Nuclear and Particle Physics Problem Solving

This final lesson brings together everything from the course into a problem-solving toolkit. Exam questions on nuclear and particle physics often combine multiple concepts — you might need to calculate binding energy AND explain a Feynman diagram, or work out an activity AND identify a quark composition. The key to success is having a systematic approach and being confident with the mathematics.

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Decay Calculations: A Systematic Approach

When faced with radioactive decay problems, follow this framework:

Step 1: Identify what you know and what you need

Common quantities:

• N₀ = initial number of nuclei
• N = remaining nuclei at time t
• A₀ = initial activity (Bq)
• A = activity at time t
• λ = decay constant (s⁻¹)
• t½ = half-life
• t = elapsed time

Step 2: Use the correct equation

If you need...	Use...
Half-life from decay constant	t½ = ln 2 / λ
Decay constant from half-life	λ = ln 2 / t½
Remaining nuclei	N = N₀e^(−λt)
Remaining activity	A = A₀e^(−λt)
Activity from N	A = λN
Time to reach a certain N or A	t = −(1/λ) ln(N/N₀) = −(1/λ) ln(A/A₀)
Number of atoms from mass	N = (m/M_r) × N_A

Step 3: Check for whole numbers of half-lives

If the time elapsed is a whole number of half-lives, you can use repeated halving instead of exponentials — this is faster and avoids calculator errors.

Worked Example: Complete Decay Problem

Problem: A sample of phosphorus-32 (t½ = 14.3 days) has an initial activity of 4.0 × 10⁶ Bq. Find: (a) the decay constant in s⁻¹, (b) the initial number of P-32 atoms, (c) the activity after 42.9 days, (d) the time for the activity to fall to 1.0 × 10⁵ Bq.

(a) λ = ln 2 / t½ = 0.693 / (14.3 × 24 × 3600) = 0.693 / 1,235,520 = 5.61 × 10⁻⁷ s⁻¹

(b) N₀ = A₀/λ = 4.0 × 10⁶ / 5.61 × 10⁻⁷ = 7.13 × 10¹² atoms

(c) 42.9 days = 3 half-lives. A = 4.0 × 10⁶ / 2³ = 4.0 × 10⁶ / 8 = 5.0 × 10⁵ Bq

(d) t = −(1/λ) ln(A/A₀) = −(1/5.61 × 10⁻⁷) × ln(1.0 × 10⁵ / 4.0 × 10⁶) = −(1.783 × 10⁶) × ln(0.025) = −(1.783 × 10⁶) × (−3.689) = 6.577 × 10⁶ s = 76.1 days

Check: 76.1/14.3 = 5.32 half-lives. After 5 half-lives: 1/32 = 0.03125, and 0.03125 × 4.0 × 10⁶ = 1.25 × 10⁵ Bq. After 6 half-lives: 6.25 × 10⁴ Bq. Since 1.0 × 10⁵ is between these, 5.3 half-lives is reasonable. ✔

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Binding Energy Calculations

Mass Defect Method

1. Calculate total mass of separate nucleons: (Z × mₚ) + (N × mₙ)
2. Subtract the nuclear mass: Δm = separate mass − nuclear mass
3. Convert to energy: E = Δm × 931.5 MeV (if Δm in atomic mass units)
4. Divide by A for binding energy per nucleon

Worked Example: Binding Energy of Nickel-62

Nickel-62: Z = 28, N = 34, nuclear mass = 61.91283 u

Total separate mass = (28 × 1.00728) + (34 × 1.00866) = 28.20384 + 34.29444 = 62.49828 u

Mass defect = 62.49828 − 61.91283 = 0.58545 u

Total binding energy = 0.58545 × 931.5 = 545.5 MeV

Binding energy per nucleon = 545.5 / 62 = 8.80 MeV

This is the highest binding energy per nucleon of any nucleus, making Ni-62 the most tightly bound nucleus (slightly higher than Fe-56 at 8.79 MeV per nucleon).

