r) Force = −∇V(r)

Repulsive core

~2 fm r

B/A~8 MeV

expect V0 ~few 10 MeV

Study detailed features using interactions between two nucleons:

The deuteron and nucleon-nucleon scattering.

Lent Term 2004 Nuclear / V.Gibson 66 The Deuteron

The deuteron is the only two nucleon (n-p) bound state (no p-p or n-n bound states).

2H or 2D

Property summary: B = 2.23 MeV JP = 1+ z µ = +0.857 µN Q = +2.82x10-31 m2 Prolate R = 2.1 fm Q > 0 No excited states observed

Deductions: n-p state 3 S1 (l=0, S=1 ↑↑) : µ = µP + µn = 0.88 µN 1 S0 (l=0, S=0 ↑↓) : µ = µP - µn = 4.71 µN

Experimental value µ = +0.857 µN

⇒ n=1, no orbital contributions to µ (l=0) 3 Deuteron is a S1 state.

Lent Term 2004 Nuclear / V.Gibson 67 Assume simple V(r) V(r) Square well

0 b r

-V0

Consider radial Schrödingers equation (l=0): ⎡− 2 d2 ⎤ h + V(r) R(r)= ER(r) ⎢ 2M 2 ⎥ ⎣ dr ⎦

m pm n where M = reduced mass = m p + m n

Let u ( r ) = r R ( r ) r = internucleon distance Probability particle between r and r+dr= 2 2 r2 R (r) dr= u(r) dr

For bound state E < 0 = -Binding energy E B

Lent Term 2004 Nuclear / V.Gibson 68 Two regions:

1) r < b V = -V0 , E = -B 2) r > b V = 0, E = -B 1) r < b 2 − 2 d u(r) h + (B − V0)u(r)= 0 2M dr2 General solution u(r)= A sinαr+ C cosαr 2M α2 = (V − B) 2 0 h u(r) R (r)= Require r finite for r→0 ⇒ C=0 2 (i.e. don’t want infinite density R (r) at centre of nucleus.) ∴u(r)= A sinαr r < b

2) r > b 2 − 2 d u(r) h + Bu(r)= 0 2M dr2 General solution u(r)= De−βr+ Fe+βr 2MB β2 = 2 h r→∞, e+βr →∞ ⇒ F=0 r ∴u(r)= De−β r > b

Lent Term 2004 Nuclear / V.Gibson 69 r=b u(r) and du(r)/dr continuous

−βb u(r) A sinαb = De −βb du(r)/dr αA cosαb = −βDe 1/2 β ⎛ B ⎞ Ratio cotαb = − = −⎜ ⎟ α ⎝ V0 − B ⎠

Assume V0 > B : 2 unknowns b,V0 π 3π 5π cotαb ≈ 0 αb = , , , 2 2 2 K 2 2M 2 ⎛ π ⎞ lowest V0b = ⎜ ⎟ 2 ⎝ 2⎠ energy h 2 2 2 π −28 V b ≈ h ≈10 2 0 8M MeVm

For b = 2 fm ⇒ V0 ~ 25 MeV (c.f. B/A~8 MeV)

Lent Term 2004 Nuclear / V.Gibson 70 u(r) Large probability for distance between proton and neutron > b. (ii)

(i)

b 2b 3br 0.5b b V(r) r

(i)

u(r) not a strong function (ii) of reasonable V(r).

Size of the deuteron determined by Binding Energy not range of force.

Lent Term 2004 Nuclear / V.Gibson 71 Spin Dependence If there were no spin dependence in the deuteron potential, expect to observe both J=0 and J=1 bound states with same energy. Only J=1 states is observed (↑↑) Also no n-n and p-p bound states are observed which would require spins to be ↑↓ due to exclusion principle. Require angular momentum and parity to be conserved v v ⇒ scalar potential, simplest ~ s1⋅ s2 Deuteron spin v v v J = s1 + s2 v2 v v 2 J = ()s1 + s2 2 2 2 v v J(J + 1)h = s1(s1+1)h + s2(s2+1)h + 2s1 ⋅s2 v v 1 2 s1 ⋅s2 = []J(J +1) − s1(s1+1) − s2(s2+1) h 2 2 sv ⋅sv = h J=1, s1=s2=1/2 1 2 4 3 2 J=0, sv ⋅sv = − h 1 2 4 Different potentials for singlet and triplet states

Lent Term 2004 Nuclear / V.Gibson 72 Non-Central Term The deuteron has a small electric quadrupole moment, Q = +2.82x10-31 m2. Hence, the nucleon potential is not spherically symmetric.

