Nuclear Reactions

Sitio: Young Education
Curso: Nuclear and Particle Physics
Libro: Nuclear Reactions
Impreso por: ゲストユーザ
Fecha: viernes, 25 de septiembre de 2026, 02:37

1. Radioactive Decay

Learning outcomes
  • I can distinguish radioactive decay from particle interactions.
  • I can identify alpha, beta, and gamma decay.
  • I can explain how nuclei change during decay.
  • I can write nuclear decay equations.
  • I can apply conservation laws to decay equations.

What Is Radioactive Decay?

Radioactive decay is the spontaneous transformation of an unstable atomic nucleus into a more stable state.

During radioactive decay, a nucleus may emit:

  • an alpha particle
  • a beta particle
  • a gamma-ray photon
  • other particles in more advanced decay processes

The original unstable nucleus is called the parent nucleus.

The nucleus formed after the decay is called the daughter nucleus.

A general decay can be represented as:

Parent nucleus → daughter nucleus + emitted radiation

Radioactive decay happens naturally and does not need to be triggered by another particle.

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5

Radioactive Decay vs Particle Interactions

Radioactive decay is different from a particle collision or other particle interaction.

In radioactive decay:

one unstable nucleus spontaneously changes

For example:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

In a particle interaction, two or more particles may interact because they meet or collide:

A + B → C + D

The key difference is:

Radioactive decay begins with an unstable nucleus on its own.

Particle interactions involve particles interacting with one another.


Comparing the Two

Radioactive Decay Particle Interaction
Begins with an unstable nucleus Usually involves two or more particles
Occurs spontaneously Requires particles to interact
Produces daughter nucleus and radiation Can produce many different particles
Governed by conservation laws Governed by conservation laws
Example: alpha decay Example: high-energy collision

Both processes obey fundamental conservation principles.


Why Do Nuclei Undergo Radioactive Decay?

Some nuclei are unstable because their arrangement of protons and neutrons has too much energy or an unfavorable balance.

Factors affecting nuclear stability include:

  • neutron-to-proton ratio
  • electrostatic repulsion between protons
  • nuclear shell structure
  • binding energy
  • overall nuclear size

An unstable nucleus can transform into a lower-energy, more stable configuration.

The excess energy may be released as particles or electromagnetic radiation.


The Three Main Types of Radioactive Decay

The three radioactive processes most commonly introduced are:

alpha decay

beta decay

gamma decay

Each affects the nucleus differently.


Alpha Decay

An alpha particle consists of:

2 protons + 2 neutrons

It is the same nuclear composition as a helium-4 nucleus.

Its symbol is:

⁴₂He

or:

α

Alpha decay is common in very heavy nuclei.


Example of Alpha Decay

Uranium-238 can undergo alpha decay:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

Let's examine what changed.

Mass number:

238 → 234

Difference:

−4

Atomic number:

92 → 90

Difference:

−2

Therefore, during alpha decay:

mass number decreases by 4

atomic number decreases by 2

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5

General Alpha Decay Equation

Alpha decay can be represented generally as:

ᴬ_ZX → ᴬ⁻⁴_Z₋₂Y + ⁴₂He

where:

A = mass number

Z = atomic number

After alpha decay:

A → A − 4

Z → Z − 2

Because the atomic number changes, the nucleus becomes a different element.


Example: Completing an Alpha Decay Equation

Suppose:

²²⁶₈₈Ra → ? + ⁴₂He

Mass number of daughter:

226 − 4 = 222

Atomic number:

88 − 2 = 86

Element 86 is radon.

Therefore:

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He


Beta-Minus Decay

In beta-minus decay, a neutron in the nucleus changes into a proton.

At the nucleon level:

n → p + e⁻ + ν̄ₑ

The electron is emitted from the nucleus as a beta-minus particle.

An electron antineutrino is also produced.


What Changes During Beta-Minus Decay?

Because a neutron becomes a proton:

number of protons increases by 1

number of neutrons decreases by 1

The total number of nucleons does not change.

Therefore:

mass number stays the same

atomic number increases by 1


Example of Beta-Minus Decay

Carbon-14 undergoes beta-minus decay:

¹⁴₆C → ¹⁴₇N + e⁻ + ν̄ₑ

Before:

Carbon has:

6 protons

8 neutrons

After:

Nitrogen has:

7 protons

7 neutrons

The mass number remains:

14

but the atomic number changes:

6 → 7

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4

General Beta-Minus Equation

The general form is:

ᴬ_ZX → ᴬ_Z₊₁Y + e⁻ + ν̄ₑ

Sometimes beta-minus particles are written using nuclear-style notation as:

⁰₋₁e

or:

⁰₋₁β

So the equation may also appear as:

ᴬ_ZX → ᴬ_Z₊₁Y + ⁰₋₁β + ν̄ₑ


Why Does the Mass Number Stay the Same?

Beta decay does not remove a nucleon from the nucleus.

Instead:

one neutron changes into one proton

The nucleus still contains the same total number of nucleons.

Therefore:

A remains unchanged


Beta-Plus Decay

Another type of beta decay is beta-plus decay.

In a suitable proton-rich nucleus, a proton effectively changes into a neutron:

p → n + e⁺ + νₑ

The emitted positron is called a beta-plus particle.


What Changes During Beta-Plus Decay?

A proton becomes a neutron.

Therefore:

proton number decreases by 1

neutron number increases by 1

Mass number remains unchanged.

So:

A stays the same

Z decreases by 1


Example of Beta-Plus Decay

Carbon-11 can undergo beta-plus decay:

¹¹₆C → ¹¹₅B + e⁺ + νₑ

Mass number:

11 → 11

Atomic number:

6 → 5

Carbon becomes boron.


General Beta-Plus Equation

ᴬ_ZX → ᴬ_Z₋₁Y + e⁺ + νₑ

or:

ᴬ_ZX → ᴬ_Z₋₁Y + ⁰₊₁β + νₑ


Gamma Decay

A nucleus can sometimes have the correct number of protons and neutrons but still contain excess energy.

Such a nucleus is said to be in an excited state.

The nucleus can release this energy by emitting a gamma-ray photon.

Gamma radiation is high-energy electromagnetic radiation.

A gamma photon is represented as:

γ


Gamma Decay Equation

A simple representation is:

X → X + γ*

The asterisk indicates that the original nucleus is excited.

For example:

⁹⁹ᵐTc → ⁹⁹Tc + γ

During gamma decay:

mass number does not change

atomic number does not change

Only the nuclear energy state changes.

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5

Comparing Alpha, Beta, and Gamma Decay

Decay Emitted Particle/Radiation Change in A Change in Z
Alpha ⁴₂He −4 −2
Beta-minus e⁻ + ν̄ₑ 0 +1
Beta-plus e⁺ + νₑ 0 −1
Gamma γ 0 0

This table is extremely useful when writing nuclear equations.


A Useful Memory Pattern

Alpha

A − 4

Z − 2

Beta-minus

A unchanged

Z + 1

Beta-plus

A unchanged

Z − 1

Gamma

A unchanged

Z unchanged


Writing Nuclear Decay Equations

A nuclear decay equation must balance important quantities.

For introductory nuclear equations, always check:

mass number

and

atomic number

These reflect conservation of nucleon number and electric charge in the nuclear notation.

For more complete particle descriptions, also consider:

  • electric charge
  • baryon number
  • lepton number
  • energy
  • momentum

Example: Balancing Alpha Decay

Consider:

²¹⁰₈₄Po → ? + ⁴₂He

Mass numbers must balance:

210 = A + 4

Therefore:

A = 206

Atomic numbers must balance:

84 = Z + 2

Therefore:

Z = 82

Element 82 is lead.

So:

²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He


Checking Conservation in Alpha Decay

For:

²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He

Mass number:

210 = 206 + 4

Atomic number:

84 = 82 + 2

Both balance.


Example: Balancing Beta-Minus Decay

Suppose:

³H → ? + e⁻ + ν̄ₑ

Tritium is hydrogen-3:

³₁H

During beta-minus decay:

A stays 3

Z increases from 1 to 2

Element 2 is helium.

Therefore:

³₁H → ³₂He + e⁻ + ν̄ₑ


Checking Charge More Carefully

For:

³₁H → ³₂He + e⁻ + ν̄ₑ

The nuclear charge before is:

+1

After:

Helium nucleus:

+2

Electron:

−1

Antineutrino:

0

Total:

+2 − 1 = +1

Charge is conserved.


Applying Lepton Number

Consider beta-minus decay:

n → p + e⁻ + ν̄ₑ

Before:

L = 0

After:

Electron:

L = +1

Antineutrino:

L = −1

Total:

L = 0

Lepton number is conserved.

This is one reason the antineutrino must appear in the complete decay equation.


Example: Why This Equation Is Incomplete

Suppose someone writes:

n → p + e⁻

Charge is conserved:

0 = +1 − 1

Baryon number is conserved:

+1 = +1

But lepton number is not.

Before:

L = 0

After:

L = +1

The missing particle is the electron antineutrino:

ν̄ₑ

Therefore the complete equation is:

n → p + e⁻ + ν̄ₑ


Conservation Laws in Alpha Decay

An alpha particle contains:

2 protons + 2 neutrons

Therefore its baryon number is:

B = 4

At the nuclear level, baryon number is equivalent to keeping track of the total number of nucleons.

For example:

²³⁸U → ²³⁴Th + ⁴He

Baryon number:

238 = 234 + 4

So baryon number is conserved.


Conservation Laws in Gamma Decay

Gamma decay:

X → X + γ*

The photon has:

charge = 0

baryon number = 0

lepton number = 0

The nucleus keeps the same number of protons and neutrons.

Energy and momentum are also conserved.

The photon carries away some of the excess energy and momentum.


Radioactive Decay Releases Energy

Radioactive decay occurs when the total mass-energy of the final state is lower than that of the initial state.

The difference appears as energy carried by:

  • emitted particles
  • daughter nucleus recoil
  • electromagnetic radiation

The released energy can be calculated using:

Q = (minitial − mfinal)c²

If:

Q > 0

the decay is energetically allowed.


Recoil of the Daughter Nucleus

Suppose an initially stationary nucleus emits an alpha particle.

The alpha particle travels in one direction.

Momentum must remain conserved.

Therefore the daughter nucleus recoils in the opposite direction.

So:

momentum before = 0

and:

momentum after = palpha + pdaughter = 0

This is why the daughter nucleus cannot simply remain completely stationary after particle emission.


Radioactive Decay Is Random

It is impossible to predict exactly when one particular unstable nucleus will decay.

A specific nucleus might decay:

  • almost immediately
  • much later
  • after an extremely long time

However, for a large number of identical unstable nuclei, the statistical behavior is very predictable.

