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Mass-energy Relation

Key idea: Use E = mc^2 to link mass defect to energy release in nuclear reactions, and write exam-ready explanations of mass–energy conservation (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–energy relation

The mass–energy relation is: E = mc² where c = 3.00 × 10⁸ m s⁻¹.

B. Mass defect and energy released

In nuclear processes, we often use changes: Δ E = Δ m c²

If the products have smaller total mass than the reactants, Δ m > 0 and energy is released.

Key phrasing

“In nuclear processes, nucleon number and charge are conserved, and mass–energy is conserved.”

2. Key Ideas (What Earns Marks)

  • A tiny mass corresponds to a huge energy because of the factor c².
  • Energy released in nuclear reactions comes from a decrease in total mass of products vs reactants: Δ E = Δ m c²
  • This underpins:
    • binding energy,
    • energy release in fission and fusion,
    • Q-value calculations (if used).

3. Detailed Explanations

A. What “mass is converted to energy” really means

Mass is not “destroyed”; instead, mass–energy is conserved. If the system releases energy (e.g. as kinetic energy of products and radiation), the rest mass of the products can be smaller by an amount Δ m such that: Δ E = Δ m c²

B. Where the released energy shows up

In many nuclear reactions, released energy appears mainly as:

  • kinetic energy of the reaction products,
  • gamma radiation.

4. Common Mistakes

  • Using E = mc² with m in u but c in SI without converting units.
  • Using the wrong sign for Δ m (define Δ m = m_reactants-m_products to keep it positive when energy is released).
  • Saying “mass isn’t conserved” without clarifying “mass–energy is conserved”.

5. Exam Tips

  • State your definition clearly: Δ m = m_reactants-m_products
  • Then: E_released = Δ m c²
  • If masses are in u, either convert to kg first or use 1u c² ≈ 931 MeV (if allowed/given).

6. Worked Examples

Modelled example 1

Energy equivalent of a small mass

Core

Problem

Find the energy equivalent of m = 2.0 × 10⁻²⁸ kg.
Study the worked solution

  1. Choose the relation

    Method

    Use the mass–energy relation E = mc².

    Reason

    The question asks for the energy equivalent of a stated mass, not an energy change from two system masses.

    Working

    E = mc²

  2. Substitute SI values

    Method

    Insert the mass in kilograms and the speed of light in metres per second.

    Reason

    These SI quantities produce energy in joules.

    Working

    E = (2.0 × 10⁻²⁸)(3.00 × 10⁸)²

  3. Evaluate and round

    Method

    The energy equivalent is 1.8 × 10⁻¹¹ J.

    Reason

    The mass is given to two significant figures.

    Working

    E = 1.8 × 10⁻¹¹ J

Guided practice 2

Energy released from mass defect

About 4 min

Problem

A reaction has mass defect Δ m = 3.0 × 10⁻²⁹ kg. Find the energy released.

Try this before viewing the solution

Unit: J

Hints

Hint 1: select the change relation
Use E_released = Δ m c².
Hint 2: check the power of ten
(10⁸)² = 10¹⁶, so combine 10⁻²⁹ with 10¹⁶.
View solution step by step

  1. Link mass decrease to released energy

    Method

    Use E_released = Δ m c² with the positive mass defect.

    Reason

    The mass defect is defined as reactant mass minus product mass for an energy-releasing reaction.

    Working

    Δ m = m_reactants-m_products > 0

  2. Substitute

    Method

    Insert the SI mass defect and c = 3.00 × 10⁸ m s⁻¹.

    Reason

    No mass-unit conversion is needed because Δ m is already in kilograms.

    Working

    E = (3.0 × 10⁻²⁹)(3.00 × 10⁸)²

  3. State the result

    Method

    The released energy is 2.7 × 10⁻¹² J.

    Reason

    The calculation is reported in joules to two significant figures.

    Working

    E = 2.7 × 10⁻¹² J

Common misconception 3

Energy released from mass defect (in atomic mass units)

Find and correct the mistake

Learner claim

A reaction has mass defect Δ m = 0.015 u. A learner substitutes 0.015 directly into E = mc² with c = 3.00 × 10⁸ m s⁻¹. Diagnose the unit error and estimate the energy released in MeV using 1u c² ≈ 931 MeV.

Try this before viewing the solution

Unit: MeV

View solution step by step

  1. Identify the mismatch

    Method

    The number 0.015 is measured in atomic mass units, not kilograms.

