UNIVERSITY OF CAPE TOWN EEE4106Z Introductory Nuclear ...
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UCT EEE4106Z 2014 Nuclear Physics 20150405.1830
UNIVERSITY OF CAPE TOWNEEE4106Z
Introductory Nuclear Physics02-01-00 Nuclear Physics
March 2015
Emeritus Professor David AschmanRoom 4T7, Physics DepartmentUniversity of Cape Townmailto:[email protected]
http://www.phy.uct.ac.za/courses/eee4106z/current/ . *
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UCT EEE4106Z 2014 Nuclear Physics CT Gaunt *
EEE4106Z SyllabusUCT EEE4106Z 2015 Introductory nuclear physics and radia-tion for power supply 16 credits, NQF level 08
Course aims: To develop strong concepts of nuclear physicsand radiation in the context of nuclear power reactors.Course outcomes: An understanding of nuclear fission pro-cess and ability to calculate reaction energy.Course outline: Nuclear physics and radiation in the contextof nuclear power reactors: atomic nature of matter; binding en-ergy; radioactive decay; nuclear fission; neutron efficiency; ion-izing radiation; radiation detection and measurement; effects ofradiation on matter and biological systems.
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UCT EEE4106Z 2014 Nuclear Physics *
02-01-00 Nuclear PhysicsNuclear systematics: size, stability, binding energyThe nuclear force and the strong interaction.
Nuclear structure: binding energy
Nuclear decay and radiation
Nuclear reactions
. . .
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Reading and referencesEEE4106Z Reading list:Reading list on course website
. . .
Martin
Meyerhof
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UCT EEE4106Z 2014 Nuclear Physics Martin Ch 2 *
Martin Ch 2: Nuclear Phenomenology2.1 Mass spectroscopy and binding energies2.2 Nuclear shapes and sizes2.3 Nuclear instability2.4 Radioactive decay2.5 Semi-empirical mass formula: the liquid drop model2.6 β-decay phenomenology2.8 γ-decays2.9 Nuclear reactions
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Mass spectroscopy and binding energiesNuclide A
ZXN specified by atomic number Z, neutron number Nand mass number A = N + Z. Isotopes have same Z; isotoneshave same N; and isobars have same A.
Use atomic masses to define mass deficit
∆M(Z, A) = M(Z, A) − Z(Mp + Me) − NMn
and hence binding energy ∆Mc2 = −B
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Binding energy in atom and nucleus
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Mass spectrometer
Crossed E and B fields give velocity.Curvature gives momentum. q/m = E/B2ρ.
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Nuclear mass and binding energyNucleus is bound, so mass of nucleus is less than sum of con-stituent masses. Binding energy, B, is important.
Mnuc = Zmp + Nmn − B/c2 or B = (Zmp + Nmn −Mnuc)c2
In terms of atomic massesMat = Mnuc + Zme − Bat/c2
where the atomic binding energy Bat of the Z electrons is negli-gible (Bat ∼ 10−6Matc2). So
B = (ZMH + Nmn − Mat)c2
Curve of binding energy B/A versus A. Fission. Fusion.Neutron separation energy isS n(A,Z) = [M(A − 1,Z) + mn − M(A,Z)]c2
Neutron capture, followed by β decay often occurs.20150405.1830 uct-physics-dga 9
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Curve of nuclear binding energy
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.1
Curve of binding energy
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UCT EEE4106Z 2014 Nuclear Physics Lilley 1.6.3
Cross-sectionWe define cross-section (Lilley, Nuclear Physics p 24) in termsof the reaction rate R, the beam flux Φ and the number of targetnuclei exposed to the beam N,
R = Φ N σ = (I)(ntt)(σ)
for a beam of area S , the beam intensity is I = Φ/S and a targetof thickness t and number density of target nuclei nt = N/(S t)