This leads to the concept of half-life, which describes how quickly a radioactive sample decays.

We will treat half-life separately because it focuses on the statistical rate of radioactive decay rather than the nuclear transformations themselves.


Penetrating Ability

Alpha, beta, and gamma radiation interact with matter differently.

Alpha

Alpha particles are relatively massive and carry charge +2.

They interact strongly with matter and lose energy quickly.

They have:

  • high ionizing ability
  • low penetrating ability

Beta

Beta particles are electrons or positrons.

They are much lighter than alpha particles.

They typically have:

  • moderate ionizing ability
  • moderate penetrating ability

Gamma

Gamma rays are photons.

They carry no electric charge.

They typically have:

  • lower ionizing ability per interaction
  • high penetrating ability

Thick materials may be required to substantially reduce gamma radiation.

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4

Ionizing Radiation

Alpha, beta, and gamma radiation can all ionize matter.

Ionization occurs when enough energy is transferred to remove electrons from atoms or molecules.

Ionizing radiation can therefore alter chemical structures and damage biological molecules.

The degree of biological effect depends on factors including:

  • radiation type
  • energy
  • absorbed dose
  • exposure time
  • tissue exposed

Radioactive Decay Chains

Sometimes one radioactive decay does not immediately produce a stable nucleus.

The daughter nucleus may itself be radioactive.

It may then decay again.

This creates a decay chain.

For example:

unstable parent → radioactive daughter → another daughter → … → stable nucleus

Heavy radioactive elements such as uranium can undergo long decay chains containing several alpha and beta decays.


Example: Tracking Changes Through Several Decays

Suppose a nucleus undergoes:

one alpha decay

followed by:

two beta-minus decays

Start with:

A = 238

Z = 92

After alpha:

A = 234

Z = 90

After first beta-minus:

A = 234

Z = 91

After second beta-minus:

A = 234

Z = 92

Final:

A = 234

Z = 92

This method allows us to follow changes through a decay chain without memorizing every isotope.


Worked Example: Identify the Decay Type

Consider:

²¹⁴₈₂Pb → ²¹⁴₈₃Bi + ?

Compare the nuclei.

Mass number:

214 → 214

No change.

Atomic number:

82 → 83

Increase of 1.

This is beta-minus decay.

Therefore the missing products include:

e⁻ + ν̄ₑ

So:

²¹⁴₈₂Pb → ²¹⁴₈₃Bi + e⁻ + ν̄ₑ


Worked Example: Identify the Missing Daughter

Consider:

²²²₈₆Rn → X + ⁴₂He

Mass number:

222 − 4 = 218

Atomic number:

86 − 2 = 84

Element 84 is polonium.

Therefore:

²²²₈₆Rn → ²¹⁸₈₄Po + ⁴₂He


Worked Example: Gamma Emission

Suppose an excited cobalt nucleus emits gamma radiation:

⁶⁰Co → ⁶⁰Co + γ*

Mass number:

60 → 60

Atomic number:

27 → 27

No element change occurs.

Only the nuclear energy decreases.


A Reliable Method for Nuclear Decay Equations

When completing a radioactive decay equation:

Step 1: Identify the decay type.

Is it:

  • alpha
  • beta-minus
  • beta-plus
  • gamma

Step 2: Write the known particles.

Include the parent nucleus and any emitted particles.

Step 3: Balance mass number.

Check that:

total A before = total A after

Step 4: Balance atomic number.

Check that:

total Z before = total Z after

Step 5: Identify the element.

Use the atomic number to determine the daughter element.

Step 6: Check additional conservation laws.

For beta decay especially, consider:

  • electric charge
  • baryon number
  • lepton number

Step 7: Remember energy and momentum.

Every physical decay must conserve both.


Common Mistakes

Mistake 1: Changing mass number in beta decay

Beta decay changes a neutron into a proton or vice versa.

The total number of nucleons remains the same.

So:

A does not change.


Mistake 2: Treating gamma radiation as a particle with mass number

A gamma photon has:

A = 0

Z = 0

Gamma decay does not change the element.


Mistake 3: Forgetting the neutrino

Complete beta-decay equations require the appropriate neutrino or antineutrino.

Beta-minus:

e⁻ + ν̄ₑ

Beta-plus:

e⁺ + νₑ


Mistake 4: Thinking alpha decay removes only two particles

An alpha particle contains four nucleons:

2 protons + 2 neutrons

So mass number decreases by 4.


Did You Know?

The word radioactive does not mean that an object is continuously firing out all of its nuclear energy at once.

Each unstable nucleus has a probability of decaying during a given interval.

Some radioactive isotopes decay very rapidly, while others have half-lives of millions or even billions of years.

This is why radioactive materials can be useful for very different purposes, from medical imaging to determining the age of ancient rocks.


Connecting the Ideas

Radioactive decay brings together many concepts from this course:

Unstable nucleus

↓

Nuclear transformation

↓

Alpha, beta, or gamma emission

↓

Daughter nucleus formed

↓

Mass number and atomic number balanced

↓

Charge, baryon number, and lepton number conserved

↓

Energy and momentum carried by products

↓

More stable nuclear configuration

Radioactive decay is therefore both a nuclear process and an application of the fundamental conservation laws used throughout particle physics.


Key Terms

Radioactive decay – The spontaneous transformation of an unstable atomic nucleus.

Parent nucleus – The original unstable nucleus.

Daughter nucleus – The nucleus formed after radioactive decay.

Alpha particle – A helium-4 nucleus containing two protons and two neutrons.

Beta-minus particle – An electron emitted during beta-minus decay.

Beta-plus particle – A positron emitted during beta-plus decay.

Gamma ray – A high-energy photon emitted by an excited nucleus.

Beta decay – A weak-interaction process that changes the proton-neutron balance of a nucleus.

Excited nucleus – A nucleus containing more energy than its lowest-energy state.

Decay chain – A sequence of radioactive decays leading eventually toward a stable nucleus.

Ionizing radiation – Radiation capable of removing electrons from atoms or molecules.


Key Takeaways

  • Radioactive decay is a spontaneous nuclear transformation.
  • It differs from a particle collision because it does not require another particle to initiate the process.
  • The three main introductory forms are alpha, beta, and gamma decay.
  • Alpha decay emits a helium-4 nucleus.
  • In alpha decay, A decreases by 4 and Z decreases by 2.
  • In beta-minus decay, a neutron changes into a proton, so A stays constant and Z increases by 1.
  • In beta-plus decay, a proton effectively changes into a neutron, so A stays constant and Z decreases by 1.
  • Gamma decay releases excess nuclear energy without changing A or Z.
  • Nuclear equations must conserve mass number and electric charge.
  • Complete beta-decay equations also conserve lepton number through the production of neutrinos or antineutrinos.
  • Energy and momentum are conserved in every radioactive decay.
  • The daughter nucleus may also be radioactive, creating a decay chain.
  • Radioactive decay allows an unstable nucleus to move toward a lower-energy, more stable configuration.

2. Nuclear Fission

Learning outcomes
  • I can describe nuclear fission.
  • I can explain chain reactions.
  • I can calculate simple energy changes.
  • I can describe how nuclear reactors control fission.
  • I can evaluate the advantages and disadvantages of fission.

What Is Nuclear Fission?

Nuclear fission is the splitting of a heavy atomic nucleus into two smaller nuclei.

This process usually also releases:

  • neutrons
  • gamma radiation
  • a large amount of energy

A common example involves uranium-235.

One possible reaction is:

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + energy

The exact fission products can vary, but the general pattern is:

heavy nucleus + neutron → two smaller nuclei + neutrons + energy

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5

Why Can Heavy Nuclei Undergo Fission?

Very heavy nuclei contain many protons.

All protons repel one another because of the electromagnetic force.

The strong nuclear interaction holds nucleons together, but it acts only over very short distances.

As nuclei become very large:

  • proton-proton repulsion becomes increasingly important
  • the nucleus can become easier to deform
  • some heavy nuclei become susceptible to splitting

A nucleus such as uranium-235 can absorb a neutron and become highly excited.

That excitation can cause the nucleus to stretch and split.


Fission of Uranium-235

Uranium-235 is especially important because it can undergo fission after absorbing a slow neutron.

The first step can be represented as:

²³⁵U + n → ²³⁶U*

The asterisk means that uranium-236 is in an excited state.

The excited nucleus can then split:

²³⁶U → fission fragments + neutrons + energy*

For example:

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + energy


Checking the Nuclear Equation

Mass numbers:

235 + 1 = 141 + 92 + 3

236 = 236

Atomic numbers:

92 = 56 + 36

92 = 92

So nucleon number and charge are conserved.


Where Does the Energy Come From?

The energy released during fission comes from a difference in nuclear binding energy.

Very heavy nuclei generally have a lower binding energy per nucleon than medium-mass nuclei.

When a heavy nucleus splits, the products move toward the more tightly bound middle region of the binding-energy curve.

Therefore:

fission products have greater binding energy per nucleon than the original heavy nucleus

The total mass of the products is slightly less than the total mass of the starting system.

This mass defect appears as released energy.


Mass-Energy in Fission

The energy released is calculated using:

E = Δmc²

where:

Δm = mass before − mass after

Even a very small mass difference can produce a large amount of energy because:

c² ≈ 9 × 10¹⁶ m²/s²


Example 1: Energy from a Mass Change

Suppose a fission reaction has a mass defect of:

3.5 × 10⁻²⁸ kg

Use:

E = Δmc²

E = (3.5 × 10⁻²⁸)(3.0 × 10⁸)²

E = (3.5 × 10⁻²⁸)(9.0 × 10¹⁶)

E = 3.15 × 10⁻¹¹ J

This is the energy from only one reaction.

That may seem tiny, but an enormous number of nuclei can undergo fission in a macroscopic sample.


Using Atomic Mass Units

Particle and nuclear masses are often given in atomic mass units.

Recall:

1 u c² ≈ 931.5 MeV

So:

Energy released in MeV = mass defect in u × 931.5


Example 2: Energy in MeV

Suppose the mass defect is:

0.210 u

Then:

E = 0.210 × 931.5

E ≈ 196 MeV

A single uranium-235 fission typically releases energy on the order of about 200 MeV.


Where Does the Fission Energy Go?

The released energy appears mainly as:

  • kinetic energy of the fission fragments
  • kinetic energy of emitted neutrons
  • gamma radiation
  • later radioactive decay energy from unstable fission products

Most of the immediate energy is carried by the two large fission fragments.

As these fragments move through surrounding material, collisions convert their kinetic energy into thermal energy.


Chain Reactions

One of the most important features of nuclear fission is that it can produce additional neutrons.

