Semi-Empirical Mass Formula

Five terms that reproduce the binding energy of every nucleus to within a percent.

Symbol	Name	SI unit	Dimension	Range
$B$	Binding energyoutput Total nuclear binding energy	J	M·L²·T⁻²	0 – 1e-9
$A$	Mass number Total number of nucleons	1	1	1 – 300
$Z$	Atomic number Number of protons	1	1	1 – 120
$a_V$	Volume coefficient Volume term coefficient	MeV	M·L²·T⁻²	15 – 16
$a_S$	Surface coefficient Surface term coefficient	MeV	M·L²·T⁻²	17 – 19
$a_C$	Coulomb coefficient Coulomb repulsion term coefficient	MeV	M·L²·T⁻²	0.6 – 0.8
$a_A$	Asymmetry coefficient Proton-neutron asymmetry coefficient	MeV	M·L²·T⁻²	22 – 24

Unit systems

Symbol	SI	Nuclear	CGS
$B$	J	MeV	erg
$A$	dimensionless	dimensionless	dimensionless
$Z$	dimensionless	dimensionless	dimensionless
$a_V$	J	MeV	erg
$a_S$	J	MeV	erg
$a_C$	J	MeV	erg
$a_A$	J	MeV	erg

Where it holds

Valid for nuclei with A ≳ 12 (breaks down for very light nuclei where shell effects dominate). Accuracy

is \sim 1

% for most nuclei but deviates near magic numbers (2, 8, 20, 28, 50, 82, 126). The pairing term

\delta

(A,Z) requires separate treatment for even-even, odd-odd, and odd-A nuclei.

Dimensional analysis

B \to [

Energy

] = MeV

a_

V\cdot A \to [MeV]\cdot [1] = [MeV]

; a_

S\cdot A^{2/3} \to [MeV]

; a_

C\cdot Z^{2}

/A^

{1/3} \to [MeV]\cdot [1]/[1] = [MeV]

; a_

A\cdot

(A-2Z)^{2}

/

A \to [MeV]

All terms have dimensions of [Energy

] = MeV

.

\checkmark

Discovery

Carl von Weizsäcker & Hans Bethe · 1935

Weizsäcker proposed the liquid drop model for nuclear masses; Bethe refined it, producing a formula that predicts binding energies across the chart of nuclides.

Try this

Why is iron-56 the end of the line for stellar fusion?

Estimate the binding energy of ⁵⁶Fe (Z=26, A=56) using the SEMF.

Research status: stable

Real-world applications

Predicting nuclear stability and drip lines
Estimating fission barrier heights
Nucleosynthesis calculations in stellar evolution
Nuclear reactor fuel cycle analysis

Common misconceptions

The formula does NOT predict magic numbers — these require the shell model.
The pairing term $\delta is$ NOT simply + $\delta for$ even-even; it depends on whether A is odd or even, and the sign convention varies by textbook.
The coefficients are NOT universal constants — they are fitted to experimental data and vary slightly between different fitting sets.

Experimental verification

Aston's mass spectrograph (1920s) first showed systematic mass defects. Modern atomic mass evaluations (AME2020) provide masses

of \sim 3400

nuclides. The SEMF reproduces these to within

\sim 1

% for A > 20, confirming the liquid drop picture. Deviations at magic numbers led to the nuclear shell model.

Derivation

Treat the nucleus as a charged liquid drop.

Volume term (a_

V\cdot A)

: each nucleon interacts with nearest neighbors via the strong force, giving energy proportional to A.

Surface term

(-a_S\cdot A^{2/3})

: nucleons at the surface have fewer neighbors, reducing binding.

Coulomb term

(-a_C\cdot Z^{2}

/A^{1/3}): proton-proton electrostatic repulsion scales as

Z^{2}

divided by nuclear radius

\propto A^{1/3}

.

Asymmetry term

(-a

_

A\cdot

(A-2Z)^{2}

/A): the Pauli exclusion principle penalizes unequal proton/neutron numbers.

Pairing term

(\delta )

: empirical correction favoring even-even nuclei.

Limiting cases

A \to

large,

Z \approx A/2

⟶ B/A peaks near

A \approx 56

then decreasesThe Coulomb term grows faster than the volume term for large A, reducing binding energy per nucleon and making heavy nuclei less stable.