Energy Released in Reactions

1. Find total mass of reactants (all particles on the left of the equation)
2. Find total mass of products (all particles on the right)
3. Energy released = (mass of reactants − mass of products) × 931.5 MeV
4. If the answer is positive, energy is released; if negative, energy must be supplied

Worked Example: Fission Energy Calculation

Problem: Calculate the energy released in: ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n

Masses: U-235 = 235.04393 u, Ba-141 = 140.91440 u, Kr-92 = 91.92616 u, n = 1.00866 u

Reactants: 235.04393 + 1.00866 = 236.05259 u Products: 140.91440 + 91.92616 + 3(1.00866) = 140.91440 + 91.92616 + 3.02598 = 235.86654 u Δm = 236.05259 − 235.86654 = 0.18605 u

Energy = 0.18605 × 931.5 = 173.3 MeV

Adding the energy from subsequent beta decay of the fragments gives the full ~200 MeV per fission.

Common Pitfall

Remember: mass defect uses individual nucleon masses (not atomic masses) compared to the nuclear mass. If you are given atomic masses, you may need to account for electron masses. In most A-Level questions, the data provided will make this clear.

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Quark Composition Problems

To find or verify a quark composition:

1. Count quarks: baryons have 3 quarks; mesons have 1 quark + 1 antiquark
2. Add up charges: must equal the particle’s total charge
3. Check baryon number: must equal +1 for baryons, 0 for mesons
4. Check strangeness: count s quarks (each gives S = −1) and s̄ antiquarks (each gives S = +1)

Quick Reference Table

Quark	Charge (e)	B	S
u	+2/3	+1/3	0
d	−1/3	+1/3	0
s	−1/3	+1/3	−1
ū	−2/3	−1/3	0
d̄	+1/3	−1/3	0
s̄	+1/3	−1/3	+1

Worked Example: Identifying an Unknown Particle

Problem: A hadron X has charge 0, baryon number 0, and strangeness +1. Identify it.

• B = 0 → meson (qq̄)
• S = +1 → contains s̄ (antistrange, S = +1). The quark partner must have S = 0.
• Charge = 0 → s̄ has charge +1/3, so the quark must have charge −1/3 → down quark (d)
• Composition: ds̄, charge = −1/3 + 1/3 = 0 ✔

This is the K⁰ (neutral kaon).

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Conservation Law Checks

When asked whether a particle interaction is allowed or forbidden:

Check each conservation law systematically:

1. Charge (Q): total charge before = total charge after (always conserved)
2. Baryon number (B): total B before = total B after (always conserved)
3. Lepton number (Lₑ, L_μ, L_τ): each generation conserved separately (always)
4. Strangeness (S): conserved in strong and electromagnetic interactions; can change by ±1 in weak interactions

Worked Example: Is This Reaction Allowed?

Reaction: p + p → p + p + π⁰

Quantity	Before	After	Conserved?
Q	+1 + 1 = +2	+1 + 1 + 0 = +2	✔
B	+1 + 1 = +2	+1 + 1 + 0 = +2	✔
L	0 + 0 = 0	0 + 0 + 0 = 0	✔
S	0 + 0 = 0	0 + 0 + 0 = 0	✔

All conserved → Allowed (strong interaction, since S is conserved)

Worked Example: Forbidden Reaction

Reaction: n → p + e⁻

Quantity	Before	After	Conserved?
Q	0	+1 + (−1) = 0	✔
B	+1	+1 + 0 = +1	✔
Lₑ	0	0 + 1 = 1	✘

Lepton number is NOT conserved → Forbidden (the actual decay n → p + e⁻ + ν̄ₑ includes the antineutrino with Lₑ = −1, which fixes the conservation law)

Worked Example: Strangeness-Changing Decay

Reaction: Λ⁰ → p + π⁻

Quantity	Before	After	Conserved?
Q	0	+1 + (−1) = 0	✔
B	+1	+1 + 0 = +1	✔
L	0	0 + 0 = 0	✔
S	−1	0 + 0 = 0	ΔS = +1

Strangeness changes by +1 → Allowed as a weak decay (but not as a strong interaction)

This explains why the Λ⁰ has a relatively long lifetime (2.6 × 10⁻¹⁰ s) compared to particles that decay via the strong force (~10⁻²³ s). Weak decays are much slower.