If the target consists of nuclei of isotopic species with atomicmass MA (in atomic mass units), so MA = A, and the targetdensity is ρ and NA is Avogadro’s number, then nt = ρNA/A,and
R = (I)(ρtNA/A)(σ)
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.2.1
Nuclear charge distributionCharge particle scattering: Born approximation gives Rutherfordcross-section (
dσdΩ
)Rutherford
=Z2α2(~c)2
4E2 sin4(θ/2)We must include electron spin(
dσdΩ
)Mott
=
(dσdΩ
)Rutherford
[1 − β2 sin2(θ/2)]
Also: target recoil and magnetic interaction(dσdΩ
)spin 1/2
=
(dσdΩ
)Mott
E′
E[1 + (−q2/4M2c2) tan2(θ/2)]
where q2 = (p − p′)2 = 2m2ec2 − 2(EE′/c2 − |p| |p′| cos θ))
So q2 ' −4EE′c2 sin2(θ/2)
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But nucleus is not pointlike; if spatial distribution is f (x) defineform factor F(q2)
F(q2) ≡1
Ze
∫e1q·x/~ f (x) d3x
where Ze =∫
f (x) d3x.For spherical symmetry
F(q2) =4π~Zeq
∫ ∞
0r ρ(r) sin
(qr~
)dr
(dσdΩ
)experimental
=
(dσdΩ
)Mott
∣∣∣F(q2)∣∣∣2
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.2.1
Electron scattering dσ/dΩ
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.2.1
Radial charge distributions of nuclei:r = 1.1 A1/3 fm
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UCT EEE4106Z 2014 Nuclear Physics Martin
Differential cross-section for d +54 Fe
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.2.1
Differential cross-section and optical model
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.3
Nuclear instabilityNot all nuclei are stable. Some decay.
α decayβ decayγ decayfission
Stable and long-lived nuclides plotted on Segre plot.
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Nuclear stabilityIt may be energetically favourable for nuclei to disintegrate, byα decay, β decay, nucleon emission, or fission. A nucleus iscompletely stable if energetics forbid disintegration.
Stable: eg. 2H, 4He, 12C, 16O, 56Fe
Metastable: eg. 40K, 14C, 22Na, 238U, 252Cf
Unstable: eg. 2He, 5X, 8B, 100Sn
Effect of pairing on stability:165 stable even-even nuclei (even A)50 stable odd-even nuclei (odd A)55 stable even-odd nuclei (odd A)4 stable odd-odd nuclei (even A)
Segre plot (Z versus N plane). Valley of stability.20150405.1830 uct-physics-dga 24
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Segre plots for odd A and even A
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.4
Radioactive decay lawλ, the decay probability per unit time is constant.Unaffected by previous history.
At t = 0 population is N(0). What is N(t)?
dN = −Nλ dt. Integrate
N(t) = N(0) exp(−λt)
where lifetime τ = 1/λ = t1/2/ ln 2
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Radioactive decay A→ B→ C
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Natural radioactivity: 235U decay chain
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Natural radioactivity: 232Th decay chain
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Natural radioactivity: 238U decay chain
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Natural radioactivity: 237Np decay chain
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Neutron separation energy systematics
S n = M(Z, A − 1) +
mn − M(Z, A)Odd-even effect in S n
in barium isotopesprovides evidence forpairing effect. Noteshell effect at N = 82.
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Neutron separation energy systematics (2)
Odd-even effect inS n in lead isotopesprovides evidence forpairing effect. Noteshell effect at N = 126.
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Semi empirical mass formula: liquid dropmodelShort range forces inspire a model similar to that of an incom-pressible liquid with surface tension.
