3.2 Particles and radiation — revision question pack

11 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.2. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.2.1.1 · Constituents of the atom

Explanation

The simple atom contains protons and neutrons in the nucleus with electrons outside it. A proton has charge $+e$ and relative mass $1$ , a neutron has zero charge and relative mass $1$ , and an electron has charge $-e$ and relative mass about $1/1836$ ; SI masses and charges may also be required.
In ${}^{A}_{Z}X$ , $Z$ is proton number, $A$ is nucleon number and neutron number is $A-Z$ .
Isotopes share $Z$ but have different neutron numbers.
Specific charge is net charge divided by total mass in $\text{C kg}^{-1}$ .
Examiners expect ion charge, nuclear composition and sign to be handled separately.

Worked example

An ion contains $12$ protons, $12$ neutrons and $10$ electrons. Find its nuclide notation, charge and approximate specific charge using nucleon mass $1.67\times10^{-27}\,\text{kg}$ .

1. $A=24$ , $Z=12$ , so the nuclide is ${}^{24}_{12}\text{Mg}$ .
2.Net charge is $+2e=+3.20\times10^{-19}\,\text{C}$ .
3.Mass $\approx24(1.67\times10^{-27})$ , so $q/m=7.98\times10^6\,\text{C kg}^{-1}$ .

Answer: ${}^{24}_{12}\text{Mg}^{2+}$ with specific charge $+8.0\times10^6\,\text{C kg}^{-1}$ .

Common mistakes

Don't use electron number as proton number $Z$ for an ion.
Don't subtract electrons when calculating nucleon number.
Don't use charge magnitude but omit the sign of specific charge.

Exam tip

Write proton, neutron and electron counts separately before calculating $A$ , $Z$ or net charge.

Tier 1 · Easy

State the numbers of protons, neutrons and electrons in a ${}^{25}_{12}\text{Mg}^{2+}$ ion.
[2 marks]
Total for this question: 2
State what is meant by isotopes of an element.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A helium nucleus has charge $+3.20\times10^{-19}\,\text{C}$ and mass $6.64\times10^{-27}\,\text{kg}$ . Calculate its specific charge.
[3 marks]
Total for this question: 3
A sample contains only isotopes ${}^{35}\text{X}$ and ${}^{37}\text{X}$ . The mass of each atom may be taken as $35\,\text{u}$ and $37\,\text{u}$ respectively. The mean mass of an atom in the sample is $35.6\,\text{u}$ . Determine the percentage of ${}^{37}\text{X}$ atoms.
[3 marks]
Total for this question: 3
Singly ionised chlorine-35 and chlorine-37 ions have charges of equal magnitude. Take ionic mass to be proportional to nucleon number. Neglect electron masses. Calculate the ratio of the magnitude of the specific charge of chlorine-35 to that of chlorine-37 and state which isotope has the greater specific charge.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An ion of an isotope with $A=40$ and $Z=20$ has specific charge $+7.19\times10^6\,\text{C kg}^{-1}$ . Use nucleon mass $1.67\times10^{-27}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ to determine the ionic charge and the number of electrons in the ion.
[5 marks]
Total for this question: 5
A nucleus is an isotope of chlorine. It has charge $+2.72\times10^{-18}\,\text{C}$ and mass $5.85\times10^{-26}\,\text{kg}$ . Use $e=1.60\times10^{-19}\,\text{C}$ and nucleon mass $1.67\times10^{-27}\,\text{kg}$ to determine its proton number, nucleon number, nuclide notation and neutron number.
[4 marks]
Total for this question: 4
A very large sample contains magnesium-24, magnesium-25 and magnesium-26 atoms in abundances $79\%$ , $10\%$ and $11\%$ respectively. Every atom is singly ionised. Assume each ion's mass is proportional to its nucleon number, and neglect electron masses. Calculate the total charge divided by the total mass of the ions, and express this specific charge as a fraction of the proton's specific charge. Use $e=1.60\times10^{-19}\,\text{C}$ and nucleon mass $1.67\times10^{-27}\,\text{kg}$ .
[5 marks]
Total for this question: 5
A sodium ion contains $11$ protons, $12$ neutrons and $10$ electrons. Calculate its specific charge using proton mass $1.673\times10^{-27}\,\text{kg}$ , neutron mass $1.675\times10^{-27}\,\text{kg}$ , electron mass $9.11\times10^{-31}\,\text{kg}$ and $e=1.602\times10^{-19}\,\text{C}$ .
[4 marks]
Total for this question: 4
A beam contains ions of two isotopes of the same element. A fraction $x$ are doubly ionised with nucleon number $63$ ; the remainder are singly ionised with nucleon number $65$ . The beam's measured specific charge is $2.09\times10^6\,\text{C kg}^{-1}$ . Neglect electron masses. Take $e=1.60\times10^{-19}\,\text{C}$ and the nucleon mass as $1.67\times10^{-27}\,\text{kg}$ . Determine $x$ .
[5 marks]
Total for this question: 5

3.2.1.2 · Stable and unstable nuclei

Explanation

The strong nuclear force stabilises nuclei: it is attractive over separations up to about $3\,\text{fm}$ but repulsive below about $0.5\,\text{fm}$ , preventing collapse. Unstable nuclei undergo alpha or beta decay.
Alpha emission lowers nucleon number by $4$ and proton number by $2$ .
In $\beta^-$ decay a neutron becomes a proton, so $A$ is unchanged and $Z$ rises by $1$ ; the equation includes an electron and electron antineutrino.
The neutrino was proposed because the continuous beta-energy distribution otherwise conflicted with energy conservation.
Examiners expect balanced $A$ and $Z$ , the correct neutrino and the strong-force range stated with femtometre scale.

Worked example

Complete ${}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+\,?$ .

1.Unchanged $A$ and an increase of one in $Z$ identify $\beta^-$ decay.
2.Include an electron to balance charge.
3.Include the electron antineutrino required by conservation laws.

Answer: ${}^{0}_{-1}e+\bar{\nu}_e$ .

Common mistakes

Don't change nucleon number during $\beta^-$ decay.
Don't omit the electron antineutrino from a $\beta^-$ equation.
Don't describe the strong force as attractive at every separation.

Exam tip

Check total nucleon number and proton number on both sides of every nuclear equation.

Tier 1 · Easy

Complete the alpha-decay equation ${}^{210}_{84}\text{Po}\rightarrow{}^{206}_{82}\text{Pb}+\,?$ .
[1 mark]
Total for this question: 1
State the particle emitted in alpha decay and the particle emitted in beta-minus decay.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write the complete equation for the $\beta^-$ decay of ${}^{14}_{6}\text{C}$ and state how the proton number changes.
[3 marks]
Total for this question: 3
Explain why the continuous range of electron energies observed in $\beta^-$ decay led to the proposal of the neutrino.
[3 marks]
Total for this question: 3
A $\beta^-$ decay makes $1.17\,\text{MeV}$ available as kinetic energy. In one event the electron receives $0.42\,\text{MeV}$ and $0.01\,\text{MeV}$ is shared among the other decay products excluding the antineutrino. Calculate the antineutrino energy, state the conservation law used and explain why the electron energy varies from event to event.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Describe how the strong nuclear force between two nucleons depends on their separation, and explain how this behaviour contributes to a stable nucleus.
[5 marks]
Total for this question: 5
A ${}^{238}_{92}\text{U}$ nucleus undergoes one alpha decay followed by two $\beta^-$ decays to form ${}^{234}_{92}\text{U}$ . Write the three separate balanced decay equations, including the electron antineutrino in each $\beta^-$ decay.
[4 marks]
Total for this question: 4
A nucleus follows the chain ${}^{214}_{82}\text{Pb}\rightarrow{}^{214}_{83}\text{Bi}\rightarrow{}^{210}_{81}\text{Tl}$ . Identify the decay at each step, write the emitted particle or particles, and state which step requires an electron antineutrino.
[5 marks]
Total for this question: 5
A ${}^{232}_{90}\text{Th}$ nucleus reaches ${}^{208}_{82}\text{Pb}$ through a sequence containing only alpha and $\beta^-$ decays. Determine the number of each type of decay and state the total number of electron antineutrinos emitted.
[4 marks]
Total for this question: 4
On a force-separation graph for two nucleons, taking repulsion as positive, the horizontal scale is $0.10\,\text{fm}$ per small division. The force curve crosses the zero-force axis $5.0$ divisions from the origin, is positive at smaller separations and negative just beyond the crossing. Determine the equilibrium separation and explain why this equilibrium is stable.
[4 marks]
Total for this question: 4

3.2.1.3 · Particles, antiparticles and photons

Explanation

Every particle has a corresponding antiparticle with the same mass and rest energy but opposite charge and additive quantum numbers. Required pairs include electron-positron, proton–antiproton, neutron–antineutron and neutrino–antineutrino.
Electromagnetic radiation is modelled as photons with $E=hf=hc/\lambda$ .
Annihilation converts a particle–antiparticle pair into photons, while pair production requires at least the pair’s combined rest energy and a nearby body for momentum conservation.
A slow electron and positron usually produce two $0.511\,\text{MeV}$ photons, relevant to PET.
Examiners expect energy to be shared correctly, wavelength and frequency units to be consistent, and neutral antiparticles not automatically to be treated as identical.

Worked example

Calculate the energy of a photon with wavelength $500\,\text{nm}$ .