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Feynman Diagram Interpretation

When analysing a Feynman diagram:

1. Identify all incoming and outgoing particles — these are the real particles before and after the interaction
2. Identify the exchange particle — the internal line connecting two vertices
3. Determine the type of interaction — photon = electromagnetic, W±/Z⁰ = weak, gluon = strong
4. Check conservation laws at each vertex — charge, baryon number, and lepton number must all be conserved at every vertex independently

Drawing Tips for Exam

• Draw time axis clearly (usually horizontal, left to right)
• Label all particles with correct symbols
• Use arrows on fermion lines (with time for particles, against time for antiparticles)
• Show the exchange particle as a wavy/dashed line between vertices
• Make sure conservation laws hold at each vertex separately

Worked Example: Drawing β⁺ Decay

Process: u → d + e⁺ + νₑ

1. Draw a horizontal time axis (left to right)
2. Incoming u quark enters from the left
3. At the first vertex, the u quark emits a W⁺ and becomes a d quark (continuing right)
4. The W⁺ (dashed line) connects to a second vertex
5. At the second vertex, the W⁺ produces e⁺ (arrow going against time, since it is an antiparticle of the electron) and νₑ (arrow with time)

Check vertex 1: Q: +2/3 = −1/3 + (+1) ✔, B: +1/3 = +1/3 + 0 ✔ Check vertex 2: Q: +1 = +1 + 0 ✔, Lₑ: 0 = −1 + 1 ✔

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Comprehensive Problem: Combining Multiple Concepts

Problem: Strontium-90 (t½ = 28.8 years) undergoes β⁻ decay to yttrium-90. A 5.0 mg sample of pure Sr-90 is prepared. (M_r of Sr-90 = 90)

(a) How many Sr-90 atoms are in the sample? N = (m/M_r) × N_A = (5.0 × 10⁻⁶ / 90 × 10⁻³) × 6.02 × 10²³ = 3.34 × 10¹⁹

(b) What is the initial activity? λ = ln 2 / t½ = 0.693 / (28.8 × 3.15 × 10⁷) = 7.63 × 10⁻¹⁰ s⁻¹ A = λN = 7.63 × 10⁻¹⁰ × 3.34 × 10¹⁹ = 2.55 × 10¹⁰ Bq = 25.5 GBq

(c) At the quark level, what happens during the β⁻ decay? A down quark in the neutron (inside the Sr-90 nucleus) changes to an up quark: d → u + W⁻. The W⁻ then decays: W⁻ → e⁻ + ν̄ₑ.

(d) What is the activity after 100 years? A = A₀ e^(−λt) = 2.55 × 10¹⁰ × e^(−7.63 × 10⁻¹⁰ × 100 × 3.15 × 10⁷) = 2.55 × 10¹⁰ × e^(−2.404) = 2.55 × 10¹⁰ × 0.0905 = 2.31 × 10⁹ Bq = 2.31 GBq

Check: 100/28.8 = 3.47 half-lives. After 3 half-lives: 25.5/8 = 3.19 GBq. After 4: 1.59 GBq. Since 2.31 GBq is between these, 3.47 half-lives is consistent. ✔

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Exam Strategy for Nuclear and Particle Physics

For Calculation Questions

1. List the given quantities with correct units
2. Identify which equation(s) are needed
3. Convert units if necessary (days to seconds, MeV to J, etc.)
4. Substitute values carefully
5. Check: does the answer have sensible magnitude and correct units?

Key Unit Conversions

From	To	Multiply by
days	seconds	86,400
years	seconds	3.15 × 10⁷
eV	joules	1.6 × 10⁻¹⁹
MeV	joules	1.6 × 10⁻¹³
u	kg	1.661 × 10⁻²⁷
u	MeV/c²	931.5

For Explanation Questions

• Always state the relevant physics principle (e.g., "binding energy per nucleon is higher for the products, so energy is released")
• Use specific numbers or data from the question
• Link cause and effect clearly

For Conservation Law Questions

• Set up a table with columns for each quantity (Q, B, Lₑ, L_μ, S)
• Calculate the total for each column before and after the interaction
• If any column doesn’t balance, state which conservation law is violated and whether the interaction is forbidden

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Summary

Successful problem solving in nuclear and particle physics requires a systematic approach. For decay problems, identify the known quantities and choose the appropriate exponential or half-life equation. For binding energy, calculate mass defect and convert using E = mc². For particle interactions, systematically check conservation of charge, baryon number, lepton number, and strangeness. For Feynman diagrams, identify the exchange particle and verify conservation at each vertex. Always show your working clearly, check units, and verify answers against reasonable estimates.

