Unit 6 · Particle Physics
The Complete Cambridge International AS & A Level Physics (9702) Guide · Cambridge International AS & A Level Physics (9702)Unit 6 · Particle Physics
- The Complete Cambridge International AS & A Level Physics (9702) Guide
- Cambridge International AS & A Level Physics (9702)
- 12 sections
Cambridge International does not publish a per-unit weighting for this exam. Exam structure (official): AS Level: Multiple Choice (75 min, 40 questions, 31%) · AS Level Structured Questions (75 min, 46%) · Advanced Practical Skills (120 min, 2 questions, 23%); A Level: Multiple Choice (75 min, 40 questions, 15.5%) · AS Level Structured Questions (75 min, 23%) · Advanced Practical Skills (120 min, 2 questions, 11.5%) · A Level Structured Questions (120 min, 38.5%) · Planning, Analysis And Evaluation (75 min, 2 questions, 11.5%). Be ready to carry the same unit skills and representations into a complete answer on every paper. Subtracting the product mass excess of 3.85 MeV/c² from 5.10 MeV/c² leaves 1.25 MeV/c². Multiplication by c² converts that mass difference to a 1.25 MeV energy release.
- How Cambridge International AS & A Level Physics (9702) assesses this Cambridge International publishes no per-unit percentage for Cambridge International A Level Physics; Paper 1 accounts for 31% of the AS Level total and 15.5% of the A Level total. · AS Level: Multiple Choice (75 min, 40 questions, 31%) · AS Level Structured Questions (75 min, 46%) · Advanced Practical Skills (120 min, 2 questions, 23%) | A Level: Multiple Choice (75 min, 40 questions, 15.5%) · AS Level Structured Questions (75 min, 23%) · Advanced Practical Skills (120 min, 2 questions, 11.5%) · A Level Structured Questions (120 min, 38.5%) · Planning, Analysis And Evaluation (75 min, 2 questions, 11.5%) · show the model with A graph of energy release (MeV) against mass difference (MeV/c²)., A formula card that explains the c² factor converts the difference in mass excess into released energy., A table linking key concept, why it matters, visible method.
- Key skills Atoms, Nuclei, Alpha-Particle Scattering and Nuclide Notation, Alpha, Beta and Gamma Radiation and Decay Equations, Fundamental Particles: Quarks, Leptons and Hadrons
- How to study for Unit 6 This page turns A graph of energy release (MeV) against mass difference (MeV/c²)., A formula card that explains the c² factor converts the difference in mass excess into released energy., A table linking key concept, why it matters, visible method. into one route: Subtract 3.85 from 5.10.
- The organizing decision Check charge, nucleon or lepton quantities as required by a reaction; matching mass–energy units alone cannot identify a missing particle.
What Cambridge International AS & A Level Physics (9702) Unit 6 covers
Use this map to connect each assessed skill to the relationship or representation that makes it visible.
Cambridge International does not publish a per-unit weighting for this exam.
Official unit name and weighting: Cambridge International syllabus.
Particle Physics
Connect the published share to the unit model
Subtracting the product mass excess of 3.85 MeV/c² from 5.10 MeV/c² leaves 1.25 MeV/c². Multiplication by c² converts that mass difference to a 1.25 MeV energy release.
The alternative joule form uses 1.60 × 10⁻¹⁹ J for each electronvolt and therefore gives 2.00 × 10⁻¹³ J. Mixing joules and electronvolts within one subtraction would break the energy chain.
Momentum conservation still governs how the 1.25 MeV is shared among products. A balanced numerical release does not replace the reaction-level conservation check.
Identify the interaction before balancing any reaction, then apply every relevant conservation rule rather than charge alone. Include a missing particle where the conserved quantities require it, and keep mass and energy units consistent if a calculation follows. The completed symbolic reaction should end with a statement of what the interaction or decay implies.
The decision that organizes this unit
Define the system and choose the route before calculating
Check charge, nucleon or lepton quantities as required by a reaction; matching mass–energy units alone cannot identify a missing particle.
Subtract 3.85 from 5.10
Mechanism route and repair branches
Relationships to preserve
- Mass difference: Initial excess exceeds product excess by 1.25 MeV/c².
- Energy release: The c² factor cancels the denominator.
- Joule form: One electronvolt equals 1.60 × 10⁻¹⁹ J here.
Representations to read
- A graph of energy release (MeV) against mass difference (MeV/c²).
- A formula card that explains the c² factor converts the difference in mass excess into released energy.
- A table linking key concept, why it matters, visible method.
