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Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles

The standard model and particle interactions

40 free, high-quality past-paper-style questions on The standard model and particle interactions for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 147 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1IALP-U4-8-A-01 Unit 4 · A2Easy1 mark

Which particle is a lepton?

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AnswerMarkNotes
Leptons are fundamental particles that are not made of quarks: the electron, muon, tau and the three neutrinos. The neutron and proton are baryons (three quarks) and the pion is a meson (quark and antiquark), so the answer is the muon.✓ 1Answer B

Answer: B

Common error: The muon is sometimes confused with a meson because of its name. Ask “is it made of quarks?” — the muon is not.

2IALP-U4-8-A-03 Unit 4 · A2Easy1 mark

The charges of the up, down and strange quarks are +2/3 e, −1/3 e and −1/3 e.
Which combination of quarks forms a baryon with a charge of −1 e?

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AnswerMarkNotes
Add the charges: d d s = −1/3 − 1/3 − 1/3 = −1. The others give 0 (udd), 0 (uds) and +1 (uus).✓ 1Answer A

Answer: A

Exam tip: Write the charge under each quark and add as fractions. A hadron's total is always a whole number.

3IALP-U4-8-A-05 Unit 4 · A2Easy1 mark

Which quantity is not always conserved in particle interactions?

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AnswerMarkNotes
Charge, baryon number and lepton number are conserved in every interaction. The number of hadrons can change: for example a proton and an antiproton (two hadrons) can annihilate into three pions.✓ 1Answer D

Answer: D

Exam tip: The three conservation checks you need are charge, baryon number and lepton number (plus energy and momentum).

4IALP-U4-8-B-01 Unit 4 · A2Easy4 marks

A muon has a rest mass of 1.88 × 10−28 kg.

(a)

Calculate the rest energy of the muon in joules.

(2)
(b)

Express this energy in MeV.

(1)
(c)

State the mass of the muon in MeV/c2.

(1)
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AnswerMarkNotes
Part (a)
E = mc2 = 1.88 × 10−28 × (3.00 × 108)2✓ 1
= 1.69 × 10−11 J✓ 1
Part (b)
1.69 × 10−11 ÷ 1.60 × 10−13 = 106 MeV✓ 1
Part (c)
106 MeV/c2 — allow error carried forward from (b)✓ 1

Exam tip: A mass in MeV/c2 has the same number as the rest energy in MeV.

5IALP-U4-8-B-02 Unit 4 · A2Easy5 marks

Particles in the standard model can be grouped as baryons, mesons, leptons and photons. The charges of the up and down quarks are +2/3 e and −1/3 e.

(a)

State the difference in quark structure between a baryon and a meson.

(2)
(b)

State the quark composition of the proton and show that it gives the correct charge.

(2)
(c)

The antiproton has the same mass as the proton. State its charge and its quark composition.

(1)
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AnswerMarkNotes
Part (a)
a baryon consists of three quarks✓ 1
a meson consists of a quark and an antiquark✓ 1
Part (b)
u u d✓ 1
+2/3 + 2/3 − 1/3 = +1 (e)✓ 1
Part (c)
charge −1 (e) and u u d✓ 1

Common error: Writing the antiproton as “uud with a negative charge”. Every quark is replaced by its antiquark.

6IALP-U4-8-B-05 Unit 4 · A2Easy6 marks

Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0. Baryon numbers: proton and neutron +1; antiproton −1; pions 0.
For each interaction below, state whether it is possible, giving a reason based on conservation laws.

(a)

p + p → p + p + p + p

(2)
(b)

e− + p → n + π0

(2)
(c)

π− + p → n + γ

(2)
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AnswerMarkNotes
Part (a)
charge 2 → 2 and baryon number 2 → 2 (lepton number 0 → 0)✓ 1
all conserved, so possible✓ 1
Part (b)
charge 0 → 0 and baryon number 1 → 1 conserved✓ 1
lepton number 1 → 0 is not conserved, so not possible✓ 1
Part (c)
charge −1 + 1 = 0 → 0; baryon number 1 → 1✓ 1
lepton number 0 → 0; all conserved, so possible✓ 1

Exam tip: Show the totals before and after for each quantity; “it doesn't balance” earns nothing without the numbers.

7IALP-U4-8-FA-01 Unit 4 · A2Easy1 mark

The charges of the up, down and strange quarks are +2/3 e, −1/3 e and −1/3 e. Figure 1 shows the quark content of four particles W, X, Y and Z.
Which particle is a meson with zero charge?

udparticle Wuusparticle Xuudparticle Ydsparticle Zquark symbols with a bar (shaded) are antiquarks
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AnswerMarkNotes
A meson is one quark and one antiquark, so only W and Z are mesons. W = u d: +2/3 + 1/3 = +1. Z = d s: −1/3 + 1/3 = 0. The answer is Z. (X = uus is a baryon, charge +1; Y is three antiquarks, an antibaryon of charge −1.)✓ 1Answer D

Answer: D

Common error: Giving an antiquark the same charge as its quark. d and s are +1/3 e, not −1/3 e.

8IALP-U4-8-FA-06 Unit 4 · A2Easy1 mark

Figure 1 shows the pattern of fundamental particles known before the top quark was discovered. The symmetry of the pattern led physicists to predict a particle to fill the box marked ?.
Which describes the predicted particle?