1.Convert $\lambda=5.00\times10^{-7}\,\text{m}$ .
2.Use $E=hc/\lambda$ .
3. $E=(6.63\times10^{-34})(3.00\times10^8)/(5.00\times10^{-7})$ .

Answer: $E=3.98\times10^{-19}\,\text{J}$ .

Common mistakes

Don't assign $1.022\,\text{MeV}$ to each annihilation photon.
Don't assume every neutral particle is its own antiparticle.
Don't use wavelength in nanometres without converting to metres.

Exam tip

For annihilation, compare total photon energy with the combined initial rest and kinetic energy.

Tier 1 · Easy

Calculate the energy of a photon of frequency $6.0\times10^{14}\,\text{Hz}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ .
[2 marks]
Total for this question: 2
Compare the rest mass and electric charge of an antineutron with those of a neutron.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A slow electron and a slow positron annihilate to produce two identical photons. Each electron has rest energy $0.511\,\text{MeV}$ . Calculate the wavelength of either photon. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[3 marks]
Total for this question: 3
A gamma photon has wavelength $8.5\times10^{-13}\,\text{m}$ . Calculate its energy in $\text{MeV}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[3 marks]
Total for this question: 3
An electron and a positron approach with equal and opposite momenta and then annihilate. Explain why a single photon cannot be the only product and why two photons travelling in opposite directions can conserve both energy and momentum.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A photon of wavelength $3.00\times10^{-13}\,\text{m}$ produces an electron-positron pair near a nucleus. Neglecting the nucleus's recoil energy, calculate the total kinetic energy of the pair in $\text{MeV}$ and explain why the nucleus is required. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ , $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ and electron rest energy $0.511\,\text{MeV}$ .
[5 marks]
Total for this question: 5
A photon produces a proton-antiproton pair near a massive nucleus. Neglect the nucleus's recoil energy. Each new particle has rest energy $938\,\text{MeV}$ . Calculate the minimum photon frequency that can produce the pair. Use $h=6.63\times10^{-34}\,\text{J s}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[5 marks]
Total for this question: 5
Two photons of wavelength $1.00\times10^{-12}\,\text{m}$ travel in opposite directions after an electron-positron pair annihilates. Before the event, the particles had momenta of equal magnitude in opposite directions. Deduce the kinetic energy of each particle. Use electron rest energy $0.511\,\text{MeV}$ , $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[4 marks]
Total for this question: 4
An electron-positron pair is moving slowly before annihilation. The products are two oppositely directed photons, each with energy $0.511\,\text{MeV}$ . Calculate the frequency and wavelength of either photon. Use $h=4.14\times10^{-15}\,\text{eV s}$ and $c=2.998\times10^8\,\text{m s}^{-1}$ . State why two photons are produced rather than one.
[4 marks]
Total for this question: 4
Evaluate two proposals for producing an electron-positron pair in empty space. Proposal A uses one photon of energy $1.30\,\text{MeV}$ . Proposal B uses two photons, each of energy $0.700\,\text{MeV}$ , travelling in opposite directions. Use electron rest energy $0.511\,\text{MeV}$ .
[4 marks]
Total for this question: 4

3.2.1.4 · Particle interactions

Explanation

The four fundamental interactions are gravitational, electromagnetic, weak and strong. Exchange particles describe force transfer between elementary particles.
The electromagnetic interaction uses virtual photons. At this level, weak interactions include $\beta^-$ and $\beta^+$ decay, electron capture and electron-proton collisions, using $W^+$ or $W^-$ bosons; gluons, $Z^0$ and gravitons are not tested.
Simple interaction diagrams must identify incoming and outgoing particles and the exchanged particle while conserving charge at each vertex.
For $\beta^-$ decay, $n\rightarrow p+W^-$ followed by $W^-\rightarrow e^-+\bar{\nu}_e$ .
Examiners expect the interaction and exchange particle, not merely a named decay.

A simple weak-interaction diagram for beta-minus decay through a W-minus boson.

Worked example

State the interaction and exchange particle for neutron $\beta^-$ decay, then give the two vertices.

1.Beta decay is governed by the weak interaction.
2.The neutron vertex is $n\rightarrow p+W^-$ .
3.The boson decays as $W^-\rightarrow e^-+\bar{\nu}_e$ .

Answer: Weak interaction via a $W^-$ boson.

Common mistakes

Don't name a virtual photon as the exchange particle for beta decay.
Don't draw a vertex at which electric charge is not conserved.
Don't add untested exchange particles instead of the required $W^+$ or $W^-$ .

Exam tip

Check electric charge at each vertex separately before accepting an interaction diagram.

Tier 1 · Easy

State the fundamental interaction and exchange particle involved when a neutron undergoes $\beta^-$ decay.
[2 marks]
Total for this question: 2
State the fundamental interaction mediated by each exchange particle: a virtual photon and a $W$ boson.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Complete the weak-interaction equation $e^-+p\rightarrow n+\,?$ and identify the exchange particle involved.
[3 marks]
Total for this question: 3
State the fundamental interaction involved in each process: repulsion between two protons, $\beta^-$ decay, and binding between neighbouring nucleons.
[3 marks]
Total for this question: 3
A positron is scattered by a proton without either particle changing type. In a separate process, a neutron undergoes $\beta^-$ decay. Name the exchange particle and the fundamental interaction in each process, including the sign of the charged exchange boson.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Describe $\beta^+$ decay as two interaction vertices involving an exchange particle, and use charge at each vertex to justify the sign of that exchange particle.
[5 marks]
Total for this question: 5
Electron capture is represented by $p+e^-\rightarrow n+\nu_e$ , with the proton emitting the exchange boson. Write the two interaction-vertex equations. Explain why the exchange boson is $W^+$ and why a virtual photon cannot mediate this process.
[5 marks]
Total for this question: 5
An electron with kinetic energy $2.00\,\text{MeV}$ collides with a stationary proton. In this collision, the outgoing neutral lepton is an electron neutrino $\nu_e$ , the outgoing neutral baryon is a neutron $n$ , and the exchanged particle is a $W^-$ boson. State the fundamental interaction, justify, using charge at each vertex, why the exchanged particle must be a $W^-$ , and calculate the combined kinetic energy after the collision. Use rest energies $0.511\,\text{MeV}$ for the electron, $938.27\,\text{MeV}$ for the proton and $939.57\,\text{MeV}$ for the neutron; take the neutrino rest energy as zero.
[4 marks]
Total for this question: 4
Consider neutron $\beta^-$ decay, proton $\beta^+$ decay, electron capture with the boson drawn from the proton vertex to the electron vertex, the collision $e^-+p\rightarrow\nu_e+n$ with the boson drawn from the electron vertex to the proton vertex, and elastic electron-proton scattering in which neither particle changes type. Identify the weak processes. State the correct exchange particle for each weak process.
[5 marks]
Total for this question: 5
In an electron-proton collision the exchange boson is drawn from the electron vertex to the proton vertex, whereas in electron capture it is drawn from the proton vertex to the electron vertex. Show that charge is conserved at every vertex in both descriptions and state the exchange boson in each case.
[4 marks]
Total for this question: 4

3.2.1.5 · Classification of particles

Explanation

Hadrons experience the strong interaction. Baryons and antibaryons have baryon number $+1$ and $-1$ ; required examples are proton, neutron and their antiparticles.
Mesons, including pions and kaons, have baryon number zero. The proton is the only stable baryon, the pion mediates the strong nuclear force, and kaons can decay into pions.
Leptons do not experience the strong interaction; required families are electron, electron neutrino, muon and muon neutrino with antiparticles, and family lepton numbers are conserved separately.
Strange particles are pair-produced by strong interactions with strangeness conserved, then decay weakly with change $0$ , $+1$ or $-1$ .
Examiners expect classification plus relevant quantum numbers.

Worked example

Classify a proton, kaon and muon.

1.A proton is a three-quark hadron, so it is a baryon.
2.A kaon is a quark–antiquark hadron, so it is a meson.
3.A muon does not experience the strong interaction, so it is a lepton.

Answer: Proton: baryon; kaon: meson; muon: lepton.

Common mistakes

Don't classify a pion as a baryon rather than a meson.
Don't check only total lepton number instead of electron and muon families separately.
Don't claim strangeness must be conserved in every weak decay.

Exam tip

Build a reaction table with particle class, baryon number, both lepton numbers and strangeness.