Suppose one fission produces three neutrons.

Those neutrons may strike other uranium-235 nuclei.

Each of those nuclei may also split.

This can produce still more neutrons.

The result is a chain reaction.

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6

A Simple Chain Reaction

Imagine:

Generation 1

One fission produces:

3 neutrons

Generation 2

If all three cause fission:

3 fissions

Suppose each produces 3 more neutrons:

9 neutrons

Generation 3

Those could cause:

9 fissions

and produce:

27 neutrons

The number of reactions could increase very rapidly.

In reality, not every neutron causes another fission.

Some:

  • escape
  • are absorbed without causing fission
  • lose energy
  • interact with other nuclei

Criticality

A chain reaction can behave in three general ways.

Subcritical

On average, fewer than one neutron from each fission causes another fission.

The reaction gradually dies out.

Critical

On average, exactly one neutron from each fission causes another fission.

The reaction continues at a steady rate.

Supercritical

On average, more than one neutron from each fission causes another fission.

The reaction rate increases.

A power reactor is designed to operate close to a controlled critical condition.


The Multiplication Factor

Nuclear engineers often describe chain reactions using a multiplication factor called k.

Very simply:

k < 1 → subcritical

k = 1 → critical

k > 1 → supercritical

A reactor operating steadily aims for approximately:

k = 1

This means the fission rate remains roughly constant.


How a Nuclear Reactor Uses Fission

A nuclear reactor uses a controlled fission chain reaction to produce thermal energy.

The basic sequence is:

fission

↓

kinetic energy of fission products

↓

thermal energy in reactor fuel

↓

heat transferred to coolant

↓

steam produced, directly or indirectly

↓

turbine turns

↓

generator produces electricity

The reactor is therefore primarily a heat source.


Main Parts of a Nuclear Reactor

A simplified reactor includes:

  • nuclear fuel
  • moderator in many reactor designs
  • control rods
  • coolant
  • reactor vessel/core
  • heat exchanger or steam generator in many designs
  • turbine
  • generator
  • containment structures
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6

Nuclear Fuel

The fuel contains fissile material.

Common examples include fuels containing uranium, especially uranium-235.

The fuel is commonly formed into ceramic pellets and arranged in fuel rods.

The fuel rods are grouped into assemblies inside the reactor core.


The Moderator

In many thermal reactors, a moderator slows fast neutrons.

Why is this useful?

For uranium-235, slow or thermal neutrons are particularly effective at causing further fission.

Common moderator materials include:

  • ordinary water
  • heavy water
  • graphite

The moderator does not stop the reaction. It changes neutron energies so that the chain reaction can be sustained efficiently.


Control Rods

Control rods absorb neutrons.

They can be inserted into or withdrawn from the reactor core.

If control rods are inserted farther:

more neutrons are absorbed

↓

fewer neutrons cause fission

↓

fission rate decreases

If control rods are withdrawn:

fewer neutrons are absorbed

↓

more neutrons remain available

↓

fission rate can increase

Control rods therefore help regulate reactor power.


Common Control-Rod Materials

Control rods use materials that absorb neutrons efficiently.

Examples include materials containing:

  • boron
  • cadmium
  • hafnium

The exact materials depend on reactor design.


The Coolant

The coolant removes thermal energy from the reactor core.

Depending on reactor type, coolants may include:

  • water
  • heavy water
  • gases
  • liquid metals
  • molten salts in some designs

The coolant transports energy from the core to another part of the power-generation system.


From Heat to Electricity

A nuclear power station does not generate electricity directly from the nucleus.

Instead:

nuclear energy → thermal energy → mechanical energy → electrical energy

The heat from fission ultimately produces steam or another working fluid.

The steam turns a turbine.

The turbine turns a generator.

The generator produces electrical energy.


Why Reactors Need Continuous Control

The power output of a reactor depends strongly on the neutron population.

If too many neutrons cause new fissions:

power rises

If too few do:

power falls

Reactor systems use:

  • control rods
  • neutron measurements
  • temperature feedback
  • coolant systems
  • shutdown systems

to maintain safe operation.


Delayed Neutrons

Most neutrons from fission are emitted almost immediately.

However, a small fraction appear later from radioactive fission products.

These are called delayed neutrons.

Although they represent only a small fraction of the neutron population, they are extremely important because they make reactor chain reactions much easier to control over human and mechanical timescales.


Fission Products Are Often Radioactive

The smaller nuclei produced by fission are usually neutron-rich.

Many are unstable.

They undergo radioactive decay, often through beta decay.

Therefore, even after the chain reaction is stopped, radioactive fission products continue to release energy.

This is called decay heat.


Why Cooling Must Continue After Shutdown

Stopping the chain reaction does not instantly eliminate all heat production.

Radioactive fission products continue to decay.

Therefore:

reactor shutdown ≠ immediate zero heat

Cooling systems must continue removing decay heat.

This is one of the most important safety considerations in reactor design.


Advantages of Nuclear Fission

Nuclear fission has several major advantages as an energy source.

High Energy Density

A small amount of nuclear fuel can release a very large amount of energy.

This is because nuclear energy changes are much larger than ordinary chemical energy changes.


Low Direct Carbon Dioxide Emissions During Operation

A nuclear reactor does not burn fossil fuel during normal operation.

Therefore, direct carbon dioxide emissions from electricity generation are very low.

There are still lifecycle emissions from:

  • construction
  • mining
  • fuel processing
  • transport
  • decommissioning

but overall lifecycle greenhouse-gas emissions are generally low compared with fossil-fuel generation.


Reliable Power Production

Nuclear power stations can produce large amounts of electricity continuously.

They are not directly dependent on:

  • sunshine
  • wind speed
  • daily weather changes

This makes them useful for steady electricity generation.


Small Fuel Mass

Because nuclear fuel has very high energy density, relatively little fuel is required compared with coal, oil, or gas for the same amount of energy.


Disadvantages of Nuclear Fission

Fission also presents significant challenges.

Radioactive Waste

Fission produces radioactive materials.

Some waste remains hazardous for long periods.

It must be:

  • contained
  • shielded
  • transported safely
  • stored or disposed of securely

Accident Risk

Modern reactors include many safety systems, but severe accidents can have significant environmental, economic, and social consequences.

Safe reactor design therefore requires multiple independent protective systems.


High Construction Cost

Nuclear power stations are complex.

They require:

  • extensive safety systems
  • radiation shielding
  • highly engineered components
  • strict regulation

As a result, construction costs can be high.


Long Construction and Decommissioning Times

Large reactors can take years to plan and build.

At the end of their useful life, they must also be safely decommissioned.

This can be expensive and time-consuming.


Nuclear Material Security

Some nuclear materials and technologies can raise concerns involving:

  • security
  • theft
  • diversion of nuclear material
  • weapons proliferation

Civil nuclear programs therefore require strong safeguards and international oversight.


Evaluating Fission Fairly

Whether nuclear fission is considered a good energy option depends on several factors.

These include:

  • electricity demand
  • available alternatives
  • local geography
  • cost
  • climate goals
  • grid reliability
  • waste-management plans
  • reactor technology
  • public acceptance

A strong evaluation considers both advantages and disadvantages rather than treating nuclear power as simply "good" or "bad."


Comparing Fission with Fossil Fuels

Nuclear Fission Fossil Fuels
Very high energy density Lower energy density
Low direct CO₂ during operation Large CO₂ emissions when burned
Produces radioactive waste Produces greenhouse gases and air pollutants
High construction cost Often lower initial construction cost
Requires radioactive-material management Requires continuous fuel extraction and combustion
Can provide steady output Can also provide controllable steady output

Fission and Renewable Energy

Nuclear and renewable technologies are not necessarily direct opposites.

A low-carbon electricity system can potentially combine:

  • nuclear power
  • solar
  • wind
  • hydroelectricity
  • storage
  • other low-carbon technologies

The best mixture depends on the needs and resources of a particular region.


Example 3: Chain Reaction Reasoning

Suppose each fission produces an average of:

2.5 neutrons

But only:

40%

of those neutrons cause another fission.

Average successful neutrons per fission:

2.5 × 0.40 = 1.0

So:

k ≈ 1

The chain reaction would be approximately critical and could continue at a steady rate.


Example 4: Subcritical Reaction

Suppose each fission produces:

2.4 neutrons

and only:

30%

cause another fission.

Then:

2.4 × 0.30 = 0.72

Since:

k < 1

the reaction is subcritical.

The chain reaction will decrease.


Example 5: Supercritical Reaction

Suppose each fission produces:

3 neutrons

and:

50%

cause another fission.

Then:

3 × 0.50 = 1.5

Since:

k > 1

the neutron population increases.

The reaction is supercritical.


Nuclear Fission and Conservation Laws

A fission reaction must conserve:

  • electric charge
  • nucleon number
  • energy
  • momentum
  • angular momentum

Consider:

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n

Nucleon number:

235 + 1 = 141 + 92 + 3

Charge:

92 = 56 + 36

Both are conserved.

The small difference in rest mass appears as released energy.


Did You Know?

Most of the energy released in a fission event does not come directly from the emitted neutrons.

A large fraction appears as kinetic energy of the two heavy fission fragments.

These positively charged fragments move rapidly through the surrounding fuel and collide with other atoms.

Their motion is converted into thermal energy.

This is how microscopic nuclear energy ultimately becomes the heat used to generate electricity.


Connecting the Ideas

Nuclear fission links several ideas from this course:

Heavy nucleus absorbs neutron

↓

Excited compound nucleus forms

↓

Nucleus splits

↓

Smaller nuclei + neutrons produced

↓

Products have greater binding energy per nucleon

↓

Mass decreases slightly

↓

E = Δmc²

↓

Energy released

↓

Released neutrons can trigger more fissions

↓

Chain reaction

↓

Control rods and reactor systems regulate the process

Fission is therefore a direct application of nuclear stability, binding energy, mass defect, mass-energy equivalence, and conservation laws.


Key Terms

Nuclear fission – The splitting of a heavy nucleus into smaller nuclei, usually with the release of neutrons and energy.

Fission fragment – One of the smaller nuclei produced during fission.

Chain reaction – A sequence in which neutrons from one fission cause additional fissions.

Critical – A condition in which the chain reaction continues at a steady rate.

Subcritical – A condition in which the chain reaction decreases.

Supercritical – A condition in which the chain reaction increases.

Moderator – Material used in many reactors to slow neutrons.

Control rod – A neutron-absorbing component used to regulate a reactor.

Coolant – Material that transfers heat away from a reactor core.

Fissile material – Material capable of sustaining fission after absorbing appropriate neutrons.

Decay heat – Heat produced by radioactive decay of fission products after the main chain reaction has stopped.


