The Mass Defect

Key idea: Define mass defect and calculate Δm using nuclear or atomic masses; connect mass defect to binding energy via ΔE = Δm c^2 (A Level Physics).

GCE A-Level H2 Physics 2027

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Learning objectives

Use mass-energy equivalence, mass defect and binding energy.

1. Definitions (Must Know)

A. Mass defect, Δ m

The mass defect of a nucleus is: Δ m = (total mass of separated nucleons)-(mass of nucleus)

For a nucleus with Z protons and N neutrons: Δ m = Zmₚ + Nmₙ-m_nucleus

For a neutral atom (if atomic masses are used): Δ m = Zmₚ + Nmₙ + Zmₑ-mₐₜₒₘ

2. Key Ideas (What Earns Marks)

A bound nucleus has a smaller mass than the total mass of its separated nucleons.
Mass defect links directly to binding energy: E_b = Δ m c²
In many problems, using atomic masses is easier because electron masses cancel consistently (you just need to be consistent with what mass values you use).

“Mass–energy is conserved: the missing mass corresponds to binding energy released when the nucleus forms.”

3. Detailed Explanations

A. Why mass defect exists

When nucleons bind to form a nucleus, energy is released. That energy comes from a decrease in the system’s mass (mass–energy conservation), giving a positive mass defect Δ m.

B. Choosing a calculation method

Two consistent approaches:

Nuclear masses method: use m_nucleus directly with mₚ and mₙ.
Atomic masses method: use mₐₜₒₘ and include electron masses consistently, or use standard identities that cancel electrons (depending on the data given).

4. Common Mistakes

Mixing nuclear masses and atomic masses in the same calculation without accounting for electrons.
Using the wrong neutron/proton count (N = A-Z is the quick check).
Forgetting to convert u to kg (or to MeV) before using E = Δ m c².

5. Exam Tips

Write A, Z, and N clearly first: N = A-Z
Define your mass defect sign convention (recommended): Δ m = m_separated-m_bound so Δ m > 0 for stable nuclei.
If u is used, keep everything in u until the final conversion.

6. Worked Examples

Problem

A nucleus contains

Z = 10 protons and

N = 10 neutrons. The total mass of the separated nucleons is

20.180 u and the nucleus mass is

19.980 u. Find

Δ m.

Study the worked solution

Use the positive-defect convention
Method
Δ m = m_separated-m_bound.
Reason
A bound nucleus has lower mass-energy than its separated nucleons.
Working
Δ m = 20.180 u-19.980 u
Evaluate
Method
Δ m = 0.200 u.
Reason
The positive result is consistent with positive binding energy.
Working
Δ m = 0.200 u

Problem

Using

Δ m = 0.200 u from Example A, estimate the binding energy in joules. Take

1u = 1.66 × 10⁻²⁷ kg and

c = 3.00 × 10⁸ m s⁻¹.

Try this before viewing the solution

Binding energy

Unit: J

Hints

Hint 1: convert then square c

First multiply

0.200 by the kilograms per atomic mass unit.

View solution step by step

Convert mass defect
Method
Δ m = 3.32 × 10⁻²⁸ kg.
Reason
SI mass is needed for the joule form of mass–energy equivalence.
Working
Δ m = 0.200(1.66 × 10⁻²⁷) = 3.32 × 10⁻²⁸ kg
Calculate binding energy
Method
E_b = 3.0 × 10⁻¹¹ J.
Reason
The bound system’s mass reduction corresponds to released binding energy.
Working
E_b = Δ mc² = (3.32 × 10⁻²⁸)(3.00 × 10⁸)² = 3.0 × 10⁻¹¹ J

Learner claim

A nucleus has

Δ m = 0.120 u and nucleon number

A = 12. A learner calculates

0.120(931) = 112 MeV and reports this as the binding energy per nucleon. Diagnose the final step and find the requested value.

