Semi-Empirical Mass Formula from the Liquid-Drop Model

Derivation

Semi-Empirical Mass Formula from the Liquid-Drop Model

D-265 Home PU-304 Threads energy · matter · force Depends on electrostatic-energy-uniform-sphere, fermi-gas-degeneracy-pressure

Statement

Starting from the liquid-drop picture of the nucleus as an incompressible, charged, saturated fluid of \(A\) nucleons (\(Z\) protons, \(N=A-Z\) neutrons), we derive the total binding energy as a five-term algebraic sum, the semi-empirical mass formula (Bethe–Weizsäcker formula), \[ B(A,Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A-2Z)^2}{A} + \delta(A,Z), \] where each coefficient carries a definite physical origin: volume saturation, surface tension, Coulomb self-energy, isospin asymmetry from the degenerate Fermi gas, and nucleon pairing.

Why it matters

The formula is the workhorse of nuclear phenomenology: with five constants fitted once, it reproduces the binding energies of over two thousand nuclei to better than one percent, predicts the valley of \(\beta\)-stability, the \(A\simeq 56\) maximum of \(B/A\) that separates fusion from fission fuel, and — through its Coulomb and surface competition — the fissility parameter that governs whether a heavy nucleus is stable against deformation.

It is also a model lesson in how a handful of clean physical ideas — incompressible drop, surface tension, electrostatics of a charged sphere, and Pauli exclusion — combine additively to organise an otherwise intractable quantum many-body problem, and where that additivity breaks (shell structure, halo nuclei) marks the frontier of nuclear structure.

Assumptions

Charge and mass density are uniform and constant.If \(\rho\) varied with \(A\), the nuclear radius would not scale as \(R=r_0A^{1/3}\) and every term's power of \(A\) would change; saturation of the nuclear force is what makes \(\rho\) universal.

The nuclear force saturates and is short-ranged.Each nucleon interacts only with its near neighbours, so the bulk binding is proportional to the number of nucleons, not to \(A^2\); drop this and the volume term would grow like \(A^2\) as in a fully-coupled system.

Protons and neutrons fill independent Fermi seas in a common well.The asymmetry term is derived from the degeneracy energy of two Fermi gases; without the Pauli principle there would be no energy cost to converting neutrons into protons and no asymmetry term.

The Coulomb energy is that of a classical uniformly charged sphere, corrected for proton self-interaction.Treating the \(Z\) protons as a smooth charge cloud ignores exchange (a small \(Z^{4/3}\) correction) and replaces \(Z^2\) by \(Z(Z-1)\) because a proton does not repel itself; dropping the self-term overbinds light nuclei.

The surface is sharp and the drop is spherical in the ground state.Surface diffuseness and static deformation modify \(a_S\); a strongly deformed ground state (as in the actinides) requires deformation-dependent surface and Coulomb terms beyond the leading formula.

Derivation

\[ R = r_0 A^{1/3}, \qquad r_0 \approx 1.2~\text{fm} \]

Constant density \(\rho = A/(\tfrac{4}{3}\pi R^3)\) forces the volume \(\propto A\), so \(R\propto A^{1/3}\); this single scaling law fixes the \(A\)-powers of every geometric term. A

\[ B_{\text{vol}} = a_V A \]

A short-range saturating force gives each interior nucleon the same fixed number of bonds, so the bulk binding is strictly proportional to the nucleon count \(A\). A

\[ B_{\text{surf}} = -a_S\, 4\pi R^2 \cdot \frac{1}{4\pi r_0^2} = -a_S A^{2/3} \]

Nucleons in the surface layer have fewer neighbours, so they are underbound by an amount proportional to the surface area \(4\pi R^2 \propto A^{2/3}\); this is a liquid surface tension. B

\[ E_C^{\text{sphere}} = \frac{3}{5}\frac{1}{4\pi\varepsilon_0}\frac{Q^2}{R}, \qquad Q = Ze \]

Imported result: the electrostatic self-energy of a uniformly charged sphere of total charge \(Q\) and radius \(R\) (electrostatic-energy-uniform-sphere). A

\[ B_{\text{Coul}} = -\,\frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 r_0}\,\frac{Z(Z-1)}{A^{1/3}} \equiv -a_C\frac{Z(Z-1)}{A^{1/3}} \]