Key Takeaways

  • Nuclear fission is the splitting of a heavy nucleus into smaller nuclei.
  • Fission typically releases neutrons and a large amount of energy.
  • Uranium-235 can undergo fission after absorbing a neutron.
  • Fission releases energy because the products are generally more tightly bound per nucleon than the original heavy nucleus.
  • The small loss of rest mass is converted to energy according to E = Δmc².
  • One fission releases energy on the order of hundreds of MeV.
  • Neutrons released by fission can trigger further fissions, producing a chain reaction.
  • A steady reactor aims to maintain approximately k = 1.
  • Control rods absorb neutrons and help regulate the reaction.
  • A moderator slows neutrons in many reactor designs.
  • A coolant transfers thermal energy away from the reactor core.
  • Radioactive fission products continue producing decay heat even after shutdown.
  • Fission offers very high energy density and low direct operational carbon emissions, but it also produces radioactive waste and requires strict safety and security systems.
  • A fair evaluation of nuclear fission must consider both its benefits and limitations.

3. Nuclear Fusion

Learning outcomes
  • I can describe nuclear fusion.
  • I can explain why fusion releases energy.
  • I can compare fusion with fission.
  • I can describe fusion in stars.
  • I can explain challenges in producing controlled fusion.

What Is Nuclear Fusion?

Nuclear fusion is the process in which two light atomic nuclei combine to form a heavier nucleus.

A simplified example is:

light nucleus + light nucleus → heavier nucleus + energy

Fusion is the process that powers stars, including the Sun.

Unlike chemical reactions, fusion changes the nuclei of atoms.

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5

Why Can Fusion Release Energy?

Light nuclei generally have lower binding energy per nucleon than medium-mass nuclei.

When light nuclei combine, the product can be more tightly bound.

That means:

final nucleus has greater binding energy per nucleon

The total mass of the final products is slightly less than the total mass of the starting particles.

This difference is the mass defect.

The missing rest mass appears as released energy according to:

E = Δmc²


Fusion and the Binding-Energy Curve

The binding-energy-per-nucleon curve rises steeply for light nuclei and reaches a maximum near the iron-nickel region.

This means that light nuclei can release energy by moving toward more tightly bound nuclei.

So:

light nuclei → fusion → heavier, more tightly bound nuclei → energy released

For very heavy nuclei, the opposite process, fission, can release energy.

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5

A Simple Fusion Reaction

One important fusion reaction uses two isotopes of hydrogen:

deuterium + tritium

Deuterium:

²₁H

Tritium:

³₁H

They can fuse to form helium-4 and a neutron:

²₁H + ³₁H → ⁴₂He + ¹₀n + energy


Checking the Equation

Mass numbers:

2 + 3 = 4 + 1

5 = 5

Atomic numbers:

1 + 1 = 2 + 0

2 = 2

So nucleon number and charge are conserved.


Energy Released in Deuterium-Tritium Fusion

The deuterium-tritium reaction releases about:

17.6 MeV

of energy.

Most of this appears as kinetic energy of:

  • the helium nucleus
  • the neutron

A common approximate split is:

helium nucleus: 3.5 MeV

neutron: 14.1 MeV

The neutron carries most of the released energy.


Why Does the Product Have Less Mass?

Before fusion, the separated nuclei have a certain total mass.

After fusion, the products have slightly less total rest mass.

The difference is:

Δm = minitial − mfinal

That mass difference is converted into other forms of energy.

So mass has not simply vanished.

Instead:

rest mass-energy → kinetic energy + radiation + other energy


Example 1: Calculating Fusion Energy

Suppose a fusion reaction has a mass defect of:

0.0189 u

Use:

1 u c² ≈ 931.5 MeV

Then:

E = 0.0189 × 931.5

E ≈ 17.6 MeV

This is approximately the energy released in deuterium-tritium fusion.


Fusion Requires Nuclei to Get Very Close

There is a major problem.

Atomic nuclei are positively charged.

Positive charges repel one another due to the electromagnetic force.

Therefore, two nuclei approaching each other experience:

electrostatic repulsion

This is called the Coulomb barrier.

For fusion to occur, the nuclei must get close enough for the strong nuclear interaction to become important.


The Coulomb Barrier

At relatively large nuclear distances:

electromagnetic repulsion dominates

At extremely small distances:

strong nuclear interaction can bind the nuclei

Therefore, fusion requires nuclei to approach extremely closely.

That requires high particle energies.

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5

Why High Temperatures Are Needed

Temperature is related to the average kinetic energy of particles.

At extremely high temperatures:

  • nuclei move very rapidly
  • collisions are more energetic
  • some nuclei can approach closely enough to fuse

Fusion therefore requires extremely high temperatures.

For many fusion reactions, temperatures of millions of kelvin are required.


Plasma

At these temperatures, ordinary atoms cannot remain intact.

Electrons separate from nuclei.

The resulting state of matter is called plasma.

Plasma contains:

  • positively charged nuclei
  • free electrons

Most fusion research therefore involves controlling extremely hot plasma.


Quantum Tunnelling

Even at stellar temperatures, most nuclei do not classically have enough energy to completely overcome the Coulomb barrier.

Fusion is still possible because of quantum tunnelling.

Quantum mechanics gives particles a probability of passing through a barrier that they could not cross according to classical physics alone.

This effect is crucial for fusion inside stars.


Fusion in the Sun

The Sun produces energy mainly by fusing hydrogen nuclei into helium.

The overall process can be summarized as:

4 hydrogen nuclei → 1 helium-4 nucleus + other particles + energy

The actual process occurs through several stages rather than all four protons colliding at once.

The dominant sequence in the Sun is called the proton-proton chain.

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5

The Proton-Proton Chain

A simplified version begins with:

p + p → ²H + e⁺ + νₑ

Two protons interact.

One effectively changes into a neutron through the weak interaction.

This forms deuterium.

A positron and electron neutrino are also produced.


Step 2

The deuterium nucleus can then combine with another proton:

²H + p → ³He + γ

This produces helium-3 and a gamma-ray photon.


Step 3

Two helium-3 nuclei can combine:

³He + ³He → ⁴He + 2p

This produces helium-4 and returns two protons.

The overall effect is that four protons are converted into one helium-4 nucleus, along with other particles and released energy.


Overall Solar Fusion

A simplified overall reaction can be written as:

4p → ⁴He + 2e⁺ + 2νₑ + energy

The total energy released is approximately:

26.7 MeV

for the complete proton-proton chain, with some energy carried away by neutrinos.


Where Does the Sun's Energy Go?

Fusion energy in the Sun appears in several forms.

Initially it may appear as:

  • kinetic energy of particles
  • gamma radiation
  • neutrino energy

The energy undergoes many interactions inside the Sun before eventually reaching the surface.

It is then radiated into space mainly as electromagnetic radiation.


Solar Neutrinos

Neutrinos produced in fusion interact very weakly with matter.

They can escape from the Sun's core quickly.

This means neutrinos give scientists direct information about nuclear reactions occurring deep inside the Sun.

Detecting solar neutrinos provides strong evidence that fusion powers the Sun.


Gravity Makes Stellar Fusion Possible

Stars contain enormous amounts of matter.

Gravity pulls this matter inward.

This creates:

  • extremely high pressure
  • extremely high temperature

in the core.

These conditions allow nuclei to collide frequently and make fusion possible.


Hydrostatic Equilibrium

A stable star is approximately in hydrostatic equilibrium.

Gravity pulls material inward.

Pressure produced by the hot interior pushes outward.

So:

inward gravitational force ↔ outward pressure

Fusion helps maintain the high temperature and pressure needed to support the star.


What Happens When Fusion Changes?

As a star uses its nuclear fuel, the types of fusion reactions occurring in its core can change.

A star's evolution depends strongly on its mass.

Stars can eventually fuse increasingly heavy nuclei under suitable conditions.

Massive stars can progress through fusion stages involving elements such as:

  • hydrogen
  • helium
  • carbon
  • neon
  • oxygen
  • silicon

Why Fusion Stops Releasing Energy Near Iron

Fusion of light nuclei releases energy because it moves nuclei toward greater binding energy per nucleon.

This trend continues toward the iron-nickel region.

Beyond this region, fusing nuclei into still heavier nuclei generally requires energy rather than releasing it.

Therefore, ordinary stellar fusion cannot continue releasing energy indefinitely by building heavier and heavier nuclei.

This is crucial in the evolution of massive stars.


Comparing Fusion and Fission

Nuclear Fusion Nuclear Fission
Combines light nuclei Splits heavy nuclei
Common in stars Used in present nuclear power reactors
Requires extremely high temperatures Can be triggered by neutron absorption
Releases energy from increased binding Releases energy from increased binding
Produces fewer long-lived fission-type products Produces many radioactive fission products
Difficult to control on Earth Commercially controlled in reactors
Fuel can include hydrogen isotopes Fuel can include uranium-235 or plutonium-239

Both processes release energy because the products move toward more tightly bound nuclear configurations.


Fusion and Fission on the Binding-Energy Curve

Fusion moves:

light nuclei → toward medium-mass nuclei

Fission moves:

very heavy nuclei → toward medium-mass nuclei

Both processes can therefore move nuclei toward the high-binding-energy region.

This is why both can release nuclear energy.


Why Fusion Is Attractive as an Energy Source

Controlled fusion could offer several potential advantages.

High Energy Density

Fusion reactions release enormous amounts of energy compared with chemical reactions.


Abundant Fuel Sources

Deuterium can be obtained from water.

Tritium is much rarer, but proposed fusion systems can produce tritium from lithium using neutrons.


No Carbon Combustion

Fusion itself does not require burning fossil fuels.

Therefore, operational carbon dioxide emissions could be very low.


No Self-Sustaining Fission Chain Reaction

Fusion does not rely on the same type of neutron-multiplying chain reaction used in fission reactors.

If the required plasma conditions are lost:

fusion rapidly decreases or stops


Is Fusion Free of Radioactive Waste?

No.

This is an important misconception.

Fusion can produce less long-lived radioactive waste than conventional fission, depending on reactor design, but it does not produce zero radioactive material.

For example, high-energy neutrons can strike reactor structures and make some materials radioactive.

This is called neutron activation.


Why Controlled Fusion Is Difficult

Fusion is easy to describe but extremely difficult to sustain.

Scientists must create conditions in which nuclei:

  • have enough energy to fuse
  • collide frequently enough
  • remain confined long enough

These requirements are often summarized in terms of:

  • temperature
  • plasma density
  • confinement time

The Fusion Challenge

A useful way to think about controlled fusion is:

heat the plasma enough

  •  

keep enough particles together

  •  

hold them together long enough

=

significant fusion

If any of these conditions are inadequate, the fusion rate is too low.


