Nuclear Binding Energy

Equivalent forms

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

Weigh a nucleus carefully enough, and you can read its stability straight off the scale.

Symbol	Name	SI unit	Dimension	Range
$B$	Binding energyoutput Energy required to separate the nucleus into free nucleons	J	M·L²·T⁻²	0 – 1e-10
$Δm$	Mass defect Difference between nucleon mass sum and nuclear mass	kg	M	0 – 1e-27
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Nuclear
$B$	J	erg	MeV
$Δm$	kg	g	u
$c$	m/s	cm/s	1 (with 1 u·c² = 931.494 MeV)

Where it holds

Valid for any bound nuclear system. The mass defect must be computed using precise atomic or nuclear masses. Applies to all nuclei from deuterium

(A=2)

to the heaviest known elements.

Dimensional analysis

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

\Delta m\cdot c^{2} \to [M]\cdot [L\cdot T^{-1}]^{2} = [M\cdot L^{2}\cdot T^{-2}]

Both sides are

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

= Joules.

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Discovery

Francis Aston · 1920

Aston's mass spectrograph revealed that nuclei weigh less than the sum of their nucleons, confirming Einstein's mass-energy equivalence on nuclear scales.

Try this

Why does a helium atom weigh less than its parts?

A helium-4 nucleus has a mass defect of 0.0304 u. What is its binding energy in MeV?

Research status: stable

Real-world applications

Nuclear power generation (fission of uranium releases $\sim 0.9\,\mathrm{MeV}$ /nucleon of binding energy difference)
Stellar nucleosynthesis (fusion of hydrogen to helium releases 6.7 MeV/nucleon)
Medical isotope production (understanding which nuclei are stable enough to be useful)
Nuclear weapons design (maximizing energy release from binding energy differences)

Common misconceptions

Binding energy is not energy stored in the nucleus — it is energy you must ADD to break the nucleus apart. A nucleus with high binding energy is MORE stable, not less.
The mass defect is not 'lost' mass — it is the mass equivalent of the energy radiated away when the nucleus formed.
Binding energy per nucleon, not total binding energy, determines nuclear stability. Iron-56 has maximum B/A, not maximum B.

Experimental verification

Aston's mass spectrograph (1920s) first revealed nuclear mass defects. Modern Penning trap mass spectrometry measures atomic masses to parts in

10^{11}

, confirming binding energies. The binding energy curve (B/A vs A, peaking near iron-56) is reproduced across thousands of nuclides.

Derivation

Consider Z protons and

N = A - Z

neutrons combining to form a nucleus of mass M(A,Z).

The mass defect

is \Delta m = Z\cdot m_p + N\cdot m_n - M

(A,Z).

By Einstein's mass-energy equivalence, the energy released in forming the nucleus (equal to the energy needed to disassemble it) is

B = \Delta m\cdot c^{2}

.

This binding energy accounts for the strong nuclear force overcoming Coulomb repulsion at short range.

Limiting cases

\Delta m \to 0

⟶

B \to 0

Zero mass defect means the nucleus weighs exactly as much as its free nucleons — no binding energy and the nucleus is unbound.

\Delta m

large (e.g., iron-56)⟶

B \approx 492\,\mathrm{MeV}

Iron-56 has one of the largest binding energies per nucleon

(\sim 8.8\,\mathrm{MeV}

/nucleon), making it the most tightly bound common nucleus.

\Delta m

< 0⟶ B < 0 (unphysical for stable nuclei)A negative mass defect would mean the nucleus is heavier than its parts — it would spontaneously disassemble.

What if…

What if the strong force were 2% weaker?

Deuterium would be unbound (B < 0), hydrogen fusion in stars would fail, and the universe would contain almost no elements heavier than hydrogen.

What if we could measure mass defect to infinite precision?

We could predict nuclear reaction energies exactly. In practice, Penning trap measurements at $10^{-11}$ precision already constrain binding energies to sub-keV accuracy.

What if binding energy per nucleon kept increasing with A?

Fusion would be energetically favorable for all elements, not just light ones. Stars would fuse all the way to the heaviest elements instead of stopping at iron.

1

Binding energy of Helium-4

Given ·

Δm u:: 0.0304
c:: 299792458

Find · B

Steps

Step 1: Mass defect $\Delta m = 2(1.007825) + 2(1.008665) - 4.002602 = 0.03038\,\mathrm{u}$ .
Step 2: Convert to energy using $1 u\cdot c^{2} = 931.494\,\mathrm{MeV}$ : $B = 0.03038 \times 931.494 = 28.3\,\mathrm{MeV}$ .
Step 3: Binding energy per nucleon $= 28.3/4 = 7.07\,\mathrm{MeV}$ /nucleon.
Step 4: In SI: $\Delta m = 0.03038 \times 1.66054 \times 10^{-27} = 5.04 \times 10^{-29} kg$ , $B = 5.04 \times 10^{-29} \times (299792458)^{2} = 4.53 \times 10^{-12} J$ .

Result ·

B = 0.0304\,\mathrm{u} \times 931.494\,\mathrm{MeV/u} = 28.3\,\mathrm{MeV}

2

Binding energy of Deuterium

Given ·

Δm u:: 0.002388
c:: 299792458

Find · B

Steps

Step 1: Mass defect $\Delta m = m_p + m_n - m_d = 1.007825 + 1.008665 - 2.014102 = 0.002388\,\mathrm{u}$ .
Step 2: Convert: $B = 0.002388 \times 931.494 = 2.224\,\mathrm{MeV}$ .
Step 3: This is the minimum photon energy needed to photodisintegrate deuterium (observed experimentally as the deuterium photodisintegration threshold).

Result ·

B = 0.002388\,\mathrm{u} \times 931.494\,\mathrm{MeV/u} = 2.224\,\mathrm{MeV}

Mass–Energy Equivalence→Semi-Empirical Mass Formula→Q-Value of Nuclear Reaction→