Tier 1 · Easy

Classify the proton, pion and muon.
[2 marks]
Total for this question: 2
State the two classes of hadron.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Complete the decay $\mu^-\rightarrow e^-+\,?+\,?$ using an electron-type neutrino and a muon-type neutrino, and state the particle class shared by all four particles.
[3 marks]
Total for this question: 3
A $\pi^-$ and a $\mu^-$ have the same electric charge. State the class of each particle and explain which one experiences the strong interaction.
[3 marks]
Total for this question: 3
The neutral particle $\Lambda^0$ experiences the strong interaction and has baryon number $+1$ and strangeness $-1$ . Classify it as a hadron or lepton and as a baryon or meson, state whether it is strange, and state the interaction by which it decays.
[3 marks]
Total for this question: 3
Two neutrons attract when they are about $1\,\text{fm}$ apart. Explain why this is not an electromagnetic interaction, name the exchange particle used to model the strong nuclear interaction between the neutrons, and state what the exchange transfers.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In the strong interaction $p+p\rightarrow p+\Lambda^0+K^+$ , the supplied data are $B(\Lambda^0)=1$ , $S(\Lambda^0)=-1$ , $B(K^+)=0$ and $S(K^+)=+1$ . Explain how the products illustrate the classification and paired production of strange particles.
[5 marks]
Total for this question: 5
A $K^+$ meson has strangeness $+1$ and decays to $\pi^++\pi^0$ . Each pion contains only up and down quarks. Determine the change in strangeness and use particle classification and strangeness to identify the interaction responsible.
[4 marks]
Total for this question: 4
A neutral hadron $X$ has baryon number $+1$ and decays as $X\rightarrow p+\pi^-$ . A student then proposes the decay $p\rightarrow\pi^++\pi^0$ . Determine the particle class of each particle and use charge and baryon number to explain why the first decay can satisfy these conservation laws but the proposed proton decay cannot.
[5 marks]
Total for this question: 5
Three particles leave tracks in a detector. Particle $X$ experiences the strong interaction and has baryon number $0$ . Particle $Y$ experiences the strong interaction and has baryon number $-1$ . Particle $Z$ does not experience the strong interaction and has muon-family lepton number $+1$ . State the class of each particle and state which could be a pion, an antiproton and a muon.
[4 marks]
Total for this question: 4
In the reaction $\Sigma^-+p\rightarrow\Lambda^0+X$ , the particles have $\Sigma^-:(Q,B,S)=(-1,+1,-1)$ , $p:(+1,+1,0)$ and $\Lambda^0:(0,+1,-1)$ . Use conservation of charge, baryon number and strangeness to deduce the unknown product $X$ .
[4 marks]
Total for this question: 4

3.2.1.6 · Quarks and antiquarks

Explanation

Only up, down and strange quarks and their antiquarks are required. Their charges are $+\tfrac23e$ , $-\tfrac13e$ and $-\tfrac13e$ ; antiquarks have opposite charge and quantum numbers.
Each quark has baryon number $+\tfrac13$ , each antiquark $-\tfrac13$ ; strange quark has strangeness $-1$ and antistrange $+1$ . Required baryons are $p=uud$ and $n=udd$ , with corresponding three-antiquark antibaryons.
Mesons are quark–antiquark pairs, including pions and kaons.
Neutron beta decay changes a down quark to an up quark.
Examiners expect charge, baryon number and strangeness to be obtained by adding the constituent values.

Worked example

Show that $u\bar{s}$ has the quantum numbers of a $K^+$ meson.

1.Charge is $+\tfrac23e+\tfrac13e=+e$ .
2.Baryon number is $+\tfrac13-\tfrac13=0$ .
3.The antistrange quark gives strangeness $+1$ .

Answer: $u\bar{s}$ has charge $+e$ , baryon number $0$ and strangeness $+1$ .

Common mistakes

Don't give an antiquark the same charge as its corresponding quark.
Don't use three quarks for a meson rather than a quark–antiquark pair.
Don't assign strangeness $+1$ to the strange quark.

Exam tip

Add constituent charges in thirds of $e$ before naming or checking a hadron.

Tier 1 · Easy

State the quark composition of a proton and show that it has charge $+e$ .
[2 marks]
Total for this question: 2
State the electric charge and baryon number of an anti-up quark.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A meson has quark composition $u\bar{s}$ . Determine its charge, baryon number and strangeness, and identify the meson.
[3 marks]
Total for this question: 3
Determine the quark composition, baryon number and strangeness of a $\pi^-$ meson.
[3 marks]
Total for this question: 3
A neutral meson has strangeness $+1$ and contains a down quark. Deduce its antiquark, write its quark composition and state the particle.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Describe neutron $\beta^-$ decay in terms of a quark change and a $W$ boson. Verify charge conservation at both vertices.
[5 marks]
Total for this question: 5
An antineutron has zero electric charge. Determine its quark composition and baryon number, then explain why it is distinct from a neutron even though both particles are neutral.
[4 marks]
Total for this question: 4
Show that the products in $p+n\rightarrow n+n+\pi^+$ can be formed by creating a down quark-antidown quark pair, and show that charge and baryon number are conserved.
[5 marks]
Total for this question: 5
A meson $X$ has charge $-e$ and strangeness $-1$ . Deduce its quark composition and show that it has the stated charge and strangeness. State the quark composition and strangeness of its antiparticle.
[4 marks]
Total for this question: 4
A baryon contains only up, down and strange quarks. It has charge $+e$ , baryon number $+1$ and strangeness $-1$ . Determine the number of each type of quark and write its quark composition.
[5 marks]
Total for this question: 5

3.2.1.7 · Applications of conservation laws

Explanation

Charge, baryon number, electron-family lepton number, muon-family lepton number, energy and momentum are conserved in particle interactions. Strangeness is conserved in strong interactions but may change by $0$ , $+1$ or $-1$ in weak interactions.
In $\beta^-$ decay a down quark changes to an up quark; in $\beta^+$ decay an up quark changes to a down quark.
A reaction satisfying charge conservation alone may still be forbidden.
Data are supplied for particles outside the required set.
Examiners expect initial and final totals to be tabulated for each relevant quantum number and energy–momentum feasibility to be considered rather than inferred from one successful check.

Worked example

Test charge, baryon number and lepton number for $p+p\rightarrow p+n+\pi^+$ .

1.Charge: initial $+2e$ ; final $+e+0+e=+2e$ .
2.Baryon number: initial $2$ ; final $1+1+0=2$ .
3.All lepton-family numbers are zero on both sides.

Answer: Charge, baryon number and lepton number are all conserved; energy and momentum must also be possible.

Common mistakes

Don't accept a reaction after checking charge alone.
Don't combine electron and muon lepton numbers into one total.
Don't require strangeness conservation in a weak interaction.

Exam tip

Use one row per conserved quantity and total every particle on both sides.

Tier 1 · Easy

Show that charge, baryon number and lepton number are conserved in $p+p\rightarrow p+n+\pi^+$ .
[2 marks]
Total for this question: 2
Other than energy and momentum, state three quantities that are conserved in every particle interaction.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Use conservation of charge, baryon number and electron lepton number to determine $X$ in $n\rightarrow p+e^-+X$ .
[4 marks]
Total for this question: 4
In the annihilation $p+\bar{p}\rightarrow\pi^++X$ , use conservation laws to determine $X$ , given that it is a pion.
[4 marks]
Total for this question: 4
For $\beta^+$ decay, a proton changes into a neutron, a positron and an electron neutrino. State the quark change and show that charge, baryon number and electron lepton number are conserved.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two proposed reactions are $\nu_e+n\rightarrow p+e^-$ and $\bar{\nu}_e+n\rightarrow p+e^-$ . Determine which can occur by checking charge, baryon number, electron lepton number, energy and momentum.
[6 marks]
Total for this question: 6
A proposed strong interaction is $K^-+p\rightarrow\Omega^-+K^++K^0$ . The particle data are: $K^-$ has $(Q,B,S)=(-1,0,-1)$ , $p$ has $(+1,+1,0)$ , $\Omega^-$ has $(-1,+1,-3)$ , $K^+$ has $(+1,0,+1)$ and $K^0$ has $(0,0,+1)$ . Determine whether the reaction is allowed by charge, baryon-number and strangeness conservation.
[4 marks]
Total for this question: 4
A negative muon at rest is captured by a stationary proton in the reaction $\mu^-+p\rightarrow n+X$ . The rest energies of the muon, proton and neutron are $106\,\text{MeV}$ , $938\,\text{MeV}$ and $940\,\text{MeV}$ respectively. Particle $X$ is a neutral muon-type neutrino or antineutrino with negligible rest energy. Determine $X$ , show that charge, baryon number and muon lepton number are conserved, state the interaction and calculate the total kinetic energy of the products.
[6 marks]
Total for this question: 6
An electron and a positron collide head-on with equal and opposite momenta. Each has total energy $118\,\text{MeV}$ . They produce a muon and an antimuon, each of rest energy $106\,\text{MeV}$ . Show that charge and both family lepton numbers are conserved, and calculate the total kinetic energy of the products.
[5 marks]
Total for this question: 5
A free neutron at rest decays as $n\rightarrow p+e^-+\bar\nu_e$ . Use rest energies $939.57\,\text{MeV}$ for the neutron, $938.27\,\text{MeV}$ for the proton and $0.511\,\text{MeV}$ for the electron. Deduce whether the decay is energetically allowed, calculate the total kinetic energy available, and show that electron-family lepton number is conserved.
[4 marks]
Total for this question: 4

3.2.2.1 · The photoelectric effect

Explanation

In the photon model, one photon transfers all its energy $hf$ to one surface electron.
Emission occurs only when $hf\geq\phi$ , where $\phi$ is work function, so threshold frequency is $f_0=\phi/h$ .
Above threshold, $E_{k,\max}=hf-\phi$ and stopping potential satisfies $eV_s=E_{k,\max}$ .
Raising frequency increases maximum kinetic energy; raising intensity at fixed frequency increases photon arrival rate and therefore emission rate, but not maximum kinetic energy.
Examiners expect work function converted consistently between eV and J, immediate emission explained by one-photon transfer, and a negative calculated kinetic energy interpreted as no emission.

One incident photon transferring energy to one electron at a metal surface.