3 However, S1 l=0, J=1 ↑↑ is symmetric Require some angular dependence in the deuteron wavefunction, ψ. For J=1, other possible states: L S 1 1 0 P1 3 l 1 1 P1 P = (-1) 3 2 1 D1 Overall state must have definite parity 3 P + (same as S1 J =1 ). 3 ψ has 5% D1 state. 2 2 (ψ=a1ψS+a2ψD, |a1| =0.95, |a2| =0.05) i.e. 5% of time L switches 0→2 and hence nuclear force must apply a torque. Potential is a function of θ as well as r. θ ⇒ non-central force (tensor force). r Also explains small difference in expected

µ = 0.88 µN and measured µ = +0.857 µN .

Lent Term 2004 Nuclear / V.Gibson 73 Deuteron Summary In addition to the general features of the nucleon-nucleon interaction, the properties of the deuteron imply:

 Depth of nucleon potential,

V0 ~25 MeV for nuclear radius (b) =2 fm.

 Nuclear force is spin dependent

 Non-central terms in potential

Two nucleon potential v v v v V = VC(r) +VS(s1 ⋅s2) +VT (r,s) +K central spin tensor

Limitations:  only one state to study  not enough information to determine all parameters required  no information for l≠0and excited states ⇒ Study nucleon-nucleon scattering

Lent Term 2004 Nuclear / V.Gibson 74 Nucleon-Nucleon Scattering

Consider n-p scattering Elastic scattering k k from near centre plane wave of nucleus (l=0) E≤10 MeV Nucleus z

ikz ikrcosϑ 1 ψIN = e = e , k = 2mE h Expand ψ I N in spherical harmonics Spherical ikrcosϑ ∞ harmonic where ψIN = e = B (r)Y ,φ(ϑ) ∑l=0 l l B (r)= il(4π(2 +1))1/2j(kr) l l l Spherical Bessel function sinkr sinkr coskr j0 = , j1 = − ,K kr ()kr 2 kr Each term is a solution of the Schrödinger equation in spherical coordinates for a constant potential energy and with scattering centre as the origin. Terms are called Partial Waves

For low energies (E ≤ 10 MeV) only need consider l=0 term. Lent Term 2004 Nuclear / V.Gibson 75 Example: A nucleon of 10 MeV (5 MeV cms)

940 k = 2ME = 2⋅ ⋅5 M = reduced mass≈ mn/2 2 m = 940 MeV/c2 = 68.6 MeV n h=1, hc=197 MeV fm = 0.348 fm-1

2 Coefficient |B (r)|2 in |Bl(r)| l l=0 partial wave expansion.

r = 2 fm kr

For a range of 2 fm for the nuclear force only the l=0 partial wave important for beam energies ≤ 10 MeV.

Angular momentum conserved, l does not change in scattering processes.

Lent Term 2004 Nuclear / V.Gibson 76 Consider l=0: Free particle wavefunction, eikr − e−ikr ψ → ψ = V=0 IN 0 2ikr e−ikr represents a spherical wave going towards the origin.

represents a spherical wave going e+ikr away from the origin.

In presence of nucleon potential,

e−ikr is not affected for r > range of potential (i.e. before particle gets to scattering centre)

for elastic scattering, amplitude must +ikr ikr e be same as e − part i.e. no particles created or destroyed.

