University Physics V · Nuclear Physics · 14.6
Isobaric Parabolas & Beta Stability
Beta decay changes Z but never A, so a whole decay chain lives on one slice of the mass surface. This is how you cut that slice, read the Q of every step straight off a parabola, and see why pairing lets two even-A isobars survive where only one odd-A isobar can.
Build the model
Connect the measurement to the mechanism.
Beta decay moves a nucleus one step in Z and leaves A untouched, so an entire beta chain lives on a single line of the nuclide chart. Take the semi-empirical mass formula, hold A fixed, and that line becomes a curve you can write down: expanding the asymmetry term as (A − 2Z)²/A = A − 4Z + 4Z²/A leaves the atomic mass as a plain quadratic, M(Z)c² = α − βZ + γZ² ± δ, with γ = aC A(−1/3) + 4aA/A the only curvature and δ = aP A(−1/2) the pairing offset. Everything the model can say about beta stability now follows from three numbers.
The vertex Zₘᵢₙ = β/2γ sits below A/2 because the Coulomb term grows as A²⁄³ while the symmetry cost per nucleon does not, which is exactly why the valley bends to N > Z. The curvature turns each decay step into Q = 2γ(Zₘᵢₙ − Z − ½), a Q that falls by 2γ every rung until it changes sign — that, and nothing else, is what stops a chain. And for even A the ±δ splits one parabola into two, the odd-odd branch raised 2δ above the even-even branch, which is why an even-A slice can hold two or three nuclides that no single beta decay can reach.
What it costs is smoothness: fitted to a liquid drop, the parabola knows nothing of closed shells, so it places Zₘᵢₙ to about a unit in mid-mass and can be several MeV wrong beside a magic number.
- Simple definition
- An isobaric mass parabola is the semi-empirical mass formula evaluated at fixed mass number A, where it reduces to a quadratic in proton number Z, split into two offset branches by the pairing term whenever A is even.
- Example
- At A = 100 the fitted coefficients give curvature γ = 1.101 MeV and a minimum at Zₘᵢₙ = 43.40, so the nearest integer is Tc (Z = 43) — odd-odd, and therefore lifted 2δ = 2.24 MeV onto the upper branch, above both of its neighbours.
One curve per isobar. Every beta Q value on that isobar is a difference of two points on it.
α, β, γ, δ in MeV; sign + for odd-odd, − for even-even, and δ = 0 for odd A
γ is the sole curvature — Coulomb repulsion plus symmetry energy — and it alone sets how fast Q falls along a chain.
(mₙ − mH)c² = 0.782 MeV; with aC = 0.711 and aA = 23.7 MeV, γ = 1.101 MeV at A = 100
That A²⁄³ is why the valley bends to N > Z: 43.0 at A = 100 and 82.7 at A = 209, against A/2 = 50 and 104.5.
0.782 MeV term dropped here; aC/4aA = 0.0075, so the bracket is 1.16 at A = 100 and 1.26 at A = 209
Capture stays open across the 1.022 MeV window where positron emission is shut — the usual route just off stability.
Atomic masses throughout; 2mec² = 1.022 MeV; QEC is the same bracket with nothing subtracted
Q drops by exactly 2γ each rung and changes sign at Zₘᵢₙ − ½, not at Zₘᵢₙ. At A = 101, 2γ = 2.18 MeV.
Odd A only, where δ = 0. γ is in MeV per unit Z².
The entire reason an even-A slice can hold two or three β-stable nuclides instead of one.
aP ≈ 11.2 MeV, giving 2δ = 2.24 MeV at A = 100 and 1.77 MeV at A = 160
Fixing A turns the mass formula into a quadratic in Z
Write the atomic mass as M(Z, A)c² = Z m(¹H)c² + (A − Z)mₙ c² − B(Z, A) and put in the semi-empirical B = aV A − aS A²⁄³ − aC Z² A(−1/3) − aA (A − 2Z)²/A ± δ. At fixed A the volume and surface terms are constants. The asymmetry term is the only awkward one, and it expands cleanly: (A − 2Z)²/A = A − 4Z + 4Z²/A. Collecting powers of Z gives M(Z)c² = α − βZ + γZ² ± δ, with β = (mₙ − mH)c² + 4aA, γ = aC A(−1/3) + 4aA/A, and α holding every term with no Z in it. Two things deserve notice. The quadratic is exact within the model — nothing was expanded about Zₘᵢₙ, so it holds as far out on the slice as the mass formula itself does. And γ has two competing pieces: Coulomb repulsion that grows with Z, and symmetry energy that punishes any departure from N = Z. Using the aC Z(Z − 1)A(−1/3) form of the Coulomb term instead leaves γ untouched and adds aC A(−1/3) to β, moving Zₘᵢₙ by 0.07 at A = 100.
