understand the equivalence between energy and mass as represented by E = mc2 and recall and use this equation

Mass–Energy Equivalence, Mass Defect & Nuclear Binding Energy (Cambridge 9702)

Learning Objectives (mapped to Cambridge A‑Level Assessment Objectives)

  • AO1 – Knowledge & understanding: Explain Einstein’s mass‑energy relation \(E=mc^{2}\); define mass defect, nuclear binding energy and the main modes of radioactive decay (α, β⁻, β⁺, γ).
  • AO2 – Application: Calculate mass defects from atomic masses, convert them to binding energies, evaluate Q‑values for nuclear reactions and decays, and use the binding‑energy‑per‑nucleon curve to predict whether fusion or fission releases energy.
  • AO3 – Practical/experimental skills: Design, analyse and evaluate an experiment (e.g. magnetic mass spectrometer) to determine a nucleus’s mass defect.


0. Prerequisite Recap (AS‑Level Foundations)

QuantitySymbolSI unitKey formula
EnergyEJ (or eV)E=mc^{2},\;E=qV,\;E=\tfrac12mv^{2}
Massmkg (or u)m=E/c^{2}
Speed of lightcm s⁻¹c=3.00 × 10⁸ m s⁻¹
Atomic mass unitukg1 u=1.660539 × 10⁻²⁷ kg
Energy–mass conversion1 u c²=931.5 MeV  or  1 MeV c⁻²=1.073 × 10⁻³ u

Key Conservation Laws (relevant to nuclear processes)

  • Energy, linear momentum, electric charge and nucleon number (A) are conserved.
  • Mass is not conserved; the “missing” mass appears as binding energy via \(E=mc^{2}\).


1. Einstein’s Mass–Energy Relation

For any system the total energy \(E\) and its invariant mass \(m\) are linked by

\(E = mc^{2}\)

In nuclear physics we normally use:

  • Energy in mega‑electron‑volts (MeV) or giga‑electron‑volts (GeV).
  • Mass in atomic mass units (u).

Exact conversion (to four significant figures):

1 u c² = 931.5 MeV  or  1 MeV c⁻² = 1.073 × 10⁻³ u

Why the factor matters

When nucleons bind to form a nucleus the total mass of the bound system is less than the sum of the masses of the free nucleons. The difference – the mass defect – multiplied by 931.5 MeV u⁻¹ gives the binding energy released during formation.


2. Mass Defect

Definition: the amount of mass lost when free protons and neutrons combine to form a nucleus.

\(\Delta m = \bigl(Z\,m{p}+N\,m{n}\bigr)-M_{\text{nucl}}\)

  • \(Z\) – number of protons, \(m_{p}=1.007276\;\text{u}\)
  • \(N\) – number of neutrons, \(m_{n}=1.008665\;\text{u}\)
  • \(M_{\text{nucl}}\) – atomic mass of the nuclide (includes electrons).

    For a pure nuclear mass subtract the electron contribution: \(Z\times0.00054858\;\text{u}\).

Important points

  • Masses in the syllabus are given to at least four significant figures.
  • Always check whether you are using atomic or nuclear masses – the electron mass cancels when you compare like‑for‑like.
  • \(\Delta m\) is always positive; the corresponding binding energy is the energy released when the nucleus forms.


3. Nuclear Binding Energy

Binding energy is the energy required to separate a nucleus into its constituent nucleons:

\(E_{b}= \Delta m\,c^{2}= \Delta m \times 931.5\;\text{MeV}\)

Binding energy per nucleon (a measure of stability):

\(\displaystyle \frac{E_{b}}{A}= \frac{\text{total binding energy}}{\text{mass number }A}\)

Stability trend – the binding‑energy‑per‑nucleon curve

  • Rises steeply for light nuclei, peaks at \(A\approx 56\) (iron‑56), then falls slowly for heavier nuclei.
  • Consequences:

    • For \(A<56\) (light nuclei) fusion moves nuclei toward the peak → energy released.
    • For \(A>56\) (heavy nuclei) fission** moves nuclei toward the peak → energy released.

Suggested sketch for the classroom (label the axes, indicate the peak at Fe‑56, and shade the regions where fusion or fission is exothermic).


4. Worked Examples

4.1 Alpha particle – \(^{4}\)He

ParticleMass (u)Number in \(^{4}\)He
Proton (p)1.0072762
Neutron (n)1.0086652
Atomic mass of \(^{4}\)He (incl. 2 e⁻)4.002603

  1. Separate nucleon mass: \(M_{\text{sep}}=2(1.007276)+2(1.008665)=4.031882\;\text{u}\)
  2. Mass defect: \(\Delta m =4.031882-4.002603=0.029279\;\text{u}\)
  3. Binding energy: \(E_{b}=0.029279\times931.5=27.2\;\text{MeV}\)
  4. Binding energy per nucleon: \(E_{b}/A=27.2/4=6.8\;\text{MeV nucleon}^{-1}\)

4.2 Carbon‑12 – \(^{12}\)C (mid‑mass example)

Data: \(m{p}=1.007276\;\text{u},\; m{n}=1.008665\;\text{u},\; M(^{12}\text{C})=12.000000\;\text{u}\) (by definition).

  1. \(M{\text{sep}}=6m{p}+6m_{n}=12.099882\;\text{u}\)
  2. \(\Delta m=12.099882-12.000000=0.099882\;\text{u}\)
  3. \(E_{b}=0.099882\times931.5=93.1\;\text{MeV}\)
  4. \(E_{b}/A=93.1/12=7.8\;\text{MeV nucleon}^{-1}\)

4.3 Iron‑56 – the peak of the curve

Atomic mass \(M(^{56}\text{Fe})=55.934937\;\text{u}\).