⇒ Phase change only

Lent Term 2004 Nuclear / V.Gibson 77 Attractive potential raises K.E. within range of force ⇒ λ decreases. free particle rR (r)

δ0

Attractive potential ⇒ +ve phase shift δ0>0 Repulsive potential ⇒ -ve phase shift δ0<0

Introducing V≠0 changes the phase of the outgoing wave:

i(kr+2δ ) −ikr iδ 0 e 0 − e e sin(kr+ δ ) ≠ ψ′ = = 0 V 0 2ikr kr Convention: 2δ0 phase shift in outgoing partial wave δ0 “ “ in l=0 scattered wave

Probability of scattering given by the amplitude of the scattered wave: i(kr+δ ) e 0 ψ = ψ′-ψ = sinδ scat 0 kr 0

Lent Term 2004 Nuclear / V.Gibson 78 Differential cross-section: dσ = Number of particles/sec scattered into dΩ dΩ Incident flux . dΩ Area r2 dΩ dΩ v Number particles/sec through area r2 dΩ 2 2 = |ψscat| r dΩ v v=velocity of particles Flux= Number of particles through unit area/sec=v 2 2 2 dσ ψscat r dΩ v 2 2 sin δ0 2 = = ψscat r = r dΩ v dΩ k2r2 dσ sin2δ = 0 dΩ k2 4π sin2δ In centre of mass, =0, isotropic σ = 0 l k2 Low E, k → 0, δ0 → 0 ei(kr+δ0 ) (e2iδ0 −1)eikr ψ = sinδ = scat kr 0 2ik r δ eikr aeikr 0 = 2 k r r σ = 4πa k → 0, δ0 → 0

ais the amplitude of ψscat, often called the “scattering length”.

Lent Term 2004 Nuclear / V.Gibson 79 n-p scattering cross-section (barns) σ

neutron K.E. (eV)

Low energy σ ∼ constant ≈ 20 barns

Extract phase shifts, δ0 , from experimental measurements of the differential cross-section and compare to predicted phase shifts to determine the nucleon potential.

Need to relate the phase shift, δ0, to the parameters of nucleon potential.

Lent Term 2004 Nuclear / V.Gibson 80 Phase Shift δ0 Solve Schrödingers equation in interaction region. Particles can collide in ↑↓ or ↑↑ n p n p Consider a square well potential: V(r) E > 0 0 b r I II -V0 Radial wave equation: u(r)=rR(r) 2 − 2 d u(r) h + V (r)u(r)= E u(r) 2 2M dr M=reduced mass

d2u(r) 2M Region I + (V + E)u(r)= 0 2 2 0 r < b dr h k = 2M (E + V ) uI(r)= A sinkr 0 h

d2u(r) 2M Region II + E u(r)= 0 2 2 r > b dr h uII(r)= B sin(k′r+ δ0) k′ = 2ME h

Lent Term 2004 Nuclear / V.Gibson 81 Boundary conditions: u(r), du(r)/dr continuous

u(r) A sinkb = B sin(k′b+ δ0) du(r)/dr kA coskb = k′B cos(k′b+ δ0) ratio kcotkb = k′cot(k′b+ δ0)

k δ = cot−1( cotkb)− k′b 0 k′

k = 2M (E + V0) h k′ = 2ME h

Given a set of potential well parameters V0, b, δ0 can be compared to the measured value 2 2 extracted from σ = 4 π si n δ 0 k ′ as a function of energy. Example: Using triplet parameters for deuteron 2 -28 2 V0b ≈ 10 MeV m b = 2.1 fm and V0 = 25 MeV

⇒ σ ∼ 5 barns (c.f. 20 barns experimentally) Need to consider spin dependence

Lent Term 2004 Nuclear / V.Gibson 82 Repulsive Core

Nucleon scattering λ≤range of δ nuclear force.