Mass has six terms:constituentsvolume, surface, Coulomb, asymmetry, pairing
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UCT EEE4106Z 2014 Nuclear Physics Meyerhof figure 2.08
Curve of binding energy
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UCT EEE4106Z 2014 Nuclear Physics Blin-Stoyle
Liquid-drop modelSemi-empirical mass formula (Weizsacker) based on liquid-dropanalogy: constituents, binding energy (volume, surface, coulomb,asymmetry, pairing)M(A,Z) = Zmp + (A − Z)mn + Zme − B(A,Z)/c2
where the binding energy
B(A,Z) = [+aVA−aSA2/3−aCZ2/A1/3−aAs(A−2Z)2/A−δ(A,Z)]c2
with δ(A,Z) = −aoeA−3/4 for A even, Z even,0 for A odd, and +aoeA−3/4 for A even, Z odd
Typical values (in MeV): aV = 15.76, aS = 17.81, aC = 0.71,aAs = 23.70, aoe = 34.0
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UCT EEE4106Z 2014 Nuclear Physics Meyerhof figure 2.14
Model for the asymmetry term
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UCT EEE4106Z 2014 Nuclear Physics Meyerhof figure 2.16
Summary of B in liquid-drop model
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UCT EEE4106Z 2014 Nuclear Physics Lilley figure 2.3
Semi-empirical mass formula vs data
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.5
Nuclear energy level near highest filledlevels
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Nuclear binding energy per nucleon B/Aand SEMF
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SEMF
M(A,Z) =Zmp + (A − Z)mn + Zme
+ (− aVA
+ aSA2/3
+ aCZ2/A1/3
+ aa(Z − A/2)2/A
+ apA−1/2[−1, 0,+1]
) c2
with (in MeV): aV = 15.56, aS = 17.23, aC = 0.697, aa = 93.14,ap = 12.020150405.1830 uct-physics-dga 46
UCT EEE4106Z 2014 Nuclear Physics Martin 2.6
β decay phenomenology
M(Z, A) = αA − βZ + γZ2 +δ
A1/2
whereα = Mn − aV + aS
A1/3 + aa4
β = aa + (Mn − Mp − me)γ = aa
A + aCA1/3
δ = ap
Energetically possible ifβ− : M(Z, A) > M(Z + 1, A)β+ : M(Z, A) > M(Z − 1, A) + 2me
EC : M(Z, A) > M(Z − 1, A) + ε
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Mass parabola for A = 111 isobars
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Mass parabolae for A = 102 isobars
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.7
Fission
Volume preserving deformation, parameter ε. Large deforma-tion can lead to fission. a = R(1 + ε) and b = R/(1 + ε)1/2
ES = aSA2/3(1 + (2ε2/5)) + . . .)EC = aCZ2A−1/3(1 − ε2/5) + . . .)∆E = (ES + EC) − (ES + EC)SEMF = (ε2/5)(2aSA2/3 − aCZ2A−1/3)Spontaneous fission for ∆E < 0, thus Z2/A ≥ 2aS/aC ≈ 49,ie for Z > 116 and A ≥ 270.20150405.1830 uct-physics-dga 50
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The fission barrier
Fission barrier ≈ 6 MeV. Tunneling or activation.
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Spontaneous fission half-lives
Spontaneous fission half life versus Z2/A. For Z/A ≥ 50 coulombterm dominates surface term in binding energy.
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Fission barrier
There is a potential energy barrier to fission. As A increasesthe fission barrier decreases. For A ' 236 the fission barrier isabout 6 MeV.
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Neutron induced fission of 235U
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Fission cross-sections for 235U and 238U
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UCT EEE4106Z 2014 Nuclear Physics Martin 2.8
Gamma decay of a nuclear levelEnergy typically ∼ 100 keV or ∼ 1 MeV.Lifetimes typically ∼ 1 × 10−12 s, and depend on energy and an-gular momentum
J = Si − S f or S i + S f ≤ J ≤ |S i − S f |
and mi = M + m f
Also parity change (−1)J.
Classification by multipolarity: E1, M1, E2, M2, E3, M3, . . .
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Nuclear reactionsNotation: a + X → b + Y or X(a, b)Y
Direct reactions: reaction time ∼ time to cross a nucleus.Pickup, stripping and knockout reactions.
Compound nucleus reactions: reaction time long, compoundnucleus forms, with no memory of initial state. Decay by par-ticle emission or gamma emission. Resonances.
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Excitation and decay of compound nucleus
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Cross section for neutron on 12C
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Direct and compound nucleus reactions
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Nucleons emitted from nucleon-nucleuscollisions
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Cross-section for neutrons on 238U
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Ground and excited states of nuclei
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Nuclear SpinNucleons are fermions, s = +1/2.
Eigenvalue of s2 is (1/2)(1/2 + 1)~2
eigenvalue of sz is ±(1/2)~
Orbital angular momentum of nucleons l = 0, 1, 2, . . . for eachnucleon.
Add all spins vectorially to get total spin.Add all orbital angular momenta vectorially to get total orbitalangular momentum.Add spin and orbital angular momentum to get total angular mo-mentum, nuclear spin J.