    Reason

    Using SI c directly requires mass in kilograms; otherwise the numerical result does not have joule units.

    Working

    0.015 u ≠ 0.015 kg

  2. Use the compatible conversion

    Method

    Multiply the mass defect in u by the given 931 MeV per u c².

    Reason

    The conversion already incorporates c² and returns the energy in MeV.

    Working

    E = 0.015(931) MeV

  3. Evaluate

    Method

    The estimated energy released is 14.0 MeV.

    Reason

    The calculation retains the source’s requested energy unit.

    Working

    E = 13.965 MeV ≈ 14.0 MeV

Examiner practice 4

Mass defect from released energy

3 marks

Examination question

A nuclear reaction releases E = 8.0 MeV. Find the mass defect Δ m in u. Use 1u c² ≈ 931 MeV. [3 marks]

Try this before viewing the solution

Unit: u

View solution step by step

  1. Rearrange the conversion

    1 mark

    Method

    Divide the released energy by 931 MeV per atomic mass unit.

    Reason

    The supplied equivalence makes energy proportional to mass defect in u.

    Working

    Δ m = E/(931 MeV u⁻¹)

  2. Substitute

    1 mark

    Method

    Use the energy value in the same MeV unit as the conversion.

    Reason

    The MeV units cancel without an SI conversion.

    Working

    Δ m = 8.0/931 u

  3. Report the mass defect

    1 mark

    Method

    The mass defect is 8.59 × 10⁻³ u, or approximately 0.0086 u.

    Reason

    The final unit is an atomic mass unit and the rounded value matches the given data.

    Working

    Δ m = 8.59 × 10⁻³ u ≈ 0.0086 u

Challenge 5

Energy released per kilogram of fuel (order of magnitude)

Minimal support

Independent transfer

If a reaction releases 200 MeV per event, estimate the energy released per kilogram of fuel (order of magnitude). Take 1 MeV = 1.60 × 10⁻¹³ J and assume one event per nucleus with mass m ≈ 4u (helium-scale). Take 1u = 1.66 × 10⁻²⁷ kg.

Try this before viewing the solution

Hints

Hint 1: build a per-kilogram route
Convert the energy per event to joules and find the mass per event in kilograms.
View solution step by step

  1. Convert energy per event

    Method

    One event releases 3.2 × 10⁻¹¹ J.

    Reason

    The final energy per kilogram is required in SI units.

    Working

    Eₑᵥₑₙₜ = 200(1.60 × 10⁻¹³) = 3.2 × 10⁻¹¹ J

  2. Find the fuel mass per event

    Method

    The assumed nucleus has mass 6.64 × 10⁻²⁷ kg.

    Reason

    The prompt assigns one event to each nucleus of mass 4u.

    Working

    mₑᵥₑₙₜ = 4(1.66 × 10⁻²⁷) = 6.64 × 10⁻²⁷ kg

  3. Scale to one kilogram

    Method

    One kilogram contains about 1.5 × 10²⁶ such event masses.

    Reason

    Divide the total fuel mass by the mass used per event.

    Working

    n = 1/(6.64 × 10⁻²⁷) ≈ 1.5 × 10²⁶ kg⁻¹

  4. Combine the factors

    Method

    The estimated energy release is 4.8 × 10¹⁵ J kg⁻¹.

    Reason

    Energy per event multiplied by events per kilogram gives energy per kilogram under the stated one-event-per-nucleus assumption.

    Working

    E/m ≈ (3.2 × 10⁻¹¹)(1.5 × 10²⁶) ≈ 4.8 × 10¹⁵ J kg⁻¹

7. Mind Stretchers

Mind stretcher 1: Why is fission energy per reaction so large compared to chemical reactions?Extension

Show Answer

Nuclear reactions involve changes in binding energy associated with the strong nuclear force, producing much larger mass defects than chemical bond energy changes. Since E = Δ m c², even a tiny mass defect gives a very large energy release.

Mind stretcher 2: Why does a nucleus have a mass defect?Extension

Explain, in terms of energy, why a bound nucleus has a smaller mass than its separated nucleons.

Show Answer

When nucleons bind, energy is released (binding energy). The bound system has lower total energy than the separated nucleons.

Since mass–energy is equivalent, the lower energy corresponds to a smaller mass: the mass defect represents the binding energy via E_b = Δ m c².

8. Optional (Enrichment)

A. Video intuition (optional)

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027