Mass-Energy — AQA A-level Physics

SPEC 3.8.1.6 • Nuclear Physics

Mass and Energy

Mass defect, binding energy, fission and fusion

Learning Objectives

By the end of this lesson, you will be able to:

Objectives

Apply

E = m c^{2}

to calculate energy from mass changes

Use the atomic mass unit (u) and convert using 1 u = 931.5 MeV

Calculate mass defect and binding energy for nuclei

Interpret the binding energy per nucleon curve

Calculate energy released in fission and fusion reactions

Mass-Energy Equivalence

Einstein's famous equation shows that mass and energy are equivalent and interchangeable:

Key Equation

E = m c^{2}

where

c = 3.00 \times 1 0^{8}

m s

^{- 1}

Key Point

This equation applies to all energy changes, not just nuclear reactions. Even heating water increases its mass (by an immeasurably small amount).

Why Nuclear?

Nuclear reactions involve significant mass changes because nuclear binding energies are millions of times larger than chemical bond energies.

The Atomic Mass Unit

Nuclear masses are inconveniently small in kilograms. We use the atomic mass unit (u) instead:

Definition

The atomic mass unit (u) is defined as exactly one-twelfth of the mass of a carbon-12 atom.

1 u = 1.661 \times 1 0^{- 27} kg

ParticleMass (u)Mass (kg)

Proton1.007276 u

1.673 \times 1 0^{- 27}

kg

Neutron1.008665 u

1.675 \times 1 0^{- 27}

kg

Electron0.000549 u

9.109 \times 1 0^{- 31}

kg

Mass-Energy Conversion

Using

E = m c^{2}

, we can convert the atomic mass unit to energy:

Key Conversion

1 u = 931.5 MeV

This is derived as follows:

Derivation

E = m c^{2} = (1.661 \times 1 0^{- 27}) \times (3.00 \times 1 0^{8})^{2}

E = 1.494 \times 1 0^{- 10}

J

Converting to MeV:

\frac{1.494 \times 1 0 ^{- 10}}{1.60 \times 1 0 ^{- 13}} = 931.5

MeV

Exam Tip

Use 1 u = 931.5 MeV directly to avoid lengthy conversions through joules. This is much faster in calculations!

Mass Defect

When nucleons bind together, the nucleus has less mass than the sum of its parts:

Definition

Mass defect (Δm) is the difference between the total mass of the separate nucleons and the mass of the nucleus.

Δ m = Z m_{p} + N m_{n} - M_{nucleus}

Why is mass "lost"?

When nucleons bind, energy is released. By

E = m c^{2}

, this energy corresponds to a decrease in mass. The "missing" mass has become binding energy.

Binding Energy

Definition

Binding energy (BE) is the energy required to completely separate all the nucleons in a nucleus to infinity.

BE = Δ m \times c^{2}

or equivalently:

BE = Δ m \times 931.5

MeV (when Δm is in u)

Common Misconception

Binding energy is NOT "energy stored in the nucleus". It is the energy that was released when the nucleus formed, and would need to be supplied to break it apart.

Key Insight

Higher binding energy = nucleus is more tightly bound = more stable

Exam Tip

In these calculations, more significant figures than usual must be used because of the tiny mass changes. Keep 5-6 s.f. in working, then round final answer.

Worked Example 1

01Mass-Energy Conversion

Show that the energy equivalent of 1 u is 931.5 MeV.

Data: 1 u = 1.661 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.60 × 10⁻¹⁹ J

1

Use

E = m c^{2}

2

E = (1.661 \times 1 0^{- 27}) \times (3.00 \times 1 0^{8})^{2}

3

E = 1.661 \times 1 0^{- 27} \times 9.00 \times 1 0^{16}

4

E = 1.4949 \times 1 0^{- 10}

J

5

Convert to eV:

\frac{1.4949 \times 1 0 ^{- 10}}{1.60 \times 1 0 ^{- 19}} = 9.343 \times 1 0^{8}

eV

6

Convert to MeV:

9.343 \times 1 0^{8} \div 1 0^{6} = 934

MeV ≈ 931.5 MeV

Answer1 u = 931.5 MeV (verified)

Worked Example 2

02Mass Defect Calculation

Calculate the mass defect of a helium-4 nucleus (

_{2}^{4}

He).

Data: m_p = 1.007276 u, mₙ = 1.008665 u, M(He-4 nucleus) = 4.001506 u

1

Identify constituents: He-4 has Z = 2 protons and N = 2 neutrons

2

Total mass of separate nucleons:

•

m_{sep} = 2 (1.007276) + 2 (1.008665)

•

m_{sep} = 2.014552 + 2.017330 = 4.031882

u

3

Mass defect:

Δ m = m_{sep} - M_{nucleus}

•

Δ m = 4.031882 - 4.001506 = 0.030376

u

AnswerΔm = 0.0304 u (3 s.f.)