/. . .
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Nucleus has wavefunction ψJM
where J = 0, 1, 2, . . .and M = −J,−J + 1, . . . , Jand
J2ψJM = J(J + 1)~2 ψJM
JzψJM = M~ψJM
Even A nuclei must have J = 0, 1, 2, . . .Odd A nuclei must have J = 1/2, 3/2, 5/2, . . .
All even-even nuclei have ground state with J = 0
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Nuclear ParityParity operator, P, reflects each point through origin. Clearly ifP is a symmetry operator P2 leaves physics unchanged.|ψ|2 = |P2ψ|2 so Pψ = ±ψP has eigenvalues P = ±1
It seems strong and electromagnetic interaction Hamiltonian con-serve parityH(r1, . . . , rA) = H(−r1, . . . ,−rA) = PH(r1, . . . , rA)or [P,H] = 0
So nuclear states have either even or odd parity. Label as JP.Even-even nuclei g.s. are 0+.
Weak interaction violates parity maximally.
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Nuclear Magnetic MomentsL gives a magnetic moment µL = −(e/2m)Lspin gives an intrinsic magnetic moment µS = −g(e/2m)sg-factor predicted: Dirac→ g = 2; QED→ g = 2.0023192 . . .
Value of electron magnetic moment, µ, is eigenvalue of µS z whenin ms = +1/2 substate. µS z = −g(e/2m)sz
where sz has eigenvalues ms~ with ms = ±(1/2). ThusµS = −(1/2)g(e/2m) = −(1/2)gµB
where µB = e~/2m = 9.274 × 10−24 J T−1 (Bohr magneton)
All charged particles with spin have magnetic momentµp = gp(e/2mp)sp and µn = gn(e/2mp)sn
Nuclear magneton is µN = (e~/2mp) = 5.55078 × 10−27 J T−1.
/. . .
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Actual nucleon g-factors and magnetic moments aregp = +5.5856 and gn = −3.8262µp = +2.7928 µN and µn = −1.9131 µN
These anomalous magnetic moments differ fromDirac values (µp = 1.0, µn = 0) due to finite nucleon size andinternal (quark) motion.Note µp/µn ' −3/2.
Magnetic moments of nuclei have spin and orbital component,and are indicator of structure.
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Nuclear electric quadrupole momentsDeviation from spherical→ electric quadrupole momentQ0 =
∫ρch(r) (3z2 − r2) dV
with z-axis along symmetry axis defined by nuclear spin.Q0 = Z(3 < z2 > − < r2 >). (Z in units of e, Q0 has units of m2 orbarns. (1 barn = 10−28 m2).
For sphere, Q0 = 0. If < z2 >, (1/3) < r2 >Q0 > 0 is prolate ellipsoid, Q0 < 0 is oblate ellipsoid.
Quantum mechanics: for J = 0, no rotation, so Q = 0.Q = [(2J − 1)/2(J + 1)] Q0 so Q = 0 for J = 1/2. As J → ∞,Q→ Q0.
Large |Q| for deformed nuclei (studied by Afrodite at NAC) be-tween magic numbers.
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Electric quadrupole moments of odd A nu-clei
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UCT EEE4106Z 2014 Nuclear Physics Blin Stoyle §3.1
Form of Internucleon PotentialElectrostatic and gravitational potential is long range (V ∼ 1/r).Near constancy of nuclear binding energy per nucleon B/A meansthat each nucleon feels only the effect of a few neighbours. Thisis called saturation. It implies the strong internucleon potential isshort range. Range is of order of the 1.8 fm internucleon sepa-ration. Since volume ∼ A, nuclei do not collapse, there is a veryshort range repulsive component. Reminiscent of interatomicpotential in molecule. Is nuclear physics just quark chemistry?Depth of potential is of order of binding energy, perhaps tens ofMeV.
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DeuteronDeuteron, 2H, is bound state of a neutron and a proton.
Deuteronbinding energy B = 2.23 MeVspin J = 1magnetic moment µ = 0.86 µN
elec quad moment Q = +2.82 × 10−31 m2
Approximate V(r) by a square well and solve Schrodinger eq[−~2
2m∇2 + V(r)
]u(r, θ, φ) = E u(r, θ, φ)
where m = mpmn/(mp + mn) ' mp/2.