Where the minimum sits, and why it is not A/2
Set dM/dZ = 0 at fixed A: −β + 2γZ = 0, so Zₘᵢₙ = β/2γ. Divide top and bottom by 4aA and the structure appears: Zₘᵢₙ = (A/2)[1 + (mₙ − mH)c²/4aA] / [1 + (aC/4aA)A²⁄³]. With aC = 0.711 and aA = 23.7 MeV, aC/4aA = 0.0075, so the denominator is 1.163 at A = 101 and 1.259 at A = 209. That single A²⁄³ is the whole reason the valley of stability bends away from N = Z: the Coulomb cost of packing protons grows with the size of the nucleus, while the symmetry cost per nucleon does not. The numerator correction is the neutron-hydrogen mass difference, (mₙ − mH)c² = 0.782 MeV, worth a factor 1.0082. It shifts Zₘᵢₙ by 0.36 at A = 101 — well under one unit, as promised, but there it straddles a half-integer and moves the nearest integer from 43 to 44, the difference between Tc-101 (14.2 min) and stable Ru-101. At A = 209 the formula returns 83.35 against A/2 = 104.5, and Bi-209 has Z = 83. Zₘᵢₙ is a real number; no nuclide sits exactly on it.
Reading decay energies off the parabola
The parabola is built from atomic masses, so β− and electron-capture Q values are plain vertical differences on it: Qβ− = [M(Z) − M(Z+1)]c², and QEC = [M(Z) − M(Z−1)]c² up to the keV-scale binding of the captured electron. Positron emission is the exception. Two electron masses go unaccounted for in the atomic bookkeeping, so Qβ+ = [M(Z) − M(Z−1)]c² − 2mec², with 2mec² = 1.022 MeV — and there is a 1.022 MeV window in which capture proceeds and β+ cannot, which is why proton-rich nuclides close to stability decay by capture alone. For odd A the pairing term vanishes and the step is exact: Qβ−(Z → Z+1) = γ[(Z − Zₘᵢₙ)² − (Z + 1 − Zₘᵢₙ)²] = 2γ(Zₘᵢₙ − Z − ½). Q therefore falls by exactly 2γ per rung. At A = 101, where γ = 1.091 MeV and Zₘᵢₙ = 43.79, the chain gives 5.01, 2.82 and 0.64 MeV at Z = 41, 42 and 43, then −1.54 MeV at Z = 44. The chain stops at Ru-101, and the sign change sits at Zₘᵢₙ − ½, not at Zₘᵢₙ.
Pairing cuts an even-A slice into two parabolas
For odd A one of Z and N is even and the other odd, the pairing term is zero, and there is one parabola. For even A both are even or both odd, and the term is ∓δ with δ = aP A(−1/2): even-even nuclides drop by δ, odd-odd rise by δ, and the two branches stay 2δ = 2aP A(−1/2) apart at every Z — 2.24 MeV at A = 100, 1.77 MeV at A = 160. Successive Z along an even-A isobar therefore alternate branches, and the mass sequence zigzags rather than descending smoothly to a single floor. Two consequences do all the work. An odd-odd nuclide near the vertex can lie above both of its even-even neighbours, so β− and capture are both open and it decays in both directions; Cu-64 does exactly this, 39% by β− to Zn-64 and 61% by capture and β+ to Ni-64. And an even-even nuclide two steps from the vertex can lie below the odd-odd nuclide beside it while still lying above the even-even nuclide beyond that — β-stable, but not the most bound isobar on its own slice.
