University Physics IV · Nuclear Physics · 13.3
Nuclear Force & the Shell Model
Two nucleons barely bind, yet a hundred and twenty-six of them fall into shells as clean as an atom's. This lesson runs from the two-body evidence to a well you can fill, and to the one number a shell-model assignment really stakes: the ground-state spin and parity of an odd-A nucleus.
Build the model
Connect the measurement to the mechanism.
The two-nucleon data say the force is barely strong enough and violently short-ranged. The deuteron binds by 2.224 MeV in a well about 34 MeV deep; flip one spin to make the singlet and it does not bind at all. Saturation supplies the rest: B/A pinned near 8 MeV and a density fixed near 0.17 nucleons per fm³ mean each nucleon binds only to its neighbours, with a repulsive core underneath to stop the collapse.
None of that predicts shells — a force that strong ought to scatter any orbiting nucleon within a fermi. What rescues the orbits is Pauli blocking: the states a deep nucleon could scatter into are already filled, so it drifts almost freely in the average well the others build. Make that well Woods-Saxon and you recover 2, 8 and 20, then miss every higher magic number.
Mayer and Jensen's repair was a single term, Vₗₛ(r) l⋅s, carrying the opposite sign to the atomic one and a strength measured in MeV rather than meV, so j = l + 1/2 falls while j = l − 1/2 rises and the gap between them grows as 2l + 1. The high-l orbital of each shell is driven out of it, and 28, 50, 82 and 126 appear. The cost is that the model follows one nucleon at a time: it nails the spin and parity of odd-A nuclei near closed shells, and it is silent or plainly wrong wherever deformation takes over.
- Simple definition
- The nuclear shell model treats each nucleon as moving independently in a smooth average well built by all the others — a Woods-Saxon potential plus an inverted spin-orbit term — whose level ordering gives the magic numbers and the ground-state spin and parity.
- Example
- ⁴¹Ca has 20 protons and 21 neutrons. Twenty neutrons fill through 1d3/2 and the twenty-first sits alone in 1f7/2, so the model predicts J = 7/2 with parity (−1)³ = −1, that is 7/2⁻ — the measured ground state.
The depth that fits is about 34 MeV, so binding is 6% of it: the force clears the bar and no more.
μ = mN/2 = 469.5 MeV/c², so π²ℏ²/8μ = 102 MeV fm² and R = 2.1 fm needs V₀ > 23.2 MeV
Flat where the density is flat, with a 2-3 fm skin. On its own it gives 2, 8 and 20 and nothing above.
V₀ ≈ 50 MeV, R = 1.25 A¹⁄³ fm, a ≈ 0.52 fm; r and a in fm, V in MeV
Angular-momentum algebra alone: the two partners move in opposite directions and by unequal amounts.
s = 1/2, so ⟨l⋅s⟩ = +(ℏ²/2)l for j = l + 1/2 and −(ℏ²/2)(l+1) for j = l − 1/2
Linear in l, so a high-l orbital splits far more than a low one — and that is what makes intruders.
C is the spin-orbit strength in MeV: about 2 MeV in the sd shell, under 1 MeV in heavy nuclei
A check on any level scheme you draw: what one partner gains in binding, the other pays for.
Weights are the occupancies 2j + 1; both energies in MeV, measured from the unsplit level
Turns a filling diagram into a falsifiable number for several hundred odd-A nuclei.
