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Derivation

The Valley of Beta Stability

Statement

For a fixed mass number \(A\), the nuclear mass \(M(Z,A)\) predicted by the semi-empirical mass formula (SEMF) is a quadratic function of the proton number \(Z\). Minimising this parabola at constant \(A\) gives the most stable charge, \(Z_*(A) \simeq A\big/\!\left(2 + \tfrac{a_C}{2a_A}A^{2/3}\right)\); the locus of these minima traced over all \(A\) is the valley (line) of beta stability, and the residual mass excess of a neighbouring isobar is \(M(Z) c^2 = M(Z_*)c^2 + \gamma\,(Z-Z_*)^2\) with \(\gamma = a_C A^{-1/3} + 4a_A A^{-1}\).

Why it matters

The single formula \(Z_*(A)\) explains, with one line of algebra, why light nuclei sit near \(N=Z\) while heavy nuclei carry a large neutron excess, and why every unstable isobar beta-decays in a definite direction — toward the bottom of its mass parabola. It is the quantitative backbone of the chart of nuclides.

The parabolic shape also predicts the energetics of beta decay: the curvature \(\gamma\) sets the \(Q\)-value scale, and for even \(A\) the pairing term splits the single parabola into two, immediately accounting for the existence of stable even–even isobar pairs and for double beta decay.

Assumptions
The binding energy is given by the SEMF (liquid-drop) with smooth \(A\)-dependence.Drop it and the mass surface acquires shell corrections at magic \(N,Z\); the smooth parabola is then only an average and \(Z_*\) can be off by one to two units.
The Coulomb term is that of a uniformly charged sphere, \(a_C Z^2 A^{-1/3}\), with \(Z(Z-1)\to Z^2\).Restore \(Z(Z-1)\) and an exchange correction \(\propto Z^{4/3}\) and both \(\gamma\) and the linear term shift slightly, moving \(Z_*\) by a fraction of a unit.
The asymmetry term is \(a_A (A-2Z)^2/A\), quadratic in the neutron excess.This quadratic form is what makes \(M(Z)\) a parabola at all; a non-quadratic Fermi-gas correction would bend the isobar and displace the minimum.
\(Z\) may be treated as a continuous variable when differentiating.\(Z\) is an integer; the continuous minimum \(Z_*\) must be rounded to the nearest physically allowed (and, for even \(A\), correct-parity) integer, so \(Z_*\) is a guide, not a literal count.