300 MeV E (MeV)

At 300 MeV: Phase shift becomes negative ⇒ repulsive force

V = +∞ r < Rcore = -V0 Rcore< r < R = 0 r > R

1 1 1 D = = = p 2mE 2×940×300 m= 940 MeV/c2 = 1.3×10-3 MeV-1 h=1, hc=197 MeV fm ⇒ Rcore ≈ 0.5 fm

Lent Term 2004 Nuclear / V.Gibson 83 Spin Dependence The total cross-section for n-p scattering is made up of a fixed mixture of n-p interactions in the n-p states: 1 S=0 S0 ↑↓-↓↑ 3 and S=1 S1 ↑↑, ↓↓, ↑↓+↓↑ If the orientations of the neutrons in the incident beam and protons in the target are random, then 3 1 t=triplet σ = σt+ σs 4 4 s=singlet

To separate the contributions of σt and σs, scatter very low-energy neutrons (E<1KeV) from ortho- and para-hydrogen (H2):

ortho-H2 ↑↑ SH2 = 1 p p para-H ↑↓ S = 0 2 p p H2

Low neutron E: λ >> separation of protons in H2 2 Get coherent scattering σ = (∑ amplitudes) 2 (c.f. incoherent σ = ∑ ()amplitudes )

Lent Term 2004 Nuclear / V.Gibson 84 Total amplitude for scattering neutron from one proton aˆp = asπˆs+ atπˆt where πˆ s and πˆ t are operators that project singlet and triplet parts of the n-p wavefn:

↑p ↑n πˆs ψnp = 0, πˆt ψnp =1, ap = at ˆ 1, ˆ 0, a a ↑p ↓n πs ψnp = πt ψnp = p = s

For coherent scattering of neutron from two protons in H2 aˆ= aˆ + aˆ = a (πˆ + πˆ )+ a (πˆ + πˆ ) p1 p2 s s1 s2 t t1 t2

2 Total cross-section σ = 4πa 2 ⎛ 1 3 ⎞ see question σpara = 4π⎜ as+ at⎟ para-H2 ↑↓ ⎝ 2 2 ⎠ sheet 2 ⎛ 2 2 1⎛ 3 1 ⎞ ⎞ ortho-H ↑↑σ = 4π⎜ ()2a + a + a ⎟ 2 ortho ⎜ 3 t 3⎜ 2 s 2 t⎟ ⎟ ⎝ ⎝ ⎠ ⎠ ↑↑↑↑↓↓ n o-H2 n o-H2 S=3/2 S=1/2

Lent Term 2004 Nuclear / V.Gibson 85 σpara can be measured using H2 at 20K. H2 at 20K is all para-hydrogen.

If the nuclear force were independent of spin,

σt = σs and at = as, thus σpara and σortho would be the same.

Experimentally, σpara = 4 b and σortho= 130 b

⇒ Nuclear force is spin-dependent

The large difference between the measured values shows that at ≠ as and that at and as must have different signs to make σpara small.

at = 5.4 fm and as = -23.7 fm

⇒ ↑↓ singlet state is unbound ↑↑ triplet state is bound

Lent Term 2004 Nuclear / V.Gibson 86 Sketch of the radial wavefunction solution to Schrödingers equation u(r)

b as at r I II -Vs

-Vt r < b uI(r)= A sinkr k = 2M (E + V0) h r > b uII(r)= B sin(k′r+ δ0) k′ = 2ME h a is where uII(r) crosses r axis for k’→0 δ u (r)= 0 at r = − 0 = a II k′ sinδ Convention: a= − lim 0 k′→0 k′ a negative → unbound state a positive → bound state

Lent Term 2004 Nuclear / V.Gibson 87 Charge Dependence

Study charge dependence of nuclear force by comparing p-p and n-n scattering.

Important difference to n-p scattering:

→ Identical particles

 Total wavefunction antisymmetric

∴ l=0 (i.e. low energy) scattering only possible in singlet state ↑↓

 Cannot distinguish

ϑ ϑ p p p p

Must include interference between 2 possibilities.