l and j label that one nucleon's orbital; even-even nuclei are 0⁺ without exception
Two nucleons, one bound state, and barely that
The deuteron is the whole bound two-nucleon spectrum. It binds by 2.224 MeV with J = 1⁺, and there is nothing else: no excited deuteron, no dineutron, no diproton. Model it as a square well of radius R = 2.1 fm and a first bound state needs V₀R² > π²ℏ²/8μ = 102 MeV fm², so V₀ > 23.2 MeV; the depth that actually reproduces 2.224 MeV is about 34 MeV. Binding is 6% of the well depth — the force clears the bar and stops. Now flip the two spins antiparallel. The spin-singlet np state does not bind at all, and its scattering length aₛ = −23.7 fm, large and negative, says it misses by a hair; the triplet's is aₜ = +5.42 fm. So the force is strongly spin dependent. It is also not central: the deuteron carries a quadrupole moment of +0.286 fm², which a pure 3S1 state cannot have, and the roughly 4% 3D1 admixture from a tensor term is also why its magnetic moment is 0.857 μN rather than the 0.880 μN you get by adding the free proton and neutron moments.
Saturation, a hard core, and charge independence
If every nucleon attracted every other, binding would follow the pair count A(A−1)/2 and B/A would climb with A. Instead B/A sits near 8 MeV from A ≈ 20 to A ≈ 240, and R = 1.2 A¹⁄³ fm holds the density fixed near 0.17 nucleons per fm³ however many you add. That is saturation, and it forces a range of a fermi or two: a nucleon binds only to its neighbours. Short range alone would still let matter collapse to any density, so something must push back — the 1S0 nucleon-nucleon phase shift is attractive at low energy but changes sign near 250 MeV, which is a repulsive core of radius about 0.5 fm. Charge independence completes the picture. Strip the Coulomb term out of proton-proton scattering and the singlet scattering lengths line up: aₚₚ ≈ −17.3 fm, aₙₙ ≈ −18.9 fm, aₙₚ ≈ −23.7 fm. To the strong force a proton and a neutron are one particle in two charge states, which is why mirror pairs such as ¹⁵N and ¹⁵O differ in binding by little more than their Coulomb energies.
Pauli blocking is what licenses a mean field
How can independent orbits survive a force this strong? A nucleon crossing a nucleus ought to scatter within a fermi, and then no orbit would last long enough to quantise. The exclusion principle is the answer: a collision needs an empty final state for each partner, and deep in the Fermi sea every nearby state is occupied, so the scattering is blocked. The mean free path stretches beyond the nuclear radius and each nucleon moves smoothly in the average potential the others build. Take that potential to be Woods-Saxon, V(r) = −V₀/[1 + exp((r − R)/a)] with V₀ ≈ 50 MeV, R = 1.25 A¹⁄³ fm and a ≈ 0.52 fm: flat where the density is flat, falling across a 2-3 fm skin, and lying between the infinite square well and the harmonic oscillator, whose solvable orderings bracket it. Fill it and closed shells appear at 2, 8 and 20 — ⁴He, ¹⁶O and ⁴⁰Ca, all anomalously bound. Then it fails.
One term, the wrong sign, twenty times too big
The oscillator's next closures are 40, 70 and 112, the square well's are 34, 58 and 92, and the data insist on 28, 50, 82 and 126 — read off high separation energies, small quadrupole moments, extra stable isotopes and low neutron-capture cross-sections. No choice of V₀, R or a repairs that, because the well has no term distinguishing the two ways l and s can couple. Mayer and Jensen (1949) added Vₗₛ(r) l⋅s. Its expectation follows from j = l + s: ⟨l⋅s⟩ = ½[j(j+1) − l(l+1) − 3/4]ℏ², which is +(ℏ²/2)l for j = l + 1/2 and −(ℏ²/2)(l+1) for j = l − 1/2. Choose the coefficient so that j = l + 1/2 is pushed down, the opposite of the atomic case, and the pair separates by ΔE = C(l + 1/2), linear in l. C runs from about 2 MeV in the sd shell to under 1 MeV in the heaviest nuclei, so the 1d pair in ¹⁷O is split by 5.08 MeV — comparable to the oscillator spacing ℏω ≈ 41 A(−1/3) MeV rather than a correction to it.