Substitute \(R=r_0A^{1/3}\); replace \(Z^2\to Z(Z-1)\) because each proton contributes to the field felt by the other \(Z-1\) protons but not to its own self-repulsion, removing the spurious self-energy of the smoothed cloud. B

\[ E_{\text{FG}}(n) = \frac{3}{5} n\,\varepsilon_F, \qquad \varepsilon_F = \frac{\hbar^2}{2m}\left(3\pi^2 \frac{n}{V}\right)^{2/3} \]

Imported result: the total kinetic energy of a degenerate Fermi gas of \(n\) spin-\(\tfrac12\) fermions is \(\tfrac{3}{5}n\varepsilon_F\) (fermi-gas-degeneracy-pressure). Protons and neutrons form two independent gases in the same volume \(V=\tfrac{4}{3}\pi R^3\propto A\). B

\[ E_{\text{kin}}(N,Z) = \frac{3}{5}\frac{\hbar^2}{2m}\left(3\pi^2\right)^{2/3} V^{-2/3}\left(N^{5/3}+Z^{5/3}\right) \]

Each gas contributes \(\tfrac{3}{5}n\varepsilon_F \propto n^{5/3}/V^{2/3}\); summing the two seas and using a common mass \(m\) and volume \(V\) gives the total degeneracy energy as a function of \(N\) and \(Z\). C

\[ N = \frac{A}{2}\!\left(1+x\right),\quad Z=\frac{A}{2}\!\left(1-x\right),\quad x=\frac{N-Z}{A} \]

Introduce the asymmetry variable \(x\) and expand \(N^{5/3}+Z^{5/3}=\left(\tfrac{A}{2}\right)^{5/3}\big[(1+x)^{5/3}+(1-x)^{5/3}\big]\) to isolate the symmetric part from the cost of imbalance. C

\[ (1+x)^{5/3}+(1-x)^{5/3} = 2\left(1+\frac{5}{9}x^2 + \mathcal{O}(x^4)\right) \]

Taylor-expand to second order; the linear terms cancel by symmetry, leaving the leading energy cost of asymmetry quadratic in \(x=(N-Z)/A\). C

\[ \Delta E_{\text{kin}} = E_{\text{kin}}-E_{\text{kin}}^{\text{sym}} \propto \frac{1}{V^{2/3}}A^{5/3}x^2 \propto \frac{(N-Z)^2}{A} = \frac{(A-2Z)^2}{A} \]

Since \(V\propto A\), the \(V^{-2/3}A^{5/3}\) prefactor reduces to \(A\); multiplying by \(x^2=(N-Z)^2/A^2\) gives \((N-Z)^2/A\). This kinetic asymmetry, plus the symmetry energy of the strong force, defines the phenomenological \(a_A\). C

\[ B_{\text{asym}} = -a_A\frac{(A-2Z)^2}{A} \]

Converting more of a symmetric nucleus into one species raises the top of that Fermi sea and lowers binding; the penalty is quadratic and minimised at \(N=Z\), giving the asymmetry term. B

\[ \delta(A,Z) = \begin{cases} +a_P A^{-1/2} & Z,N \text{ even}\\ 0 & A \text{ odd}\\ -a_P A^{-1/2} & Z,N \text{ odd} \end{cases} \]

Nucleons of like kind pair into time-reversed \(0^+\) states with extra binding; the empirical pairing term is added by hand (it is a quantum, not liquid-drop, effect), with the \(A^{-1/2}\) scaling fixed by fit. A

\[ B(A,Z) = a_V A - a_S A^{2/3} - a_C\frac{Z(Z-1)}{A^{1/3}} - a_A\frac{(A-2Z)^2}{A} + \delta(A,Z) \]

Sum the volume, surface, Coulomb, asymmetry and pairing contributions; additivity is the central liquid-drop assumption that each mechanism perturbs a common incompressible drop independently. A

Result

\[ B(A,Z) = a_V A - a_S A^{2/3} - a_C\frac{Z(Z-1)}{A^{1/3}} - a_A\frac{(A-2Z)^2}{A} + \delta(A,Z) \]

Reading. The binding energy is a competition between a bulk volume gain \(a_V A\) that wants nuclei large, and three penalties: the surface term (fewer bonds at the edge), the Coulomb term (proton–proton repulsion, growing like \(Z^2/A^{1/3}\)), and the asymmetry term (Pauli cost of \(N\neq Z\)). The pairing term is a small even–odd correction. Their balance produces the peak of \(B/A\approx 8.8\) MeV near iron and the drift of the stability line toward neutron excess in heavy nuclei. A representative fit (in MeV): \(a_V\approx 15.8\), \(a_S\approx 18.3\), \(a_C\approx 0.71\), \(a_A\approx 23.2\), \(a_P\approx 12\).