Worked example

A metal has work function $2.0\,\text{eV}$ and receives $3.5\,\text{eV}$ photons. Find maximum kinetic energy and stopping potential.

1. $E_{k,\max}=3.5-2.0=1.5\,\text{eV}$ .
2.Use $eV_s=E_{k,\max}$ .
3.An electron energy of $1.5\,\text{eV}$ corresponds to $V_s=1.5\,\text{V}$ .

Answer: $E_{k,\max}=1.5\,\text{eV}$ and $V_s=1.5\,\text{V}$ .

Common mistakes

Don't claim increased intensity raises maximum photoelectron energy.
Don't subtract a work function in eV from photon energy in joules.
Don't report negative kinetic energy instead of no emission.

Exam tip

State first whether $f$ exceeds $f_0$ ; only then calculate kinetic energy.

Tier 1 · Easy

A metal has work function $2.4\,\text{eV}$ . Calculate its threshold frequency. Use $h=6.63\times10^{-34}\,\text{J s}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[2 marks]
Total for this question: 2
Explain why increasing the intensity of radiation below a metal's threshold frequency does not cause photoelectron emission.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Radiation of frequency $8.2\times10^{14}\,\text{Hz}$ illuminates a metal of work function $2.1\,\text{eV}$ . Calculate the maximum kinetic energy in $\text{eV}$ and the stopping potential. Use $h=6.63\times10^{-34}\,\text{J s}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3
A metal has threshold wavelength $540\,\text{nm}$ . Determine its work function in electronvolts. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[3 marks]
Total for this question: 3
For one metal, increasing the radiation frequency by $1.2\times10^{14}\,\text{Hz}$ raises the stopping potential from $0.45\,\text{V}$ to $0.95\,\text{V}$ . Calculate an experimental value for the Planck constant. Use $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Light of wavelength $420\,\text{nm}$ illuminates a metal with work function $2.30\,\text{eV}$ . Calculate the stopping potential and state the effect on emission rate and stopping potential when the light intensity is doubled at the same wavelength. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
Radiation of wavelength $365\,\text{nm}$ illuminates a metal of work function $2.15\,\text{eV}$ . Calculate the maximum speed of an emitted electron. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ , $e=1.60\times10^{-19}\,\text{C}$ and electron mass $9.11\times10^{-31}\,\text{kg}$ .
[5 marks]
Total for this question: 5
Metals A and B are illuminated by radiation of the same frequency. Their stopping potentials are $0.80\,\text{V}$ and $0.25\,\text{V}$ respectively. Metal A has threshold frequency $6.0\times10^{14}\,\text{Hz}$ . Calculate the threshold frequency of metal B and the ratio $v_A/v_B$ of the maximum photoelectron speeds. Use $h=6.63\times10^{-34}\,\text{J s}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
For one metal, stopping potentials of $0.38\,\text{V}$ and $1.42\,\text{V}$ are measured at frequencies $7.1\times10^{14}\,\text{Hz}$ and $9.6\times10^{14}\,\text{Hz}$ respectively. Use $e=1.60\times10^{-19}\,\text{C}$ to determine an experimental value of the Planck constant, the work function in electronvolts and the threshold frequency.
[4 marks]
Total for this question: 4
Monochromatic radiation of wavelength $310\,\text{nm}$ and power $2.8\,\text{mW}$ illuminates a metal of work function $2.65\,\text{eV}$ . $18\%$ of incident photons release one electron each, and all emitted electrons are collected. Calculate the stopping potential and the resulting photoelectric current. Use $hc=1.989\times10^{-25}\,\text{J m}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[6 marks]
Total for this question: 6

3.2.2.2 · Collisions of electrons with atoms

Explanation

Excitation raises an atomic electron to a higher bound level, while ionisation removes it from the atom.
In an inelastic collision, the incident electron must transfer an allowed discrete excitation energy or at least the ionisation energy; remaining energy stays as kinetic energy.
An electron accelerated through potential difference $V$ gains $eV$ , numerically $V\,\text{eV}$ .
In a fluorescent tube, accelerated electrons excite mercury atoms; de-excitation produces ultraviolet photons that excite the coating, whose later de-excitation emits visible light.
Examiners expect excitation to remain bound, ionisation to produce a free electron, eV–J conversion, and collision energy to be shared without inventing a partly allowed atomic transition.

Worked example

An electron with $12.0\,\text{eV}$ excites an atom by $8.5\,\text{eV}$ . Find its remaining kinetic energy in eV and J.

1.Subtract the allowed excitation energy: $12.0-8.5=3.5\,\text{eV}$ .
2.Convert using $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
3. $E=3.5(1.60\times10^{-19})=5.6\times10^{-19}\,\text{J}$ .

Answer: $3.5\,\text{eV}=5.6\times10^{-19}\,\text{J}$ .

Common mistakes

Don't call an electron in a higher bound level ionised.
Don't assume all incident kinetic energy must transfer to the atom.
Don't treat a potential difference in volts as energy in joules without multiplying by charge.

Exam tip

Compare incident energy with each discrete gap before deciding whether excitation or ionisation can occur.

Tier 1 · Easy

Convert an electron energy of $18\,\text{eV}$ into joules. Use $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[1 mark]
Total for this question: 1
Starting from rest, an electron is accelerated across a potential difference of $12\,\text{V}$ . State the kinetic energy it gains in electronvolts.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An electron with kinetic energy $12.0\,\text{eV}$ collides with an atom in its ground state. Excited states are $4.0\,\text{eV}$ and $10.5\,\text{eV}$ above the ground state, and the ionisation energy is $13.6\,\text{eV}$ . Determine the greatest possible excitation and the electron kinetic energy immediately afterwards.
[3 marks]
Total for this question: 3
Explain the difference between excitation and ionisation of an atom in terms of its electrons and the energy transferred.
[3 marks]
Total for this question: 3
An electron with kinetic energy $18.0\,\text{eV}$ ionises a stationary atom whose ionisation energy is $13.6\,\text{eV}$ . Neglect atomic recoil. Calculate the total kinetic energy of the two outgoing electrons and explain why their individual kinetic energies cannot be determined from the information given.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Mercury atoms in a fluorescent tube have an excitation energy of $7.7\,\text{eV}$ . Determine the minimum accelerating potential needed for an electron to cause this excitation, then explain how the collision ultimately produces visible light from the tube coating.
[5 marks]
Total for this question: 5
An electron with kinetic energy $20.0\,\text{eV}$ collides successively with two identical atoms. Their allowed excitation energies are $6.0\,\text{eV}$ and $10.5\,\text{eV}$ . In each collision the electron causes the greatest possible excitation and does not ionise the atom. Determine the excitation in each collision and the electron's final kinetic energy.
[5 marks]
Total for this question: 5
An electron starts from rest and is accelerated through a potential difference $V$ . It then excites a stationary atom by $6.7\,\text{eV}$ and emerges at $1.50\times10^6\,\text{m s}^{-1}$ . Neglect atomic recoil. Determine $V$ . Use electron mass $9.11\times10^{-31}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[4 marks]
Total for this question: 4
An electron accelerated from rest through $28.0\,\text{V}$ ionises a stationary atom. After the collision, the two electrons have speeds $1.70\times10^6\,\text{m s}^{-1}$ and $1.10\times10^6\,\text{m s}^{-1}$ . Neglect atomic recoil. Determine the atom's ionisation energy in electronvolts. Take $e=1.60\times10^{-19}\,\text{C}$ and electron mass $9.11\times10^{-31}\,\text{kg}$ .
[5 marks]
Total for this question: 5
A fluorescent coating absorbs one ultraviolet photon of wavelength $254\,\text{nm}$ . It then emits two photons as it returns to its ground state. One emitted photon has wavelength $430\,\text{nm}$ . Assuming all the absorbed energy is carried by the two emitted photons, calculate the wavelength of the other photon and explain the excitation and de-excitation sequence.
[5 marks]
Total for this question: 5

3.2.2.3 · Energy levels and photon emission

Explanation

Atomic electrons occupy discrete energy levels. A line emission spectrum is evidence that only particular downward transitions and photon energies occur.
For a transition from higher $E_1$ to lower $E_2$ , $hf=E_1-E_2$ ; the positive energy gap may also be used in $\lambda=hc/\Delta E$ .
Upward absorption requires a photon whose energy exactly matches an available gap, rather than partial absorption of an intermediate energy.
Energy levels may be quoted in joules or electronvolts.
Examiners expect the correct initial and final levels, a positive gap converted to joules before using $h$ , and each spectral line linked to one permitted transition.

A photon emitted when an electron falls between two discrete atomic energy levels.

Worked example

An electron falls from $-1.5\,\text{eV}$ to $-3.4\,\text{eV}$ . Find the emitted photon energy and wavelength.

1. $\Delta E=(-1.5)-(-3.4)=1.9\,\text{eV}$ .
2.Convert $\Delta E=1.9(1.60\times10^{-19})=3.04\times10^{-19}\,\text{J}$ .
3. $\lambda=hc/\Delta E=6.54\times10^{-7}\,\text{m}$ .

Answer: $1.9\,\text{eV}$ and $6.5\times10^{-7}\,\text{m}$ .

Common mistakes

Don't subtract negative energy levels in the wrong order and report negative photon energy.
Don't use an energy gap in eV directly with $h$ in joule seconds.
Don't claim a photon can be partly absorbed when its energy misses every level gap.