Filling the levels: magic numbers and intruders
Order the split levels and fill them. 1s1/2 closes at 2; 1p3/2 then 1p1/2 close at 8; 1d5/2, 2s1/2 and 1d3/2 close at 20. Now the new physics. Because ΔE grows as 2l + 1, the j = l + 1/2 member of the highest-l orbital in each shell is driven out of that shell and down beside the one below. 1f7/2 goes first, holding 8 and closing 28. 1g9/2 follows, holding 10 and closing 50 — and it carries even parity into an otherwise odd-parity neighbourhood, which is why ⁹³Nb has a 9/2⁺ ground state. 1h11/2 gives 82 and 1i13/2 gives 126. Two book-keeping checks keep this honest. The occupancy-weighted centre of gravity never moves, since 2l + 2 states drop by Cl/2 while 2l states rise by C(l+1)/2 and the two products are equal. And each partner keeps the parity of its l, so an intruder is always the odd one out in its new company.
Spin, parity, and where the mean field gives out
Even-even nuclei are 0⁺ without exception, because pairing binds identical nucleons in time-reversed pairs coupled to zero. For odd A the single unpaired nucleon carries everything: J = j of its orbital and π = (−1)l. ¹⁷O comes out 5/2⁺, ⁴¹Ca 7/2⁻, ²⁰⁷Pb a 3p1/2 hole at 1/2⁻, ²⁰⁹Bi a 1h9/2 proton at 9/2⁻ — all measured, all right. Then look at ¹⁹F, where the model says 5/2⁺ and the ground state is 1/2⁺, or ²³Na, where it says 5/2⁺ and the answer is 3/2⁺. Both sit mid-shell, where a spherical mean field is simply the wrong starting point: the nucleus is deformed, the last nucleon moves in a non-spherical well, and l and j stop being good labels. Two signatures give it away. Quadrupole moments in the rare earths and actinides run about ten times the single-particle estimate, and those nuclei show rotational bands with E proportional to J(J+1) — collective motion the independent-particle model has no coordinate for. Near closed shells the mean field is excellent; away from them it needs the deformed Nilsson basis.
Change one variable at a time
Make the relationship visible.
Keep c negative, the nuclear sign, and drag l from 2 up to 6: the solid branch dives, the dashed one climbs, and by l = 5 the solid branch has sunk past the shell below, taking 2l + 2 states with it. Flip c positive, the atomic sign, and the order reverses - no intruder, no new gap.
SPIN-ORBIT SPLITTING5.00 MeV
j = l + 1/2 LEVEL-3.20 MeV
j = l - 1/2 LEVEL1.80 MeV
STATES IN j = l + 1/26
Live interpretationSPIN-ORBIT SPLITTING: 5.00 MeV. j = l + 1/2 LEVEL: −3.20 MeV. j = l - 1/2 LEVEL: 1.80 MeV. STATES IN j = l + 1/2: 6
Catch the common trap
Explain before calculating.
²⁰⁷Pb has Z = 82 and N = 125, one neutron short of the magic 126. Between 82 and 126 the neutron orbitals fill in the order 1h9/2, 2f7/2, 1i13/2, 2f5/2, 3p3/2, 3p1/2. What ground-state spin and parity does the shell model predict?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAssign the ground-state spin and parity of ⁴¹Ca (Z = 20, N = 21) and of ¹⁵N (Z = 7, N = 8), using the level order 1s1/2, 1p3/2, 1p1/2, 1d5/2, 2s1/2, 1d3/2, 1f7/2.
- Paired nucleons contribute nothing. In ⁴¹Ca the 20 protons and 20 of the neutrons pair to 0⁺, so only the odd neutron matters; in ¹⁵N the 8 neutrons pair off and the odd proton is the seventh.
- Count the ⁴¹Ca neutrons into the levels: 1s1/2 takes 2, 1p3/2 takes 4 (total 6), 1p1/2 takes 2 (8), 1d5/2 takes 6 (14), 2s1/2 takes 2 (16), 1d3/2 takes 4 (20). The twenty-first neutron opens 1f7/2.