Worked Example 3

03Binding Energy Calculation

Calculate the binding energy of helium-4 in MeV, using the mass defect from Example 2.

Data: Δm = 0.030376 u, 1 u = 931.5 MeV

1

Use

BE = Δ m \times 931.5

MeV

2

BE = 0.030376 \times 931.5

3

BE = 28.30

MeV

AnswerBE = 28.3 MeV

Binding Energy per Nucleon

To compare stability of different nuclei, we calculate binding energy per nucleon:

Definition

BE per nucleon = \frac{BE}{A}

where A = nucleon number (mass number)

Key Principle

Higher BE/A = more stable nucleus

This means more energy would be needed (per nucleon) to break it apart.

NucleusABE (MeV)BE/A (MeV)

Helium-4428.37.08

Iron-56564928.79

Uranium-23523517847.59

Worked Example 4

BE per Nucleon Comparison04

Calculate the binding energy per nucleon for (a) helium-4 (BE = 28.30 MeV) and (b) iron-56 (BE = 492.26 MeV). Which is more stable?

1

(a) Helium-4:

•

\frac{BE}{A} = \frac{28.30}{4} = 7.08

MeV per nucleon

2

(b) Iron-56:

•

\frac{BE}{A} = \frac{492.26}{56} = 8.79

MeV per nucleon

3

Iron-56 has higher BE/A, so it is more stable.

Answer(a) 7.08 MeV, (b) 8.79 MeV. Iron-56 is more stable.

The Binding Energy per Nucleon Curve

Key Features

Peak at Fe-56: Iron-56 has the highest BE/A (~8.8 MeV) — most stable nucleus

Light nuclei (A < 56): Can release energy by fusion (moving up the curve)

Heavy nuclei (A > 56): Can release energy by fission (moving up the curve)

Fission and Fusion

Fission

Heavy nucleus splits into two smaller nuclei (plus neutrons).

Products have higher BE/A → energy released.

Fusion

Light nuclei combine to form a heavier nucleus.

Product has higher BE/A → energy released.

Energy Release Principle

In both processes, products are more stable (higher BE/A). Energy =

Σ B E_{products} - Σ B E_{reactants}

Why is fusion difficult?

Fusion needs extremely high temperatures (millions °C) to overcome electrostatic repulsion between positive nuclei.

Worked Example 5

05Energy from Fission

Calculate the energy released when uranium-235 undergoes fission:

_{92}^{235} U +_{0}^{1} n \to_{56}^{141} Ba +_{36}^{92} Kr + 3_{0}^{1} n

Data: U-235: 235.0439 u, Ba-141: 140.9144 u, Kr-92: 91.9262 u, n: 1.0087 u

1

Total mass of reactants:

•

m_{react} = 235.0439 + 1.0087 = 236.0526

u

2

Total mass of products:

•

m_{prod} = 140.9144 + 91.9262 + 3 (1.0087) = 235.8667

u

3

Mass difference:

•

Δ m = 236.0526 - 235.8667 = 0.1859

u

4

Energy released:

•

E = 0.1859 \times 931.5 = 173.2

MeV

AnswerE = 173 MeV

Worked Example 6

06Energy from Fusion

Calculate the energy released in deuterium-tritium fusion:

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n

Data: D (H-2): 2.0141 u, T (H-3): 3.0161 u, He-4: 4.0026 u, n: 1.0087 u

1

Total mass of reactants:

•

m_{react} = 2.0141 + 3.0161 = 5.0302

u

2

Total mass of products:

•

m_{prod} = 4.0026 + 1.0087 = 5.0113

u

3

Mass difference:

•

Δ m = 5.0302 - 5.0113 = 0.0189

u

4

Energy released:

•

E = 0.0189 \times 931.5 = 17.6

MeV

AnswerE = 17.6 MeV

Worked Example 7

Comparing Fission and Fusion07

Compare the energy released per nucleon in (a) U-235 fission (E = 173 MeV, A = 236 nucleons involved) and (b) D-T fusion (E = 17.6 MeV, A = 5 nucleons involved). Which process releases more energy per nucleon?

1

(a) Fission:

•

Energy per nucleon =

\frac{173}{236} = 0.73

MeV per nucleon

2

(b) Fusion:

•

Energy per nucleon =

\frac{17.6}{5} = 3.5

MeV per nucleon

3

Fusion releases ~5× more energy per nucleon than fission.