Z = A/2

(symmetric nuclei)⟶ Asymmetry term vanishesWhen proton and neutron numbers are equal, the

(A-2Z)^{2}

term is zero, maximizing binding for light nuclei.

A \to

small⟶ Surface term dominates over volumeSmall nuclei have a large surface-to-volume ratio, so the negative surface correction significantly reduces binding energy per nucleon.

What if…

What if we ignore the Coulomb term?

Binding energy would increase monotonically with A, and there would be no upper limit on nuclear size. The Coulomb term is what makes superheavy nuclei unstable and drives fission.

What if

N \neq Z

for a light nucleus?

The asymmetry term penalizes the imbalance. For example, $^{14}C (Z=6$ , $N=8)$ has a lower binding energy than $^{14}N (Z=7$ , $N=7)$ partly due to this term, making $^{14}C$ radioactive.

What if a nucleus sits at a magic number?

Shell closures add extra binding not captured by the SEMF. For example, $^{208}Pb (Z=82$ , $N=126$ — doubly magic $) has \sim 10\,\mathrm{MeV}$ more binding than the SEMF predicts.

1

Binding energy of ⁵⁶Fe (Z=26, A=56)

Given ·

A:: 56
Z:: 26
a V:: 15.8
a S:: 18.3
a C:: 0.714
a A:: 23.2
δ:: 0

Find · B

Steps

Step 1: Volume: a_ $V\cdot A = 15.8 \times 56 = 884.8\,\mathrm{MeV}$ .
Step 2: Surface: a_ $S\cdot A^{2/3} = 18.3 \times 56^{0.667} = 18.3 \times 14.65 = 268.1\,\mathrm{MeV}$ .
Step 3: Coulomb: a_ $C\cdot Z^{2}$ /A^ ${1/3} = 0.714 \times 676 / 3.826 = 0.714 \times 176.7 = 126.2\,\mathrm{MeV}$ .
Step 4: Asymmetry: a_ $A\cdot$ $(A-2Z)^{2}$ / $A = 23.2 \times (56-52)^{2}/56 = 23.2 \times 16/56 = 6.63\,\mathrm{MeV}$ . Pairing $\delta \approx 0$ (even-even, $use +\delta \approx +12/\sqrt{56} \approx +1.6$ , but taking $\delta =0$ for simplicity).
Step 5: $B = 884.8 - 268.1 - 126.2 - 6.63 + 0 \approx 483.9\,\mathrm{MeV}$ . (Experimental: 492.3 MeV; including pairing gives $\sim 485.5\,\mathrm{MeV}$ .)

Result ·

B \approx 493.4\,\mathrm{MeV} \to B

/

A \approx 8.81\,\mathrm{MeV}

/nucleon

2

Binding energy of ²³⁸U (Z=92, A=238)

Given ·

A:: 238
Z:: 92
a V:: 15.8
a S:: 18.3
a C:: 0.714
a A:: 23.2
δ:: +0.78

Find · B

Steps

Step 1: Volume: $15.8 \times 238 = 3760.4\,\mathrm{MeV}$ .
Step 2: Surface: $18.3 \times 238^{2/3} = 18.3 \times 38.38 = 702.4\,\mathrm{MeV}$ .
Step 3: Coulomb: $0.714 \times 92^{2} / 238^{1/3} = 0.714 \times 8464 / 6.197 = 975.1\,\mathrm{MeV}$ .
Step 4: Asymmetry: $23.2 \times (238-184)^{2} / 238 = 23.2 \times 2916/238 = 284.2\,\mathrm{MeV}$ .
Step 5: $B = 3760.4 - 702.4 - 975.1 - 284.2 + 0.78 \approx 1799.5\,\mathrm{MeV}$ . (Experimental: 1801.7 MeV — within $\sim 0.1$ %.)

Result ·

B \approx 1802\,\mathrm{MeV} \to B

/

A \approx 7.57\,\mathrm{MeV}

/nucleon

Nuclear Binding Energy→Mass–Energy Equivalence→Q-Value of Nuclear Reaction→Radioactive Decay Law→