A = 100, and why only double β decay is left
Take A = 100 with aC = 0.711, aA = 23.7 and aP = 11.2 MeV. Then γ = 1.101 MeV, δ = 1.120 MeV and Zₘᵢₙ = 43.40. Measuring mass upward from the unpaired vertex, E(Z) = γ(Z − Zₘᵢₙ)² ± δ puts Mo-100 (Z = 42, even-even) at +1.04 MeV, Tc-100 (Z = 43, odd-odd) at +1.30 MeV and Ru-100 (Z = 44, even-even) at −0.72 MeV. Read the decays off. Tc-100 tops both neighbours, so it goes both ways: the model gives Qβ− = 2.02 MeV and QEC = 0.26 MeV, against measured 3.20 and 0.17 MeV. Mo-100 cannot reach Tc-100 at all, since Qβ− = 1.04 − 1.30 = −0.26 MeV; the measured value is −0.17 MeV. Single beta decay is shut, and the only open channel is the second-order Mo-100 → Ru-100 + 2e⁻ + 2ν̄, which skips the odd-odd rung as a virtual intermediate rather than populating it. The model puts that Q at 1.76 MeV against a measured Qββ of 3.03 MeV — and Mo-100 is a headline double-β source for precisely this reason.
Where a smooth five-parameter model has to fail
The parabola is a smooth function of Z fitted to a liquid drop, so it cannot know about closed shells, and beside one it fails loudly rather than gently. At A = 136 the formula gives γ = 0.835 MeV and Zₘᵢₙ = 57.2, which leaves Xe-136 (Z = 54) 3.2 units up the even-even branch and predicts Qβ− = +2.6 MeV to Cs-136. The measured value is −0.09 MeV: Xe-136 is β-stable, and the model misses that by nearly 3 MeV, because N = 82 is closed and Xe-136 is far more bound than any smooth surface allows. Residuals of that size stacked at N or Z = 28, 50, 82 and 126 are the fingerprint of shell structure, not fit noise. Two further limits are worth saying out loud. The parabola gives energetics only: a positive Q permits a decay and fixes no rate, which needs matrix elements and phase space. And the whole construction is about ground states — an isomer sits on a rung of its own, with its own Q.
Change one variable at a time
Make the relationship visible.
Drag the pairing coefficient to zero: the two branches merge into the single odd-A parabola with one stable isobar. Restore 11.2 MeV and the split reopens — at A = 100 both open markers drop below the filled one, so the odd-odd nuclide decays both ways while its two neighbours are stable.
MINIMUM Zₘᵢₙ43.40 protons
CURVATURE γ1.101 MeV
PAIRING SPLIT 2δ2.24 MeV
A/2 − Zₘᵢₙ6.60 protons
Live interpretationMINIMUM Zₘᵢₙ: 43.40 protons. CURVATURE γ: 1.101 MeV. PAIRING SPLIT 2δ: 2.24 MeV. A/2 − Zₘᵢₙ: 6.60 protons
Catch the common trap
Explain before calculating.
An A = 100 isobaric parabola fitted with aC = 0.711 MeV, aA = 23.7 MeV and aP = 11.2 MeV has curvature γ = 1.101 MeV, its minimum at Zₘᵢₙ = 43.40, and a pairing split 2δ = 2.24 MeV. Which nuclides on this slice are stable against single beta decay?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor A = 101 use aC = 0.711 MeV, aA = 23.7 MeV and (mₙ − mH)c² = 0.782 MeV to locate the minimum of the isobaric mass parabola, first with the 0.782 MeV term dropped and then with it kept. Which integer Z does each version pick, and which nuclide is actually the stable isobar?
- Curvature first: γ = aC A(−1/3) + 4aA/A. With 101¹⁄³ = 4.6570, γ = 0.711/4.6570 + 94.8/101 = 0.1527 + 0.9386 = 1.0913 MeV.
- Dropped form: Zₘᵢₙ = (A/2)/[1 + (aC/4aA)A²⁄³]. Here aC/4aA = 0.711/94.8 = 0.00750 and A²⁄³ = 4.6570² = 21.688, so the bracket is 1 + 0.1627 = 1.1627 and Zₘᵢₙ = 50.5/1.1627 = 43.43.
- Full form: Zₘᵢₙ = β/2γ with β = (mₙ − mH)c² + 4aA = 0.782 + 94.8 = 95.582 MeV, so Zₘᵢₙ = 95.582/(2 × 1.0913) = 95.582/2.1826 = 43.79.
- The two differ by 0.36, well under one unit — but they straddle 43.5, so the nearest integer moves from 43 to 44.
- A = 101 is odd, so there is a single parabola and a single stable isobar. Ru-101 (Z = 44) is stable; Tc-101 (Z = 43) β− decays with a 14.2 min half-life. Keeping the 0.782 MeV term picks the right one.