Exam tip

Mark the downward transition first, then calculate the positive level difference.

Tier 1 · Easy

An electron falls from an energy level at $-1.5\,\text{eV}$ to one at $-3.4\,\text{eV}$ . State the photon energy.
[1 mark]
Total for this question: 1
Explain why an atomic emission spectrum consists of discrete lines.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Calculate the wavelength emitted when an electron falls from $-1.51\,\text{eV}$ to $-3.40\,\text{eV}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[3 marks]
Total for this question: 3
An electron falls between two atomic levels whose energy difference is $3.30\times10^{-19}\,\text{J}$ . Calculate the frequency and period of the emitted radiation. Use $h=6.63\times10^{-34}\,\text{J s}$ .
[3 marks]
Total for this question: 3
An atom has energy levels $0$ , $2.1\,\text{eV}$ and $4.9\,\text{eV}$ above its ground state. It receives photons of energy $2.8\,\text{eV}$ . Determine which initial atomic level permits absorption and state the final level.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An atom has energy levels $0$ , $-1.2$ , $-3.0$ and $-5.0\,\text{eV}$ . Electrons are raised to the $0\,\text{eV}$ level and can return by any sequence. Determine the number of distinct emission lines and calculate the longest wavelength. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[5 marks]
Total for this question: 5
An atom emits photons of wavelength $450\,\text{nm}$ and $650\,\text{nm}$ in a two-step cascade from an upper level to the ground state. Calculate the wavelength of the photon emitted by a direct transition between the same initial and final levels.
[5 marks]
Total for this question: 5
Two atomic transitions have energy differences $2.10\,\text{eV}$ and $3.40\,\text{eV}$ . Their measured emission wavelengths are $590\,\text{nm}$ and $365\,\text{nm}$ respectively. Calculate the Planck constant from each transition and calculate their percentage difference relative to their mean. Use $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[5 marks]
Total for this question: 5
An atom has energy levels $0$ , $1.65$ , $3.90$ and $6.20\,\text{eV}$ above its ground state. Atoms are excited to the highest level and every possible downward transition occurs. Deduce the number of distinct photon energies, list them, and calculate the longest emitted wavelength. Use $hc=1240\,\text{eV nm}$ .
[5 marks]
Total for this question: 5
An atom has a ground-state energy of $-6.80\,\text{eV}$ . A $3.10\,\text{eV}$ photon raises a ground-state atom to an intermediate level. A further $1.90\,\text{eV}$ photon raises it from that level to an upper level. Determine both excited-state energies and the wavelength emitted in a direct transition from the upper level to the ground state. Use $hc=1.989\times10^{-25}\,\text{J m}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[5 marks]
Total for this question: 5

3.2.2.4 · Wave-particle duality

Explanation

Electron diffraction shows that matter particles possess wave properties, while the photoelectric effect shows particle-like photon transfer by electromagnetic radiation.
The de Broglie wavelength is $\lambda=h/p=h/(mv)$ for a non-relativistic particle.
Increasing momentum shortens wavelength and reduces diffraction for unchanged crystal spacing or aperture.
For an electron accelerated from rest through potential $V$ , energy $eV$ gives momentum $\sqrt{2meV}$ and hence $\lambda=h/\sqrt{2meV}$ .
Examiners expect momentum in $\text{kg m s}^{-1}$ , voltage converted through $eV$ rather than treated as joules, and evidence to be evaluated as scientific models change through peer review and community validation.

Electron diffraction from a polycrystalline target produces concentric rings on a screen.

Worked example

Calculate the de Broglie wavelength of a particle with momentum $3.0\times10^{-24}\,\text{kg m s}^{-1}$ .

1.Use $\lambda=h/p$ .
2. $\lambda=(6.63\times10^{-34})/(3.0\times10^{-24})$ .
3.Express the result in metres and nanometres.

Answer: $\lambda=2.2\times10^{-10}\,\text{m}=0.22\,\text{nm}$ .

Common mistakes

Don't state that increasing momentum increases de Broglie wavelength.
Don't use voltage directly as energy in joules.
Don't cite photoelectric emission as evidence only for the wave nature of light.

Exam tip

When momentum changes, use $\lambda\propto1/p$ before discussing the amount of diffraction.

Tier 1 · Easy

Calculate the de Broglie wavelength of a particle with momentum $3.0\times10^{-24}\,\text{kg m s}^{-1}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ .
[2 marks]
Total for this question: 2
State the observation from electron diffraction and the conclusion it supports about electrons.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Electrons travel at $2.5\times10^6\,\text{m s}^{-1}$ . Calculate their de Broglie wavelength and state how doubling their speed affects the wavelength. Use electron mass $9.11\times10^{-31}\,\text{kg}$ and $h=6.63\times10^{-34}\,\text{J s}$ .
[3 marks]
Total for this question: 3
A neutron beam has de Broglie wavelength $0.180\,\text{nm}$ . Calculate the momentum and speed of each neutron. Use $h=6.63\times10^{-34}\,\text{J s}$ and neutron mass $1.67\times10^{-27}\,\text{kg}$ .
[3 marks]
Total for this question: 3
In the same electron-diffraction apparatus, the diameter of a diffraction ring changes from $42\,\text{mm}$ to $30\,\text{mm}$ . The ring diameter is proportional to the de Broglie wavelength. Calculate the ratio of the final electron momentum to the initial electron momentum.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Electrons are accelerated from rest through $150\,\text{V}$ and then diffract from a crystal. Derive an expression for their de Broglie wavelength in terms of $V$ , calculate it, and explain how increasing $V$ changes the diffraction. Use $e=1.60\times10^{-19}\,\text{C}$ , electron mass $9.11\times10^{-31}\,\text{kg}$ and $h=6.63\times10^{-34}\,\text{J s}$ .
[5 marks]
Total for this question: 5
A non-relativistic electron and proton have equal kinetic energies. Calculate the ratio $\lambda_e/\lambda_p$ of their de Broglie wavelengths and state which particle produces the larger diffraction angle from the same crystal. Use electron mass $9.11\times10^{-31}\,\text{kg}$ and proton mass $1.67\times10^{-27}\,\text{kg}$ .
[4 marks]
Total for this question: 4
A student claims that electrons temporarily become waves only while passing through a crystal, and that the first observed diffraction pattern is enough to prove this claim. Explain why the observations do not justify this interpretation and how a change to the scientific model should be validated.
[4 marks]
Total for this question: 4
A proton and an alpha particle have the same de Broglie wavelength of $0.090\,\text{nm}$ . Calculate their common momentum, their speeds and their kinetic energies in electronvolts. Use $h=6.63\times10^{-34}\,\text{J s}$ , proton mass $1.67\times10^{-27}\,\text{kg}$ , alpha-particle mass $6.68\times10^{-27}\,\text{kg}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[6 marks]
Total for this question: 6
Electrons accelerated from rest through $(275\pm3)\,\text{V}$ have a measured de Broglie wavelength $(7.50\pm0.15)\times10^{-11}\,\text{m}$ . Use $h=\lambda\sqrt{2meV}$ to determine an experimental value of the Planck constant and its percentage uncertainty. Deduce whether it is consistent with $6.63\times10^{-34}\,\text{J s}$ . Treat $m=9.11\times10^{-31}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ as exact.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.2.1.1 · Constituents of the atom

Tier 1 · Easy

Mark scheme for 3.2.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$12$ protons, $13$ neutrons and $10$ electrons	The proton number gives $12$ protons. The neutron number is $25-12=13$ . A $2+$ ion has lost two electrons, so it has $12-2=10$ electrons.	2
02.1	Atoms with the same proton number but different numbers of neutrons.	Isotopes belong to the same element, so they have the same proton number $Z$ . Their neutron numbers, and therefore their nucleon numbers $A$ , are different.	2

Tier 2 · Standard

Mark scheme for 3.2.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+4.82\times10^7\,\text{C kg}^{-1}$	Specific charge is $q/m$ . Therefore $q/m=(3.20\times10^{-19})/(6.64\times10^{-27})=4.82\times10^7\,\text{C kg}^{-1}$ . It is positive because the nucleus has positive charge.	3
02.1	$30\%$	Let the fraction of ${}^{37}\text{X}$ be $x$ . The weighted mean is $37x+35(1-x)=35.6$ . Hence $2x=0.6$ and $x=0.30$ , so ${}^{37}\text{X}$ makes up $30\%$ of the atoms.	3
03.1	ratio $=1.06$ ; chlorine-35 has the greater specific charge	For equal ionic charge, specific charge is inversely proportional to nuclear mass and hence approximately to $1/A$ . Therefore $s_{35}/s_{37}=37/35=1.057$ , which is $1.06$ to three significant figures. The lighter chlorine-35 ion has the greater specific charge.	3

