University Physics V · Nuclear Physics · 14.7

Stability, Drip Lines & Decay Maps

Stability is not a property you look up; it is a sign you compute. This topic turns the mass table into a map — which channels a nuclide has open, where the last bound isotope sits on each side, and why "stable" is only a statement about how hard anyone has looked.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Nothing on a chart of nuclides is measured directly except mass. Stability is derived: collect the atomic masses into a vector and every decay channel becomes one fixed integer-coefficient contraction of it — Sₚ = B(Z, N) − B(Z−1, N) for proton emission, S₂n for two-neutron emission, Q_α for α decay, an isobaric mass difference for β. A decay-mode map is then the sign pattern of half a dozen such functionals evaluated at every (N, Z), and it inherits exactly the errors of the table beneath it.

Two structural facts shape that map. Pairing puts an odd-even sawtooth into Sₙ, so the neutron edge is located by S₂n = 0, where the pairing term cancels identically, not by Sₙ = 0. And the Coulomb term grows as Z²A^(−1/3) against a symmetry term growing only as (N−Z)²/A, so the valley floor bends steadily to N > Z and Q_α crosses zero near A = 150.

What the construction cannot give you is a rate: Q > 0 opens a channel, and the transmission through the barrier behind it varies by thirty orders of magnitude at fixed sign. "Stable" is therefore not a computed property but an experimental bound, and every one of them carries a date.

Simple definition: A nuclide is particle-bound when every one- and two-nucleon separation energy computed from the mass table is positive, and β-stable when no isobaric neighbour lies lower in mass; the drip lines are the loci where those separation energies pass through zero.
Example: ²⁴O has S₂n = +6.93 MeV and is the last bound oxygen; ²⁶O has Sₙ = +0.75 MeV yet S₂n = −0.02 MeV, so it is bound against emitting one neutron and unbound against emitting two.

Nucleon separation energiesSₚ = B(Z, N) − B(Z−1, N) · Sₙ = B(Z, N) − B(Z, N−1)

First differences of one measured surface: no model, no fit, and no rate.

B and S in MeV, read off the mass table. Sₚ = 0 defines the proton drip line.

Two-neutron separation energyS₂n = B(Z, N) − B(Z, N−2) = Sₙ(N) + Sₙ(N−1)

The operative neutron-side criterion, with no odd-even sawtooth left to misread.

N and N−1 have opposite parity, so the two pairing contributions cancel identically.

Alpha Q valueQ_α = [Mₐₜ(A, Z) − Mₐₜ(A−4, Z−2) − Mₐₜ(⁴He)]c²

Turns positive near A = 150, where the α's own 28.296 MeV beats 4 × (B/A).

Atomic masses in u, × 931.494 MeV/u; the Z electrons balance exactly, so no correction.

Beta Q values in atomic massesQ_β⁻ = [Mₐₜ(Z) − Mₐₜ(Z+1)]c² · Q_β⁺ = Q_EC − 2mₑ c²

A 1.022 MeV window exists in which electron capture runs and positron emission cannot.

2mₑ c² = 1.022 MeV. Q_EC = [Mₐₜ(Z) − Mₐₜ(Z−1)]c², using atomic not nuclear masses.

Β-stability line from the mass formulaZₘᵢₙ = (A/2)(1 + Δ/4a_A) ÷ [1 + (a_C/4a_A) A²⁄³]

The A²⁄³ in the denominator is the valley bending to N > Z as A grows.

Δ = (mₙ − m_H)c² = 0.782 MeV, a_C ≈ 0.70 MeV, a_A ≈ 23.2 MeV; Zₘᵢₙ is dimensionless.

What a null counting experiment provesT₁/2 > ln2 · ε N t / nᵤₚ

One mole for one year with no counts buys only T₁/2 > 1.8 × 10²³ yr. That is all "stable" asserts.

N nuclei watched for time t at efficiency ε; nᵤₚ = 2.3 is the 90 %-CL limit on zero counts.

Q is a linear functional on the mass vector

Load the atomic masses into a vector m and every question about a decay channel becomes one fixed integer-coefficient contraction of it: Q_β⁻ = m(Z, A) − m(Z+1, A); Q_α = m(A, Z) − m(A−4, Z−2) − m(⁴He); Sₙ = m(Z, N−1) + mₙ − m(Z, N). No dynamics enters — nothing in these expressions knows about barriers, matrix elements or phase space. A decay-mode map on the (N, Z) plane is therefore the sign pattern of a handful of such functionals, and in NumPy the whole map is one integer matrix times one mass vector. Two consequences follow. Uncertainties propagate linearly, so a nuclide whose Q comes out at 30 ± 40 keV has no determined mode, only a determined ignorance. And the AME flags extrapolated masses with a #, so whole regions of any published map rest on a model of the mass surface rather than on a measurement of it — which is exactly where new facilities keep moving the lines.