Units check. Every coefficient carries units of energy (MeV) and every \(A,Z\) factor is dimensionless, so each term is an energy. The Coulomb coefficient \(a_C=\tfrac{3}{5}e^2/(4\pi\varepsilon_0 r_0)\): with \(e^2/(4\pi\varepsilon_0)=1.44~\text{MeV\,fm}\) and \(r_0=1.2~\text{fm}\), \(a_C=\tfrac{3}{5}(1.44/1.2)=0.72~\text{MeV}\), matching the fitted value and confirming the electrostatic origin of that term.

Limiting cases

Large \(A\), fixed \(Z/A\): \(B/A \to a_V - a_S A^{-1/3} - \dots\); the surface correction vanishes as \(A^{-1/3}\), so \(B/A\) approaches the bulk value \(a_V\) reduced by Coulomb and asymmetry.
\(Z=0\) (pure neutron matter): Coulomb term vanishes but the asymmetry term \(-a_A A\) is maximal, so free neutron drops are unbound — consistent with the non-existence of bound multineutrons.
Light nuclei \(A\to 1\): surface and volume terms become comparable, the \(A^{2/3}\) expansion loses accuracy, and the formula fails for \(A\lesssim 20\).
Fissility \(x_f=\dfrac{a_C Z^2}{2a_S A}\to 1\): Coulomb energy overwhelms surface tension, the spherical drop becomes unstable to deformation, marking the fission limit near \(Z^2/A\approx 49\).
\(N=Z\), even–even: asymmetry term vanishes and pairing is maximally positive, giving the most tightly bound light nuclei (e.g. \(^{4}\text{He}\), \(^{12}\text{C}\), \(^{16}\text{O}\)).

Breaks when

Magic numbers. Near closed shells (\(Z\) or \(N=2,8,20,28,50,82,126\)) the smooth liquid drop misses the extra binding from shell gaps; \(^{208}\text{Pb}\) is bound about 10 MeV more than the formula predicts. A Strutinsky shell correction must be added.
Very light nuclei (\(A\lesssim 20\)). The surface-to-volume ratio is order unity and the \(A^{2/3}\) surface expansion and continuum liquid picture both fail; few-body quantum structure dominates.
Halo and drip-line nuclei. Weakly bound nuclei like \(^{11}\text{Li}\) have diffuse, extended matter distributions that violate the sharp-surface, constant-density assumption entirely.
Strongly deformed ground states. Actinides and rare-earth nuclei are not spherical; the leading surface and Coulomb terms need deformation-dependent corrections (the basis of the deformed liquid-drop / fission-barrier calculation).

Failure modes

Using \(Z^2\) instead of \(Z(Z-1)\) in the Coulomb term for light nuclei — this double-counts a proton's self-repulsion and noticeably overbinds \(^{4}\text{He}\).
Writing the asymmetry term as \((N-Z)^2\) instead of \((N-Z)^2/A\) — forgetting the \(1/A\) from \(x^2=(N-Z)^2/A^2\) times the \(A\) prefactor; the term must be intensive-per-nucleon-scaled.
Sign errors: writing \(B\) with the surface, Coulomb and asymmetry terms as positive contributions. They reduce binding and carry minus signs; only \(a_V A\) and (sometimes) \(\delta\) are positive.
Confusing binding energy \(B\) with the nuclear mass \(M c^2 = Z m_p c^2 + N m_n c^2 - B\); students sometimes maximise \(M\) instead of minimising it when finding the stable \(Z\).
Applying the pairing sign backwards — even–even is more bound (\(+\)), odd–odd less bound (\(-\)); reversing this predicts the wrong \(\beta\)-decay chains for odd-\(A\) versus even-\(A\) isobars.
Forgetting the \(3/5\) geometric factor in \(a_C\), which comes specifically from the uniform-sphere self-energy, not a point charge.