Tier 3 · Hard

Mark scheme for 3.2.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	charge $+3e$ ; $17$ electrons	Approximate the ion mass by its $40$ nucleons: $m=40(1.67\times10^{-27})=6.68\times10^{-26}\,\text{kg}$ . Its charge is $q=(7.19\times10^6)(6.68\times10^{-26})=4.80\times10^{-19}\,\text{C}$ . Dividing by $e$ gives $q/e=(4.80\times10^{-19})/(1.60\times10^{-19})=3$ , so the ion has charge $+3e$ . A neutral atom with $Z=20$ has $20$ electrons; losing three leaves $17$ electrons.	5
02.1	proton number $17$ ; nucleon number $35$ ; ${}^{35}_{17}\text{Cl}$ ; $18$ neutrons	The proton number is $Z=q/e=(2.72\times10^{-18})/(1.60\times10^{-19})=17$ . The nucleon number is $A\approx(5.85\times10^{-26})/(1.67\times10^{-27})=35.0$ , so $A=35$ . Element $Z=17$ is chlorine, giving ${}^{35}_{17}\text{Cl}$ . Its neutron number is $A-Z=35-17=18$ .	4
03.1	$3.94\times10^6\,\text{C kg}^{-1}$ ; $0.0411$ of the proton's specific charge	The mean nucleon number is $0.79(24)+0.10(25)+0.11(26)=24.32$ . Each ion has charge $+e$ , so the total-charge to total-mass ratio is $e/(24.32m_n)=(1.60\times10^{-19})/[24.32(1.67\times10^{-27})]=3.94\times10^6\,\text{C kg}^{-1}$ . The proton's specific charge is $e/m_n$ , so the ratio is $[e/(24.32m_n)]/(e/m_n)=1/24.32=0.0411$ .	5
04.1	$+4.16\times10^6\,\text{C kg}^{-1}$	The ion has one fewer electron than protons, so its charge is $+e=+1.602\times10^{-19}\,\text{C}$ . Its mass is $11(1.673\times10^{-27})+12(1.675\times10^{-27})+10(9.11\times10^{-31})=3.851211\times10^{-26}\,\text{kg}$ . Therefore $q/m=(1.602\times10^{-19})/(3.851211\times10^{-26})=4.1597\times10^6\,\text{C kg}^{-1}$ , giving $+4.16\times10^6\,\text{C kg}^{-1}$ to three significant figures.	4
05.1	$x=0.40$ , so $40\%$ of the ions have nucleon number $63$ (accept $0.39$ to $0.41$ )	For an arbitrary total of $N$ ions, the charge is $Ne[2x+(1-x)]=Ne(1+x)$ . The mass is $Nm_n[63x+65(1-x)]=Nm_n(65-2x)$ . Hence $2.09\times10^6=e(1+x)/[m_n(65-2x)]$ . Substitution gives $(2.09\times10^6)(1.67\times10^{-27})(65-2x)=(1.60\times10^{-19})(1+x)$ . Expanding gives $1+x=1.4179-0.04363x$ , so $1.04363x=0.41794$ and $x=0.400$ (the small difference from exactly $0.400$ depends on retaining the rounded measured ratio), so $40\%$ is the appropriate result.	5

3.2.1.2 · Stable and unstable nuclei

Tier 1 · Easy

Mark scheme for 3.2.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	${}^{4}_{2}\text{He}$	The missing particle must carry nucleon number $210-206=4$ and proton number $84-82=2$ , so it is an alpha particle, ${}^{4}_{2}\text{He}$ .	1
02.1	alpha particle; electron	Alpha decay emits an alpha particle, while beta-minus decay emits an electron.	2

Tier 2 · Standard

Mark scheme for 3.2.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	${}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+e^-+\bar{\nu}_e$ ; proton number increases by $1$	In $\beta^-$ decay a neutron changes into a proton, an electron and an electron antineutrino. The nucleon number therefore stays $14$ , while the proton number increases from $6$ to $7$ : ${}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+e^-+\bar{\nu}_e$ .	3
02.1	A fixed nuclear energy change did not appear entirely as electron kinetic energy; an unseen neutral particle carrying variable energy was proposed so that energy is conserved.	The parent and daughter nuclear states have a fixed energy difference, but emitted electrons have a continuous range of energies. A second emitted particle can share the available energy by varying amounts. The undetected particle was proposed to be the neutrino, preserving energy conservation.	3
03.1	$0.74\,\text{MeV}$ ; conservation of energy; the available energy is shared in varying proportions between the electron, antineutrino and recoiling daughter nucleus	Conservation of energy gives the antineutrino energy as $1.17-0.42-0.01=0.74\,\text{MeV}$ . The available energy is shared among the decay products. The share carried by the antineutrino can differ between events, so the electron has a continuous range of possible energies.	3

Tier 3 · Hard

Mark scheme for 3.2.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	It is repulsive below about $0.5\,\text{fm}$ , attractive from about $0.5\,\text{fm}$ to $3\,\text{fm}$ , and negligible beyond about $3\,\text{fm}$ ; attraction binds nearby nucleons while short-range repulsion prevents collapse.	Award the marking-point chain: below roughly $0.5\,\text{fm}$ the force is repulsive; from roughly $0.5\,\text{fm}$ to $3\,\text{fm}$ it is attractive; beyond roughly $3\,\text{fm}$ it is negligible. The attractive region binds neighbouring nucleons and can overcome proton-proton electrostatic repulsion at nuclear separations. The very-short-range repulsion prevents nucleons from collapsing into the same position, giving a stable separation.	5
02.1	${}^{238}_{92}\text{U}\rightarrow{}^{234}_{90}\text{Th}+{}^{4}_{2}\text{He}$ ${}^{234}_{90}\text{Th}\rightarrow{}^{234}_{91}\text{Pa}+{}^{0}_{-1}e+\bar{\nu}_e$ ${}^{234}_{91}\text{Pa}\rightarrow{}^{234}_{92}\text{U}+{}^{0}_{-1}e+\bar{\nu}_e$	Alpha decay lowers $A$ by $4$ and $Z$ by $2$ , giving ${}^{234}_{90}\text{Th}$ . Each $\beta^-$ decay leaves $A$ unchanged and raises $Z$ by $1$ : thorium-234 becomes protactinium-234 and then uranium-234. Writing the electron as ${}^{0}_{-1}e$ balances proton number, and each beta equation includes an electron antineutrino.	4
03.1	${}^{214}_{82}\text{Pb}\rightarrow{}^{214}_{83}\text{Bi}+{}^{0}_{-1}e+\bar{\nu}_e$ ${}^{214}_{83}\text{Bi}\rightarrow{}^{210}_{81}\text{Tl}+{}^{4}_{2}\text{He}$ The first step requires the electron antineutrino.	In the first step, $A$ stays at $214$ while $Z$ increases from $82$ to $83$ , so the decay is $\beta^-$ . It emits ${}^{0}_{-1}e$ and an electron antineutrino, giving ${}^{214}_{82}\text{Pb}\rightarrow{}^{214}_{83}\text{Bi}+{}^{0}_{-1}e+\bar{\nu}_e$ . In the second step, $A$ falls by $4$ and $Z$ by $2$ , so it is alpha decay: ${}^{214}_{83}\text{Bi}\rightarrow{}^{210}_{81}\text{Tl}+{}^{4}_{2}\text{He}$ . Only the $\beta^-$ step requires $\bar{\nu}_e$ .	5
04.1	$6$ alpha decays, $4$ $\beta^-$ decays and $4$ electron antineutrinos	Each alpha decay lowers nucleon number by $4$ . The decrease $232-208=24$ therefore requires $24/4=6$ alpha decays. These lower the proton number from $90$ to $90-2(6)=78$ . Each $\beta^-$ decay raises proton number by $1$ without changing nucleon number, so $82-78=4$ $\beta^-$ decays are required. Every $\beta^-$ decay emits one electron antineutrino, giving four antineutrinos in total.	4
05.1	$0.50\,\text{fm}$ ; the force is zero there, repulsive at smaller separations and attractive at slightly larger separations, so a displacement produces a restoring force.	The zero crossing is at $r=5.0(0.10)=0.50\,\text{fm}$ , so the force vanishes at this separation. If the nucleons move closer than $0.50\,\text{fm}$ , the positive repulsive force pushes them apart. If they move slightly farther apart, the negative attractive force pulls them together. In either direction the force acts back towards $0.50\,\text{fm}$ , so the equilibrium is stable.	4

3.2.1.3 · Particles, antiparticles and photons

Tier 1 · Easy

Mark scheme for 3.2.1.3 Tier 1 · Easy
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01.1	$4.0\times10^{-19}\,\text{J}$	Use $E=hf$ : $E=(6.63\times10^{-34})(6.0\times10^{14})=3.978\times10^{-19}\,\text{J}$ , which is $4.0\times10^{-19}\,\text{J}$ to two significant figures.	2
02.1	They have the same rest mass and both have zero charge.	A particle and its antiparticle have equal rest mass. The neutron and antineutron are both electrically neutral, although their additive quantum numbers are opposite.	2

Tier 2 · Standard

Mark scheme for 3.2.1.3 Tier 2 · Standard
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01.1	$2.43\times10^{-12}\,\text{m}$	Two photons share the two rest energies, so each photon has $0.511\,\text{MeV}$ . In joules this is $E=0.511\times10^6\times1.60\times10^{-19}=8.176\times10^{-14}\,\text{J}$ . Hence $\lambda=hc/E=(6.63\times10^{-34})(3.00\times10^8)/(8.176\times10^{-14})=2.43\times10^{-12}\,\text{m}$ .	3
02.1	$1.5\,\text{MeV}$ (2 s.f.) or $1.46\,\text{MeV}$ (3 s.f.)	Use $E=hc/\lambda$ : $E=(6.63\times10^{-34})(3.00\times10^8)/(8.5\times10^{-13})=2.34\times10^{-13}\,\text{J}$ . Since $1\,\text{MeV}=1.60\times10^{-13}\,\text{J}$ , this is $(2.34\times10^{-13})/(1.60\times10^{-13})=1.4625\,\text{MeV}$ , which is $1.5\,\text{MeV}$ to two significant figures or $1.46\,\text{MeV}$ to three significant figures.	3
03.1	The initial total momentum is zero, but one photon must have non-zero momentum; two photons can carry energy while their equal and opposite momenta sum to zero.	Equal and opposite initial momenta give total momentum zero. A photon carrying the annihilation energy also has momentum, so a single photon would leave non-zero final momentum and violate momentum conservation. Two photons of equal energy travelling in opposite directions have momenta of equal magnitude that cancel, while together carrying the initial total energy.	3