The neutron–proton mass difference \((m_n-m_p)c^2\approx1.29\,\text{MeV}\) is negligible beside \(4a_A\approx93\,\text{MeV}\).Keep it and the numerator of \(Z_*\) gains a term \((m_n-m_p)c^2\); this shifts \(Z_*\) down by \(\sim0.007A\), a real but small neutron-favouring correction.
Derivation
1
\[ M(Z,A)c^2 = Z\,m_p c^2 + (A-Z)\,m_n c^2 - B(Z,A) \]
Definition of the nuclear mass as constituent masses minus binding energy. A
2
\[ B(Z,A) = a_V A - a_S A^{2/3} - a_C\frac{Z^2}{A^{1/3}} - a_A\frac{(A-2Z)^2}{A} + \delta(A,Z) \]
Insert the assumed SEMF; volume, surface, Coulomb, asymmetry and pairing terms. A
3
\[ M(Z)c^2 = \Big[a_C A^{-1/3} + 4a_A A^{-1}\Big]Z^2 - \Big[(m_n-m_p)c^2 + 4a_A\Big]Z + \alpha(A) \]
Substitute \(N=A-Z\), expand \((A-2Z)^2\), and collect powers of \(Z\) at fixed \(A\); all \(Z\)-independent pieces are absorbed into \(\alpha(A)\). This exposes the parabola. B
4
\[ M(Z)c^2 \equiv \gamma Z^2 - \beta Z + \alpha,\qquad \gamma = \frac{a_C}{A^{1/3}} + \frac{4a_A}{A},\quad \beta = (m_n-m_p)c^2 + 4a_A \]
Name the coefficients. \(\gamma>0\) guarantees a minimum (upward parabola). A
5
\[ \left.\frac{\partial (M c^2)}{\partial Z}\right|_A = 2\gamma Z - \beta = 0 \quad\Longrightarrow\quad Z_* = \frac{\beta}{2\gamma} \]
Stationarity at fixed \(A\); treating \(Z\) as continuous (assumption). Since \(\partial^2/\partial Z^2 = 2\gamma>0\), this is the minimum. B
6
\[ Z_* = \frac{(m_n-m_p)c^2 + 4a_A}{\,2a_C A^{-1/3} + 8a_A A^{-1}\,} = \frac{A\big[(m_n-m_p)c^2 + 4a_A\big]}{2a_C A^{2/3} + 8a_A} \]
Substitute \(\beta,\gamma\) and multiply numerator and denominator by \(A\) to clear the fractional powers. Exact within the SEMF. B
7
\[ \boxed{\,Z_* \simeq \frac{A}{\,2 + \dfrac{a_C}{2a_A}A^{2/3}\,}\,} \]
Neglect \((m_n-m_p)c^2\ll 4a_A\) (assumption) so \(\beta\to 4a_A\), then divide top and bottom of step 6 by \(4a_A\). A
8
\[ M(Z)c^2 = M(Z_*)c^2 + \gamma\,(Z-Z_*)^2 \]
Complete the square in step 4: \(\gamma Z^2-\beta Z+\alpha = \gamma(Z-\beta/2\gamma)^2 + (\alpha-\beta^2/4\gamma)\). The isobaric mass parabola with curvature \(\gamma\). B
Result
\[ Z_*(A) \simeq \frac{A}{\,2 + \dfrac{a_C}{2a_A}A^{2/3}\,},\qquad M(Z)c^2 = M(Z_*)c^2 + \gamma\,(Z-Z_*)^2,\quad \gamma = \frac{a_C}{A^{1/3}} + \frac{4a_A}{A} \]