Branches to reject
- Reversing 5.10 MeV/c² and 3.85 MeV/c²
- Treating MeV as though it were eV
- Using the total mass excess instead of the difference
| Key concept | Why it matters | Visible method |
|---|---|---|
| Mass difference | Initial excess exceeds product excess by 1.25 MeV/c² | Subtract 3.85 from 5.10 |
| Energy release | The c² factor cancels the denominator | State Q = 1.25 MeV |
| Joule form | One electronvolt equals 1.60 × 10⁻¹⁹ J here | Convert to 2.00 × 10⁻¹³ J |
| Interaction first | The interaction determines which conserved quantities apply. | Identify it before completing the symbolic reaction. |
| Missing-particle check | Charge balance alone may not complete the reaction. | Account for all required particle quantities on both sides. |
How Cambridge International AS & A Level Physics (9702) assesses Particle Physics
What a complete response must make visible
Match the task to evidence that a reader can audit, then check the most likely reasoning failure before finalizing the response.
| Task | Evidence to show | Hurdle |
|---|---|---|
| Atoms, Nuclei, Alpha-Particle Scattering and Nuclide Notation | mass excess; binding energy; nucleon number | Reversing 5.10 MeV/c² and 3.85 MeV/c² |
| Alpha, Beta and Gamma Radiation and Decay Equations | mass excess; binding energy; nucleon number | Treating MeV as though it were eV |
| Fundamental Particles: Quarks, Leptons and Hadrons | mass excess; binding energy; nucleon number | Using the total mass excess instead of the difference |
Worked example: nuclear notation, conservation and mass–energy
Read each question as it appears on the exam paper, choose your answer, then tap it to see the feedback.
Practice 1: Energy released in a decay
A decay changes the total mass excess of a system.
A nucleus with mass excess 5.10 MeV/c² decays into products with total mass excess 3.85 MeV/c². Calculate the released energy in MeV and joules.
[4]Show all necessary working.
A. Q = 5.10 − 3.85 = 1.25 MeV = 2.00 × 10⁻¹³ J.
✅ Correct! 😊 Correct. The decrease in mass excess becomes released energy, and 1.25 × 10⁶ eV converts to 2.00 × 10⁻¹³ J.
B. Q = 3.85 − 5.10 = −1.25 MeV, so 1.25 MeV is absorbed.
❌ Not quite 😭 This reverses the before-minus-after subtraction. The products have lower mass excess, so the stated decay releases energy.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
C. Q = 5.10 + 3.85 = 8.95 MeV.
❌ Not quite 😭 Mass–energy release depends on the decrease between initial and final totals, not their sum.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
D. Q = 1.25 × 10⁻¹⁹ J because 1 MeV equals 10⁶ J.
❌ Not quite 😭 The MeV-to-joule conversion is wrong. Convert MeV to eV, then use 1 eV = 1.60 × 10⁻¹⁹ J.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
Full explanation
- Step 1Use Q = (mass excess before − mass excess after)c².
- Step 2The c² cancels the /c² in MeV/c², giving Q = 5.10 − 3.85 = 1.25 MeV.
- Step 3Convert 1.25 MeV to 1.25 × 10⁶ eV.
- Step 4Calculate Q = (1.25 × 10⁶)(1.60 × 10⁻¹⁹) = 2.00 × 10⁻¹³ J.
Practice 2: Conservation in beta-minus decay
A nucleus undergoes beta-minus decay.
Which change in the nucleus and emitted particles is consistent with beta-minus decay?
[1]Choose the best answer.
A. The proton number decreases by one, and a positron and neutrino are emitted.
❌ Not quite 😭 This describes the charge pattern of beta-plus decay, not beta-minus decay. The emitted charged particle is also wrong.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
B. The proton number increases by one, and an electron and electron antineutrino are emitted.
✅ Correct! 😊 Correct. A neutron changes into a proton while the electron and antineutrino preserve charge and lepton number.
C. The nucleon number decreases by four, and an electron is emitted.
❌ Not quite 😭 Beta decay leaves nucleon number unchanged. A decrease of four belongs to emission of a helium nucleus, not an electron.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
D. The proton number is unchanged, and only a photon is emitted.
❌ Not quite 😭 Photon emission does not produce the nuclear charge change characteristic of beta-minus decay. It is an electromagnetic de-excitation.
Prep with Sia Agent
Free download for Mac and Windows. Choose your computer:
Full explanation
- Step 1In beta-minus decay, a neutron in the nucleus changes into a proton.
- Step 2The nucleon number stays constant, while the proton number increases by one.
- Step 3An electron carries charge −1, balancing the extra positive nuclear charge.
- Step 4An electron antineutrino is also emitted so lepton number is conserved.