Tier 3 · Hard

Mark scheme for 3.2.1.3 Tier 3 · Hard
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01.1	$3.12\,\text{MeV}$ ; the nucleus recoils to conserve momentum	The photon energy is $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(3.00\times10^{-13})=6.63\times10^{-13}\,\text{J}$ . This is $(6.63\times10^{-13})/(1.60\times10^{-13})=4.14\,\text{MeV}$ . Creating the two rest masses requires $2(0.511)=1.022\,\text{MeV}$ , leaving $4.14-1.022=3.12\,\text{MeV}$ as total kinetic energy. A single photon has momentum, so the nearby nucleus must recoil and take momentum for momentum to be conserved.	5
02.1	$4.53\times10^{23}\,\text{Hz}$	At the threshold, the photon supplies the combined rest energy $2(938)=1876\,\text{MeV}$ . In joules this is $1876\times10^6\times1.60\times10^{-19}=3.0016\times10^{-10}\,\text{J}$ . Hence $f=E/h=(3.0016\times10^{-10})/(6.63\times10^{-34})=4.527\times10^{23}\,\text{Hz}$ , which is $4.53\times10^{23}\,\text{Hz}$ to three significant figures.	5
03.1	$0.732\,\text{MeV}$	Each photon has energy $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(1.00\times10^{-12})=1.989\times10^{-13}\,\text{J}$ . This is $(1.989\times10^{-13})/(1.60\times10^{-13})=1.243\,\text{MeV}$ . The equal and opposite initial momenta and the two equal photon energies make each particle's initial total energy equal to one photon energy. Therefore the kinetic energy of each particle is $1.243-0.511=0.732\,\text{MeV}$ .	4
04.1	$f=1.23\times10^{20}\,\text{Hz}$ ; $\lambda=2.43\times10^{-12}\,\text{m}$ ; two photons are needed because the initial momentum is (almost) zero and a single photon could not conserve momentum.	Convert the photon energy to electronvolts: $E=0.511\times10^6=5.11\times10^5\,\text{eV}$ . From $E=hf$ , $f=(5.11\times10^5)/(4.14\times10^{-15})=1.234\times10^{20}\,\text{Hz}$ , or $1.23\times10^{20}\,\text{Hz}$ to three significant figures. Keeping the unrounded frequency, $\lambda=c/f=(2.998\times10^8)/(1.234\times10^{20})=2.43\times10^{-12}\,\text{m}$ . The electron and positron are slow, so the total momentum before annihilation is approximately zero; a single photon would carry non-zero momentum, so at least two photons moving in opposite directions are required to conserve momentum.	4
05.1	A is impossible because one photon cannot conserve momentum in empty space; B can conserve momentum and has $0.378\,\text{MeV}$ available as total kinetic energy.	Both proposals exceed the combined rest energy $2(0.511)=1.022\,\text{MeV}$ . Proposal A nevertheless fails in empty space: its single initial photon has momentum, whereas a two-particle final state cannot simultaneously match the photon energy and momentum without another body taking recoil. In proposal B the two equal, opposite photon momenta sum to zero, so the electron and positron can emerge with equal and opposite momenta. Their total initial energy is $2(0.700)=1.400\,\text{MeV}$ , leaving $1.400-1.022=0.378\,\text{MeV}$ as their combined kinetic energy.	4

3.2.1.4 · Particle interactions

Tier 1 · Easy

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01.1	weak interaction; $W^-$ boson	Beta decay is caused by the weak interaction. In $\beta^-$ decay the neutron emits a negatively charged exchange boson, so the exchange particle is $W^-$ .	2
02.1	virtual photon: electromagnetic interaction; $W$ boson: weak interaction	A virtual photon mediates the electromagnetic interaction. A charged $W^+$ or $W^-$ boson mediates the weak interaction.	2

Tier 2 · Standard

Mark scheme for 3.2.1.4 Tier 2 · Standard
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01.1	$e^-+p\rightarrow n+\nu_e$ ; a $W^-$ boson	Initial charge is $-1+1=0$ , so the missing particle is neutral. Electron lepton number starts at $+1$ , so the outgoing neutral lepton must be $\nu_e$ , also with electron lepton number $+1$ . The electron emits $W^-$ as it becomes $\nu_e$ , and the proton absorbs $W^-$ to become a neutron. Thus $e^-+p\rightarrow n+\nu_e$ is mediated by $W^-$ .	3
02.1	electromagnetic; weak; strong, respectively	Like charged protons repel through the electromagnetic interaction. Beta decay is governed by the weak interaction. The short-range attraction binding neighbouring nucleons is the strong interaction.	3
03.1	positron-proton scattering: virtual photon, electromagnetic interaction; neutron $\beta^-$ decay: $W^-$ boson, weak interaction	The charged positron and proton scatter electromagnetically by exchanging a virtual photon, with neither particle changing type. Neutron $\beta^-$ decay is a weak interaction. At its first vertex $n\rightarrow p+W^-$ , so charge conservation requires the exchanged boson to be $W^-$ .	3

Tier 3 · Hard

Mark scheme for 3.2.1.4 Tier 3 · Hard
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01.1	$p\rightarrow n+W^+$ followed by $W^+\rightarrow e^++\nu_e$ ; charge is conserved at both vertices	At the first vertex a proton of charge $+1$ becomes a neutron of charge $0$ , so the emitted exchange particle must carry charge $+1$ : $p\rightarrow n+W^+$ . At the second vertex the $W^+$ produces a positron of charge $+1$ and a neutral electron neutrino: $W^+\rightarrow e^++\nu_e$ . Charge is therefore $+1=0+(+1)$ at the first vertex and $+1=(+1)+0$ at the second; the complete reaction is $p\rightarrow n+e^++\nu_e$ .	5
02.1	$p\rightarrow n+W^+$ and $e^-+W^+\rightarrow\nu_e$ ; charge conservation requires a $W^+$ from the proton vertex, and a virtual photon cannot change particle type.	At the proton vertex $p\rightarrow n+W^+$ , for which charge is $+e=0+(+e)$ . The electron absorbs the boson: $e^-+W^+\rightarrow\nu_e$ , for which $-e+e=0$ . The exchanged boson emitted by the proton therefore carries charge $+e$ and is $W^+$ . Electron-to-neutrino and proton-to-neutron changes occur through the weak interaction; a virtual photon mediates the electromagnetic interaction and cannot cause either change of particle type.	5
03.1	weak interaction; at the electron vertex $-1\to0$ so the boson carries charge $-1$ , and at the proton vertex $+1+(-1)\to0$ balances, so it must be a $W^-$ ; total product kinetic energy $=1.21\,\text{MeV}$	The electron-to-neutrino and proton-to-neutron changes identify the weak interaction. Charge at the electron vertex falls from $-1$ to $0$ , so the exchanged boson must carry charge $-1$ ; the proton vertex then balances as $+1+(-1)\to0$ , confirming a $W^-$ . Energy conservation gives the total product kinetic energy as $(0.511+2.00+938.27)-939.57=1.211\,\text{MeV}$ , or $1.21\,\text{MeV}$ .	4
04.1	The first four processes are weak; elastic electron-proton scattering is electromagnetic. The exchange particles are $W^-$ for $\beta^-$ decay, $W^+$ for $\beta^+$ decay, $W^+$ for electron capture, and $W^-$ for $e^-+p\rightarrow\nu_e+n$ .	The four reactions that change a nucleon and a lepton are weak interactions; elastic scattering between charged particles without a change of particle type is electromagnetic and uses a virtual photon. In $\beta^-$ decay, $n\rightarrow p+W^-$ , whereas in $\beta^+$ decay, $p\rightarrow n+W^+$ . With the stated direction for electron capture, $p\rightarrow n+W^+$ and $e^-+W^+\rightarrow\nu_e$ ; reversing the direction of the exchange line gives the equivalent $W^-$ description. With the stated direction for the collision, $e^-\rightarrow\nu_e+W^-$ and $p+W^-\rightarrow n$ .	5
05.1	$e^-\rightarrow\nu_e+W^-$ then $p+W^-\rightarrow n$ (a $W^-$ boson); $p\rightarrow n+W^+$ then $e^-+W^+\rightarrow\nu_e$ (a $W^+$ boson). Charge is conserved at every vertex.	For the electron-proton collision, at the electron vertex $e^-\rightarrow\nu_e+W^-$ charge is $-e=0+(-e)$ , and at the proton vertex $p+W^-\rightarrow n$ charge is $+e-e=0$ , so the boson drawn from the electron is $W^-$ . For electron capture, at the proton vertex $p\rightarrow n+W^+$ charge is $+e=0+(+e)$ , and at the electron vertex $e^-+W^+\rightarrow\nu_e$ charge is $-e+e=0$ , so the boson drawn from the proton is $W^+$ . Both descriptions conserve charge at every vertex; the two processes differ in which particle emits the boson.	4