1st generation2nd generation3rd generationquarksu+2/3upc+2/3charm?d−1/3downs−1/3strangeb−1/3bottomleptonse⁻−1electronμ⁻−1muonτ⁻−1tauνe0e-neutrinoνμ0μ-neutrinoντ0τ-neutrinocharge in units of e shown at the top right of each box
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AnswerMarkNotes
Each generation contains a pair of quarks, one of charge +2/3 e and one of charge −1/3 e, and a pair of leptons. The third generation had the bottom quark (−1/3 e) but no partner, so the model predicted a quark of charge +2/3 e — the top quark.✓ 1Answer C

Answer: C

Exam tip: “Symmetry of the model” means every generation has the same pattern: two quarks (+2/3, −1/3) and two leptons (−1, 0).

9IALP-U4-8-FA-07 Unit 4 · A2Easy1 mark

Figure 1 shows the tracks of five particles that enter a uniform magnetic field at the same speed. Track P is made by a proton.
Which track could be made by an antiproton?

P(proton)ABCDB into page (×)the uniform field fills the shaded region
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AnswerMarkNotes
An antiproton has the same mass as a proton but the opposite charge. At the same speed, r = p/BQ gives the same radius, while the force is in the opposite direction. The track curves the other way with the same radius: B.✓ 1Answer B

Answer: B

Exam tip: Particle and antiparticle: same mass, opposite charge. In the same field at the same speed their tracks are mirror images.

10IALP-U4-8-FB-04 Unit 4 · A2Easy5 marks

Figure 1 shows the charges of the up, down and strange quarks and of their antiquarks.

−1−2/3−1/30+1/3+2/3+1charge / eudsudsquarksantiquarks
(a)

State the charge of the anti-strange quark.

(1)
(b)

The Δ− particle is a baryon with a charge of −1 e. It contains no strange quarks. Deduce its quark composition.

(2)
(c)

A student suggests a particle made of one up quark and one down quark. Use Figure 1 to explain why this particle cannot exist.

(2)
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AnswerMarkNotes
Part (a)
+1/3 e✓ 1
Part (b)
a baryon is made of three quarks✓ 1
−1/3 − 1/3 − 1/3 = −1, so Δ− = d d d✓ 1
Part (c)
its charge would be +2/3 − 1/3 = +1/3 e, which is not a whole number of e✓ 1
a two-quark hadron (meson) must be a quark and an antiquark, e.g. u d✓ 1

Exam tip: Every hadron has a charge that is a whole number of e. Use this to check any combination of quarks you write down.

11IALP-U4-8-FB-07 Unit 4 · A2Easy5 marks

Figure 1 is a chart used to classify particles. P, Q and R are three groups of particles.

all particleshadronsmade of quarksfundamentalnot made of quarksPthree quarks(or three antiquarks)Qone quark andone antiquarkRe.g. electron,neutrinophotoncarries thee.m. force
(a)

Name the groups P, Q and R.

(3)
(b)

A particle Z consists of a strange quark and an anti-down quark (s d). State which box Z belongs in.

(1)
(c)

The muon has a rest mass close to that of the pion. Explain why the muon belongs in R but the pion does not.

(1)
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AnswerMarkNotes
Part (a)
P: baryons✓ 1
Q: mesons✓ 1
R: leptons✓ 1
Part (b)
Q — it is one quark and one antiquark, a meson✓ 1
Part (c)
the muon is fundamental (not made of quarks), whereas the pion is made of a quark and an antiquark; the groups depend on structure, not on mass✓ 1

Common error: Classifying particles by mass. The muon is a lepton even though it is heavier than some mesons.

12IALP-U4-8-A-02 Unit 4 · A2Medium1 mark

A particle has a mass of 1.00 GeV/c2.
What is this mass in kilograms?

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AnswerMarkNotes
Rest energy = 1.00 × 109 × 1.60 × 10−19 = 1.60 × 10−10 J. m = E/c2 = 1.60 × 10−10 ÷ 9.00 × 1016 = 1.78 × 10−27 kg.✓ 1Answer A

Answer: A

Exam tip: Two steps: GeV to joules (× 1.60 × 10−10), then divide by c2. A mass near 1 GeV/c2 should be close to the proton's mass.

13IALP-U4-8-A-04 Unit 4 · A2Medium1 mark

Mass of electron = 9.11 × 10−31 kg.
What is the minimum energy of a photon that can produce an electron–positron pair?

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AnswerMarkNotes
The photon must supply the rest energy of both particles: 2mec2 = 2 × 9.11 × 10−31 × 9.00 × 1016 = 1.64 × 10−13 J = 1.025 MeV, so 1.03 MeV (rounded up: 1.0 MeV is not enough).✓ 1Answer C

Answer: C

Common error: Using the rest energy of one particle. Pair production creates two particles, so the photon needs at least 2mc2.

14IALP-U4-8-A-07 Unit 4 · A2Medium1 mark

The positive pion π+ consists of an up quark and an anti-down quark (ud̄).
What is the quark composition of its antiparticle?

