advanced30 min studyLesson 572 in Physics

Binding energy and the semiempirical mass formula

Calculate binding and separation energies with consistent mass conventions, then interpret a fitted liquid-drop approximation.

Before you begin

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The question for today

What you’ll understand

Convert atomic to nuclear mass with the correct electron-binding sign.
Separate total binding from one- and two-nucleon separation energies.
Evaluate and interpret a declared semiempirical mass model.

The mass of a bound nucleus is smaller than the sum of its separated nucleon masses. That difference records the energy released during assembly, but its interpretation depends on which masses were actually measured. Tabulations commonly report neutral atoms, not bare nuclei. After establishing consistent bookkeeping, smooth mass formulae reveal systematic trends in nuclear binding. Their success is empirical and approximate: a global fit need not resolve a small separation energy near a shell closure or particle threshold.

Keep atomic electrons in the bookkeeping

Let the electron binding energy be positive: removing all electrons from a neutral atom requires that energy. The atomic mass therefore equals the nuclear mass plus electron rest masses minus the electron binding divided by light speed squared. Solving for nuclear mass gives a positive binding correction. Subtracting electron rest masses alone is often adequate at coarse nuclear precision, but its remaining correction must be estimated before a high-precision comparison.

Nuclear binding is positive for a bound ground state when the reference is separated free protons and neutrons. Rest masses expressed as energy must be multiplied by light speed squared exactly once. Atomic masses can also be combined with neutral-hydrogen masses to cancel electron rest terms, but electronic binding differences remain. A measured mass excess is another convention: it subtracts nucleon number times the atomic mass unit, rather than directly measuring the nuclear binding. Mixing these definitions creates large artificial offsets.

M_{\rm nuc}=M_{\rm at}-Zm_e+\frac{B_e}{c^2},\qquad B=[Zm_p+Nm_n-M_{\rm nuc}]c^2

M values are masses, B and positive B-e are energies, and A=N+Z. Electronic and nuclear binding must use consistent energy units; B-e refers to all atomic electrons.↔ Scroll wide equations horizontally.

Separate total binding from removal thresholds

A neutron separation energy is the energy required to remove one neutron while leaving the daughter in its specified ground state. It is a difference of neighboring total bindings at fixed proton number. The two-neutron separation energy compares nuclei two neutrons apart; it is the sum of two consecutive one-neutron values, not generally twice either one. Pairing and shell effects make those consecutive values differ.

Positive total binding does not guarantee stability against every breakup. A nucleus can lie below separated nucleons yet above a particular bound daughter plus emitted particles. Similarly, a positive reaction Q value expresses energy release, not the absence of a Coulomb barrier or a selection-rule suppression. For radiative neutron capture into the ground state, the available energy is the corresponding neutron separation plus incoming kinetic energy, with recoil sharing the final energy rather than disappearing.

S_n(A,Z)=B(A,Z)-B(A-1,Z),\qquad S_{2n}(A,Z)=B(A,Z)-B(A-2,Z)

The daughter and parent bindings use the same bare-nucleon reference. The separation energies have energy units and the definitions do not assume a smooth mass formula.↔ Scroll wide equations horizontally.

Use the liquid drop as a fitted baseline

The semiempirical formula balances a volume attraction against a surface penalty, electrostatic repulsion, and neutron-proton imbalance. Saturation motivates a volume term proportional to nucleon number rather than its square. Missing neighbors at the surface motivate an area term. The asymmetry term reflects the energy cost of unequal proton and neutron populations, including the fermionic filling pressure and interaction effects. An additional signed pairing term captures broad odd-even trends.

Coefficients depend on the chosen fit, data range, and even the detailed algebraic convention of the formula. A Coulomb factor Z squared and one using Z times Z minus one require different fitted interpretations. Shell corrections, deformation, weak binding, and local correlations are absent from the simplest expression. Numerical examples should therefore label toy coefficients explicitly and avoid presenting an approximate mass prediction as a measured value. Error correlations among neighboring predictions also matter when their difference is small.

B_{\rm LD}=a_vA-a_sA^{2/3}-a_c\frac{Z(Z-1)}{A^{1/3}}-a_a\frac{(A-2Z)^2}{A}+\delta_{\rm pair}

All coefficients and delta-pair have energy units. This is a phenomenological liquid-drop fit form, with positive pairing contribution for even-even nuclei in the usual convention, not an exact binding theorem.↔ Scroll wide equations horizontally.

Work it through

Evaluate a declared toy liquid drop

Use toy coefficients $(a_v,a_s,a_c,a_a)=(16,18,0.75,24)\,\mathrm{MeV}$ and set pairing to zero. Evaluate the formula for $A=27,Z=13$ and find binding per nucleon.

Since $A^{1/3}=3$ and $A^{2/3}=9$ , the volume contribution is $16(27)=432\,\mathrm{MeV}$ and the surface penalty is $18(9)=162\,\mathrm{MeV}$ .
The Coulomb penalty is $0.75(13)(12)/3=39\,\mathrm{MeV}$ . The imbalance is $A-2Z=1$ , giving asymmetry penalty $24/27=8/9\,\mathrm{MeV}$ . Each term has energy units.
Subtract the three penalties to obtain $B_{\rm LD}=432-162-39-8/9=230.111\ldots\,\mathrm{MeV}$ . Divide by 27 to find $B/A\approx8.523\,\mathrm{MeV}$ . This is a toy-model result, not an evaluated mass.

The result

The declared liquid-drop model gives total binding approximately 230.111 MeV and binding per nucleon approximately 8.523 MeV.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A local quadratic mass penalty

A fixed-A asymmetry penalty is quadratic in displacement from equal populations. As a dimensionless analogy, Reset and set $a=1,x=0.5$ in the quadratic lab; predict y equal to 0.25. Compare a fresh Reset run at $a=1,x=1$ ; predict y equal to one. This does not evaluate a mass formula, electrons, pairing, separation energies, or a nuclear stability line.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA quadratic imbalance penalty grows faster than the displacement, while its fitted physical coefficient must come from a declared mass model.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Atomic binding lowers the atom mass relative to a nucleus plus free electrons. Removing the electron rest masses therefore requires adding back the positive electronic binding-energy equivalent.

Your turn

Try it yourself

In a fictional bookkeeping example, $Z=2$ , $M_{\rm at}c^2=3750\,\mathrm{MeV}$ , $Zm_ec^2=1\,\mathrm{MeV}$ , and $B_e=0.001\,\mathrm{MeV}$ . Free nucleons sum to $3777\,\mathrm{MeV}$ . Find nuclear binding. Separately, given bindings 100, 94, and 85 MeV for A, A−1, A−2 at fixed Z, find both neutron separations.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First convert the atomic rest energy; the separate neighboring-binding data need no electron calculation.

Compare with the solution

The nuclear rest energy is $3750-1+0.001=3749.001\,\mathrm{MeV}$ . Binding is $3777-3749.001=27.999\,\mathrm{MeV}$ . Omitting electronic binding would give 28 MeV, too high by one keV. For the separate binding sequence, $S_n=100-94=6\,\mathrm{MeV}$ and $S_{2n}=100-85=15\,\mathrm{MeV}$ . The other consecutive one-neutron separation is nine MeV, confirming that two-neutron separation need not equal twice the first.

Keep the idea

Take it with you

Atomic-to-nuclear mass conversion retains an electronic binding correction.
Separation energy is a neighboring-mass difference rather than binding per nucleon.
Semiempirical coefficients and local prediction errors belong to a specified fit.

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