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Since V(r) is spherically symmetric
unlm(r, θ, φ) = Rnl(r)Ylm(θ, φ)
with n = 1, 2, 3, . . .; l = 0, 1, . . . , n − 1; m = −l,−l + 1, . . . , l.Energy eigenvalues are Enl. Note use of radial quantum numbernr = np + l, where nr − 1 is number of nodes in R.
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Deuteron ground state configurationDeuteron has very small binding energy, and no excited states.Ground state has n = 1, l = 0, m = 0. l = 0 denotes a S -state.Two ways to add the spin 1/2 of n and p, viz s = 1 or s = 0. ThenJ = l + s. Using notation 2S +1(l)J, we could get a 3S1 triplet or a1S0 singlet. But experiment says J = 1, so deuteron is in tripletstate.
Check with magnetic moments. l = 0 so no orbital motion con-tribution. Evaluate µ for M = J = 1. Spins are aligned.
µ = µn + µp = (−1.91 + 2.79)µN = 0.88µN
Good agreement with experimental µ = 0.86µN
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Deuteron: tensor forceThe non-central tensor forceVT = fT(r)
[3(s1 · r)(s2 · r)/r2 − s1 · s2
]causes ∼ 5% 3D1 admixture in deuteron wave function, giving aprolate shape with Q > 0.
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Deuteron ground state energySolve Schrodinger equation for n = 1, l = 0. Y00 constant; onlyradial part. Put R10 = v/r
−~2
2md2v
dr2 + V(r) = E10v
Solve in two regions, and match.
For r ≤ a, V(r) = −V0
−~2
2md2v
dr2 = (V0 − B)v
with general solution v< = A sinαr + C cosαr with α = [(2m(V0 −
B)/~2]1/2. Take C = 0 otherwise v/r → ∞ as r → 0.
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For r > a, V(r) = 0
−~2
2md2v
dr2 = −Bv
with general solution v> = De−βr + Fe+βr with β = [(2mB)/~2]1/2.Take F = 0 otherwise v→ ∞ as r → ∞. /. . .
Now match v and dv/dr at r = a
A sinαa = De−βa
αA cosαa = −βDe−βa
Dividingcotαa = −[B/(V0 − B)]1/2
Solve numerically, or approximate. Since B << V0, neglect B, socotαa = 0. where α ' (2mV0/~
2)1/2. Hence αa = π/2, 3π/2, 5π/2, . . ..20150405.1830 uct-physics-dga 78
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For ground state take state with fewest oscillations (lowest en-ergy), ie αa = π/2. Then
V0a2 'π2~2
8m' 10−28 MeV m2
Size of wavefunction β−1 = 4.3 fm. So β−1 a means nucleonsoften outside range of potential. If a ' 2 fm, then V0 ' 25 MeV.Exact calc gives V0 ' 35 MeV. /. . .
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No singlet state observed degenerate with triplet, so potential isspin dependent. Analysis of scattering data givestriplet (J = 1): V0t ' 31.3 MeV and at ' 2.21 fmsinglet (J = 0): V0s ' 13.4 MeV and as ' 2.65 fmSinglet is unbound by only 0.1 MeV.
V = VC + V§ = fC(r) + f§(r)s1 · s2
Quadrupole moment→ anisotropic admixture:3P1 (§ = 1,L = 1, J = 1)1P1 (§ = 0,L = 1, J = 1)3D1 (§ = 1,L = 2, J = 1)But P state has negative parity ((−1)l factor).So only D state admixture allowedΨ = (1 − p)1/2Ψ(3§1) + p1/2Ψ(3D1), p ' 5%. Tensor componentto NN potential aligns spin along relative separation vector
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VT = fT(r)[3(s1 · r)(s2 · r)/r2 − s1 · s2)
]Also spin orbit potential VL§ = fL§L · (s1 + s2), and exchangepotential terms. Very complicated, yet models do account forscattering data.