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AnswerMarkNotes
The antiparticle has every quark replaced by its antiquark: u → ū and d̄ → d, giving ū d, charge −2/3 − 1/3 = −1 (the π−).✓ 1Answer A

Answer: A

Exam tip: To form an antiparticle, swap every quark for its antiquark and every antiquark for its quark. Charge and baryon number both change sign.

15IALP-U4-8-A-08 Unit 4 · A2Medium1 mark

Muons are produced about 12 km above the Earth's surface and travel downwards at a speed close to c. Their mean lifetime at rest is 2.2 µs, in which time they would travel less than 1 km. Large numbers of muons are detected at sea level.
Which statement explains this?

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AnswerMarkNotes
At speeds close to c the lifetime of an unstable particle, measured by observers it is moving past, is greatly increased. Measured from the Earth, the muons' lifetime is much longer than 2.2 µs, so many survive the journey.✓ 1Answer D

Answer: D

Exam tip: The relativistic increase in lifetime only becomes significant at speeds that are a large fraction of c.

16IALP-U4-8-A-10 Unit 4 · A2Medium1 mark

The top quark was predicted many years before it was first detected.
Which reason led physicists to predict it?

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AnswerMarkNotes
The leptons form three pairs (e, νe), (μ, νμ), (τ, ντ). The symmetry of the model required three quark pairs too; once the bottom quark was found, a partner of charge +2/3 was needed to complete the third pair.✓ 1Answer C

Answer: C

Exam tip: “Symmetry” in the standard model means the matching pattern of quark pairs and lepton pairs across three generations.

17IALP-U4-8-B-03 Unit 4 · A2Medium7 marks

In a hospital scanner, a radioactive tracer emits positrons inside the patient's body. Each positron slows down and then annihilates with an electron. Assume both particles are at rest when they annihilate. Mass of electron = 9.11 × 10−31 kg.

annihilationγγring of detectors
Figure: two photons from one annihilation reaching opposite detectors (not to scale)
(a)

State what is meant by the antiparticle of a particle.

(1)
(b)

Explain why the annihilation produces two photons travelling in opposite directions.

(2)
(c)

Calculate the energy of each photon in joules and its wavelength.

(3)
(d)

Detectors surround the patient. Suggest how recording two photons at the same moment, on opposite sides, helps to locate where the annihilation took place.

(1)
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AnswerMarkNotes
Part (a)
a particle with the same mass but opposite charge (and opposite baryon/lepton number)✓ 1
Part (b)
the total momentum before annihilation is (approximately) zero✓ 1
a single photon would carry momentum, so two photons with equal and opposite momenta are needed to conserve momentum✓ 1
Part (c)
E = mec2 = 9.11 × 10−31 × 9.00 × 1016 = 8.20 × 10−14 J✓ 1
λ = hc/E = 6.63 × 10−34 × 3.00 × 108 ÷ 8.20 × 10−14✓ 1
= 2.43 × 10−12 m✓ 1
Part (d)
the annihilation must have happened somewhere on the straight line joining the two detectors (many such lines cross at the tracer's location)✓ 1

Exam tip: In annihilation at rest, each photon has the rest energy of one particle, mc2 — not 2mc2.

18IALP-U4-8-B-08 Unit 4 · A2Medium7 marks

A student places a source of positrons next to a gamma-ray detector. The positrons annihilate with electrons in the surrounding material, almost at rest. The detector records a sharp peak in the number of photons at an energy of 0.511 MeV. The uncertainty in this energy reading is ±0.004 MeV.

(a)

Explain why the photons from annihilation all have the same energy.

(2)
(b)

Use the peak energy to calculate a value for the mass of the electron.

(3)
(c)

Calculate the percentage uncertainty in the energy reading and state whether the student's value of the electron mass is consistent with 9.11 × 10−31 kg.

(2)
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AnswerMarkNotes
Part (a)
the electron and positron (almost at rest) have a fixed total energy 2mec2✓ 1
it is shared equally between two photons emitted in opposite directions (to conserve momentum), so each has mec2✓ 1
Part (b)
E = 0.511 × 1.60 × 10−13 = 8.18 × 10−14 J✓ 1
m = E/c2 = 8.18 × 10−14 ÷ 9.00 × 1016✓ 1
= 9.08 × 10−31 kg✓ 1
Part (c)
0.004 ÷ 0.511 × 100 = 0.8%✓ 1
the difference from 9.11 × 10−31 kg is about 0.3%, less than 0.8%, so the value is consistent✓ 1

Exam tip: To judge agreement, compare the percentage difference from the accepted value with the percentage uncertainty.

19IALP-U4-8-B-09 Unit 4 · A2Medium6 marks

A detector on a mountain, 2500 m above sea level, counts 600 muons per minute arriving vertically. An identical detector at sea level counts 410 per minute. Assume all these muons travel vertically at 0.995c and that no muons are produced between the two heights. For muons at rest, half of any sample decays every 1.52 µs.

(a)

Calculate the time taken by the muons to travel from the mountain detector to sea level, as measured from the Earth.

(1)
(b)

Calculate the number of muons per minute that would reach sea level if the muons decayed as they do at rest.

(3)
(c)

Explain what the measured count of 410 per minute shows.