Discussion

The power of the formula lies in its additivity: each term isolates one physical mechanism with a distinct \(A\)-scaling — volume \(\propto A\), surface \(\propto A^{2/3}\), Coulomb \(\propto A^{-1/3}Z^2\), asymmetry \(\propto A^{-1}(N-Z)^2\). Because these scale differently, their competition produces the characteristic curve of \(B/A\): rising steeply for light nuclei as the surface penalty shrinks, peaking near \(^{56}\text{Fe}\)–\(^{62}\text{Ni}\), then falling as Coulomb repulsion accumulates. That single curve is why fusion releases energy below iron and fission above it, and it underlies stellar nucleosynthesis and reactor physics alike.

The valley of stability follows from minimising the mass \(Mc^2 = Zm_pc^2 + (A-Z)m_nc^2 - B\) at fixed \(A\). Setting \(\partial M/\partial Z=0\) balances the Coulomb term (which prefers fewer protons) against the asymmetry term (which prefers \(N=Z\)), giving the most-stable charge \(Z^* \approx \dfrac{A}{2}\dfrac{1}{1+\tfrac{a_C}{4a_A}A^{2/3}}\). For light nuclei \(Z^*\approx A/2\); for heavy nuclei the \(A^{2/3}\) Coulomb growth pushes \(Z^*\) well below \(A/2\), explaining the neutron excess of stable heavy nuclei.

The Coulomb–surface competition is made precise by the fissility parameter. A small quadrupole deformation \(R(\theta)=R_0[1+\alpha_2 P_2(\cos\theta)]\) at fixed volume lowers the Coulomb energy by \(\tfrac{1}{5}\alpha_2^2 E_C^{(0)}\) but raises the surface energy by \(\tfrac{2}{5}\alpha_2^2 E_S^{(0)}\). The drop is unstable when the Coulomb gain exceeds the surface cost, i.e. when \(E_C^{(0)}>2E_S^{(0)}\), giving the fissility \(x_f = E_C^{(0)}/(2E_S^{(0)}) = a_C Z^2/(2a_S A)\). At \(x_f=1\) the fission barrier vanishes; this is the liquid-drop prediction of the limit of nuclear stability, refined by shell corrections into the observed island of the heaviest elements.

Common misconceptions. The formula is often mistaken for a first-principles derivation; in fact only the surface, Coulomb and (partly) asymmetry terms have transparent classical or Fermi-gas origins, while \(a_V\) and \(a_P\) are genuinely phenomenological, absorbing the strong-force dynamics that the model does not compute. It is also not a nuclear-structure theory: it knows nothing of individual orbitals, so it cannot explain magic numbers, deformation, or spectroscopy — those require the shell model layered on top.

Worked examples

\[ \text{Binding energy of } ^{56}\text{Fe} \quad (A=56,\ Z=26,\ N=30) \]

Compute each term with \(a_V=15.8\), \(a_S=18.3\), \(a_C=0.71\), \(a_A=23.2\), \(a_P=12\) MeV; \(^{56}\text{Fe}\) is even–even so \(\delta=+a_P A^{-1/2}\). A

\[ B_{\text{vol}} = 15.8\times 56 = 884.8~\text{MeV} \]

Volume term \(a_V A\). A

\[ B_{\text{surf}} = -18.3\times 56^{2/3} = -18.3\times 14.64 = -267.9~\text{MeV} \]

\(56^{2/3}=(56^{1/3})^2=(3.826)^2=14.64\). A

\[ B_{\text{Coul}} = -0.71\times\frac{26\times 25}{56^{1/3}} = -0.71\times\frac{650}{3.826} = -120.6~\text{MeV} \]

Coulomb term with \(Z(Z-1)=650\), \(56^{1/3}=3.826\). A

\[ B_{\text{asym}} = -23.2\times\frac{(56-52)^2}{56} = -23.2\times\frac{16}{56} = -6.63~\text{MeV} \]

Asymmetry term with \(A-2Z=56-52=4\). A

\[ \delta = +\frac{12}{\sqrt{56}} = +\frac{12}{7.483} = +1.60~\text{MeV} \]

Even–even pairing bonus. A

\[ B = 884.8 - 267.9 - 120.6 - 6.63 + 1.60 = 491.3~\text{MeV} \]

Reading. \(B/A = 491.3/56 = 8.77~\text{MeV}\), within 0.3% of the experimental \(8.79\) MeV — \(^{56}\text{Fe}\) sits at the peak of the binding curve, which is exactly why it is the ash of stellar burning.