Key terms for Unit 6: Particle Physics
Models, uses, and boundaries
- Particle Physics
- The c² factor converts the difference in mass excess into released energy. Use this relationship when the c² factor converts the difference in mass excess into released energy..
- Worked example: nuclear notation, conservation and mass–energy
- Subtracting 3.85 from 5.10 gives the positive 1.25 MeV available to the products. Use this relationship when subtracting 3.85 from 5.10 gives the positive 1.25 MeV available to the products..
- Particle Physics: change one condition
- Along this unit-slope relation, increasing mass difference from 1.0 to 1.5 raises Q by the same factor. Use this relationship when along this unit-slope relation, increasing mass difference from 1.0 to 1.5 raises Q by the same factor..
- Particle Physics: test proposed results
- A viable option must preserve both the MeV scale and the sign of an energy release. Use this relationship when a viable option must preserve both the MeV scale and the sign of an energy release..
Cambridge International AS & A Level Physics (9702) Unit 6 FAQ
How much of Cambridge International AS & A Level Physics (9702) does Unit 6 carry?
Cambridge International does not publish a per-unit weighting for this exam.
What is the first move on a Particle Physics problem?
Subtract 3.85 from 5.10
Which relationships should I preserve?
Mass difference: Initial excess exceeds product excess by 1.25 MeV/c². Energy release: The c² factor cancels the denominator. Joule form: One electronvolt equals 1.60 × 10⁻¹⁹ J here.
Which representations should I practice?
Practice moving among A graph of energy release (MeV) against mass difference (MeV/c²)., A formula card that explains the c² factor converts the difference in mass excess into released energy., A table linking key concept, why it matters, visible method..
What error should I check before submitting an answer?
Check for Reversing 5.10 MeV/c² and 3.85 MeV/c²; Treating MeV as though it were eV; Using the total mass excess instead of the difference.
Evidence workshop
Continue from the free model into complete practice
The full unit guide continues with the chapter’s worked examples, figures, scoring tables, and answer checks.
- Worked example: nuclear notation, conservation and mass–energy
- Particle Physics: change one condition
- Particle Physics: test proposed results
- Check the worked-example evidence for Particle Physics
Full unit practice. Open the complete guide for the full evidence workshop and synthesis.
Related Cambridge International AS & A Level Physics (9702) unit guides
Cambridge International AS & A Level Physics (9702) Exam Guide & Review
The whole exam and its official unit sequence.01Physical Quantities and Units
Cambridge International does not publish a per-unit weighting for this exam.02Kinematics and Dynamics
Cambridge International does not publish a per-unit weighting for this exam.03Forces, Energy and Deformation of Solids
Cambridge International does not publish a per-unit weighting for this exam.04Waves and Superposition
Cambridge International does not publish a per-unit weighting for this exam.05Electricity and D.C. Circuits
Cambridge International does not publish a per-unit weighting for this exam.07Circular Motion, Gravitational Fields and Oscillations
Cambridge International does not publish a per-unit weighting for this exam.08Thermal Physics
Cambridge International does not publish a per-unit weighting for this exam.09Electric Fields and Capacitance
Cambridge International does not publish a per-unit weighting for this exam.10Magnetic Fields and Alternating Currents
Cambridge International does not publish a per-unit weighting for this exam.11Quantum and Nuclear Physics
Cambridge International does not publish a per-unit weighting for this exam.12Medical Physics, Astronomy and Cosmology
Cambridge International does not publish a per-unit weighting for this exam.How to study Cambridge International AS & A Level Physics (9702) Unit 6
Start with the organizing decision
Before solving, restate the decision in operational terms: Check charge, nucleon or lepton quantities as required by a reaction; matching mass–energy units alone cannot identify a missing particle.. Your first written move should be to Subtract 3.85 from 5.10.
Practice the same idea in several representations
Rotate through A graph of energy release (MeV) against mass difference (MeV/c²)., A formula card that explains the c² factor converts the difference in mass excess into released energy., A table linking key concept, why it matters, visible method.. Use each representation to practice Atoms, Nuclei, Alpha-Particle Scattering and Nuclide Notation, Alpha, Beta and Gamma Radiation and Decay Equations, Fundamental Particles: Quarks, Leptons and Hadrons, and explain what stays invariant when the surface form changes.
Turn each error into a repair check
After every attempt, audit the response for Reversing 5.10 MeV/c² and 3.85 MeV/c²; Treating MeV as though it were eV; Using the total mass excess instead of the difference. Then redo only the first step that made the reasoning diverge, keeping units, direction, and model conditions visible.
Confirm current course details in the official Cambridge International syllabus for the 2025-2027 administration.
Where to go next
Return to the Cambridge International AS & A Level Physics (9702) to see every unit and plan your revision.