Lent Term 2004 Nuclear / V.Gibson 88 p-p scattering Exclusive study of singlet interaction. However, both Coulomb and nuclear interactions are present. Theoretical expression for dσ/dΩ for p-p scattering: 2 dσ ⎛ e2 ⎞ 1 ⎪⎧ 1 =⎜ ⎟ dΩ ⎜ 4πε ⎟ 2 ⎨ 4 ⎝ 0⎠ 4T ⎩⎪sin ()ϑ/2 Rutherford scattering Mott Scattering 1 cosηLntan2()ϑ/2 + − [] cos4()ϑ/2 sin2()ϑ/2 cos2 ()ϑ/2 Rutherford Wave-mechanical classical term interference term Corrections for two identical particles 2 ⎛cosδ +ηLnsin2()ϑ/2 cosδ +ηLncos2()ϑ/2 ⎞ − sinδ ⎜ []0 + []0 ⎟ η 0⎜ 2 2 ⎟ ⎝ sin ()ϑ/2 cos ()ϑ/2 ⎠ Wave intereference cross-terms between Coulomb and nuclear potential scattering 4 ⎪⎫ + sin2δ 2 0⎬ T = laboratory K.E. η ⎭⎪ ϑ = scattering angle in c.m.s system 2 −1 Pure nuclear η = (e /4πε0hc)β β=v/c potential scattering δ0 = l=0 phase shift

Lent Term 2004 Nuclear / V.Gibson 89 δ0 only unknown

From dσ/dΩ find sign and magnitude of δ0

Interference allows sign determination:

dσ/dΩ

Total

Interference

Mott

0 90 ϑ (cms)

Find a -ve → no pp bound states

σpp = 36.7 ± 0.1 b

Lent Term 2004 Nuclear / V.Gibson 90 n-n scattering

Difficult as no neutron only targets.

Use reactions that create 2 neutrons within nuclear range as separate (comparable to a scattering experiment) e.g. π- + d → 2n + γ γ detector

π- beam d target  π stop in target  Look for coincidences of 2n and γ Liquid scintillator neutron detector If 2n bound ⇒ γ monochromatic, 2 body final state If no n-n ⇒ energy shared interaction between 3 particles

σnn = 33.8 ± 1.8 b

(c.f. σpp = 36.7 ± 0.1 b) ⇒ Nuclear force is charge independent

Lent Term 2004 Nuclear / V.Gibson 91 Spin-Orbit Potential A momentum dependent force can be represented by a spin-orbit term in the potential v v ~ Vso (r)L ⋅S Atomic electrons experience a spin-orbit coupling arising from the interaction of the electrons spin and the internal magnetic field of the atom. Nucleons experience a spin-orbit coupling arising from the interaction of the nucleon spin and the strong nuclear force. 20x strength and opposite sign c.f. atomic spin-orbit term

Evidence for a nucleon spin-orbit term from the polarization of scattered nucleons. Polarized: magnetic substates not equally populated N(↑) −N(↓) Polarization = N(↑) + N(↓) P = ±1 100% polarization, P=0 unpolarized

Lent Term 2004 Nuclear / V.Gibson 92 Observe polarization of scattered nucleon when beam and target unpolarized. v v Assume V ~ −Vso (r)L ⋅S

3 possibilities:

(i) Beam nucleon spin ↑, target nucleon spin ↑ Total S=1

= vr× pv ↑ l 1 ↑ ↑ 2 v v l = r× p

v v v v Nucleon 1: l = r× p into plane ∴ l ⋅ s -ve ⇒ V is +ve i.e. repulsive. v v v v Nucleon 2: l = r× p out of plane ∴ l ⋅ s +ve ⇒ V is -vei.e. attractive.

All spin ↑ incident on spin ↑ (target) deflected in same direction due to spin-orbit potential.

Lent Term 2004 Nuclear / V.Gibson 93 (ii) Beam nucleon spin ↓, target nucleon spin ↓ Total S=1 = vr× pv ↓ l 1 ↓ ↓ 2 v v l = r× p

v v v v Nucleon 1: l = r× p into plane ∴ l ⋅ s +ve ⇒ V is -vei.e. attractive. v v v v Nucleon 2: l = r× p out of plane ∴ l ⋅ s -ve ⇒ V is +ve i.e. repulsive.