Try this before viewing the solution

Binding energy per nucleon

Unit: MeV

View solution step by step

Identify the total
Method
E_b = 112 MeV is the total binding energy.
Reason
The full nuclear mass defect belongs to all 12 nucleons together.
Working
E_b = 0.120(931) = 112 MeV
Divide by nucleon number
Method
E_b/A = 9.33 MeV per nucleon.
Reason
Binding energy per nucleon is the total divided by A.
Working
E_b/A = 112/12 = 9.33 MeV nucleon⁻¹

Examination question

A nucleus has total binding energy

E_b = 28.3 MeV. Find the mass defect in

u. Use

1u c² = 931 MeV. [2 marks]

Try this before viewing the solution

Conversion and mass defect

View solution step by step

Use the energy equivalent
1 mark
Method
Divide the energy by 931 MeV per u.
Reason
1u corresponds to 931 MeV/c² of mass.
Working
Δ m = 28.3/931 u
Evaluate
1 mark
Method
Δ m = 3.04 × 10⁻² u ≈ 0.030 u.
Reason
The result is the positive separated-minus-bound mass difference.
Working
Δ m = 3.04 × 10⁻² u

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark the energy-to-mass conversion and mass defect.

Mark points

1 mark: Divides by 931 MeV per u.
1 mark: Obtains about 0.030 u.

Independent transfer

A balanced nuclear reaction is evaluated using neutral-atom masses, which include electron masses. Explain why the nuclear mass difference remains valid when atomic masses are used consistently on both sides.

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Consistency and cancellation explanation

Hints

Hint 1: count electrons from charge balance

Compare the total proton number, and hence the electron count for neutral atoms, on each side.

View solution step by step

Use the balanced charge
Method
The total number of electrons included in the neutral-atom masses is the same on both sides.
Reason
Conservation of total proton number gives equal electron totals when all listed atoms are neutral.
Working
balanced total Z → balanced electron count
Cancel the common mass
Method
Electron masses cancel in the reactant-minus-product mass difference.
Reason
The same total electron rest mass is included on both sides.
Working
Δ m = m_reactants-m_products
Interpret the remainder
Method
The remaining difference represents the relevant nuclear mass-energy change.
Reason
Consistent mass conventions preserve the physical difference even though each tabulated atomic mass includes electrons.
Working
consistent atomic masses → valid nuclear energy difference

Next step

Attempt why mass defect differs among nuclei independently.

7. Mind Stretchers

Mind stretcher 1: Why is Δ m not the same for all nuclei?Extension

Show Answer

Different nuclei have different binding energies (strength of binding depends on nucleon number and nuclear structure), so the mass defect varies from nucleus to nucleus.

Mind stretcher 2: Why does a stable nucleus have Δ m > 0?Extension

Using E_b = Δ m c², explain why a stable nucleus has a positive mass defect (under the sign convention Δ m = m_separated-m_bound).

Show Answer

A stable nucleus requires energy input to separate its nucleons to infinity, so its binding energy E_b is positive.

Since E_b = Δ m c² and c² > 0, this implies Δ m > 0 under the convention Δ m = m_separated-m_bound.

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Course and syllabus information

Course: GCE A-Level H2 Physics
Edition: GCE A-Level H2 Physics 2027

The Mass Defect

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Learning objectives

1. Definitions (Must Know)

A. Mass defect, Δ m

2. Key Ideas (What Earns Marks)

3. Detailed Explanations

A. Why mass defect exists

B. Choosing a calculation method

4. Common Mistakes

5. Exam Tips

6. Worked Examples

Mass defect from given masses (in u)

Use the positive-defect convention

Evaluate

Binding energy from mass defect

Convert mass defect

Calculate binding energy

Binding energy per nucleon from mass defect (MeV form)

Identify the total

Divide by nucleon number

Mass defect from binding energy

Use the energy equivalent

Evaluate

Why atomic masses can be used (calculation check)

Use the balanced charge

Cancel the common mass

Interpret the remainder

7. Mind Stretchers

Mind stretcher 1: Why is Δ m not the same for all nuclei?Extension

Mind stretcher 2: Why does a stable nucleus have Δ m > 0?Extension