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A-Level Deep Dive: Nuclear and Particle Physics Problem Solving

Spec mapping

Edexcel 9PH0 specification, Topics 8 (Nuclear and particle physics) and 11 (Nuclear radiation) are examined together as a synoptic capstone — most often on Paper 3, where unfamiliar contexts and AO3 problem-solving dominate (refer to the official specification document for exact wording). Topic 8 supplies the Standard Model framework: classification of hadrons and leptons, quark composition, exchange particles, and the conservation rules (charge, baryon number, lepton number, strangeness) that govern interactions. Topic 11 supplies the nuclear toolkit: $N$N–$Z$Z stability, alpha/beta/gamma decay, the exponential decay law $N = N_0 e^{-\lambda t}$N=N0​e−λt, half-life, mass defect, binding energy per nucleon, and $E = mc^2$E=mc2 applied to fission and Q-value calculations. Synoptic links extend to Topic 5 (electric and magnetic fields, used for accelerator and bubble-chamber problems) and Topic 13 (oscillations, used in some neutrino-oscillation contexts). The Edexcel A-Level data and formulae booklet provides $c$c, $e$e, $u$u, $N_A$NA​, and the decay-law equations; quark charges, lepton numbers, and exchange-particle assignments must be memorised.

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Worked example with full mark scheme

Question (12 marks):

A neutral kaon $K^0$K0 (quark content $d\bar{s}$dsˉ) decays at rest into two charged pions:

$K^0 \rightarrow \pi^+ + \pi^-$K0→π++π−

Rest masses: $m(K^0) = 497.6 \text{ MeV}/c^2$m(K0)=497.6 MeV/c2, $m(\pi^\pm) = 139.6 \text{ MeV}/c^2$m(π±)=139.6 MeV/c2.

(a) Show that charge and baryon number are conserved in this decay. (3)

(b) Determine whether strangeness is conserved, and state what this implies about the interaction responsible. (3)

(c) Calculate the Q-value of the decay and explain its physical meaning. (3)

(d) Sketch a Feynman diagram consistent with the decay, identifying the exchange particle. (3)

Solution with mark scheme:

(a) Charge: $Q(K^0) = 0$Q(K0)=0; $Q(\pi^+) + Q(\pi^-) = (+1) + (-1) = 0$Q(π+)+Q(π−)=(+1)+(−1)=0. Charges balance.

M1 — quoting charges of all three particles correctly. A1 — explicit statement that LHS = RHS = 0.

Baryon number: Mesons (one quark, one antiquark) have $B = 0$B=0 by definition. $B(K^0) = 0$B(K0)=0; $B(\pi^+) + B(\pi^-) = 0 + 0 = 0$B(π+)+B(π−)=0+0=0. Baryon number conserved.

B1 — correct identification of all three particles as mesons with $B = 0$B=0.

(b) Strangeness: $K^0 = d\bar{s}$K0=dsˉ, so $S(K^0) = +1$S(K0)=+1 (the antistrange quark carries $S = +1$S=+1; strange quark carries $S = -1$S=−1 by convention). Pions contain only up and down quarks, so $S(\pi^+) = S(\pi^-) = 0$S(π+)=S(π−)=0. Total strangeness: LHS = +1, RHS = 0. Strangeness is not conserved ($\Delta S = -1$ΔS=−1).

M1 — assigning $S = +1$S=+1 to $K^0$K0 using quark content. M1 — assigning $S = 0$S=0 to pions. A1 — concluding $\Delta S \neq 0$ΔS=0, so the decay must proceed via the weak interaction (the only interaction that violates strangeness conservation by $\Delta S = \pm 1$ΔS=±1).

(c) Q-value: $Q = [m(K^0) - 2 m(\pi^\pm)] c^2 = [497.6 - 2(139.6)]$Q=[m(K0)−2m(π±)]c2=[497.6−2(139.6)] MeV $= 497.6 - 279.2 = 218.4$=497.6−279.2=218.4 MeV.

M1 — correct mass-difference expression. A1 — numerical value 218.4 MeV (allow 218 MeV).

Q is positive, so the decay is exothermic — the rest-energy difference appears as kinetic energy of the two pions, shared equally by momentum conservation since $K^0$K0 decays at rest.

A1 — physical interpretation: positive Q means the products carry kinetic energy equal to Q (shared between the two pions).