(2)
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AnswerMarkNotes
Part (a)
t = 2500 ÷ (0.995 × 3.00 × 108) = 8.4 × 10−6 s (8.38 µs)✓ 1
Part (b)
number of halving times = 8.38 ÷ 1.52 = 5.5✓ 1
fraction left = (½)5.5 = 0.022✓ 1
600 × 0.022 = 13 per minute✓ 1
Part (c)
far more muons survive than predicted, so fewer have decayed during the journey✓ 1
the lifetime/half-life of the fast-moving muons measured from the Earth is much longer than at rest (relativistic increase in lifetime)✓ 1

Exam tip: Work out the number of halving times first, then the fraction remaining; keep the unrounded value until the final step.

20IALP-U4-8-B-10 Unit 4 · A2Medium6 marks

A free neutron decays into a proton, an electron and an antineutrino. Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0. Rest energies: neutron 939.57 MeV; proton 938.27 MeV; electron 0.511 MeV; the antineutrino's rest energy is negligible.

(a)

Write the particle equation for this decay.

(1)
(b)

Show that charge, baryon number and lepton number are conserved in this decay.

(2)
(c)

Describe the change that happens to the quarks in this decay.

(1)
(d)

Calculate the total kinetic energy, in MeV, shared by the products, assuming the neutron is at rest.

(2)
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AnswerMarkNotes
Part (a)
n → p + e− + ν✓ 1
Part (b)
charge: 0 → +1 − 1 + 0 = 0; baryon number: 1 → 1 + 0 + 0 = 1✓ 1
lepton number: 0 → 0 + 1 − 1 = 0✓ 1
Part (c)
a down quark changes into an up quark (udd → uud)✓ 1
Part (d)
939.57 − 938.27 − 0.511✓ 1
= 0.79 MeV (0.789 MeV)✓ 1

Common error: Writing a neutrino instead of an antineutrino. Only the antineutrino (lepton number −1) balances the electron's +1.

21IALP-U4-8-FA-02 Unit 4 · A2Medium1 mark

Figure 1 shows the rest energies of five particles. A gamma photon of energy 250 MeV passes close to a nucleus, which takes negligible energy.
Which is the most massive particle that this photon can create together with its antiparticle?

01002003004005006007008009001000electron0.511 MeVmuon106 MeVpion π⁺140 MeVkaon K⁺494 MeVproton938 MeVrest energy / MeV
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AnswerMarkNotes
Pair production needs at least the rest energy of both particles, 2mc2: electron 1.02 MeV, muon 212 MeV, pion 280 MeV, kaon 988 MeV. A 250 MeV photon can make a muon–antimuon pair but not a pion pair, so the answer is the muon.✓ 1Answer B

Answer: B

Common error: Comparing the photon energy with the rest energy of one particle. The photon must create the particle and its antiparticle.

22IALP-U4-8-FA-03 Unit 4 · A2Medium1 mark

Photons of different energies produce pairs of a particle X and its antiparticle near nuclei, which take negligible energy. Figure 1 shows the total kinetic energy of X and its antiparticle against the energy of the photon.
What is the rest mass of X?

0100200300400500600050100150200250300350photon energy / MeVtotal kinetic energy of X and anti-X / MeVno pair is producedbelow the threshold
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AnswerMarkNotes
The line meets the photon-energy axis at 300 MeV. At this threshold all of the photon's energy becomes rest energy: 2mc2 = 300 MeV, so mc2 = 150 MeV and m = 150 MeV/c2. Check: at 480 MeV the kinetic energy is 480 − 300 = 180 MeV, as plotted.✓ 1Answer B

Answer: B

Exam tip: On a kinetic energy against photon energy graph, the intercept on the photon-energy axis is the total rest energy of the pair, 2mc2.

23IALP-U4-8-FA-08 Unit 4 · A2Medium1 mark

Figure 1 shows how the kinetic energy of protons varies with distance along a linear accelerator.
What is the kinetic energy gained by a proton between 20 m and 50 m from the start?

0102030405060020406080100120distance along accelerator / mkinetic energy of protons / MeV
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AnswerMarkNotes
From the graph: 40 MeV at 20 m and 100 MeV at 50 m, a gain of 60 MeV. 60 × 106 × 1.60 × 10−19 = 9.6 × 10−12 J.✓ 1Answer D

Answer: D

Exam tip: 1 MeV = 1.60 × 10−13 J. Find the change in energy from the graph first, then convert.

24IALP-U4-8-FA-09 Unit 4 · A2Medium1 mark

Figure 1 shows the tracks in a magnetic field when a negative kaon K− (a meson) comes to rest and decays into a negative muon μ− and a neutral particle X, which leaves no track. Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0.
Which particle is X?

K⁻ (comes to rest)μ⁻X (no track)B into page (×)
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AnswerMarkNotes
Charge: −1 → −1 + 0, so X is neutral. Baryon number: 0 → 0, so X is not a baryon (rules out the neutron). Lepton number: 0 → +1 + LX, so LX = −1: X is an antineutrino.✓ 1Answer A

Answer: A

Common error: Choosing the neutrino. A negative muon has lepton number +1, so the particle produced with it must have −1.

25IALP-U4-8-FA-10 Unit 4 · A2Medium1 mark

A particle Y and its antiparticle, both at rest, annihilate and produce two photons. Figure 1 shows the energy spectrum recorded by a photon detector; the sharp peak is caused by these annihilations.
What is the mass of particle Y?