Reading. At fixed \(A\) the isobars lie on a parabola in \(Z\); the vertex \(Z_*\) is the most stable charge and the valley of stability is the curve \(Z_*(A)\). Because \(a_C/2a_A>0\), the denominator exceeds \(2\), so \(Z_*<A/2\): heavier nuclei are pushed to a neutron excess by the Coulomb energy, which the asymmetry term can only partly resist. Any isobar with \(Z<Z_*\) lowers its mass by \(\beta^-\) decay (\(n\to p\), \(Z\to Z+1\)); any with \(Z>Z_*\) does so by \(\beta^+\)/electron capture (\(Z\to Z-1\)).

Units check. \(a_C, a_A\) are energies (MeV), so the ratio \(a_C/2a_A\) is dimensionless; \(A^{2/3}\) is dimensionless, hence the denominator and \(Z_*\) are pure numbers, as a particle count must be. The curvature \(\gamma\) has units of MeV (energy per \((\Delta Z)^2\)), so \(\gamma(Z-Z_*)^2\) is an energy — the mass excess above the vertex.

Limiting cases
  • Light nuclei \(A^{2/3}\to\) small: the denominator \(\to 2\), so \(Z_*\to A/2\), i.e. \(N=Z\). The Coulomb term is negligible and symmetry alone rules.
  • Coulomb switched off \(a_C\to 0\): denominator \(\to 2\) for all \(A\), giving \(Z_*=A/2\) exactly — the whole neutron excess of heavy nuclei is a Coulomb effect.
  • Heavy nuclei \(a_C A^{2/3}/2a_A \gg 1\): \(Z_*\to 2a_A A^{1/3}/a_C\), scaling as \(A^{1/3}\); the neutron excess \(N-Z = A-2Z_*\) grows roughly linearly with \(A\).
  • Curvature limit large \(A\): \(\gamma\approx 4a_A/A\to 0\), so the parabola flattens and more isobars become quasi-stable/long-lived near the valley floor.
Breaks when
  • Near shell closures (magic \(N\) or \(Z\): 2, 8, 20, 28, 50, 82, 126). The SEMF has no shell structure; real minima are pinned to magic numbers, so the observed stable \(Z\) can differ from \(Z_*\) by 1–2 (e.g. \(A=40\): SEMF gives \(\approx18.4\), yet doubly-magic \(^{40}\mathrm{Ca}\) with \(Z=20\) is stable).
  • Very light nuclei \(A\lesssim 20\). Surface and Coulomb terms are not small perturbations, clustering (alpha structure) dominates, and the smooth liquid-drop parabola loses meaning.
  • Near the drip lines. Far from \(Z_*\) the parabola eventually crosses the particle-emission threshold; nuclei then decay by prompt neutron/proton emission, not by \(\beta\), and "most bound isobar" ceases to describe the ground-state landscape.
  • Superheavy region. Spontaneous fission competes with and outpaces \(\beta\) processes, so proximity to \(Z_*\) no longer implies stability.
Failure modes
  • Rounding \(Z_*\) blindly. For even \(A\) the ground state must have even-parity structure split by pairing; naively rounding to an odd \(Z\) can name an unstable odd–odd nuclide as "most stable".
  • Forgetting the double parabola for even \(A\). Setting \(\delta=0\) for even \(A\) hides the even–even/odd–odd split and predicts a single stable isobar where two (or more) actually exist.
  • Using \(Z_*=A/2\) for heavy nuclei. Ignoring the \(a_C A^{2/3}\) term predicts \(^{238}\mathrm{U}\) at \(Z=119\) instead of \(92\).
  • Confusing the asymmetry coefficient convention. Writing the term as \(a_{\text{sym}}(N-Z)^2/A\) vs \(\tfrac{a_{\text{sym}}}{4}\) forms changes the factor in \(a_C/2a_A\) by four; mixing conventions gives a badly wrong slope of the valley.
  • Differentiating the binding energy but forgetting the constituent-mass linear term \(-(m_n-m_p)c^2 Z\). This drops a genuine (if small) neutron-favouring shift in \(Z_*\).
  • Sign error in \(\partial(A-2Z)^2/\partial Z\). The chain rule gives \(-4(A-2Z)\); a lost minus sign flips the direction of predicted \(\beta\) decay.
Discussion

The parabola is a competition made visible. The asymmetry term \(a_A(A-2Z)^2/A\) is a quadratic well centred on \(N=Z\): it comes from the Pauli principle, since converting a neutron near the Fermi surface into a proton (or vice versa) costs kinetic energy once the two Fermi seas are unequally filled. The Coulomb term \(a_C Z^2/A^{1/3}\) is a rising cost in \(Z\) alone. Their sum is still quadratic, so a minimum always exists; the minimum sits below \(A/2\) by exactly the amount the Coulomb slope pulls the vertex. That is the entire content of the valley of stability.

The curvature \(\gamma\) is physically the "stiffness" of an isobaric chain: the mass penalty for straying one unit of charge from the floor is \(\gamma\), which for mid-mass nuclei is of order \(1\,\text{MeV}\) — precisely the scale of observed \(\beta\)-decay \(Q\)-values. Because \(\gamma\) decreases with \(A\), heavy isobaric chains are shallow and host several long-lived species clustered near the valley, whereas light chains are steep with a single sharply-favoured member.