(iii) Beam nucleon spin ↓ or ↑, target nucleon spin ↓ or ↑. Total S=0 v v l ⋅ s= 0 ∴ No deflection due to spin-orbit

The spin-orbit interaction deflects the spin ↑ component of the incident beam left and the spin ↓ component of the incident beam right if total S=1.

Lent Term 2004 Nuclear / V.Gibson 94 Any individual nucleon passing through the interior of a nucleus will on average pass an equal number of nucleons with spin ↑ and spin ↓, hence there will be no net spin-orbit interaction.

A net spin-orbit interaction is obtained from those nucleons passing near the surface of the nucleus. p-p scattering

N(↑) −N(↓) P = N(↑) + N(↓)

ϑ (cms) Only see effect when incident beam energy is high enough for l > 0. Polarization increases with energy.

Lent Term 2004 Nuclear / V.Gibson 95 Yukawa Potential

Consider the electromagnetic interaction:

 Classically: E-M forces arise from action at Ev Bv a distance of the v and fields. F v q q F = 1 2 rˆ v 2 q1 r q2 r  Quantum mechanically: Forces arise due to exchange of virtual field quanta (second quantization) Fv

q1 pv q2

The field strength at any point is uncertain ∆p∆r ~ ∆t = ∆r h c

Number of quanta emitted and absorbed~q1q2 v dp q q ⇒ F = = 1 2 rˆ dt r2 Massless particle e.g. photon, force has infinite range. Lent Term 2004 Nuclear / V.Gibson 96 Nucleons start to experience the strong interaction at a distance of ~ 2fm. 2 ∆E∆t ~ h E = mc c ⇒ mc2 ~ h ~ h ∆t r 1 h=c=1 m ~ h = rc r For r=2 fm, predict existence of a particle of mass ≈ 100 MeV/c2 (i.e. m ≈ 200 me) ⇒ Meson (e.g. pion)

The pion (π) was discovered in cosmic ray photographic plates in 1947.

3 types: π+ π- π0

u d u d u u

mass(π+)=mass(π-)=139.6 MeV/c2 mass(π0)=135 MeV/c2 JP = 0+

Lent Term 2004 Nuclear / V.Gibson 97 Relativistic wave equation: 2 2 2 E = p + m c = 1 ∂ v Operator E → i p→ −i∇ ∂t ∂2ψ − = −∇2ψ + m 2ψ ∂t2 2 2 2 ∂ ψ Klein-Gordon ∇ ψ − m ψ − = 0 Equation ∂t2 Static solution g2 e−mr ψ = − 4π r Assume the pion wavefunction can be represented equivalently by a potential in the vicinity of the nucleon: Yukawa g2 e−mr V (r)= − Potential 4π r g= coupling constant (dimensionless) Note: m→0, V(r) → 1/r g2 e2 1 α = ~ O (1) c.f. α = ≈ s 4π 4π 137 Strong coupling Fine structure constant constant (EM)

Lent Term 2004 Nuclear / V.Gibson 98 Yukawa interaction between 2 nucleons:

nnπ0 ppπ0 npπ0 g g n n p p n p

pnπ- ppπ+ n p n n Quark model: (see Particles course) ppπ0 p p u u u u d d

d u uu d u u u

Lent Term 2004 Nuclear / V.Gibson 99 Summary Nucleon-Nucleon Potential v v V(r) V = VC(r)+VS(s1 ⋅s2) v v v v +VT (r,s) +Vso(r) L ⋅S+K Repulsive core 0.5 1 2 r (fm) V0~few 10 MeV Yukawa Potential g2 e−mr V(r)= − 4π r The nucleon force is  strong, (existence of nuclei, nucleon scattering,

strength αs~O(1))  short range, (nuclei size → range ~ 2 fm)  attractive, (existence of nuclei)  has a repulsive core, (nucleon scattering r < 0.5 fm)  saturates, (B/A~constant)  charge independent, (mirror nuclei, pp vs nn vs np scattering)  spin dependent, (deuteron J=1, np and nH2 scattering)  spin-orbit interaction (polarization in nucleon scattering at high E)  non-central terms (deuteron Q) Lent Term 2004 Nuclear / V.Gibson 100