- So ⁴¹Ca has J = 7/2, and f means l = 3, giving π = (−1)³ = −1: the prediction is 7/2⁻.
- For ¹⁵N count protons instead: 1s1/2 takes 2, 1p3/2 takes 4 (total 6), and the seventh proton opens 1p1/2. So J = 1/2, and p means l = 1, giving π = −1.
Answer⁴¹Ca is 7/2⁻ and ¹⁵N is 1/2⁻, and both match the measured ground states. Parity comes from l alone — never from the mass number, and never from j.
Medium¹⁷O is one neutron outside doubly magic ¹⁶O. Its ground state is 5/2⁺ and a 3/2⁺ level sits 5.08 MeV above it, both single-particle states of the 1d neutron orbital. Find the spin-orbit strength C at this mass, place each level relative to the unsplit 1d level, and check that the occupancy-weighted mean is unmoved.
- The 1d orbital has l = 2, so it splits into j = 5/2 and j = 3/2. From ⟨l⋅s⟩/ℏ² = ½[j(j+1) − l(l+1) − 3/4]: for j = 5/2 that is ½(8.75 − 6 − 0.75) = +1.0, and for j = 3/2 it is ½(3.75 − 6 − 0.75) = −1.5.
- Write the shift as E = −C⟨l⋅s⟩/ℏ² with C > 0, which is the inverted nuclear sign. Then E(1d5/2) = −C and E(1d3/2) = +1.5C, so the separation is 2.5C — exactly C(l + 1/2).
- Set 2.5C = 5.08 MeV, giving C = 2.03 MeV.
- The levels therefore sit at E(1d5/2) = −2.03 MeV and E(1d3/2) = +1.5 × 2.03 = +3.05 MeV about the unsplit 1d level.
- Weighted check, with 2j + 1 = 6 and 4: 6(−2.03) + 4(+3.05) = −12.19 + 12.19 = 0. The term redistributes binding rather than adding any.
AnswerC = 2.03 MeV; 1d5/2 lies 2.03 MeV below the unsplit 1d level and 1d3/2 lies 3.05 MeV above it, with the occupancy-weighted mean unmoved at zero.
HardAbove the gap at 28 the next orbitals are 2p3/2, 1f5/2 and 2p1/2, which close a subshell at 40; the 1g orbital belongs to the oscillator shell above. Taking the spin-orbit strength near A = 90 as C = 0.90 MeV, work out how far 1g9/2 is driven down, and find the nucleon number at which the next gap opens.
- Fill upward from 28: 2p3/2 holds 4 (to 32), 1f5/2 holds 6 (to 38), 2p1/2 holds 2 (to 40).
- The 1g orbital has l = 4, so the splitting is C(l + 1/2) = 0.90 × 4.5 = 4.05 MeV, with 1g9/2 at −Cl/2 = −1.80 MeV and 1g7/2 at +C(l + 1)/2 = +2.25 MeV about the unsplit 1g level.
- Check the weights: 10(−1.80) + 8(+2.25) = −18.0 + 18.0 = 0, so the level scheme conserves the centre of gravity.
- That 1.80 MeV drop sits on top of the Woods-Saxon flat bottom, which already pulls high-l orbitals below the rest of their oscillator shell. Together they carry 1g9/2 down beside 2p1/2, while 1g7/2 stays up with the shell it came from.
- 1g9/2 holds 2j + 1 = 10 states, taking the count from 40 to 50, so the gap above it is the magic number 50. Its l = 4 gives even parity in odd-parity company, which is why ⁹³Nb, with 41 protons, has a 9/2⁺ ground state.
AnswerThe 1g pair splits by 4.05 MeV, with 1g9/2 pushed 1.80 MeV below the unsplit 1g level; it brings 10 states down and closes a shell at 50 — a positive-parity orbital sitting among negative-parity ones.