3.8.2NuclearStretch

Mass-Energy Equivalence and Binding Energy

Weigh a helium-4 nucleus and weigh its two protons and two neutrons separately, and you’ll find something strange: the assembled nucleus weighs slightly $less$ than its individual parts. That "missing" mass hasn’t vanished — it has been converted into the very energy holding the nucleus together, exactly as Einstein’s most famous equation predicts.

45 min Video by Science Shorts 3.8.2 Nuclear energy

Binding Energy — A-level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

State and use $E = m c^{2}$
Explain what is meant by mass defect
Calculate binding energy from mass defect
Calculate binding energy per nucleon and explain its significance
Interpret the binding energy per nucleon curve, including why it peaks near iron

Mass-energy equivalence

Einstein’s equation $E = m c^{2}$ states that mass and energy are two aspects of the same thing: any change in a system’s energy corresponds to a change in its mass, related by the (enormous) constant $c^{2}$ . This is why energy changes in everyday chemistry are utterly unnoticeable as mass changes — but in nuclear reactions, where energy changes are millions of times larger, the associated mass changes become measurable.

E = m c^{2} or Δ E = Δ m c^{2}

Mass–energy equivalence; c = speed of light (3.00 × 10⁸ m s⁻¹).

1Rearrange

Δ E = Δ m c^{2}

Δ m = \frac{Δ E}{c ^{2}}

Δ m = \frac{3.0 \times 1 0 ^{- 11}}{( 3.00 \times 1 0 ^{8} ) ^{2}} = \frac{3.0 \times 1 0 ^{- 11}}{9.00 \times 1 0 ^{16}}

Answer

Δ m \approx 3.33 \times 1 0^{- 28}

kg — tiny, but real and measurable with precise instruments.

Mass defect and binding energy

The mass of any nucleus is always $less$ than the sum of the masses of its separate, unbound protons and neutrons. This difference is the $mass defect$ , $Δ m$ . By mass–energy equivalence, this "missing" mass corresponds to an energy — the $binding energy$ — which is the energy that would be needed to completely separate the nucleus into individual, unbound nucleons (equivalently, the energy released if the nucleus were assembled from separate nucleons).

Δ m = (Z m_{p} + N m_{n}) - m_{nucleus} E_{B} = Δ m c^{2}

Mass defect (sum of separate nucleon masses minus the actual nuclear mass) and binding energy.

1Sum of separate nucleon masses (2 protons + 2 neutrons):

2 (1.6726 \times 1 0^{- 27}) + 2 (1.6749 \times 1 0^{- 27}) = 3.3452 \times 1 0^{- 27} + 3.3498 \times 1 0^{- 27}

2Total

= 6.6950 \times 1 0^{- 27}

kg.

3Mass defect:

Δ m = 6.6950 \times 1 0^{- 27} - 6.6447 \times 1 0^{- 27} = 5.03 \times 1 0^{- 29}

kg.

4Binding energy:

E_{B} = Δ m c^{2} = 5.03 \times 1 0^{- 29} \times (3.00 \times 1 0^{8})^{2}

AnswerMass defect

\approx 5.03 \times 1 0^{- 29}

kg; binding energy

\approx 4.53 \times 1 0^{- 12}

J (about 28 MeV).

Tip — Binding energy is not stored "in" the nucleus as extra energy waiting to escape — it is the energy that must be SUPPLIED to break the nucleus apart, which is exactly why more tightly bound (higher binding energy) nuclei are more stable.

Binding energy per nucleon

Total binding energy naturally increases with a nucleus’s size (more nucleons means more bonds), so it isn’t a fair way to compare the stability of different-sized nuclei. Dividing by the nucleon number gives $binding energy per nucleon$ — a genuine measure of how tightly, on average, each individual nucleon is bound, and hence how stable the nucleus is.

Plotting binding energy per nucleon against nucleon number produces a curve that rises steeply for light nuclei, peaks around iron and nickel ( $A \approx 56$ ), and then decreases slowly for heavier nuclei. Nuclei near the peak are the most stable of all; both very light and very heavy nuclei are less tightly bound per nucleon, and hence less stable.

1Use

\frac{E _{B}}{A} = \frac{4.8 \times 1 0 ^{- 11}}{12}

Answer

4.0 \times 1 0^{- 12}

J per nucleon

Tip — The peak of the binding-energy-per-nucleon curve near iron/nickel is the single most important feature of this graph — everything about fission and fusion releasing energy comes down to nuclei "moving toward" that peak.

Equation recap

Δ E = Δ m c^{2}

Mass–energy equivalence.

E_{B} = Δ m c^{2}

Binding energy from mass defect.

\frac{E _{B}}{A}

Binding energy per nucleon — a measure of nuclear stability.

Common mistakes to avoid

Thinking binding energy is energy contained within (and released spontaneously by) a stable nucleus.

Binding energy is the energy that must be SUPPLIED to pull a nucleus apart — a higher binding energy means a MORE stable, not less stable, nucleus.

Comparing total binding energy directly between nuclei of very different sizes to judge stability.

Always compare binding energy PER NUCLEON, not total binding energy, since larger nuclei naturally have more total binding energy simply from having more nucleons.

Calculating mass defect as (nucleus mass) − (sum of separate nucleon masses).

Mass defect is the sum of SEPARATE nucleon masses MINUS the actual (smaller) nuclear mass — get this the wrong way round and you get a negative, meaningless "mass defect".

Key takeaways

E=mc² relates energy and mass; any energy change corresponds to a mass change, ΔE=Δmc².
Mass defect: the actual nuclear mass is less than the sum of its separate nucleon masses.
Binding energy Eb=Δmc² is the energy needed to separate a nucleus into individual nucleons.
Binding energy per nucleon (Eb/A) measures nuclear stability, peaking near iron/nickel (A≈56).

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Mass-Energy Equivalence and Binding Energy

Weigh a helium-4 nucleus and weigh its two protons and two neutrons separately, and you’ll find something strange: the assembled nucleus weighs slightly

less

than its individual parts. That "missing" mass hasn’t vanished — it has been converted into the very energy holding the nucleus together, exactly as Einstein’s most famous equation predicts.

45 min Video by Science Shorts 3.8.2 Nuclear energy