05010015020025030035040004080120160photon energy / MeVnumber of photons detected
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AnswerMarkNotes
The peak is at 250 MeV. Each photon carries the rest energy of one particle: E = 250 × 1.60 × 10−13 = 4.00 × 10−11 J. m = E/c2 = 4.00 × 10−11 ÷ 9.00 × 1016 = 4.44 × 10−28 kg.✓ 1Answer A

Answer: A

Exam tip: Annihilation at rest: total energy 2mc2 shared by two photons, so each photon has mc2.

26IALP-U4-8-FB-01 Unit 4 · A2Medium7 marks

An antiproton slows down in a detector and annihilates with a proton at rest. The detector is in a uniform magnetic field directed into the page. Figure 1 shows the tracks. Three pions are produced; product 3 leaves no track and is shown dashed. Rest energies: proton 938 MeV; charged pion 140 MeV; neutral pion 135 MeV.

p (slows down and stops)track 1track 23: no track (shown dashed)B into page (×)
(a)

State one property of the antiproton that is the same as for the proton, and one property that is different.

(2)
(b)

Use Figure 1 to identify the pion that makes each of tracks 1, 2 and 3, and write the particle equation for the annihilation.

(2)
(c)

Show that charge and baryon number are conserved in this annihilation.

(1)
(d)

Calculate the total kinetic energy of the three pions, in joules.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
same: mass / rest energy (or size of charge)✓ 1
different: sign of charge (−1 e) or baryon number (−1)✓ 1
Part (b)
track 1 curves the same way as the (negative) antiproton, so π−; track 2 curves the opposite way, so π+; 3 leaves no track, so π0✓ 1
p + p → π− + π+ + π0✓ 1
Part (c)
charge −1 + 1 = 0 → −1 + 1 + 0 = 0 and baryon number −1 + 1 = 0 → 0 + 0 + 0 = 0✓ 1
Part (d)
2 × 938 − (140 + 140 + 135) = 1461 MeV✓ 1
1461 × 1.60 × 10−13 = 2.34 × 10−10 J✓ 1

Common error: Using the rest energy of the proton only. Both the proton and the antiproton supply rest energy, so 2 × 938 MeV is available.

27IALP-U4-8-FB-02 Unit 4 · A2Medium6 marks

A student plots the rest energy of four particles against their mass (Figure 1). The masses are: muon 1.88 × 10−28 kg, positive pion 2.49 × 10−28 kg, positive kaon 8.80 × 10−28 kg and proton 16.7 × 10−28 kg.

02468101214161801002003004005006007008009001000mass / 10⁻²⁸ kgrest energy / MeVμ⁻π⁺K⁺p
(a)

Use the graph to determine the gradient of the line, in MeV kg−1.

(2)
(b)

Convert the gradient to J kg−1 and explain what it represents.

(2)
(c)

The eta meson has a mass of 9.76 × 10−28 kg. Calculate its mass in MeV/c2.

(2)
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AnswerMarkNotes
Part (a)
large triangle from the origin to a point on the line, e.g. 900 MeV at 16 × 10−28 kg✓ 1
gradient = 900 ÷ 1.6 × 10−27 = 5.6 × 1029 MeV kg−1 (allow 5.5–5.7 × 1029)✓ 1
Part (b)
5.6 × 1029 × 1.60 × 10−13 = 9.0 × 1016 J kg−1✓ 1
this is c2 = (3.00 × 108)2 = 9.00 × 1016 m2 s−2, because E = c2m✓ 1
Part (c)
E = 9.76 × 10−28 × 9.00 × 1016 = 8.78 × 10−11 J✓ 1
÷ 1.60 × 10−13 = 549 MeV/c2 (allow 540–560 read from the graph)✓ 1

Exam tip: Take a gradient from a large triangle using points where the line crosses gridlines, not from two close data points.

28IALP-U4-8-FB-03 Unit 4 · A2Medium7 marks

In a storage ring, muons circulate at 0.9994c. Detectors record the electrons produced when the muons decay; the count rate is proportional to the number of muons remaining. Figure 1 shows how the count rate varies with time, measured in the laboratory. The half-life of muons at rest is 1.52 µs.

0153045607590105120135150165180012345678time / µscount rate / 10³ s⁻¹
(a)

Use the graph to determine the half-life of the circulating muons, as measured in the laboratory.

(2)
(b)

Calculate the ratio of the laboratory half-life to the half-life at rest.

(1)
(c)

Explain how these measurements show that the lifetime of an unstable particle depends on its speed, and state when this effect is significant.

(2)
(d)

Calculate the distance a muon travels around the ring during one laboratory half-life.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
count rate halves, e.g. from 8.0 to 4.0 (× 103 s−1), or 4.0 to 2.0✓ 1
half-life = 45 µs (4.5 × 10−5 s; allow 43–47 µs)✓ 1
Part (b)
45 ÷ 1.52 = 30 (29.6)✓ 1
Part (c)
the circulating muons take about 30 times longer to decay, measured in the laboratory, than muons at rest✓ 1
the increase in lifetime is significant only when the speed is a large fraction of c / close to c✓ 1
Part (d)
d = 0.9994 × 3.00 × 108 × 45 × 10−6✓ 1
= 1.3 × 104 m✓ 1

Exam tip: Read a half-life from two successive halvings (8 → 4 and 4 → 2) to check your reading of the graph.