Units check. All terms in MeV; \(B/A\) in MeV per nucleon, the standard measure of nuclear stability.

\[ \text{Most stable } Z \text{ for the } A=127 \text{ isobars} \]

Minimise \(Mc^2(A,Z)\) at fixed \(A\); equivalently maximise \(B\) minus the proton–neutron mass difference term, giving the analytic \(Z^*\). Use \(m_n-m_p=1.293\) MeV, and \(a_C, a_A\) as above. B

\[ \frac{\partial M c^2}{\partial Z}=0 \;\Rightarrow\; Z^* = \frac{A}{2}\,\frac{1 + \dfrac{(m_n-m_p)c^2}{4a_A}}{\,1+\dfrac{a_C}{4a_A}A^{2/3}} \]

Differentiate the Coulomb, asymmetry and mass terms; the numerator correction from \((m_n-m_p)\) is tiny and often dropped. C

\[ A^{2/3} = 127^{2/3} = (5.026)^2 = 25.26 \]

\(127^{1/3}=5.026\). A

\[ \frac{a_C}{4a_A}A^{2/3} = \frac{0.71}{4\times 23.2}\times 25.26 = \frac{0.71}{92.8}\times 25.26 = 0.1933 \]

Denominator correction. A

\[ Z^* = \frac{127}{2}\,\frac{1+\dfrac{1.293}{92.8}}{1+0.1933} = 63.5\times\frac{1.0139}{1.1933} = 63.5\times 0.8497 = 53.96 \]

Insert numbers; the small numerator term shifts \(Z^*\) up by \(\sim 0.9\). B

\[ Z^* \approx 54 \quad\Rightarrow\quad ^{127}\text{Xe (Z=54)} \]

Reading. The formula predicts \(Z^*\approx 54\); the actual stable \(A=127\) nuclide is \(^{127}\text{I}\) (\(Z=53\)), one unit away — excellent for a five-parameter model, with the small discrepancy attributable to the nearby \(N=74\) shell structure the liquid drop cannot capture.

Units check. \(Z^*\) is dimensionless; the ratio \((m_n-m_p)c^2/a_A\) is MeV/MeV, consistent inside the bracket.