AnswerFusion releases significantly more energy per nucleon (3.5 MeV vs 0.73 MeV).

Worked Example 8

08Using Binding Energies

Use binding energy per nucleon values to estimate the energy released in D-T fusion.

Data: BE/A: D = 1.11 MeV, T = 2.83 MeV, He-4 = 7.07 MeV

1

Total BE of reactants:

•

B E_{react} = 2 \times 1.11 + 3 \times 2.83 = 2.22 + 8.49 = 10.71

MeV

2

Total BE of products (He-4 only, neutron has no BE):

•

B E_{prod} = 4 \times 7.07 = 28.28

MeV

3

Energy released = increase in binding energy:

•

E = 28.28 - 10.71 = 17.6

MeV

AnswerE ≈ 17.6 MeV (consistent with mass defect method)

Practice Question 1

Calculate the mass defect and binding energy of oxygen-16.

Data: m_p = 1.007276 u, mₙ = 1.008665 u, M(O-16 nucleus) = 15.9949 u

Oxygen-16 has 8 protons and 8 neutrons.

1

Mass of separate nucleons:

•

m = 8 (1.007276) + 8 (1.008665) = 8.058208 + 8.069320 = 16.1275

u

2

Mass defect:

•

Δ m = 16.1275 - 15.9949 = 0.1326

u

3

Binding energy:

•

BE = 0.1326 \times 931.5 = 123.5

MeV

AnswerΔm = 0.133 u, BE = 124 MeV

Practice Question 2

In a fission reaction, a nucleus of mass 236.05 u splits into products with total mass 235.86 u.

Calculate the energy released in MeV and in joules.

1

Mass difference:

•

Δ m = 236.05 - 235.86 = 0.19

u

2

Energy in MeV:

•

E = 0.19 \times 931.5 = 177

MeV

3

Convert to joules:

•

E = 177 \times 1 0^{6} \times 1.60 \times 1 0^{- 19} = 2.83 \times 1 0^{- 11}

J

AnswerE = 177 MeV = 2.83 × 10⁻¹¹ J

Practice Question 3

The diagram shows the binding energy per nucleon curve.

(a) Explain why iron-56 is the most stable nucleus.

(b) Explain why energy is released in both fission of heavy nuclei and fusion of light nuclei.

(c) State why hydrogen-1 has zero binding energy.

(a)

Iron-56 has the highest binding energy per nucleon (~8.8 MeV). This means more energy would be required to separate each nucleon from the nucleus, so it is the most tightly bound and therefore most stable.

(b)

Both processes result in products with higher BE/A than reactants. Moving "up" the curve (towards Fe-56) from either direction means the products are more tightly bound. The increase in total binding energy is released.

(c)

Hydrogen-1 contains only one proton (no neutrons). There are no nucleons to bind together, so there is no binding energy. It would require zero energy to "separate" a single nucleon.

Practice Question 4

In the Sun, hydrogen nuclei fuse to form helium in a series of reactions. The overall process is:

4_{1}^{1} H \to_{2}^{4} He + 2 e^{+} + 2 ν_{e}

Calculate the energy released per reaction.

Data: m(H-1) = 1.007825 u, m(He-4) = 4.002603 u, m(e⁺) = 0.000549 u

1

Mass of reactants:

•

m_{react} = 4 \times 1.007825 = 4.031300

u

2

Mass of products (neutrinos have negligible mass):

•

m_{prod} = 4.002603 + 2 (0.000549) = 4.003701

u

3

Mass difference:

•

Δ m = 4.031300 - 4.003701 = 0.027599

u

4

Energy released:

•

E = 0.027599 \times 931.5 = 25.7

MeV

AnswerE = 25.7 MeV per fusion event

Summary

Key Equations

E = m c^{2}

— mass-energy equivalence (applies to ALL energy changes)

1 u = 931.5 MeV

— mass-energy conversion

Δ m = Z m_{p} + N m_{n} - M_{nucleus}

— mass defect

BE = Δ m \times c^{2}

or

Δ m \times 931.5

MeV — binding energy

Key Concepts

Mass defect: Mass "lost" when nucleons bind (converted to energy)

Binding energy: Energy to completely separate all nucleons

Higher BE/A = more stable (Fe-56 is most stable)

Fission: Heavy nuclei split → products have higher BE/A

Fusion: Light nuclei combine → product has higher BE/A

Exam Skills

Always use 1 u = 931.5 MeV for quick conversions. Calculate Δm first, then multiply by 931.5 to get energy in MeV. Check your mass defect is positive!