The Lawson Criterion

Fusion researchers use a concept called the Lawson criterion.

It relates important plasma conditions such as:

  • temperature
  • particle density
  • confinement time

A fusion system must achieve sufficient combinations of these quantities for the fusion energy produced to compete with energy losses.

The exact mathematical treatment is advanced, but the principle is important:

hot plasma alone is not enough.

It must also be sufficiently dense and well confined.


Magnetic Confinement Fusion

One approach is called magnetic confinement.

Because plasma contains charged particles, magnetic fields can influence its motion.

Strong magnetic fields are used to keep the hot plasma away from material walls.

A major device used for this purpose is the tokamak.

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5

The Tokamak

A tokamak uses a doughnut-shaped chamber called a torus.

Magnetic fields confine the plasma inside the chamber.

The goal is to prevent the extremely hot plasma from directly touching the walls.

Tokamaks are one of the most extensively studied fusion-reactor designs.


Why Can't the Plasma Touch the Walls?

Fusion plasma can reach temperatures of tens or even hundreds of millions of kelvin.

No ordinary solid material could remain intact if directly exposed to plasma at those temperatures.

Magnetic confinement therefore keeps the bulk plasma away from the walls.

Even so, handling heat and particle flow at the edge of the plasma remains a major engineering challenge.


Inertial Confinement Fusion

Another approach is inertial confinement fusion.

A tiny fuel pellet containing fusion fuel is compressed rapidly.

High-powered lasers or other drivers deliver energy to the pellet.

The outer material heats and expands outward, causing the inner fuel to compress strongly.

For a very short time, the fuel may reach conditions suitable for fusion.

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5

Magnetic vs Inertial Confinement

Magnetic Confinement Inertial Confinement
Uses magnetic fields Uses rapid compression
Plasma confined for relatively longer times Fuel confined for extremely short times
Often uses tokamaks or stellarators Often uses lasers
Lower density plasma Extremely compressed fuel
Continuous or long-pulse power is a goal Pulsed operation

Both approaches aim to satisfy the conditions necessary for significant fusion.


Energy Gain

One important measure in fusion research is energy gain.

In simplified terms:

energy gain = fusion energy produced ÷ energy supplied to the fuel/plasma

A scientific experiment may achieve a gain greater than 1 at the fuel or plasma level without yet representing a complete power station that produces net electricity.

A practical fusion plant must account for the energy needed for:

  • heating
  • magnets
  • lasers or other drivers
  • cooling
  • fuel processing
  • electricity generation
  • supporting equipment

Why Fusion Power Is Harder Than Demonstrating Fusion

Fusion reactions have been produced in laboratories for many decades.

The real challenge is not simply:

Can fusion happen?

It can.

The challenge is:

Can fusion be sustained, controlled, reliable, and economical while producing useful net electrical power?

That is a much more demanding engineering problem.


Major Engineering Challenges

Controlled fusion must solve problems including:

  • maintaining stable plasma
  • confining extremely hot plasma
  • controlling plasma instabilities
  • handling enormous heat loads
  • resisting neutron damage
  • producing and managing tritium
  • maintaining reactor components
  • converting fusion energy into electricity efficiently

These challenges involve both physics and engineering.


Plasma Instabilities

Plasma is not always easy to control.

It can develop:

  • waves
  • turbulence
  • sudden changes in shape
  • instabilities

These effects can allow energy and particles to escape confinement.

Modern fusion experiments use sophisticated control systems to monitor and adjust plasma conditions.


Neutron Damage

In deuterium-tritium fusion, energetic neutrons escape the magnetic confinement because they have no electric charge.

These neutrons strike the reactor walls.

Over time, they can:

  • damage materials
  • change material properties
  • create radioactive isotopes
  • weaken structural components

Developing materials that can survive this environment is a major challenge.


Tritium

Tritium is radioactive and relatively scarce in nature.

A future deuterium-tritium fusion reactor would likely need to produce much of its own tritium.

One proposed method uses lithium.

Neutrons from fusion interact with lithium-containing materials to produce tritium.

This is called tritium breeding.


Example 2: Comparing Energy Release

Suppose one reaction releases:

18 MeV

and another releases:

200 MeV

The 200 MeV reaction releases more energy per individual reaction.

However, this does not automatically mean it produces more energy per kilogram of fuel.

The masses of the reacting nuclei and the number of possible reactions must also be considered.

This is why energy comparisons should specify whether they refer to:

  • energy per reaction
  • energy per nucleon
  • energy per unit mass of fuel

Example 3: Why Fusion Stops if Temperature Falls

Suppose a fusion plasma cools significantly.

Then nuclei have lower average kinetic energy.

Fewer collisions can bring nuclei close enough together for fusion.

Therefore:

temperature falls

↓

fusion rate falls

↓

energy production falls

This is one reason fusion reactions are not self-sustaining in the same way as a fission chain reaction.


Example 4: Conservation in Fusion

Consider:

²₁H + ³₁H → ⁴₂He + ¹₀n

Check nucleon number:

2 + 3 = 4 + 1

5 = 5

Check charge:

1 + 1 = 2 + 0

2 = 2

Both are conserved.

Energy and momentum must also be conserved.


Does Fusion Produce Radiation?

Yes.

Fusion systems can produce:

  • high-energy neutrons
  • gamma radiation
  • energetic charged particles

Therefore, fusion facilities still require shielding and radiation protection.

Fusion is nuclear technology and must be treated accordingly.


Fusion in Different Stars

The proton-proton chain dominates in stars like the Sun.

In hotter, more massive stars, another fusion pathway called the CNO cycle can play a major role.

CNO stands for:

  • carbon
  • nitrogen
  • oxygen

These nuclei participate in a cycle that ultimately converts hydrogen into helium.


Fusion and Stellar Nucleosynthesis

Fusion inside stars is responsible for creating many of the elements found in the universe.

Stars can build heavier nuclei from lighter ones.

This process is called stellar nucleosynthesis.

Many elements in planets and living organisms were formed through nuclear processes in earlier generations of stars.


Did You Know?

The Sun converts a small amount of mass into other forms of energy every second.

Although this mass loss is tiny compared with the Sun's total mass, the value of c² is so large that it corresponds to an enormous energy output.

This is why the Sun can shine continuously for billions of years.


Connecting the Ideas

Nuclear fusion connects several major ideas from this course:

Light nuclei

↓

Very high temperature and energetic collisions

↓

Nuclei approach closely

↓

Strong nuclear interaction binds them

↓

More tightly bound nucleus forms

↓

Mass defect

↓

E = Δmc²

↓

Energy released

In stars:

gravity → high pressure and temperature → fusion → stellar energy

In laboratories:

heating + confinement + plasma control → fusion attempts


Key Terms

Nuclear fusion – The joining of light nuclei to form a heavier nucleus.

Coulomb barrier – The electrostatic repulsion that nuclei must overcome or tunnel through to approach closely enough to fuse.

Plasma – A state of matter containing free electrons and ions.

Proton-proton chain – The main fusion process that powers stars such as the Sun.

Quantum tunnelling – A quantum effect that allows particles to pass through barriers that classical physics would not allow them to cross.

Magnetic confinement – The use of magnetic fields to confine hot plasma.

Tokamak – A toroidal magnetic-confinement fusion device.

Inertial confinement – Fusion achieved by rapidly compressing a small amount of fuel.

Lawson criterion – A condition relating temperature, particle density, and confinement time needed for useful fusion performance.

Tritium breeding – Production of tritium inside a proposed fusion reactor, commonly using lithium.

Neutron activation – The creation of radioactive nuclei when materials absorb or interact with neutrons.

Stellar nucleosynthesis – The formation of new atomic nuclei through nuclear reactions in stars.


Key Takeaways

  • Nuclear fusion combines light nuclei to form heavier nuclei.
  • Fusion releases energy when the products are more tightly bound than the starting nuclei.
  • The energy comes from a mass defect according to E = Δmc².
  • Deuterium-tritium fusion produces helium-4, a neutron, and about 17.6 MeV of energy.
  • Fusion requires nuclei to approach extremely closely despite their electrostatic repulsion.
  • Extremely high temperatures create the energetic plasma conditions needed for fusion.
  • Quantum tunnelling helps fusion occur in stars.
  • The Sun is powered mainly by the proton-proton chain.
  • Fusion and fission both release energy by moving nuclei toward the high-binding-energy region of the binding-energy curve.
  • Fission splits heavy nuclei, while fusion combines light nuclei.
  • Controlled fusion is difficult because plasma must be sufficiently hot, dense, and confined.
  • Major approaches include magnetic confinement and inertial confinement.
  • Fusion still presents challenges involving plasma stability, neutron damage, tritium supply, heat handling, and reactor materials.
  • Fusion could become an important low-carbon energy source, but producing reliable net electrical power from controlled fusion remains a major scientific and engineering challenge.
 
 
 

4. Stellar Nucleosynthesis

Learning outcomes
  • I can explain how stars produce heavier elements.
  • I can describe hydrogen fusion.
  • I can explain the formation of elements beyond iron.
  • I can relate stellar evolution to element formation.
  • I can explain why supernovae are important.

What Is Stellar Nucleosynthesis?

Stellar nucleosynthesis is the production of new atomic nuclei inside stars.

Stars begin with large amounts of hydrogen and helium. Over their lifetimes, nuclear reactions build heavier nuclei from lighter ones.

The basic pattern is:

light nuclei → fusion → heavier nuclei + energy

This process is responsible for producing many of the elements found throughout the universe.

Massive stars can build a layered structure in which progressively heavier elements are formed as the star ages.

https://cdn.livephysics.com/cheat-sheets/astronomy-stellar-nucleosynthesis-reference.png
 
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The First Elements

Stars did not create the very first hydrogen and helium.

Those light nuclei formed shortly after the Big Bang.

The early universe contained mostly:

  • hydrogen
  • helium
  • small amounts of other light nuclei

Later, stars formed from this material.

Inside stars, nuclear reactions began producing heavier elements.

So:

Big Bang nucleosynthesis → mostly light elements

stellar nucleosynthesis → many heavier elements


Hydrogen Fusion

For most of a star's lifetime, its main energy source is hydrogen fusion.

Hydrogen nuclei are essentially protons.

In stars such as the Sun, hydrogen is gradually converted into helium through a sequence of reactions called the proton-proton chain.

A simplified overall reaction is:

4p → ⁴He + 2e⁺ + 2νₑ + energy

The complete process occurs through several stages.


Step 1 of the Proton-Proton Chain

Two protons interact:

p + p → ²H + e⁺ + νₑ

One proton effectively changes into a neutron through the weak interaction.