For even \(A\) the pairing term \(\delta=\pm a_P A^{-1/2}\) is not a smooth function of \(Z\): even–even nuclei get \(-a_P A^{-1/2}\), odd–odd get \(+a_P A^{-1/2}\), splitting the mass surface into two vertically displaced parabolas separated by \(2a_P A^{-1/2}\). An odd–odd nucleus sitting near the vertex can then have both its even–even \(\beta^-\) daughter and its even–even \(\beta^+\)/EC daughter lower in mass, so it decays in both directions (e.g. \(^{64}\mathrm{Cu}\)). Conversely two even–even isobars can straddle an intervening odd–odd nucleus that is higher than either, forbidding ordinary single \(\beta\) between them and leaving second-order double beta decay as the only channel — the observational basis of \(0\nu\beta\beta\) searches. None of this is visible without the pairing splitting.

Common misconceptions. "Stable means most tightly bound per nucleon" — no; per-nucleon binding selects the best \(A\) (the iron peak), whereas \(Z_*\) selects the best \(Z\) within a fixed \(A\). "The valley is \(N=Z\)" — only for light nuclei; Coulomb bends it steadily toward neutron excess. "\(Z_*\) is an integer prediction" — it is a continuous vertex; the real stable isobar is the nearest allowed integer, and shell/pairing effects can override it.

Worked examples
1
Most stable charge for \(A=127\) (odd \(A\)).
Odd \(A\Rightarrow\delta=0\): a single parabola, one stable isobar expected. Use \(a_C=0.71\,\text{MeV}\), \(a_A=23.2\,\text{MeV}\), so \(a_C/2a_A=0.0153\). A
2
\[ A^{2/3}=127^{2/3}=25.3,\qquad \frac{a_C}{2a_A}A^{2/3}=0.0153\times25.3=0.387 \]
Symbols first, then numbers; \(A^{2/3}\) and the ratio are dimensionless. A
3
\[ Z_* = \frac{127}{2+0.387} = \frac{127}{2.387} = 53.2 \]
Round to the nearest integer, \(Z=53\). A
\[ Z_*\approx 53 \;\Rightarrow\; {}^{127}\mathrm{I}\ \text{(iodine)} \]

Reading. \(^{127}\mathrm{I}\) is indeed the only stable \(A=127\) isobar — the prediction lands exactly. Units check. \(Z_*\) is dimensionless; \(53.2\) is a pure number, consistent with a charge count.

1
Most stable charge(s) for \(A=64\) (even \(A\)), and the fate of \(^{64}\mathrm{Cu}\).
Even \(A\Rightarrow\) two parabolas (even–even below, odd–odd above). Same coefficients. A
2
\[ A^{2/3}=64^{2/3}=16,\qquad \frac{a_C}{2a_A}A^{2/3}=0.0153\times16=0.245 \]
Numerical evaluation of the dimensionless denominator term. A
3
\[ Z_* = \frac{64}{2.245} = 28.5 \]
The continuous vertex falls between \(Z=28\) and \(Z=29\). B
4
\[ Z=28\ (^{64}\mathrm{Ni},\ \text{e–e}),\quad Z=30\ (^{64}\mathrm{Zn},\ \text{e–e}),\quad Z=29\ (^{64}\mathrm{Cu},\ \text{o–o, upper parabola}) \]
The odd–odd \(^{64}\mathrm{Cu}\) sits on the raised parabola \(2a_P A^{-1/2}\) above its two even–even neighbours. B
\[ ^{64}\mathrm{Ni}\ \text{and}\ ^{64}\mathrm{Zn}\ \text{stable};\quad ^{64}\mathrm{Cu}\xrightarrow{\ \beta^-\ }{}^{64}\mathrm{Zn},\ \ ^{64}\mathrm{Cu}\xrightarrow{\ \beta^+/\mathrm{EC}\ }{}^{64}\mathrm{Ni} \]

Reading. The vertex at \(28.5\) lies between two even–even minima; the intervening odd–odd \(^{64}\mathrm{Cu}\) is above both and so decays in both directions — a textbook double-parabola signature. Units check. \(Z_*=28.5\) is dimensionless; energies compared (the pairing gap \(2a_P A^{-1/2}\)) are all in MeV.