29IALP-U4-8-FB-05 Unit 4 · A2Medium6 marks

A beam of positive pions strikes protons at rest in a detector placed in a uniform magnetic field directed into the page. Figure 1 shows the tracks from one collision. Baryon numbers: proton +1, pions 0.

π⁺ (beam)1234proton at restB into page (×)
(a)

Use Figure 1 to deduce how many of the four outgoing particles are positively charged. Give your reasoning.

(2)
(b)

The products are one proton and three pions. Write the particle equation for this collision.

(1)
(c)

Show that charge and baryon number are conserved.

(2)
(d)

The total rest energy of the products is greater than that of the colliding particles. Explain how this is possible.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
the π+ beam (positive) curves anticlockwise; tracks 1, 2 and 3 curve the same way✓ 1
so three are positive; track 4 curves the opposite way and is negative✓ 1
Part (b)
π+ + p → p + π+ + π+ + π−✓ 1
Part (c)
charge: +1 + 1 = 2 → +1 + 1 + 1 − 1 = 2✓ 1
baryon number: 0 + 1 = 1 → 1 + 0 + 0 + 0 = 1✓ 1
Part (d)
some of the kinetic energy of the incoming pion is converted into the extra rest energy (mass) of the new particles, ΔE = c2Δm✓ 1

Exam tip: Compare each track with one whose charge you know: in the same field, curving the same way means the same sign of charge.

30IALP-U4-8-FB-10 Unit 4 · A2Medium6 marks

Figure 1 shows the charge and baryon number of several particles. J and Y are not named. Quark charges: u +2/3 e, d −1/3 e, s −1/3 e.

−2−10+1+2−10+1charge / ebaryon numberpnπ⁺π⁰π⁻JY
(a)

J is the antiparticle of one of the labelled particles. Identify it, using Figure 1 to explain your answer.

(2)
(b)

Y is a baryon that contains one strange quark and no up quarks. Use Figure 1 to deduce its quark composition.

(2)
(c)

Use Figure 1 to determine whether charge and baryon number are conserved in the interaction p + π− → Y + π+.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
J has charge −1 and baryon number −1, both opposite to those of the proton (+1, +1)✓ 1
J is the antiproton✓ 1
Part (b)
from the figure, Y has charge −1 (and baryon number +1: three quarks)✓ 1
s (−1/3) plus two quarks totalling −2/3: d d s✓ 1
Part (c)
charge: +1 − 1 = 0 → −1 + 1 = 0, conserved✓ 1
baryon number: 1 + 0 = 1 → 1 + 0 = 1, conserved✓ 1

Exam tip: A particle and its antiparticle sit opposite each other through the origin of a charge–baryon number grid.

31IALP-U4-8-A-06 Unit 4 · A2Hard1 mark

Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0.
Which interaction is possible?

Show mark scheme
AnswerMarkNotes
Check charge, baryon number and lepton number. π− + p → n + π0: Q 0 → 0, B 1 → 1, L 0 → 0, so C is possible. A fails on charge (2 → 1); B fails on lepton number (0 → 2); D fails on lepton number (1 → −1).✓ 1Answer C

Answer: C

Exam tip: Check all three quantities for each option; a single failure rules an interaction out.

32IALP-U4-8-A-09 Unit 4 · A2Hard1 mark

A charged lepton has a rest mass of 1.78 GeV/c2. It annihilates with its antiparticle when both are at rest, producing two photons.
What is the wavelength of each photon?

Show mark scheme
AnswerMarkNotes
Each photon carries the rest energy of one particle: E = 1.78 × 109 × 1.60 × 10−19 = 2.85 × 10−10 J. λ = hc/E = 6.63 × 10−34 × 3.00 × 108 ÷ 2.85 × 10−10 = 6.98 × 10−16 m.✓ 1Answer B

Answer: B

Exam tip: The total energy 2mc2 is shared between two photons, so each has mc2.

33IALP-U4-8-B-04 Unit 4 · A2Hard8 marks

A gamma photon of energy 3.20 MeV passes close to a lead nucleus in a detector and produces an electron–positron pair. The detector is in a uniform magnetic field directed into the page. Mass of electron = 9.11 × 10−31 kg.

××××××××××××××××××××××××γ, 3.20 MeVlead nucleuse⁺e⁻B into page
Figure: the electron and positron tracks in the detector (not to scale)
(a)

Show that the minimum photon energy needed to produce an electron–positron pair is about 1 MeV.

(2)
(b)

Calculate the kinetic energy of each particle, assuming the energy is shared equally and the nucleus gains negligible energy.

(2)
(c)

Calculate the wavelength of the gamma photon.