First differences of B, with a pairing sawtooth

Sₙ(Z, N) = B(Z, N) − B(Z, N−1) is the discrete derivative of the binding surface along N, and the pairing term makes it alternate. Add a neutron to an odd-N core and it pairs with the unpaired one, so Sₙ is large; add it to an even-N core and it stays unpaired, so Sₙ is small. The full swing is 2Δ with Δ ≈ 12/√A MeV — about 2.4 MeV at A = 24, about 0.8 MeV at A = 208. The measured oxygen chain shows it plainly: Sₙ = 6.85 MeV at ²²O (N = 14), 2.74 MeV at ²³O (N = 15), 4.19 MeV at ²⁴O (N = 16). The sequence is not monotonic, and the 1.45 MeV rise from ²³O to ²⁴O is pairing, not extra resistance to the downward trend. Any drip-line criterion built on "the first N where Sₙ stops falling" is reading the parity of N, not the edge of binding.

The two drip lines, and why they are not mirror images

On the neutron side the criterion is S₂n = 0, because S₂n = B(N) − B(N−2) compares two nuclides of the same neutron parity and the pairing term drops out exactly. Beyond that line the pair cannot be held at all: an l = 0 neutron sees no barrier, the state is a resonance of width Γ of order 1 MeV, and ħ/Γ = 6.58 × 10⁻²² MeV s ÷ 1 MeV ≈ 10⁻²¹ s — gone before it is a nucleus. The proton side behaves completely differently, because the emitted proton must tunnel through a Coulomb-plus-centrifugal barrier of order 10 MeV. Sₚ can go a full MeV negative and the nuclide still lives long enough to be identified: ¹⁵¹Lu emits a 1.23 MeV proton from its ground state with T½ = 80.6 ms. The edges also sit at wildly different distances. For tin the proton drip line lies within a couple of nucleons of the doubly magic ¹⁰⁰Sn, and the proton line is mapped across most of the chart; the tin neutron drip line lies far beyond the heaviest tin yet made, and experimentally the neutron line is established only as far as neon, Z = 10.

Why the valley bends to N > Z

At fixed A the mass formula is an exact quadratic in Z, so setting dM/dZ = 0 gives Zₘᵢₙ = (A/2)(1 + Δ/4a_A) ÷ [1 + (a_C/4a_A)A²⁄³] with Δ = (mₙ − m_H)c² = 0.782 MeV. Read the competition term by term instead. Pushing one more unit of charge in costs 2a_C Z A^(−1/3) of Coulomb energy — 19.50 MeV at ²⁰⁹Bi — and returns 4a_A(N−Z)/A from the symmetry term, 19.09 MeV there. They balance only once N − Z has grown to 43, i.e. N/Z = 1.52. Since the Coulomb slope grows as Z A^(−1/3) while the symmetry restoring slope is bounded by 4a_A, the required excess N − Z keeps growing with A, and the valley floor pulls away from the N = Z line. The formula delivers 83.34 at A = 209 (²⁰⁹Bi observed) and 93.14 at A = 238, which sits neatly between the two β-stable isobars ²³⁸U and ²³⁸Pu.

Why Q_α turns positive above A ≈ 150

Write B = A b(A) with b the binding per nucleon. Then Q_α = B(A−4) + 28.296 − B(A) ≈ 28.296 − 4[b(A) + A b′(A)], where 28.296 MeV is the α particle's own binding. Past the ⁶²Ni peak b′ is negative, so the −4A b′ piece is a positive contribution that grows with A. Take b falling linearly from 8.60 MeV at A = 100 to 7.57 MeV at A = 238, giving b′ = −0.00746 MeV: Q_α crosses zero where b + A b′ = 7.074 MeV, at A = 152. Below that the α is not worth the four nucleons it removes; above it, it is. The observed onset is earlier — ¹⁴⁴Nd with Q_α = 1.90 MeV and ¹⁴⁷Sm with 2.31 MeV — because ¹⁴⁴Nd's daughter ¹⁴⁰Ce closes the N = 82 shell, a several-MeV bonus a smooth curve has no way to carry, and ¹⁴⁷Sm sits in the same shell-driven island.