Problems
  1. (Easy) Show from \(Z_*=A/(2+\tfrac{a_C}{2a_A}A^{2/3})\) that light nuclei satisfy \(N=Z\).
    Solution As \(A\to\) small, \(A^{2/3}\to\) small, so \(\tfrac{a_C}{2a_A}A^{2/3}\to0\) and the denominator \(\to2\), giving \(Z_*\to A/2\). Then \(N=A-Z_*\to A/2 = Z_*\), i.e. \(N=Z\). Physically the Coulomb energy is negligible for small \(Z\), leaving only the symmetry term, whose minimum is at \(N=Z\).
  2. (Easy) Compute \(Z_*\) for \(A=40\) with \(a_C/2a_A=0.0153\) and compare with the stable isobars \(^{40}\mathrm{Ar}\ (Z=18)\) and \(^{40}\mathrm{Ca}\ (Z=20)\).
    Solution \(40^{2/3}=11.7\); \(0.0153\times11.7=0.179\); \(Z_*=40/2.179=18.4\). Nearest integer \(Z=18=\ ^{40}\mathrm{Ar}\), which is stable. However \(^{40}\mathrm{Ca}\ (Z=20)\) is also stable because \(Z=20\) and \(N=20\) are both magic — a shell effect the SEMF cannot capture, illustrating a "breaks when" regime.
  3. (Medium) For even \(A=128\), find \(Z_*\) and predict the stable isobar(s), noting the parity split.
    Solution \(128^{2/3}=25.4\); \(0.0153\times25.4=0.389\); \(Z_*=128/2.389=53.6\). The vertex lies between \(Z=52\) and \(Z=54\), both even–even: \(^{128}\mathrm{Te}\ (Z=52)\) and \(^{128}\mathrm{Xe}\ (Z=54)\) are stable, while the intervening odd–odd \(^{128}\mathrm{I}\ (Z=53)\) sits on the upper parabola and is unstable (it \(\beta^-\)-decays to Xe and EC-decays to Te). \(^{128}\mathrm{Te}\to{}^{128}\mathrm{Xe}\) is a double-beta-decay candidate.
  4. (Medium) Compute the neutron excess \(N-Z\) predicted for \(A=208\) and compare with doubly-magic \(^{208}\mathrm{Pb}\ (Z=82)\).
    Solution \(208^{2/3}=35.1\); \(0.0153\times35.1=0.537\); \(Z_*=208/2.537=82.0\), so \(N-Z=A-2Z_*=208-164=44\). This matches \(^{208}\mathrm{Pb}\) (\(Z=82,\ N=126\), \(N-Z=44\)) essentially exactly — here the smooth SEMF vertex happens to coincide with the doubly-magic nucleus.
  5. (Hard) Derive the isobaric curvature \(\gamma\) and use it to estimate the mass penalty for being one charge unit off the vertex at \(A=127\) (\(a_C=0.71,\ a_A=23.2\,\text{MeV}\)). Comment on the \(\beta\)-decay \(Q\)-value scale.
    Solution From \(M(Z)c^2=\gamma Z^2-\beta Z+\alpha\), \(\gamma=\tfrac12\partial^2(Mc^2)/\partial Z^2 = a_C A^{-1/3}+4a_A A^{-1}\). At \(A=127\): \(A^{1/3}=5.03\), so \(a_C/A^{1/3}=0.71/5.03=0.141\,\text{MeV}\) and \(4a_A/A=92.8/127=0.731\,\text{MeV}\); thus \(\gamma=0.87\,\text{MeV}\). The mass excess of an isobar one unit off is \(M(Z_*\pm1)c^2-M(Z_*)c^2=\gamma(1)^2\approx0.87\,\text{MeV}\). This is the order of magnitude of the difference driving \(\beta\) decay along the chain, consistent with observed \(Q_\beta\sim\text{MeV}\); note it grows with \(|Z-Z_*|\) as \(\gamma(2|Z-Z_*|-1)\) between successive neighbours, so isobars far from the floor decay with large \(Q\) and short half-lives.