university of cape town eee4106z introductory nuclear

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/ . *

20150405.1830 uct-physics-dga 1

mailto:[email protected]

http://www.phy.uct.ac.za/courses/eee4106z/current/

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.


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

. . .



Reading and referencesEEE4106Z Reading list:Reading list on course website

. . .

Martin

Meyerhof


http://www.phy.uct.ac.za/courses/eee4101f/current/eee4106zr14.htm

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


UCT EEE4106Z 2014 Nuclear Physics Martin 2.1 *

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



Binding energy in atom and nucleus


UCT EEE4106Z 2014 Nuclear Physics Martin 2.1 *

Mass spectrometer

Crossed E and B fields give velocity.Curvature gives momentum. q/m = E/B2ρ.



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


Curve of nuclear binding energy


UCT EEE4106Z 2014 Nuclear Physics Martin 2.1

Curve of binding energy


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)



Cross-section


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)



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



Electron scattering dσ/dΩ



Radial charge distributions of nuclei:r = 1.1 A1/3 fm


UCT EEE4106Z 2014 Nuclear Physics Martin

Differential cross-section for d +54 Fe



Differential cross-section and optical model



Nuclear instabilityNot all nuclei are stable. Some decay.

α decayβ decayγ decayfission

Stable and long-lived nuclides plotted on Segre plot.



Segre plot



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


Segre plot



Segre plot 2



Segre plots for odd A and even A



Valley of stability



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



Radioactive decay A→ B→ C



Natural radioactivity: 235U decay chain



Natural radioactivity: 232Th decay chain



Natural radioactivity: 238U decay chain



Natural radioactivity: 237Np decay chain



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.



Neutron separation energy systematics (2)

Odd-even effect inS n in lead isotopesprovides evidence forpairing effect. Noteshell effect at N = 126.



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


UCT EEE4106Z 2014 Nuclear Physics Meyerhof figure 2.08

Curve of binding energy


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



Model for the asymmetry term



Summary of B in liquid-drop model


UCT EEE4106Z 2014 Nuclear Physics Lilley figure 2.3

Semi-empirical mass formula vs data



Nuclear energy level near highest filledlevels



Nuclear binding energy per nucleon B/Aand SEMF



Contributions to B/A



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


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



Mass parabola for A = 111 isobars



Mass parabolae for A = 102 isobars



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


The fission barrier

Fission barrier ≈ 6 MeV. Tunneling or activation.



Spontaneous fission half-lives

Spontaneous fission half life versus Z2/A. For Z/A ≥ 50 coulombterm dominates surface term in binding energy.



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.



Neutron induced fission of 235U



Fission cross-sections for 235U and 238U



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, . . .



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.



Excitation and decay of compound nucleus



Cross section for neutron on 12C



Direct and compound nucleus reactions



Nucleons emitted from nucleon-nucleuscollisions



Cross-section for neutrons on 238U



Ground and excited states of nuclei



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.

/. . .



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



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.



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.

/. . .



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.



Quadrupole deformed nuclei



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.



Electric quadrupole moments of odd A nu-clei


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.



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.



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.



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



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.



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.



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


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. /. . .



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



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.



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



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


UCT EEE4106Z 2014 Nuclear Physics Martin fig 7.4

Single particle levels with spin-orbit


UCT EEE4106Z 2014 Nuclear Physics Blin-Stoyle fig 4.3



UCT EEE4106Z 2014 Nuclear Physics Krane fig 5.6




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)



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.



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)+



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.



End EEE4106Z 02-01-00 Nuclear Physics

*


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