Permission is not a rate

²⁰⁹Bi has Q_α = +3.137 MeV; ²¹²Po has Q_α = +8.954 MeV. The Q values differ by a factor 2.85, and the half-lives by 2.0 × 10¹⁹ yr against 0.29 µs — 33.3 decades. The Geiger-Nuttall exponent, which goes as Z_d/√Q, accounts for about 30 of them; a sign test on the mass table accounts for none. So a decay-mode map states which channels are open, and ranking their rates is a different calculation entirely. This is also what "stable" means operationally. Watch one mole for one year, see nothing, and at 90 % confidence you have shown only T½ > ln2 × 6.02 × 10²³ ÷ 2.3 = 1.8 × 10²³ yr. ²⁰⁹Bi was printed as stable for a century with a positive Q_α in plain sight, until a bolometric measurement in 2003 counted its α decays; ¹²⁸Te's double beta decay at 2.25 × 10²⁴ yr sits near the practical edge of the technique.

Change one variable at a time

Make the relationship visible.

Interactive model

Trend slope s0.8 MeV/neutron

Pairing gap Δ1.4 MeV

Cursor neutron number N20

Keep the slope at 0.8 and step the cursor outward from N = 18: Sₙ dips below zero at 19, climbs back above it at 20, then drops again — while the dashed S₂n/2 crosses zero once. N = 20 is bound against one neutron and unbound against two, the ²⁶O case.

Sₙ AT CURSOR0.80 MeV

S₂n AT CURSOR-0.40 MeV

LAST 2n-BOUND N19

ODD-EVEN JUMP 2Δ − s2.00 MeV

Live interpretationSₙ AT CURSOR: 0.80 MeV. S₂n AT CURSOR: −0.40 MeV. LAST 2n-BOUND N: 19. ODD-EVEN JUMP 2Δ − s: 2.00 MeV

Catch the common trap

Explain before calculating.

Misconception check

The drip line is not where nuclides start being radioactive. Almost everything between the valley floor and the drip line is radioactive — much of it β-decays in milliseconds — and is still perfectly particle-bound, with Sₚ, Sₙ and S₂n all comfortably positive. The drip line marks something much narrower: the point where the mass surface can no longer hold one more nucleon, so the extra nucleon leaves in about 10⁻²¹ s and there is no nucleus left to name. The mirror error is to read the black squares on a chart of nuclides as a computed property. Stable is an observational claim with a date on it: ²⁰⁹Bi was black for a century with Q_α = +3.14 MeV sitting in plain view in the mass table, until a bolometer counted long enough to return 2.0 × 10¹⁹ yr. Q > 0 opens a channel; it says nothing about whether anything comes through it.

²⁰⁹Bi has Q_α = +3.14 MeV and was printed as a stable nuclide until a 2003 measurement returned T½ = 2.0 × 10¹⁹ yr. ²¹²Po has Q_α = +8.95 MeV and T½ = 0.29 µs. Taken together, what do these two nuclides establish?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUsing atomic masses Mₐₜ(¹⁵O) = 15.003066 u, Mₐₜ(¹⁵N) = 15.000109 u, Mₐₜ(¹⁶O) = 15.994915 u, Mₐₜ(¹H) = 1.007825 u and mₙ = 1.008665 u, find Sₙ and Sₚ for ¹⁶O and say how far it sits from either drip line.

Sₙ = [Mₐₜ(¹⁵O) + mₙ − Mₐₜ(¹⁶O)]c². Both sides carry 8 atomic electrons, so they cancel with no correction.
15.003066 + 1.008665 − 15.994915 = 0.016816 u, and 0.016816 × 931.494 MeV/u = 15.664 MeV.
Sₚ = [Mₐₜ(¹⁵N) + Mₐₜ(¹H) − Mₐₜ(¹⁶O)]c². Use ¹H, not the bare proton, so the electron count balances: 7 + 1 = 8.
15.000109 + 1.007825 − 15.994915 = 0.013019 u, and 0.013019 × 931.494 MeV/u = 12.127 MeV.
Both are strongly positive, so ¹⁶O is far inside both edges: oxygen's proton drip line is at ¹³O (five neutrons lighter) and its neutron drip line at ²⁴O (eight neutrons heavier).