AnswerZₘᵢₙ = 43.43 without the 0.782 MeV term and 43.79 with it; only the full value rounds to Z = 44, which is stable Ru-101.
MediumFor A = 64 with aC = 0.711, aA = 23.7 and aP = 11.2 MeV, locate Zₘᵢₙ, place Ni-64, Cu-64 and Zn-64 on the split parabola, and find the Q value of every beta channel open to Cu-64. Compare with the measured 0.579 MeV for β− and 1.675 MeV for capture.
- 64¹⁄³ = 4.000 exactly, so γ = 0.711/4 + 94.8/64 = 0.1778 + 1.4813 = 1.6590 MeV, and δ = aP A(−1/2) = 11.2/8 = 1.400 MeV, a branch separation of 2δ = 2.80 MeV.
- Zₘᵢₙ = (0.782 + 94.8)/(2 × 1.6590) = 95.582/3.3180 = 28.807, so the nearest integer is 29 — copper, and odd-odd.
- Measure mass upward from the unpaired vertex with E(Z) = γ(Z − Zₘᵢₙ)² ± δ, taking −δ for even-even and +δ for odd-odd. Ni-64: 1.6590(−0.807)² − 1.400 = 1.081 − 1.400 = −0.319 MeV. Cu-64: 1.6590(0.193)² + 1.400 = 0.062 + 1.400 = +1.462 MeV. Zn-64: 1.6590(1.193)² − 1.400 = 2.361 − 1.400 = +0.961 MeV.
- Cu-64 lies above both neighbours, so both directions are open. Qβ− = 1.462 − 0.961 = 0.501 MeV. The atomic-mass difference to Ni-64 is 1.462 − (−0.319) = 1.781 MeV, which is QEC directly; subtract 2mec² = 1.022 MeV for the positron channel, giving Qβ+ = 0.759 MeV.
- Against measurement — 0.579, 1.675 and 0.653 MeV — the five-parameter fit is good to about 0.1 MeV in every channel, and it reproduces the qualitative point: Cu-64 really does split 39% β− against 61% capture plus β+.
AnswerZₘᵢₙ = 28.81; Cu-64 sits 0.501 MeV above Zn-64 and 1.781 MeV above Ni-64 in atomic mass, so Qβ− = 0.50 MeV, QEC = 1.78 MeV and Qβ+ = 0.76 MeV — all three channels open.
HardFor A = 100 with the same coefficients, show that Mo-100 cannot β decay to Tc-100, find the Q of the double β decay to Ru-100, and compare both with the measured −0.17 MeV and 3.03 MeV. Say where the model's error comes from.
- 100¹⁄³ = 4.6416, so γ = 0.711/4.6416 + 94.8/100 = 0.1532 + 0.9480 = 1.1012 MeV; δ = 11.2/10 = 1.120 MeV; Zₘᵢₙ = 95.582/(2 × 1.1012) = 43.400.
- E(Z) = γ(Z − Zₘᵢₙ)² ± δ. Mo-100 (Z = 42, even-even): 1.1012(−1.400)² − 1.120 = 2.158 − 1.120 = +1.038 MeV. Tc-100 (Z = 43, odd-odd): 1.1012(−0.400)² + 1.120 = 0.176 + 1.120 = +1.296 MeV. Ru-100 (Z = 44, even-even): 1.1012(0.600)² − 1.120 = 0.396 − 1.120 = −0.724 MeV.
- Single step: Qβ− = 1.038 − 1.296 = −0.258 MeV. Negative, so Mo-100 → Tc-100 is forbidden; the measured value is −0.17 MeV, the same sign and the same order.
- Two steps at once: Qββ = 1.038 − (−0.724) = 1.762 MeV, positive, so the second-order process is open although every single step is not. Tc-100 need not be reachable — it enters as a virtual intermediate, never as a populated state.
- The measured Qββ is 3.03 MeV, so the model is 1.27 MeV low, a 42% error, while the single-step Q was good to 0.09 MeV. Differences between neighbours cancel most of a smooth model's error; a two-step difference does not, and A near 100 sits in the Zr–Mo region where deformation sets in and the liquid drop is furthest from the truth.
AnswerQβ−(Mo-100 → Tc-100) = −0.26 MeV, so single beta decay is closed; Qββ(Mo-100 → Ru-100) = +1.76 MeV against a measured 3.03 MeV, the shortfall being deformation and shell structure the liquid drop omits.