Problems

Compute \(B/A\) for \(^{208}\text{Pb}\) (\(A=208, Z=82\)) using the fitted coefficients and compare to the experimental \(7.87\) MeV. Comment on the sign of the discrepancy.
Solution
Even–even. \(B_{\text{vol}}=15.8\times208=3286.4\). \(208^{1/3}=5.925\), \(208^{2/3}=35.10\). \(B_{\text{surf}}=-18.3\times35.10=-642.3\). \(Z(Z-1)=82\times81=6642\); \(B_{\text{Coul}}=-0.71\times6642/5.925=-795.9\). \(A-2Z=208-164=44\); \(B_{\text{asym}}=-23.2\times44^2/208=-23.2\times9.31=-216.0\). \(\delta=+12/\sqrt{208}=+0.83\). Sum \(B=3286.4-642.3-795.9-216.0+0.83=1633.0\) MeV, \(B/A=1633.0/208=7.85\) MeV. Experiment is \(7.87\); the formula slightly underbinds because \(^{208}\text{Pb}\) is doubly magic (\(Z=82, N=126\)) and gains extra shell binding the liquid drop omits.
Show that the asymmetry term follows from a two-Fermi-gas model. Given \(E_{\text{kin}}\propto V^{-2/3}(N^{5/3}+Z^{5/3})\) at fixed \(A=N+Z\), expand to second order in \((N-Z)/A\) and identify the \((N-Z)^2/A\) dependence.
Solution
Write \(N=\tfrac{A}{2}(1+x)\), \(Z=\tfrac{A}{2}(1-x)\), \(x=(N-Z)/A\). Then \(N^{5/3}+Z^{5/3}=(\tfrac{A}{2})^{5/3}[(1+x)^{5/3}+(1-x)^{5/3}]\). Taylor: \((1\pm x)^{5/3}=1\pm\tfrac53 x+\tfrac{5}{9}x^2\pm\dots\); the sum \(=2(1+\tfrac59 x^2+\dots)\). So \(E_{\text{kin}}=E_0(1+\tfrac59 x^2)\) with \(E_0\propto V^{-2/3}A^{5/3}\). Since \(V\propto A\), \(V^{-2/3}A^{5/3}\propto A^{5/3-2/3}=A\). Thus \(\Delta E_{\text{kin}}\propto A\cdot x^2=A\cdot(N-Z)^2/A^2=(N-Z)^2/A\), which is the asymmetry term. The kinetic contribution accounts for roughly half of the empirical \(a_A\); the rest comes from the isospin dependence of the strong interaction.
Estimate the fissility parameter \(x_f=a_C Z^2/(2a_S A)\) for \(^{238}\text{U}\) (\(Z=92, A=238\)). Is it stable against small deformations?
Solution
\(x_f = 0.71\times92^2/(2\times18.3\times238) = 0.71\times8464/8710.8 = 6009.4/8710.8 = 0.690\). Since \(x_f<1\), \(^{238}\text{U}\) is (meta)stable against small deformations — it has a finite fission barrier — but with \(x_f\approx0.69\) the barrier is modest, consistent with its spontaneous-fission half-life and its readiness to fission on neutron capture. The critical value \(x_f=1\) corresponds to \(Z^2/A=2a_S/a_C\approx 51.5\); for \(^{238}\text{U}\), \(Z^2/A=35.5\).
For \(A=64\) isobars, compute \(Z^*\) from \(Z^*=\tfrac{A}{2}/(1+\tfrac{a_C}{4a_A}A^{2/3})\) (neglect the \(m_n-m_p\) term). Which stable nuclide does this pick out?
Solution
\(64^{2/3}=16\) exactly. \(\tfrac{a_C}{4a_A}A^{2/3}=\tfrac{0.71}{92.8}\times16=0.1224\). \(Z^*=32/(1.1224)=28.5\). Rounding gives \(Z^*=28\) or \(29\); the stable \(A=64\) nuclides are \(^{64}\text{Ni}\) (\(Z=28\)) and \(^{64}\text{Zn}\) (\(Z=30\)), with \(^{64}\text{Ni}\) closest to the prediction. Note \(Z=28\) is magic, enhancing \(^{64}\text{Ni}\) stability beyond the liquid-drop estimate; \(^{64}\text{Cu}\) (\(Z=29\), odd–odd) is unstable, decaying both ways — a classic even-\(A\) isobar case where the pairing term splits the odd–odd nuclide below both even–even neighbours.
Advanced. Derive the fissility condition \(E_C^{(0)}>2E_S^{(0)}\) from a quadrupole deformation \(R(\theta)=R_0[1+\alpha_2 P_2(\cos\theta)]\) at constant volume, given that to order \(\alpha_2^2\) the surface energy changes by \(+\tfrac25\alpha_2^2 E_S^{(0)}\) and the Coulomb energy by \(-\tfrac15\alpha_2^2 E_C^{(0)}\).
Solution
The total energy change to second order is \(\Delta E=\Delta E_S+\Delta E_C=\alpha_2^2\left(\tfrac25 E_S^{(0)}-\tfrac15 E_C^{(0)}\right)\). The spherical shape is a local minimum (stable) when \(\Delta E>0\) for small \(\alpha_2\), i.e. \(\tfrac25 E_S^{(0)}>\tfrac15 E_C^{(0)}\Rightarrow E_C^{(0)}<2E_S^{(0)}\). Instability (barrierless fission) sets in when \(E_C^{(0)}>2E_S^{(0)}\), i.e. \(x_f\equiv E_C^{(0)}/(2E_S^{(0)})=a_C Z^2/(2a_S A)>1\). At \(x_f=1\) the quadratic coefficient vanishes and the fission barrier disappears; beyond it the nucleus is classically unstable to spontaneous deformation and immediate fission. This is the liquid-drop upper limit on nuclear existence, giving \(Z^2/A\lesssim 2a_S/a_C\approx 51\).