The products include:

  • deuterium
  • a positron
  • an electron neutrino

Step 2

The deuterium nucleus combines with another proton:

²H + p → ³He + γ

This produces helium-3 and a gamma-ray photon.


Step 3

Two helium-3 nuclei can combine:

³He + ³He → ⁴He + 2p

The overall result is that hydrogen is converted into helium.

Energy is released because the helium nucleus is more tightly bound than the original hydrogen nuclei.


Why Hydrogen Fusion Releases Energy

The helium-4 nucleus has a greater binding energy per nucleon than separate hydrogen nuclei.

Therefore, the products have slightly less total rest mass than the starting particles.

The difference is converted into energy:

E = Δmc²

This energy helps keep the star hot and provides the outward pressure needed to resist gravitational collapse.


Hydrogen Burning and Stellar Stability

A main-sequence star exists in an approximate balance called hydrostatic equilibrium.

Gravity pulls inward.

The hot stellar interior creates outward pressure.

So:

gravity inward ↔ pressure outward

Hydrogen fusion helps maintain the temperature needed for this balance.


What Happens When Hydrogen Runs Low?

Eventually, the hydrogen in the star's core becomes depleted.

The fusion rate in the core decreases.

Gravity causes the core to contract.

As the core contracts:

  • pressure increases
  • temperature increases
  • new nuclear reactions may become possible

What happens next depends strongly on the mass of the star.


Helium Fusion

When the core becomes sufficiently hot, helium can begin to fuse.

An important reaction is the triple-alpha process.

Three helium-4 nuclei ultimately combine to produce carbon-12:

3 ⁴He → ¹²C + energy

The process occurs through intermediate steps, but the overall result is:

helium → carbon


Carbon and Oxygen Formation

Once carbon exists, helium nuclei can also combine with carbon:

¹²C + ⁴He → ¹⁶O + γ

This produces oxygen.

Therefore, helium-burning stars can produce significant amounts of:

  • carbon
  • oxygen

These elements are extremely important because they later become part of planets and living organisms.


Stellar Mass Determines What Elements Can Form

Not every star can produce the same elements.

A star's initial mass determines:

  • the temperature reached in its core
  • the pressure reached
  • which fusion reactions can occur
  • how the star eventually dies

Lower-mass stars and massive stars therefore follow different nucleosynthesis pathways.


Lower-Mass Stars

Stars with relatively low or intermediate masses do not reach the extreme core temperatures needed for all advanced fusion stages.

They can produce significant amounts of elements such as:

  • helium
  • carbon
  • oxygen

Later in their evolution, they can eject enriched material into space.

This material can become part of future generations of stars and planets.


Massive Stars

Massive stars reach much higher central temperatures.

They can continue fusion beyond carbon and oxygen.

Successive nuclear-burning stages can produce elements such as:

  • neon
  • magnesium
  • silicon
  • sulfur
  • argon
  • calcium
  • iron-group nuclei

Massive stars can therefore act as enormous element-producing factories.


The Onion-Shell Structure

Near the end of its life, a massive star can develop layers resembling an onion.

Different fusion reactions occur in different shells.

A simplified pattern is:

outer layers: hydrogen fusion

↓

helium fusion

↓

carbon fusion

↓

neon burning

↓

oxygen burning

↓

silicon burning

↓

iron-rich core

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Advanced Fusion Stages

As a massive star evolves, the required temperatures become increasingly high.

A simplified sequence is:

hydrogen → helium

helium → carbon and oxygen

carbon → neon, sodium, magnesium and other nuclei

oxygen → silicon, sulfur and other nuclei

silicon burning → iron-group nuclei

The exact reaction networks are much more complex than this simple sequence.


Why the Later Stages Become Faster

Hydrogen burning can last millions or billions of years, depending on stellar mass.

Later burning stages occur much more rapidly.

Why?

As heavier elements are fused:

  • less energy is gained per reaction
  • the star loses energy rapidly
  • higher temperatures are required
  • nuclear fuel is consumed more quickly

In massive stars, some of the final burning stages can be extremely short compared with the long hydrogen-burning stage.


Why Fusion Stops Near Iron

Fusion releases energy only when the products are more tightly bound than the starting nuclei.

The binding-energy-per-nucleon curve reaches its maximum around the iron-nickel region.

Therefore, fusing light nuclei toward this region generally releases energy.

But trying to fuse iron-group nuclei into still heavier nuclei does not provide the star with additional energy in the same way.

Instead:

fusion beyond the iron region generally requires energy

This creates a major problem for a massive star.


The Iron Core

Once an iron-rich core develops, fusion can no longer provide enough new energy to support the core against gravity.

The core grows as surrounding shells continue producing iron-group material.

Eventually, the core can become unstable.

Gravity then causes a rapid collapse.

This is the beginning of one pathway to a core-collapse supernova.


What Is a Supernova?

A supernova is an enormously energetic stellar explosion.

In the core-collapse case, the core of a massive star collapses extremely rapidly.

The collapse and subsequent explosion can eject much of the star's outer material into space.

Supernovae are important because they:

  • create conditions for additional nucleosynthesis
  • eject elements already made inside the star
  • spread those elements into interstellar space
  • enrich future generations of stars and planets

Elements Beyond Iron

Elements heavier than iron are not efficiently produced by ordinary energy-releasing fusion.

Instead, many heavy nuclei are built mainly through neutron capture.

A nucleus absorbs neutrons:

nucleus + neutron → heavier isotope

The neutron-rich nucleus may then undergo beta decay.

During beta-minus decay:

neutron → proton + electron + antineutrino

This increases the atomic number and can create a new element.


The s-Process

One important mechanism is the slow neutron-capture process, or s-process.

In the s-process:

  • neutron capture occurs relatively slowly
  • unstable nuclei often have time to beta-decay before capturing another neutron

This process occurs mainly during certain late stages of stellar evolution, especially in evolved giant stars.

It can produce many nuclei heavier than iron.


The r-Process

Another mechanism is the rapid neutron-capture process, or r-process.

In the r-process:

  • enormous numbers of neutrons are available
  • nuclei capture neutrons very rapidly
  • extremely neutron-rich nuclei are created
  • these later decay toward more stable nuclei

The r-process can produce very heavy elements.

Examples include nuclei associated with elements such as:

  • gold
  • platinum
  • uranium

Are Supernovae the Only Source of Heavy Elements?

No.

This is an important refinement.

Supernovae are extremely important for nucleosynthesis and for distributing elements, but modern astrophysics shows that different heavy elements are produced in different environments.

Important sites include:

  • evolved stars
  • core-collapse supernovae
  • neutron-star mergers
  • other explosive stellar environments

In particular, neutron-star mergers are now known to be major sites of rapid neutron-capture nucleosynthesis.

So it is better to say:

many heavy elements are produced through neutron-capture processes associated with extreme stellar events, rather than saying that every element beyond iron is produced only in supernovae.


Why Supernovae Are Still Crucial

Supernovae remain extremely important for several reasons.

First, they create extreme conditions of:

  • temperature
  • pressure
  • neutron density
  • energy

Second, they eject newly formed elements from the star.

Without this ejection, many of those elements would remain trapped in stellar remnants.

Third, the expanding material mixes with the interstellar medium.

That enriched material can later form:

  • new stars
  • planets
  • asteroids
  • living organisms

Cosmic Recycling

Stellar nucleosynthesis is part of a much larger cosmic cycle.

gas cloud

↓

star forms

↓

fusion creates heavier elements

↓

star evolves

↓

material is expelled

↓

interstellar gas becomes enriched

↓

new stars and planets form

This means later generations of stars contain more heavy elements than the earliest stars.


Where Did the Elements in Earth Come From?

The atoms that make up Earth were produced through several cosmic processes.

For example:

Hydrogen – largely from the early universe

Carbon and oxygen – produced extensively in stars

Silicon, sulfur, calcium, and iron – produced in massive stars and explosive stellar processes

Many very heavy elements – produced through neutron-capture processes in extreme astrophysical environments

The Solar System formed from gas and dust already enriched by earlier generations of stars.


We Are Made of Stellar Material

The carbon in biological molecules, oxygen in water, calcium in bones, and iron in blood were not created on Earth.

Their nuclei were produced by earlier astrophysical processes before the Solar System formed.

Stellar nucleosynthesis therefore connects nuclear physics directly to:

  • astronomy
  • planetary science
  • chemistry
  • biology

Example 1: Hydrogen to Helium

Suppose four hydrogen nuclei ultimately form one helium-4 nucleus.

Initial particles:

4 protons

Final helium nucleus:

2 protons + 2 neutrons

The process also produces other particles and energy.

The helium nucleus has a lower total rest mass than the original collection of particles.

The mass difference becomes energy.


Example 2: Triple-Alpha Process

Three helium-4 nuclei combine:

3 ⁴₂He → ¹²₆C

Check nucleon number:

3 × 4 = 12

Atomic number:

3 × 2 = 6

Therefore:

¹²₆C

is produced.


Example 3: Building Oxygen

Carbon-12 captures a helium nucleus:

¹²₆C + ⁴₂He → ¹⁶₈O + γ

Mass number:

12 + 4 = 16

Atomic number:

6 + 2 = 8

Element 8 is oxygen.


Example 4: Neutron Capture and Beta Decay

Suppose a nucleus captures a neutron.

Its:

mass number increases by 1

but:

atomic number stays the same

If the resulting nucleus later undergoes beta-minus decay:

mass number stays the same

atomic number increases by 1

A new element may therefore be produced.

This combination of neutron capture and beta decay is central to the formation of many heavy nuclei.


Stellar Evolution and Element Production

A useful summary is:

Stellar Stage Important Nuclear Process Typical Products
Main sequence Hydrogen fusion Helium
Red giant / supergiant Helium fusion Carbon, oxygen
Massive evolved star Advanced burning Neon, magnesium, silicon, sulfur, iron-group nuclei
Late-stage giant stars Slow neutron capture Many nuclei heavier than iron
Explosive events Rapid reactions and neutron capture Additional heavy nuclei
Supernova ejecta Dispersal Elements mixed into interstellar space

Why Stellar Mass Matters

Lower-Mass Star

A lower-mass star:

  • never reaches all the temperatures needed for advanced burning
  • produces fewer very heavy nuclei through core fusion
  • eventually ejects some enriched material
  • leaves a compact stellar remnant

Massive Star

A massive star:

  • reaches much higher core temperatures
  • undergoes several successive burning stages
  • develops an iron-rich core
  • may undergo core collapse
  • can explode as a supernova
  • disperses large amounts of newly synthesized material

Did You Know?

The phrase "we are made of star stuff" has a real nuclear-physics meaning.

Many of the nuclei in your body existed long before Earth formed.

