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A Level · Cambridge (CIE) · 9702

Define mass defect and nuclear binding energy

Compare a nucleus with its separated nucleons and calculate the energy equivalent of a mass defect.

Nuclear physics pathway · Cambridge 9702 A Level · 23.1

A Level · Cambridge International Physics 9702, 2025–2027 syllabus. The Topic 23 outcomes were also checked against Cambridge’s 2028–2030 syllabus, which reports no significant teaching changes. Online practice is self-checked and cannot replace supervised practical work, full timed assessment or human subject-teacher review.

Before you begin

  • Use scientific notation, SI units, proportional reasoning and the particle numbers in nuclide notation.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What does the mass defect compare?

What you will learn

  • Define mass defect
  • Define nuclear binding energy
  • Calculate energy released from a supplied mass change using E = c²Δm

The mass defect compares a bound nucleus with its nucleons

The mass defect is the difference between the sum of the masses of the separate nucleons and the mass of the nucleus. The nucleus has less mass than those same separated particles because energy was released when the bound nucleus formed.

Use only the masses specified in the question and keep their units consistent. If they are given in atomic mass units, convert the mass difference to kilograms using the supplied conversion before applying E = c²Δm.

Binding energy is the energy needed to separate a nucleus

The nuclear binding energy is the energy required to separate a nucleus completely into its constituent protons and neutrons. It is equal to the energy equivalent of the mass defect, ΔE = c²Δm.

The same magnitude is released when the separated nucleons form that nucleus. State whether a question asks for mass defect, binding energy or energy released; the numerical magnitude may be the same while the physical description differs.

Pause and explain

What is the binding energy of a nucleus physically?

Put the idea to work

Worked example

The mass defect of a nucleus is 3.32 × 10⁻²⁹ kg. Calculate its binding energy. Use c = 3.00 × 10⁸ m s⁻¹.

Show the worked solution
  1. Write ΔE = c²Δm.
  2. Square the speed of light: c² = 9.00 × 10¹⁶ m² s⁻².
  3. Multiply: ΔE = 9.00 × 10¹⁶ × 3.32 × 10⁻²⁹ = 2.99 × 10⁻¹² J.

Answer 2.99 × 10⁻¹² J (3 s.f.)

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

A supplied set of separate nucleons has total mass 4.020 × 10⁻²⁶ kg; the nucleus mass is 4.000 × 10⁻²⁶ kg. Find the mass defect.

Give me a hint

Subtract the bound-nucleus mass from the separate-nucleon total.

Compare my reasoning
  1. Δm = 4.020 × 10⁻²⁶ − 4.000 × 10⁻²⁶ kg.
  2. Keep the common power of ten.
  3. Δm = 0.020 × 10⁻²⁶ kg = 2.0 × 10⁻²⁸ kg.

2.0 × 10⁻²⁸ kg.

Look for these in your work

  • Uses separate minus bound mass
  • Subtracts the coefficients
  • Normalises the power of ten
2 · Independent

A reaction releases 4.5 × 10⁻¹² J. Calculate the corresponding mass decrease using c = 3.0 × 10⁸ m s⁻¹.

Give me a hint

Rearrange E = c²Δm to Δm = E/c².

Compare my reasoning
  1. c² = 9.0 × 10¹⁶ m² s⁻².
  2. Δm = 4.5 × 10⁻¹² / 9.0 × 10¹⁶.
  3. Δm = 5.0 × 10⁻²⁹ kg.

5.0 × 10⁻²⁹ kg.

Look for these in your work

  • Rearranges for Δm
  • Squares c
  • Reports kilograms
3 · Transfer

Explain why a bound nucleus can have less mass than the same nucleons separated.

Give me a hint

Connect the mass difference with the energy released during binding.

Compare my reasoning
  1. Binding the nucleons releases energy.
  2. Mass and energy are equivalent through E = mc².
  3. The bound system therefore has a smaller rest mass by an amount corresponding to the released energy.

The mass decrease is the mass equivalent of the binding energy released.

Look for these in your work

  • States energy is released on binding
  • Uses mass–energy equivalence
  • Connects the decrease to binding energy

Common mistakes

  • Reversing the order and reporting a negative mass defect
  • Using total nuclear mass instead of the difference in masses
  • Forgetting that the binding-energy unit is joule after conversion to kilograms
Recall without your notes

What can you explain now?

What does the mass defect compare?

Compare with the explanation

The total mass of separate nucleons with the mass of the bound nucleus

Subtract the nucleus mass from the sum for its separate nucleons.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

How are mass defect and binding energy related by E = mc²?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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