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Magic numbersEvidence for high stability of nuclei withZ = 2, 8, 20, 28, 50, 82N = 2, 8, 20, 28, 50, 82, 126
1) Actual mass of magic nuclei lower than semi-empirical massformula2) More isotones when Z magic; more isotopes when N magic3) Doubly magic nuclei are particularly stable (4He, 16O, 208Pb)4) Large abundance of 4Hein universe, and nuclides with N =
50, 82, 1265) First excited states lie higher for magic N or Z6) Reduced neutron capture cross-section for magic nuclei7) Electric quadrupole moments, Q, go through zero for magicnuclei, suggesting spherical shape
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Spin-orbit couplingCentral potential, whether square well, harmonic oscillator ordiffuse, does NOT have shell structure that gives magic num-bers.
Mayer and Jensen (1948) proposed spin orbit couplingVso = f (r)L · s
Usually f (r) ∼ (1/r) dV/dr with V(r) a Woods-Saxon shape.Complicated spin-spatial nature of NN interaction produce strongspin-orbit term. Level Enl split into two, j = l ± 1/2. Splitting∆E = E j=l−1/2 − E j=l+1/2 ' 10(2l + 1)A−2/3 MeV. Lower j level lieshigher.
Levels labelled n(l) j, e.g. 1s1/2, 1p3/2, 1p1/2
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UCT EEE4106Z 2014 Nuclear Physics Martin fig 7.4
Single particle levels with spin-orbit
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UCT EEE4106Z 2014 Nuclear Physics Blin-Stoyle fig 4.3
Single particle levels with spin-orbit
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UCT EEE4106Z 2014 Nuclear Physics Krane fig 5.6
Single particle levels with spin-orbit
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UCT EEE4106Z 2014 Nuclear Physics *
JP for closed subshellsLevel with angular momentum j has 2 j + 1 like nucleons withm = j, j− 1, . . . ,− j. If level (orbital, subshell) full, then Msubshell =
Σm = 0) so, plausibly Jsubshell = 0. Closed subshell couples toJ = 0. If all subshells full then J = 0. Since j = l ± 1/2, 2 j + 1 iseven, so filled subshells occur for even-even nuclei.Parity of a nucleon in subshell is (−1)l; if even number of them,P = +1. Hence closed subshell has even parity.
Expt: If all subshells are closed, nucleus has JP = 0+ groundstate.
e.g. doubly magic 4He, 16O,208Pb;also 12C(1s1/2, 1p3/2), 28Si(1s1/2, 1p3/2, 1p1/2)
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UCT EEE4106Z 2014 Nuclear Physics *
JP for one particle/hole outside subshellConsider JP = 0+ core plus one n or p.JP must be that of the n or p.17O has n in 1d5/2, so j = 5/2, l = 2, P = (−1)l = +1predict g.s. JP = (5/2)+. Agrees.
41Ca has n in 1 f7/2 predict g.s. JP = (7/2)−. Agrees.
Closed subshell minus nucleon is a hole. Hole has same j asparticle, opposite m.
15N has hole in 1p1/2 predict g.s. JP = (1/2)−. Agrees.
207Pbhas hole in 3p1/2
predict g.s. JP = (1/2)−. Agrees.
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UCT EEE4106Z 2014 Nuclear Physics *
Residual InteractionIf > 1 n or p outside closed subshell there is residual interaction.Basically attractive. Pairs of nucleons are as close as possible.For like nucleon pairs with same ( j,m), Pauli principle ensuresnucleons are well separated. Hence ( j,m) pairs with ( j,−m) togive (0, 0) pair. Add 0+ pair to 0+ core to give JP = 0+ for alleven-even nuclei.
Expt: all even-even nuclei have 0+ g.s.
Hence JP of odd-A nuclei, with one nucleon outside 0+ even-even core, is that of spin parity of last nucleon. Often true e.g.43Ca has 23rd n in 1 f7/2 and is (7/2)−;33S has 17th n in 1d3/2 and is (3/2)+
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UCT EEE4106Z 2014 Nuclear Physics *
Many exceptions: 1) rather pair in a higher j state, e.g. 10747 Ag60
is not (9/2)+ due to odd proton in 1g9/2, but has unpaired p inlower 2p1/2, hence (1/2)−.
2) Residual interaction mix configs, eg 23Na.
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