They were created by earlier generations of stars and explosive astrophysical events, released into space, incorporated into the cloud that formed the Solar System, and eventually became part of Earth.


Connecting the Ideas

Stellar nucleosynthesis connects many of the ideas in this course:

Hydrogen nuclei

↓

hydrogen fusion

↓

helium

↓

helium fusion

↓

carbon and oxygen

↓

advanced fusion in massive stars

↓

elements up to the iron region

↓

core collapse and extreme stellar events

↓

neutron-capture nucleosynthesis

↓

heavy elements

↓

elements dispersed through space

↓

new stars, planets, and living matter


Key Terms

Stellar nucleosynthesis – The production of atomic nuclei through nuclear reactions associated with stars.

Hydrogen fusion – The conversion of hydrogen nuclei into helium nuclei.

Proton-proton chain – A sequence of reactions responsible for most hydrogen fusion in stars such as the Sun.

Triple-alpha process – A nuclear process in which helium nuclei ultimately combine to form carbon-12.

Stellar burning – A term used for nuclear-fusion stages inside stars; it does not mean ordinary chemical combustion.

Iron core – The iron-group-rich central region that can develop late in the life of a massive star.

Supernova – A powerful stellar explosion capable of ejecting large amounts of material into space.

Neutron capture – A process in which a nucleus absorbs a neutron.

s-process – Slow neutron capture responsible for producing many nuclei heavier than iron.

r-process – Rapid neutron capture responsible for producing many very heavy neutron-rich nuclei.

Interstellar medium – The gas and dust between stars.

Stellar evolution – The sequence of changes a star undergoes during its lifetime.


Key Takeaways

  • Stellar nucleosynthesis is the production of new nuclei through nuclear reactions associated with stars.
  • Stars spend much of their lives converting hydrogen into helium.
  • Stars such as the Sun mainly use the proton-proton chain for hydrogen fusion.
  • Helium fusion can produce carbon and oxygen.
  • Massive stars can continue through advanced fusion stages and produce nuclei up to the iron-group region.
  • Fusion beyond the iron region does not normally release useful stellar energy because the binding-energy curve has already reached its maximum region.
  • Many nuclei heavier than iron are produced through neutron-capture processes rather than ordinary fusion.
  • The s-process involves relatively slow neutron capture.
  • The r-process involves extremely rapid neutron capture in highly energetic environments.
  • Supernovae are important because they provide extreme conditions and, crucially, eject newly formed elements into space.
  • Neutron-star mergers and other extreme events are also important sources of some of the heaviest elements.
  • A star's mass determines which fusion stages it can reach and therefore which elements it can help produce.
  • Material released by earlier generations of stars becomes part of later stars, planets, and living organisms.
 
 
 

5. Energy from Nuclear Reactions

Learning outcomes
  • I can compare energy released in chemical and nuclear reactions.
  • I can explain why nuclear reactions release large amounts of energy.
  • I can interpret nuclear reaction energy diagrams.
  • I can compare different nuclear energy sources.
  • I can evaluate the efficiency of nuclear energy.

Chemical Energy vs Nuclear Energy

Both chemical and nuclear reactions can release energy, but they involve very different parts of the atom.

In a chemical reaction, electrons are rearranged and chemical bonds are broken and formed.

In a nuclear reaction, the nucleus itself changes.

This difference is extremely important because nuclear binding energies are far greater than typical chemical bond energies.

As a result:

nuclear reactions can release millions of times more energy per reaction than ordinary chemical reactions.

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https://www.energy.gov/sites/default/files/Fission%20vs%20fusion%201200x627_FB%20Thumbnail.png
 

Chemical Reactions

Chemical reactions involve changes in the arrangement of electrons.

Examples include:

  • combustion
  • batteries
  • respiration
  • reactions between acids and bases

During a chemical reaction:

  • atomic nuclei remain essentially unchanged
  • atoms rearrange into new substances
  • chemical bonds are broken and formed

For example:

CH₄ + 2O₂ → CO₂ + 2H₂O + energy

The energy comes from differences in the chemical bonding of the reactants and products.


Nuclear Reactions

Nuclear reactions involve changes to the nucleus.

Examples include:

  • radioactive decay
  • nuclear fission
  • nuclear fusion

During a nuclear reaction:

  • protons and neutrons may be rearranged
  • one element may change into another
  • small changes in rest mass can occur
  • very large amounts of energy may be released

Comparing Energy Scales

Typical chemical energies are often measured in:

electronvolts per molecule

or

kilojoules per mole

Nuclear reaction energies are commonly measured in:

MeV per nucleus or reaction

Remember:

1 eV = 1.602 × 10⁻¹⁹ J

and:

1 MeV = 10⁶ eV

So an energy of:

1 MeV

is one million electronvolts.

This immediately shows why nuclear energy operates on a much larger scale than ordinary chemistry.


Typical Energy Comparison

A chemical bond may involve energy changes of a few:

eV

A nuclear reaction may involve:

millions of eV

For example:

chemical reaction: ~1–10 eV per molecular event

nuclear reaction: ~1–200 MeV per nuclear event

The exact values vary, but the difference in scale is enormous.


Why Do Nuclear Reactions Release So Much Energy?

The answer comes from nuclear binding energy.

Protons and neutrons are held together inside nuclei by the strong nuclear interaction.

The energies associated with nuclear binding are much larger than the energies associated with electron bonding in atoms and molecules.

When a nuclear reaction produces a more tightly bound arrangement of nucleons, the final system can have less total rest mass than the initial system.

That mass difference becomes energy.


Mass-Energy Equivalence

Einstein's equation is:

E = mc²

For nuclear reactions, we usually write:

ΔE = Δmc²

where:

Δm = mass difference between the initial and final systems

Because:

c ≈ 3.00 × 10⁸ m/s

then:

c² ≈ 9.00 × 10¹⁶ m²/s²

A tiny mass difference can therefore correspond to a very large amount of energy.


Example 1: Small Mass, Large Energy

Suppose a nuclear reaction has a mass defect of:

2.0 × 10⁻²⁸ kg

Then:

E = Δmc²

E = (2.0 × 10⁻²⁸)(3.0 × 10⁸)²

E = 1.8 × 10⁻¹¹ J

That is the energy from just one nuclear reaction.

A macroscopic sample contains an enormous number of nuclei, so the total energy can become very large.


Using Atomic Mass Units

Nuclear masses are often measured in atomic mass units.

Recall:

1 u c² ≈ 931.5 MeV

Therefore:

Energy released (MeV) = Δm (u) × 931.5


Example 2: Energy from Mass Defect

Suppose:

Δm = 0.150 u

Then:

E = 0.150 × 931.5

E ≈ 140 MeV

This is a typical nuclear-scale energy.


Binding Energy and Nuclear Energy

The binding energy per nucleon tells us how tightly nucleons are bound in a nucleus.

The binding-energy curve rises rapidly for light nuclei, reaches a maximum near the iron-nickel region, and gradually decreases for very heavy nuclei.

This explains both major nuclear energy processes:

fusion of light nuclei → moves toward greater binding energy per nucleon

fission of heavy nuclei → moves toward greater binding energy per nucleon

Both can therefore release energy.


Nuclear Reaction Energy Diagrams

An energy diagram shows the energy of a system before and after a reaction.

A simple exothermic nuclear reaction may look conceptually like this:

Higher energy

Reactants
────────────

↓ energy released

Products
────────

Lower energy

If the products have lower total energy than the reactants, the difference is released.


Exothermic Nuclear Reaction

If:

Eproducts < Ereactants

then energy is released.

The energy released is:

Q = Einitial − Efinal

or, using masses:

Q = (minitial − mfinal)c²

If:

Q > 0

the reaction releases energy.


Endothermic Nuclear Reaction

Some nuclear reactions require energy.

If:

Eproducts > Ereactants

then energy must be supplied.

In this case:

Q < 0

Such a reaction is endothermic.


Reading a Nuclear Energy Diagram

When interpreting an energy diagram, ask:

  1. Which level represents the reactants?
  2. Which level represents the products?
  3. Which has greater energy?
  4. What is the energy difference?
  5. Is energy released or absorbed?

If the products are lower:

energy is released

If the products are higher:

energy must be supplied


Example 3: Reading an Energy Diagram

Suppose the reactants have total energy:

150 MeV

and the products have:

132 MeV

Then:

Q = 150 − 132

Q = 18 MeV

Therefore:

18 MeV is released.


Where Does Released Nuclear Energy Go?

Released energy can appear as:

  • kinetic energy of particles
  • kinetic energy of daughter nuclei
  • gamma radiation
  • neutrino energy
  • thermal energy after particles interact with matter

In a power reactor, much of the microscopic kinetic energy eventually becomes heat.


Comparing Nuclear Energy Sources

Several nuclear processes can release usable or observable energy.

Important examples include:

  • radioactive decay
  • nuclear fission
  • nuclear fusion

Each has different characteristics.


Energy from Radioactive Decay

Radioactive decay occurs when an unstable nucleus spontaneously transforms.

Energy may be carried away by:

  • alpha particles
  • beta particles
  • gamma rays
  • neutrinos
  • recoil of the daughter nucleus

Radioactive decay is extremely important in areas such as:

  • medicine
  • dating
  • space power systems
  • scientific research

However, ordinary radioactive decay is generally not used in the same way as a large fission power reactor.


Energy from Fission

In nuclear fission, a heavy nucleus splits into smaller nuclei.

For example:

²³⁵U + n → fission products + neutrons + energy

A typical uranium-235 fission releases energy on the order of:

200 MeV

per fission event.

The fission products are more tightly bound than the original heavy nucleus, producing a mass-energy difference.


Energy from Fusion

In nuclear fusion, light nuclei combine.

For example:

²H + ³H → ⁴He + n + 17.6 MeV

The products are more tightly bound than the starting nuclei.

Fusion releases less energy per individual reaction than a uranium fission event, but the reacting nuclei are also much lighter.

Therefore, comparisons should not be based only on energy per reaction.


Energy per Reaction vs Energy per Unit Mass

This distinction is very important.

Suppose:

  • Reaction A releases 200 MeV
  • Reaction B releases 18 MeV

It may appear that A is automatically the better energy source.

But if the particles involved in B are much lighter, many more reactions can occur in the same mass of fuel.

So energy sources should also be compared using:

energy per kilogram

or:

energy per nucleon

rather than only energy per reaction.


Comparing Chemical, Fission, and Fusion Energy

A rough comparison illustrates the scale:

Process Typical Energy Scale
Chemical combustion ~10⁷ J/kg
Nuclear fission fuel ~10¹³–10¹⁴ J/kg of fissile material
D-T fusion fuel ~10¹⁴ J/kg of reacting fuel

Exact values depend on fuel composition and what is included in the calculation.

