Binding Energy per Nucleon

Equivalent forms

B(A,Z) = [Z m_p + N m_n - M(A,Z)] c^2

A single curve from H to U explains why stars fuse, why bombs fission, and why iron is the cosmic ash.

Symbol	Name	SI unit	Dimension	Range
$B̄$	Binding energy per nucleonoutput Average binding energy per nucleon	J	M·L²·T⁻²	0 – 1.5e-12
$B$	Total binding energy Energy needed to disassemble the nucleus	J	M·L²·T⁻²	0 – 3e-10
$A$	Mass number Total number of nucleons	1	1	1 – 250

Unit systems

Symbol	SI	CGS	Natural
$B̄$	J	erg	MeV
$B$	J	erg	MeV
$A$	1	1	1

Where it holds

Defined for any bound nucleus with

A \ge 2

. The shape of B̄(A) is captured semi-empirically by the SEMF (Weizsäcker formula).

Dimensional analysis

B̄

\to [M\cdot L^{2}\cdot T^{-2}]

B/A \to [M\cdot L^{2}\cdot T^{-2}]/[1] = [M\cdot L^{2}\cdot T^{-2}]

Energy units.

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Discovery

Francis W. Aston · 1920

Aston's mass spectrograph revealed the 'packing fraction' curve — the first measurement of B/A across the periodic table.

Try this

Why is iron-56 the hardest nucleus to break apart?

Given B(Fe-56) = 492.3 MeV, compute the binding energy per nucleon and explain the iron peak.

Research status: stable

Real-world applications

Stellar nucleosynthesis cutoff (silicon burning produces iron-peak elements)
Nuclear fission energy release calculations
Fusion reactor energy yield $(D-T \to He-4$ jump in B̄)
Astrophysical r-process and s-process modeling

Common misconceptions

B̄ is not the energy of a single nucleon — it is an average. Removing one nucleon costs the 'separation energy' which differs from B̄.
The iron peak is not the most stable nuclide per mass-energy; Ni-62 actually has slightly higher B̄ (8.795 MeV vs Fe-56's 8.790 MeV).
The plot is not monotonic; light nuclei show peaks at He-4, C-12, O-16 (alpha-cluster effect).

Experimental verification

Aston's mass spectrograph in the 1920s mapped the curve; modern atomic mass evaluations (AME2020) give B̄ values to keV precision

for \sim 3300

nuclides.

Derivation

Start with the mass defect

\Delta m = Zm_p + Nm_n - M

(A,Z).

Multiply by

c^{2} to

get total binding energy

B = \Delta m c^{2}

.

Divide by A: B̄

= B

/A.

Phenomenologically the SEMF gives B̄

\approx a_v - a_s

/A^

(1/3) - a_c Z(Z-1)

/A^

(4/3) - a_sym (N-Z)^{2}

/

A^{2} + \delta

/A^(3/2).

Limiting cases

A = 1

⟶ B̄

= 0

A single proton or neutron has no binding partner.

A \approx 56

⟶ B̄

\approx 8.79\,\mathrm{MeV}

Iron peak — most tightly bound region; endpoint of stellar nucleosynthesis in massive stars.

A \to 238

⟶ B̄

\approx 7.57\,\mathrm{MeV}

Heavy nuclei lose binding to Coulomb repulsion, enabling fission energy release.

What if…

What if the strong force had longer range?

Surface term a_s would shrink, B̄ would rise for light nuclei and stable nuclei could be much larger.

What if the Coulomb force were stronger?

Heavy nuclei would be unstable; iron peak would shift to lower A, fission would dominate.

What if iron's B̄ were the cosmic maximum exactly?

It nearly is; the universe's elemental abundance pattern peaks at iron because stellar fusion stops there.

1

Iron-56 per-nucleon binding energy

Given ·

B:: 7.886e-11
A:: 56

Find · B̄

Steps

Step 1: Convert $B = 492.3\,\mathrm{MeV} = 492.3 \times 1.602176634e-13 = 7.886e-11\,\mathrm{J}$ .
Step 2: B̄ $= 7.886e-11 / 56 \approx 1.408e-12\,\mathrm{J}$ .
Step 3: Convert: $1.408e-12 / 1.602176634e-13 \approx 8.79\,\mathrm{MeV}$ per nucleon.

Result · B̄

= 7.886e-11 / 56 \approx 1.408e-12\,\mathrm{J} \approx 8.79\,\mathrm{MeV}

/nucleon

2

Fusion energy gain D+T → He+n

Given ·

B:: 4.533e-12
A:: 4

Find · B̄

Steps

Step 1: B(He- $4) = 28.3\,\mathrm{MeV} = 4.533e-12\,\mathrm{J}$ .
Step 2: B̄ $= 7.07\,\mathrm{MeV}$ /nucleon.
Step 3: Per-nucleon gain $\approx 4.4\,\mathrm{MeV} \times 5$ nucleons $\approx 17.6\,\mathrm{MeV}$ release per D-T fusion.

Result · B̄(He-

4) = 4.533e-12 / 4 \approx 1.133e-12\,\mathrm{J} = 7.07\,\mathrm{MeV}

/nucleon vs D+T initial

\sim 2.7\,\mathrm{MeV}

/nucleon.

Semi-Empirical Mass Formula→q value nuclear reactionMass–Energy Equivalence→