Nuclear equations, conservation and beta decay

Key idea: A reviewed, static H2 Physics learning chain for all official Nuclear Physics outcomes, from Rutherford evidence to fusion and fission.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: Which quantities must balance in a nuclear equation?

Nuclear reactions conserve charge, nucleon number, energy and momentum. In beta-minus decay, a neutron becomes a proton while an electron and antineutrino are emitted; the antineutrino accounts for the continuous beta energy spectrum while preserving energy and momentum. Balance A and Z first, then interpret the energy.

Balance more than the visible symbols

A nuclear equation conserves nucleon number A and charge Z. It must also conserve total mass-energy and momentum. Identify an unknown nuclide from both A and Z, then check whether the proposed reaction is energetically possible.

In beta-minus decay, a neutron changes to a proton and emits an electron and electron antineutrino: n → p + e⁻ + ν̄_e. A stays constant and nuclear Z rises by one. Detailed particle classification is not required.

Check your understanding: Complete ²³⁴₉₀Th → ²³⁴₉₁Pa + X.

X is ⁰₋₁e plus an antineutrino in beta-minus decay; A is unchanged and Z balances.

Use the continuous beta spectrum as evidence

If beta decay produced only a daughter nucleus and electron from a fixed initial state, two-body energy and momentum conservation would give the electron a fixed energy apart from recoil. Experiments instead show a continuous range up to an endpoint.

The antineutrino carries a variable share of energy and momentum, restoring conservation event by event. Energy is not 'lost'; it is distributed among electron, recoil nucleus and antineutrino.

Check your understanding: What key observation motivated the antineutrino proposal?

The continuous beta-electron energy spectrum, together with the need to conserve energy and momentum.

How alpha, beta-minus and gamma affect A and ZThree quick-reference panels showing parent-to-daughter changes in nucleon number and proton number for alpha, beta-minus and gamma emissions.AlphaA -> A - 4Z -> Z - 2emit 4/2 HeBeta-minusA -> AZ -> Z + 1emit 0/-1 eGammaA -> AZ -> Zemit gamma ray
Alpha decreases A by 4 and Z by 2; beta-minus keeps A but increases Z by 1; gamma changes neither.

Key ideas to keep

  • An emitted beta electron is created in the decay; it was not orbiting inside the nucleus.
  • Beta decay changes Z but leaves A unchanged.
  • A continuous electron spectrum requires energy shared with another particle.

Worked example

Identify a reaction product using two conservation counts

Question: Balance ¹⁴₇N + ⁴₂He → ¹⁷₈O + X.

  1. Step 1: Balance nucleon number

    Why: Total A is conserved.

    Working: 14 + 4 = 17 + Aₓ, so Aₓ = 1.

  2. Step 2: Balance charge number

    Why: Total nuclear charge is conserved.

    Working: 7 + 2 = 8 + Zₓ, so Zₓ = 1.

  3. Step 3: Identify and complete the equation

    Why: A = 1 and Z = 1 identifies a proton.

    Working: X = ¹₁H; energy and momentum must also balance even though they are not shown by A and Z.

Answer: Conserving nucleon number gives A = 1 and charge gives Z = 1, so X is ¹₁H. Mass-energy and momentum must also be conserved.

Check: Both sides have total A = 18 and total Z = 9.

Practise with support

Try this

In beta-minus decay, state changes in A and Z.

Hint: A neutron becomes a proton.

Check your answer

A is unchanged and Z increases by one.

Practise independently

Your turn

Write and check a nuclear equation and explain antineutrino evidence.

Check your answer

Conserve A and charge explicitly, then conserve mass-energy and momentum. A continuous beta energy spectrum contradicts fixed two-body energy sharing; an antineutrino carries the missing variable energy and momentum.

Common mistakes

Common mistake

Only nucleon number must balance.

What is wrong with this reasoning?

Show better thinking

Charge, mass-energy and momentum must also be conserved.

Common mistake

The beta electron alone receives a fixed decay energy.

What is wrong with this reasoning?

Show better thinking

Electron, recoil and antineutrino share energy and momentum.

Exam guidance

Balance nucleon and proton numbers on both sides before calculating any energy release.

Exam-style practice [6 marks]

Complete ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + X and name conserved quantities.

Plan before you answer

  • Balance A and Z.
  • Identify the emitted particle.
  • State the other conserved quantities.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

X = ⁴₂He. Nucleon number, charge, mass-energy and momentum are conserved.

Check what stayed with you

Recall question

Why does beta-minus decay keep nucleon number unchanged?

Check the answer

A neutron changes into a proton; the total number of nucleons is unchanged.

Try this next

Continue to the next lesson in this topic.

Mass defect and binding energy

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 20 excludes knowledge of positron emission in 20(g) and detailed knowledge of the antineutrino and particle zoo in 20(o). Nuclide equations conserve nucleon number, charge, mass-energy and momentum. Count data require background correction before population-law inference. Applications must relate half-life, penetration and ionisation to benefit and hazard. The binding-energy-per-nucleon curve, not a claim that mass disappears, explains fusion and fission energy release.

  • GCE A-Level H2 PhysicsTopic 20(m) / Topic 20(n) / Topic 20(o) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027