The important conclusion is:

nuclear fuels contain vastly greater usable energy per unit mass than chemical fuels.


Why Nuclear Fuel Has High Energy Density

Energy density means the amount of energy available from a given mass or volume of fuel.

Nuclear fuels have extremely high energy density because nuclear binding-energy changes are much larger than chemical bond-energy changes.

Therefore, relatively small quantities of nuclear fuel can produce large amounts of energy.


Chemical Energy Example

Combustion of fossil fuels releases energy by rearranging electrons.

The nuclei remain unchanged.

A kilogram of fuel may release energy on the order of:

tens of megajoules

or roughly:

10⁷ J/kg


Nuclear Fission Example

A kilogram of fissile uranium undergoing complete fission can release energy on the order of:

10¹³ J/kg

or greater.

This is millions of times more energy per kilogram than typical chemical fuel.


Why Isn't All Nuclear Rest Mass Converted to Energy?

It is important not to misunderstand:

E = mc²

A nuclear reactor does not convert all of its fuel mass directly into energy.

Only a small fraction of the total mass-energy changes during the nuclear reaction.

For fission and fusion:

a small mass defect → a large energy release

Most of the original mass still remains in the reaction products.


Energy Efficiency

The word efficiency can have several meanings.

This is important when evaluating nuclear energy.

We may discuss:

  • reaction efficiency
  • fuel utilization
  • thermal efficiency
  • electrical efficiency
  • energy density

These are not the same thing.


Mass-to-Energy Conversion Efficiency

One way to describe nuclear efficiency is to ask:

What fraction of the starting rest mass becomes other forms of energy?

This can be calculated as:

Efficiency = Δm / minitial × 100%


Example 4: Mass Conversion Efficiency

Suppose:

initial mass = 5.000 u

and:

mass defect = 0.020 u

Then:

efficiency = 0.020 / 5.000 × 100%

efficiency = 0.40%

Only 0.40% of the initial rest mass has been converted into other forms of energy.

That sounds small, but because c² is enormous, the released energy can still be very large.


Fission Mass Conversion

In a typical fission reaction, roughly a small fraction of one percent of the initial rest mass appears as released energy.

Even this tiny fraction is enough to give nuclear fission its enormous energy density.


Fusion Mass Conversion

Hydrogen-to-helium fusion can convert a somewhat larger fraction of the reacting mass into energy than typical fission.

For the net hydrogen-to-helium process in stars, roughly:

0.7%

of the original mass is converted into other forms of energy.

Again, nearly all the original mass remains.


Thermal Efficiency of a Nuclear Power Station

A different concept is thermal efficiency.

A nuclear reactor produces thermal energy.

A power station then converts:

nuclear energy → heat → mechanical energy → electrical energy

Not all thermal energy becomes electricity.

Some energy is transferred to the surroundings.

Therefore:

electrical energy output < thermal energy produced


Calculating Power-Station Efficiency

Efficiency can be written:

Efficiency = useful energy output / total energy input × 100%

For a power station:

Efficiency = electrical energy output / thermal energy supplied × 100%


Example 5: Reactor Efficiency

Suppose a reactor produces:

3000 MW of thermal power

and the generators produce:

1000 MW of electrical power

Then:

Efficiency = 1000 / 3000 × 100%

Efficiency ≈ 33%

The remaining thermal energy must ultimately be transferred to the environment.


Why Is the Efficiency Not 100%?

A nuclear power station is a heat engine.

It is subject to the laws of thermodynamics.

Some thermal energy must be rejected to a lower-temperature environment.

This is why power stations often use:

  • cooling towers
  • cooling water
  • condensers

The large amount of waste heat does not mean nuclear fission itself is inefficient at releasing energy.

It reflects limitations in converting thermal energy into useful electrical energy.


Fuel Utilization

Another question is:

How much of the fuel can actually undergo the useful nuclear reaction?

Not every nucleus in a fuel assembly necessarily undergoes fission.

Reactor fuel contains mixtures of isotopes, and the fuel may be removed before every possible fissile nucleus has reacted.

Some reactor systems can also create new fissile materials from other isotopes.

Therefore, practical fuel utilization is more complicated than the theoretical energy available from one kilogram of a pure isotope.


Comparing Nuclear Fission and Fusion

Feature Fission Fusion
Basic process Heavy nucleus splits Light nuclei combine
Typical reaction energy ~200 MeV per U-235 fission 17.6 MeV for D-T fusion
Energy density Extremely high Extremely high
Commercial electricity today Yes Not yet established commercially
Chain reaction Yes No fission-type chain reaction
Long-lived waste Significant issue Generally different/lower long-lived waste challenge, but neutron activation occurs
Fuel Uranium/plutonium systems Hydrogen isotopes in major designs
Main technical challenge Safe control and waste management Achieving sustained useful controlled fusion

Comparing Nuclear and Chemical Energy

Feature Chemical Nuclear
Part of atom involved Electrons Nucleus
Typical energy scale eV MeV
Element identity changes? Usually no Often yes
Mass change Extremely tiny Small but measurable in principle
Energy density Relatively low Extremely high
Example Combustion Fission or fusion

Advantages of Nuclear Energy Efficiency

Nuclear energy has several advantages related to energy density.

Small Amounts of Fuel

A relatively small fuel mass can produce enormous amounts of energy.

Reduced Fuel Transport

Because less fuel mass is required, much less material needs to be transported compared with fossil fuels for an equivalent energy output.

Long Operating Periods

Nuclear reactors can operate for long periods between refuelling cycles.

Low Direct Carbon Emissions

Nuclear reactions themselves do not involve combustion of carbon-based fuels.


Limitations of Nuclear Energy

High energy density does not automatically make an energy source perfect.

Nuclear energy systems must also consider:

  • construction costs
  • radioactive waste
  • reactor safety
  • decommissioning
  • fuel mining and processing
  • security
  • cooling requirements
  • thermal conversion losses

Efficiency is only one part of evaluating an energy technology.


Energy Efficiency vs Environmental Impact

A process can have high energy density but still create environmental challenges.

Similarly, an energy source with lower energy density may have other advantages.

A complete evaluation should consider:

  • efficiency
  • emissions
  • land use
  • reliability
  • waste
  • resource availability
  • cost
  • safety

Therefore:

energy efficiency alone does not determine the best energy source.


Example 6: Comparing Fuel Masses

Suppose Fuel A releases:

4 × 10⁷ J/kg

and Fuel B releases:

8 × 10¹³ J/kg

Compare their energy densities:

8 × 10¹³ ÷ 4 × 10⁷

= 2 × 10⁶

Fuel B releases:

2 million times more energy per kilogram

than Fuel A.

This demonstrates why nuclear fuel requires so little mass compared with chemical fuels.


Example 7: Q-Value from Masses

Suppose a nuclear reaction has:

initial mass = 4.0350 u

final mass = 4.0120 u

Mass defect:

Δm = 4.0350 − 4.0120

Δm = 0.0230 u

Energy released:

Q = 0.0230 × 931.5

Q ≈ 21.4 MeV

Since:

Q > 0

the reaction releases energy.


Interpreting the Result

The mass did not simply disappear.

The difference in rest mass appears as other forms of energy, such as:

  • particle kinetic energy
  • radiation
  • nuclear recoil

Total mass-energy is conserved.


Example 8: Energy Diagram

Suppose an energy diagram shows:

Reactants = 80 MeV

Products = 65 MeV

Then:

ΔE = 80 − 65

ΔE = 15 MeV

Since the products are lower in energy:

15 MeV is released.

If the diagram were reversed, 15 MeV would have to be supplied.


The Sun as a Nuclear Energy Source

The Sun demonstrates the enormous energy available from nuclear reactions.

Hydrogen fusion converts a small fraction of the reacting mass into other forms of energy.

Because the Sun contains an enormous amount of hydrogen, this process can power it for billions of years.

Nuclear energy therefore operates on both:

  • microscopic particle scales
  • astronomical scales

Did You Know?

The energy difference between nuclear and chemical reactions is so large that comparing them only by fuel mass can be surprising.

A relatively small amount of nuclear fuel can contain the potential for energy comparable to enormous quantities of chemical fuel.

This does not mean all of the rest mass of the nuclear fuel is converted to energy. Only a small change in nuclear mass is required to produce the difference.


Connecting the Ideas

The energy released by nuclear reactions connects several topics from this course:

Nuclear reaction

↓

new arrangement of nucleons

↓

change in binding energy

↓

mass defect

↓

ΔE = Δmc²

↓

kinetic energy and radiation released

↓

thermal energy in practical systems

↓

electricity generation

This explains why nuclear reactions can release such enormous amounts of energy from relatively small quantities of material.


Key Terms

Chemical energy – Energy associated with the arrangement and bonding of electrons in atoms and molecules.

Nuclear energy – Energy associated with changes in atomic nuclei.

Mass defect – The difference in mass between initial and final nuclear systems or between bound nuclei and their separated nucleons.

Binding energy – Energy required to separate a nucleus completely into its individual nucleons.

Q-value – The net energy released or absorbed during a nuclear reaction.

Energy density – Energy available per unit mass or volume.

Exothermic reaction – A reaction that releases energy.

Endothermic reaction – A reaction that requires an input of energy.

Thermal efficiency – The fraction of supplied thermal energy converted into useful output.

Fuel utilization – The fraction of available fuel that is effectively used to produce energy.


Key Takeaways

  • Chemical reactions involve changes in electron arrangements, while nuclear reactions involve changes in the nucleus.
  • Nuclear reactions typically involve energy scales of MeV, compared with eV for individual chemical events.
  • Nuclear reactions release large amounts of energy because nuclear binding-energy changes are very large.
  • A small change in rest mass can release a large amount of energy through ΔE = Δmc².
  • Nuclear energy diagrams compare the energy of reactants and products.
  • If the products have lower total energy, the reaction releases energy.
  • The reaction Q-value can be calculated using Q = (minitial − mfinal)c².
  • Both fission and fusion release energy because they move nuclei toward more tightly bound configurations.
  • Fission typically releases about 200 MeV per uranium-235 event, while D-T fusion releases 17.6 MeV per reaction.
  • Energy per reaction is not the same as energy per kilogram of fuel.
  • Nuclear fuels have millions of times greater energy density than typical chemical fuels.
  • Only a small fraction of nuclear rest mass is converted into other forms of energy.
  • Power-station efficiency is lower than the theoretical nuclear-energy release because thermal energy must be converted into electricity.
  • Evaluating nuclear energy requires considering not only efficiency, but also cost, safety, waste, resource use, emissions, and reliability.