(2)
(d)

The tracks of the electron and positron are spirals that curve in opposite directions. Explain why the tracks curve in opposite directions and why their radius decreases.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
Emin = 2mec2 = 2 × 9.11 × 10−31 × 9.00 × 1016 = 1.64 × 10−13 J✓ 1
÷ 1.60 × 10−13 = 1.02 MeV✓ 1
Part (b)
total kinetic energy = 3.20 − 1.02 = 2.18 MeV✓ 1
each = 1.09 MeV✓ 1
Part (c)
E = 3.20 × 1.60 × 10−13 = 5.12 × 10−13 J✓ 1
λ = hc/E = 3.88 × 10−13 m✓ 1
Part (d)
the particles have opposite charges, so the magnetic force on each acts in the opposite direction✓ 1
they lose energy/momentum as they ionise the material, and r = p/BQ so the radius decreases✓ 1

Exam tip: In pair production the photon pays for both rest energies first; only what is left over becomes kinetic energy.

34IALP-U4-8-B-06 Unit 4 · A2Hard8 marks

A neutral particle X, at rest, decays into a proton and a negative pion: X → p + π−. Rest masses: X 1116 MeV/c2; proton 938 MeV/c2; π− 140 MeV/c2; neutron 940 MeV/c2. Quark charges: u +2/3 e, d −1/3 e, s −1/3 e.

(a)

Deduce the baryon number of X and state whether X is a baryon or a meson.

(2)
(b)

X contains one strange quark. Deduce its other two quarks.

(2)
(c)

Calculate the total kinetic energy of the proton and pion, in joules.

(2)
(d)

A student suggests that a neutron at rest could decay in the same way, n → p + π−. Explain why this decay cannot happen, even though charge and baryon number are conserved.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
baryon number after = 1 + 0 = 1, so X has baryon number 1 (by conservation)✓ 1
X is a baryon✓ 1
Part (b)
three quarks with total charge 0; s = −1/3 so the other two total +1/3✓ 1
u and d: X = u d s✓ 1
Part (c)
1116 − (938 + 140) = 38 MeV✓ 1
38 × 1.60 × 10−13 = 6.1 × 10−12 J✓ 1
Part (d)
the rest energy of the products (938 + 140 = 1078 MeV) is greater than that of the neutron (940 MeV)✓ 1
energy would not be conserved / the neutron does not have enough energy to create the products✓ 1

Common error: Thinking that a particle can decay into anything that balances charge, baryon number and lepton number. The products' total rest energy must also be less than the parent's.

35IALP-U4-8-B-07 Unit 4 · A2Hard6 marks

Muons are produced when cosmic rays strike nuclei about 15 km above sea level. They travel towards the ground at 0.998c. The mean lifetime of a muon at rest is 2.2 µs. A detector at sea level records large numbers of these muons.

(a)

Explain why the detection of large numbers of muons at sea level is evidence for the relativistic increase in the lifetime of fast-moving particles. Support your answer with a calculation.

(6)
Show mark scheme
AnswerMarkNotes
Indicative content:—
in 2.2 µs a muon at 0.998c travels about 0.998 × 3.00 × 108 × 2.2 × 10−6 ≈ 660 m✓ 1
the journey of 15 km takes about 15 × 103 ÷ 2.99 × 108 ≈ 50 µs, more than 20 lifetimes✓ 1
with a lifetime of 2.2 µs almost all would decay before reaching sea level, so very few would be detected✓ 1
large numbers are detected, so the lifetime measured in the Earth's frame must be much longer than 2.2 µs✓ 1
this increase in lifetime happens because the muons move at a speed very close to c✓ 1
at everyday speeds the increase is negligible; it is significant only at a large fraction of c✓ 1

Exam tip: Give a number for the distance a muon could travel in 2.2 µs and compare it with the height at which muons are made — then state the conclusion about the lifetime.

36IALP-U4-8-FA-04 Unit 4 · A2Hard1 mark

Figure 1 shows how the mean lifetime of charged pions, measured in a laboratory, depends on their speed. A beam of pions travels through the laboratory at 0.80c.
What is the mean distance a pion travels before it decays, as measured in the laboratory?

0.00.10.20.30.40.50.60.70.80.91.00102030405060708090100speed / cmean lifetime measured in the laboratory / ns43.3 ns at 0.80c26.0 ns at rest
Show mark scheme
AnswerMarkNotes
From the graph, the lifetime at 0.80c is 43.3 ns. Distance = speed × time = 0.80 × 3.00 × 108 × 43.3 × 10−9 = 10.4 m. The rest lifetime (26.0 ns) would give only 6.24 m: at 0.80c the increase in lifetime is significant, while below about 0.3c the graph is almost flat.✓ 1Answer C

Answer: C

Exam tip: Use the lifetime and the speed measured in the same (laboratory) frame. The graph shows the increase only matters at a large fraction of c.

37IALP-U4-8-FA-05 Unit 4 · A2Hard1 mark

Figure 1 shows the tracks produced when an antiproton collides with a proton at rest in a uniform magnetic field. The particles labelled π+ and π− are pions; particle X is not identified. Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0. Baryon numbers: proton +1, antiproton −1, pions and leptons 0.
Which particle could X be?

p (beam)π⁺π⁻π⁻Xproton at restB into page (×)
Show mark scheme
AnswerMarkNotes
Before: charge −1 + 1 = 0, baryon number −1 + 1 = 0, lepton number 0. The three labelled pions have total charge +1 − 1 − 1 = −1, so X must have charge +1 (its track curves the same way as the π+ track), baryon number 0 and lepton number 0: π+.✓ 1Answer A

Answer: A

Exam tip: Find what is missing for each conserved quantity in turn (charge, baryon number, lepton number), then pick the particle that has all three values.