  1. \(Z=26,\; N=30\)
  2. \(M_{\text{sep}}=26(1.007276)+30(1.008665)=56.10844\;\text{u}\)
  3. \(\Delta m=56.10844-55.934937=0.173503\;\text{u}\)
  4. \(E_{b}=0.173503\times931.5=161.5\;\text{MeV}\)
  5. \(E_{b}/A=161.5/56=2.89\;\text{MeV nucleon}^{-1}\) (maximum per‑nucleon value in the chart).


5. Radioactive Decay (Syllabus 23.2)

Four principal types of decay are required for the Cambridge syllabus.

Decay typeChange in (Z, A)Particle emittedTypical Q‑value (MeV)
α‑decay\((Z,A)\rightarrow(Z-2,A-4)\)\(^{4}\text{He}\) nucleus (mass 4, charge +2)4–9
β⁻‑decay\((Z,A)\rightarrow(Z+1,A)\)Electron + antineutrino0.1–5
β⁺‑decay (positron emission)\((Z,A)\rightarrow(Z-1,A)\)Positron + neutrino1.02–5
γ‑decayNo change in (Z, A)High‑energy photon0.01–10 (usually follows α or β)

General Q‑value expression

\(Q = \bigl[M{\text{parent}}-M{\text{products}}\bigr]c^{2}\)

Use atomic masses; the electron masses cancel automatically for β⁻ and β⁺ (provided the correct atomic species are used).

Example – α‑decay of \(^{238}\)U

  • \(^{238}\text{U}\rightarrow^{234}\text{Th}+^{4}\text{He}\)
  • Atomic masses: \(M(^{238}\text{U})=238.050788\;\text{u},\; M(^{234}\text{Th})=234.043601\;\text{u},\; M(^{4}\text{He})=4.002603\;\text{u}\)
  • \(Q=[238.050788-(234.043601+4.002603)]\times931.5=4.27\;\text{MeV}\)


6. Practical Skills – Determining a Mass Defect (AO3)


7. Common Pitfalls & How to Avoid Them

  • Electron mass omission: When using atomic masses, remember to subtract \(Z\times0.00054858\;\text{u}\) only if a pure nuclear mass is required.
  • Sign of \(\Delta m\): \(\Delta m\) = (mass of separate nucleons) − (mass of nucleus). It must be positive.
  • Units: Keep the conversion factor 931.5 MeV u⁻¹ separate; do not insert an extra \(c^{2}\) after multiplication.
  • Significant figures: Use four‑significant‑figure masses and the conversion factor; round the final answer to the same number of figures as the least‑precise input.
  • Energy per nucleon vs total energy: \(E_{b}/A\) is a stability indicator; the Q‑value of a reaction is the total energy change.


8. Practice Questions (with marks allocation)

  1. Mass defect & binding energy of \(^{12}\)C (6 marks)


    Given: \(m{p}=1.007276\;\text{u},\; m{n}=1.008665\;\text{u},\; M(^{12}\text{C})=12.000000\;\text{u}\).


    Calculate \(\Delta m\), \(E{b}\) (MeV) and \(E{b}/A\) (MeV nucleon⁻¹).


    [Show all steps, use 931.5 MeV u⁻¹, give answers to three significant figures.]

  2. Binding‑energy per nucleon to total energy (5 marks)


    An isotope has \(E_{b}/A=8.5\;\text{MeV}\) and \(A=56\). Find the total binding energy and the corresponding mass defect (in u).


    [Use the conversion factor, express \(\Delta m\) to four significant figures.]

  3. Q‑value of a nuclear reaction (7 marks)


    Calculate the energy released when \(^{235}\)U undergoes fission into \(^{141}\)Ba, \(^{92}\)Kr and three neutrons.


    Atomic masses:

    • \(^{235}\)U = 235.0439 u
    • \(^{141}\)Ba = 140.9144 u
    • \(^{92}\)Kr = 91.9262 u
    • n = 1.008665 u


    [Explain each step, comment on the magnitude of the result.]

  4. Conceptual – Why does the binding‑energy curve peak at iron? (4 marks)


    Explain in terms of the short‑range attractive nuclear force and the long‑range Coulomb repulsion, and state the implication for fusion and fission as energy‑producing processes.

  5. Practical – Design task (8 marks)


    Outline a method to measure the mass defect of \(^{4}\)He using a magnetic mass spectrometer. Include apparatus, procedure, data analysis, sources of error and safety considerations.

  6. Decay calculation – α‑decay of \(^{238}\)U (5 marks)


    Using the masses given in Section 5, calculate the Q‑value and state whether the decay is energetically allowed.


9. Summary

Einstein’s equation \(E=mc^{2}\) tells us that mass and energy are interchangeable. In the nucleus the binding of protons and neutrons reduces the total mass; the “missing” mass – the mass defect – appears as binding energy. By:

  • Measuring atomic masses,
  • Applying the conversion factor \(1\;\text{u}c^{2}=931.5\;\text{MeV}\),
  • Calculating \(\Delta m\,c^{2}\),

we obtain the binding energy, a key quantity for:

  • Predicting the energy yield of nuclear reactions (fusion, fission, decay).
  • Understanding nuclear stability via the binding‑energy‑per‑nucleon curve.
  • Linking nuclear physics to applications in medicine, energy production and astrophysics.

Mastery of these concepts satisfies the Cambridge A‑Level objectives AO1–AO3 and provides a solid foundation for the rest of the nuclear physics syllabus.