AnswerSₙ = 15.66 MeV and Sₚ = 12.13 MeV, both far above zero, so ¹⁶O sits nowhere near either edge — the oxygen drip lines are at ¹³O and ²⁴O.

MediumMeasured neutron separation energies along the oxygen chain are Sₙ = 6.85 MeV at ²²O (N = 14), 2.74 MeV at ²³O, 4.19 MeV at ²⁴O and −0.77 MeV at ²⁵O, and ²⁶O is measured 18 keV above the ²⁴O + 2n threshold. Locate the neutron drip line and show why Sₙ = 0 is the wrong test.

S₂n(N) = B(N) − B(N−2) = Sₙ(N) + Sₙ(N−1): the first differences telescope, and because N and N−1 have opposite parity the two pairing contributions enter with opposite signs and cancel.
Sₙ itself is not monotonic — it rises 1.45 MeV from ²³O to ²⁴O — because N = 16 is even and the added neutron pairs. That rise is parity, not stability.
Telescoping: S₂n = 2.74 + 6.85 = 9.59 MeV at ²³O, 4.19 + 2.74 = 6.93 MeV at ²⁴O, −0.77 + 4.19 = 3.42 MeV at ²⁵O. Falling by about 3 MeV a step, with no alternation left.
²⁶O sits 18 keV above threshold, so S₂n(²⁶O) = −0.02 MeV, and Sₙ(²⁶O) = S₂n(²⁶O) − Sₙ(²⁵O) = −0.02 + 0.77 = +0.75 MeV.
So ²⁶O is bound against emitting one neutron and unbound against emitting two — exactly the nuclide an Sₙ = 0 test would wrongly call bound. The last particle-bound oxygen is ²⁴O, at the N = 16 sub-shell closure.
The edge does not continue smoothly: add one proton and ³¹F (Z = 9, N = 22) is bound, six neutrons beyond the oxygen edge.

AnswerS₂n = 9.59, 6.93, 3.42 and −0.02 MeV at ²³O to ²⁶O. The last bound oxygen is ²⁴O; ²⁶O, with Sₙ = +0.75 MeV but S₂n = −0.02 MeV, is precisely the case an Sₙ = 0 test gets wrong.

HardWith a_C = 0.697 MeV, a_A = 23.2 MeV and Δ = (mₙ − m_H)c² = 0.782 MeV, locate the β-stability vertex at A = 63, say which isobar the formula predicts to be stable, then invert the measured Q_β⁻(⁶³Ni) = +67 keV to find where the vertex actually sits.

At fixed A the only Z-dependent terms are Z(m_H − mₙ)c², a_C Z²A^(−1/3) and a_A(A−2Z)²/A, so M(Z)c² is exactly quadratic in Z within one pairing parity. Setting dM/dZ = 0 gives Z₀ = (A/2)(1 + Δ/4a_A) ÷ [1 + (a_C/4a_A)A²⁄³].
Numbers: 63¹⁄³ = 3.9791 and 63²⁄³ = 15.833. a_C/4a_A = 0.697/92.8 = 0.007511, so the denominator is 1.11891; Δ/4a_A = 0.782/92.8 = 0.008427, so the numerator is 31.5 × 1.008427 = 31.765.
Z₀ = 31.765 ÷ 1.11891 = 28.39. The nearest integer is 28, so the formula predicts ⁶³Ni as the stable A = 63 isobar. Nature says ⁶³Cu, and ⁶³Ni β-decays with T½ = 101 yr.
Curvature: d²(Mc²)/dZ² = 2a_C A^(−1/3) + 8a_A/A = 0.350 + 2.946 = 3.296 MeV per unit charge squared. The parabola is shallow, so what decides the answer is the vertex position, not the depth.
Invert the measurement. For a parabola of curvature k, M(28) − M(29) = k(Z₀ − 28.5). With Q_β⁻ = +0.067 MeV, Z₀ = 28.5 + 0.067/3.296 = 28.52.
The fit misses the vertex by 28.52 − 28.39 = 0.13 charge units — invisible on a plot, and just enough to put it on the wrong side of the Ni/Cu midpoint at 28.5 and hand back the wrong stable isobar.

AnswerThe formula puts the vertex at Z₀ = 28.39 and predicts ⁶³Ni stable; the measured Q_β⁻(⁶³Ni) = +67 keV with curvature 3.30 MeV puts it at 28.52, making ⁶³Cu stable. A 0.13-unit error flips the answer.