38IALP-U4-8-FB-06 Unit 4 · A2Hard8 marks

A gamma photon passes close to a nucleus and produces a positive pion and a negative pion. The nucleus takes negligible energy. Figure 1 shows how the energy of the photon is shared after the event.

050100150200250300350400energy of the photonbeforeafterrest energy π⁺140 MeVrest energy π⁻140 MeVKE π⁺70 MeVKE π⁻50 MeVenergy / MeV
(a)

Explain, with reference to ΔE = c2Δm, how a photon can produce two particles that have mass.

(2)
(b)

Use Figure 1 to calculate the energy of the photon in joules, and its wavelength.

(3)
(c)

Calculate the minimum photon energy, in joules, that could produce this pair of pions.

(2)
(d)

Suggest why this process needs a nucleus nearby and cannot happen to a photon in empty space.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
energy is converted into mass: the photon's energy provides the rest energy (mc2) of each new particle✓ 1
the photon needs at least 2mc2; any energy above this becomes kinetic energy of the particles✓ 1
Part (b)
E = 140 + 140 + 70 + 50 = 400 MeV = 400 × 1.60 × 10−13 = 6.40 × 10−11 J✓ 1
λ = hc/E = 6.63 × 10−34 × 3.00 × 108 ÷ 6.40 × 10−11✓ 1
= 3.11 × 10−15 m✓ 1
Part (c)
2 × 140 = 280 MeV✓ 1
280 × 1.60 × 10−13 = 4.48 × 10−11 J✓ 1
Part (d)
the nucleus is needed so that momentum is conserved: it takes some of the momentum✓ 1

Common error: Treating all of the photon's energy as rest energy. Only 2mc2 becomes mass; the rest is kinetic energy.

39IALP-U4-8-FB-08 Unit 4 · A2Hard7 marks

A positive pion π+ comes to rest at A in a detector and decays into a positive muon μ+ and a neutral particle X, which leaves no track. The muon stops at B and decays into a positron and two neutral particles. Figure 1 shows the tracks in a magnetic field. Lepton numbers: electron, muon and neutrino +1; positron, antimuon and antineutrino −1; all other particles in this question 0. Rest energies: π+ 139.6 MeV; μ+ 105.7 MeV; the rest energies of neutrinos are negligible.

π⁺ (comes to rest at A)ABμ⁺e⁺B into page (×)dashed: neutral particles (no track)
(a)

Use lepton number to deduce what X is.

(2)
(b)

Write the particle equation for the decay at B and show that lepton number is conserved.

(2)
(c)

Calculate the total kinetic energy of the products of the decay at A, in joules.

(2)
(d)

In many such events the muon tracks from A to B all have the same length. Suggest why.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
lepton number: 0 → −1 + LX, so LX = +1✓ 1
X is neutral with negligible mass: a neutrino✓ 1
Part (b)
μ+ → e+ + ν + ν✓ 1
−1 → −1 + 1 − 1 = −1✓ 1
Part (c)
139.6 − 105.7 = 33.9 MeV✓ 1
33.9 × 1.60 × 10−13 = 5.42 × 10−12 J✓ 1
Part (d)
the pion is at rest and decays into two particles, so conservation of energy and momentum gives every muon the same kinetic energy, and muons with the same energy travel the same distance before stopping✓ 1

Exam tip: When a product leaves no track, use conservation of charge, baryon number and lepton number to work out its properties, then name it.

40IALP-U4-8-FB-09 Unit 4 · A2Hard7 marks

Figure 1 shows a collider in which electrons and positrons travel in opposite directions and collide head-on; the beam energies are shown. In some collisions the electron and positron annihilate and produce a meson M and its antiparticle, each of rest mass 5.28 GeV/c2.

detectore⁻ beam5.29 GeV per electrone⁺ beam5.29 GeV per positronbeams collide head-oninside the detector
(a)

Calculate the total energy of one electron and one positron from the beams, in joules.

(2)
(b)

Show that there is enough energy to produce M and its antiparticle, and calculate their total kinetic energy in MeV.

(2)
(c)

Calculate the mass of M in kg.

(2)
(d)

Explain why colliding beams of equal energy head-on allows almost all of their energy to become rest energy of new particles.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
5.29 + 5.29 = 10.58 GeV✓ 1
10.58 × 1.60 × 10−10 = 1.69 × 10−9 J✓ 1
Part (b)
rest energy of the pair = 2 × 5.28 = 10.56 GeV, which is less than 10.58 GeV✓ 1
kinetic energy = 10.58 − 10.56 = 0.02 GeV = 20 MeV✓ 1
Part (c)
5.28 × 109 × 1.60 × 10−19 = 8.45 × 10−10 J✓ 1
÷ 9.00 × 1016 = 9.39 × 10−27 kg✓ 1
Part (d)
the electron and positron have equal and opposite momenta, so the total momentum is zero and the products can be made (almost) at rest; little energy has to remain as kinetic energy✓ 1

Common error: Using the energy of one beam only. In a head-on collision the energies of both particles are available.