3.2.1.5 · Classification of particles

Tier 1 · Easy

Mark scheme for 3.2.1.5 Tier 1 · Easy
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01.1	proton: baryon; pion: meson; muon: lepton	The proton is a three-quark hadron, so it is a baryon. The pion is a quark-antiquark hadron, so it is a meson. The muon is not a hadron and belongs to the lepton family.	2
02.1	baryons and mesons	The two classes of hadron are baryons and mesons.	2

Tier 2 · Standard

Mark scheme for 3.2.1.5 Tier 2 · Standard
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01.1	$\mu^-\rightarrow e^-+\bar{\nu}_e+\nu_\mu$ ; all are leptons	Initially $L_\mu=+1$ and $L_e=0$ . The outgoing electron supplies $L_e=+1$ , so an electron antineutrino with $L_e=-1$ is also needed. A muon neutrino supplies the initial $L_\mu=+1$ . Thus $\mu^-\rightarrow e^-+\bar{\nu}_e+\nu_\mu$ , and every particle shown is a lepton or antilepton.	3
02.1	The $\pi^-$ is a meson and therefore a hadron, so it experiences the strong interaction; the $\mu^-$ is a lepton and does not.	A pion is a quark-antiquark meson, so it is a hadron and participates in the strong interaction. A muon is a lepton, and leptons do not experience the strong interaction. Equal electric charge does not determine the particle class.	3
03.1	a hadron; a baryon; a strange particle; it decays by the weak interaction	Experiencing the strong interaction makes $\Lambda^0$ a hadron rather than a lepton. Baryon number $+1$ identifies it as a baryon rather than a meson. Its non-zero strangeness makes it a strange particle, and strange particles decay by the weak interaction.	3
04.1	neutrons have no electric charge; a virtual pion; energy and momentum are transferred	A neutron has no net electric charge, so the attraction is not an electrostatic interaction mediated by a virtual photon. At this scale the attraction is the strong nuclear interaction, modelled by the exchange of a virtual pion between the neutrons. The exchanged pion transfers energy and momentum.	3

Tier 3 · Hard

Mark scheme for 3.2.1.5 Tier 3 · Hard
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01.1	$\Lambda^0$ is a strange baryon and $K^+$ a strange meson; total $B$ remains $2$ , total charge remains $+2$ , and the produced strangeness sums to $0$ .	The initial baryon number is $1+1=2$ . The final value is $1+1+0=2$ , so $\Lambda^0$ is classified as a baryon while $K^+$ is a meson. Charge is also conserved: $+2=(+1)+0+(+1)$ . Initial strangeness is $0$ , and the new particles have $-1$ and $+1$ , giving total strangeness $0$ . Their opposite strangeness therefore demonstrates paired production in a strong interaction.	5
02.1	The kaon and pions are mesons; the pions have zero strangeness, so strangeness changes from $+1$ to $0$ , giving $\Delta S=-1$ . A strong interaction would conserve strangeness, whereas this change is permitted in a weak interaction.	Kaons and pions are mesons. Because the pions contain no strange or antistrange quark, each has strangeness zero, so the total changes from $+1$ to $0$ : $\Delta S=-1$ . Strangeness is conserved in a strong interaction but may change by $0$ or $\pm1$ in a weak interaction. The decay is therefore weak.	4
03.1	$X$ and $p$ are baryons; the pions are mesons. For $X\rightarrow p+\pi^-$ , $Q:0=+1-1$ and $B:1=1+0$ . For $p\rightarrow\pi^++\pi^0$ , charge is conserved but baryon number would change from $1$ to $0$ , so it is forbidden.	A hadron with baryon number $+1$ is a baryon, so both $X$ and the proton are baryons. Pions are mesons and have baryon number zero. In the first decay, charge is $0=(+1)+(-1)$ and baryon number is $1=1+0$ , so these laws allow it if energy, momentum and any other relevant quantum numbers also permit it. In the proposed proton decay, charge is $+1=(+1)+0$ , but baryon number would be $1=0+0$ , which is false. Conservation of baryon number therefore forbids the proton from decaying only into mesons, consistent with the proton being the stable baryon.	5
04.1	$X$ is a meson and could be a pion; $Y$ is an antibaryon and could be an antiproton; $Z$ is a lepton and could be a muon. $X$ and $Y$ are hadrons, whereas $Z$ is not.	A particle that experiences the strong interaction is a hadron. A hadron with baryon number $0$ is a meson, so $X$ can be a pion. A hadron with baryon number $-1$ is an antibaryon, so $Y$ can be an antiproton. Particle $Z$ does not experience the strong interaction and carries muon-family lepton number, so it is a lepton and can be a muon. Leptons are not hadrons.	4
05.1	$X$ has $(Q,B,S)=(0,+1,0)$ , so $X$ is a neutron.	The initial totals are $Q=-1+1=0$ , $B=1+1=2$ and $S=-1+0=-1$ . Subtracting the $\Lambda^0$ values gives $Q_X=0-0=0$ , $B_X=2-1=1$ and $S_X=-1-(-1)=0$ . A particle with charge $0$ , baryon number $+1$ and strangeness $0$ is a neutron, so $X=n$ .	4

3.2.1.6 · Quarks and antiquarks

Tier 1 · Easy

Mark scheme for 3.2.1.6 Tier 1 · Easy
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01.1	$uud$ ; $+\frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e=+e$	A proton is $uud$ . Adding the quark charges gives $+\frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e=+e$ , as required.	2
02.1	charge $-\frac{2}{3}e$ ; baryon number $-\frac{1}{3}$	An up quark has charge $+\frac{2}{3}e$ and baryon number $+\frac{1}{3}$ . The corresponding antiquark has both quantum numbers reversed, giving $-\frac{2}{3}e$ and $-\frac{1}{3}$ .	2

Tier 2 · Standard

Mark scheme for 3.2.1.6 Tier 2 · Standard
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01.1	charge $+e$ , baryon number $0$ , strangeness $+1$ ; $K^+$	The charge is $+\frac{2}{3}e+\frac{1}{3}e=+e$ . Its baryon number is $+\frac{1}{3}-\frac{1}{3}=0$ . The antistrange quark gives strangeness $+1$ . A meson with composition $u\bar{s}$ is therefore $K^+$ .	3
02.1	$d\bar{u}$ ; baryon number $0$ ; strangeness $0$	The composition $d\bar{u}$ has charge $-\frac{1}{3}e-\frac{2}{3}e=-e$ , so it is $\pi^-$ . Its baryon number is $+\frac{1}{3}-\frac{1}{3}=0$ . Neither constituent is strange, so its strangeness is $0$ .	3
03.1	an anti-strange quark; $d\bar{s}$ ; $K^0$	A particle with strangeness $+1$ contains an anti-strange quark, $\bar{s}$ . Combining it with the stated down quark gives $d\bar{s}$ . Their charges $-\frac13e$ and $+\frac13e$ sum to zero, and this neutral strange meson is $K^0$ .	3

Tier 3 · Hard

Mark scheme for 3.2.1.6 Tier 3 · Hard
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01.1	$d\rightarrow u+W^-$ , changing $udd$ to $uud$ , followed by $W^-\rightarrow e^-+\bar{\nu}_e$	A neutron is $udd$ and a proton is $uud$ , so one down quark changes into an up quark: $d\rightarrow u+W^-$ . Charge at this vertex is $-\frac{1}{3}e=+\frac{2}{3}e-e$ . The remaining quarks are spectators, so $udd$ becomes $uud$ . The second vertex is $W^-\rightarrow e^-+\bar{\nu}_e$ , with charge $-e=-e+0$ . The full decay is therefore $n\rightarrow p+e^-+\bar{\nu}_e$ .	5
02.1	$\bar{u}\bar{d}\bar{d}$ ; baryon number $-1$ . It contains antiquarks and has the opposite baryon number to the neutron $udd$ , whose baryon number is $+1$ .	A neutron is $udd$ , so replacing each quark by its antiquark gives $\bar{u}\bar{d}\bar{d}$ . Its charge is $-\frac{2}{3}e+\frac{1}{3}e+\frac{1}{3}e=0$ . Three antiquarks give baryon number $3(-\frac{1}{3})=-1$ , whereas the neutron has baryon number $+1$ . Their opposite baryon numbers and quark-antiquark content distinguish them despite their shared zero charge.	4
03.1	$p+n$ contains $3u+3d$ ; $n+n+\pi^+$ contains $3u+4d+\bar d$ , so the extra constituents are $d\bar d$ . Charge is $+e$ and baryon number is $2$ on both sides.	Initially $p+n=uud+udd$ , giving $3u+3d$ . The two final neutrons contain $2u+4d$ , and $\pi^+=u\bar d$ , so all products contain $3u+4d+\bar d$ . Relative to the initial particles, the added constituents are therefore a $d\bar d$ pair. Initial charge is $+e+0=+e$ and final charge is $0+0+e=+e$ . Initial baryon number is $1+1=2$ ; the two neutrons contribute $2$ and the meson contributes $\frac13-\frac13=0$ , so baryon number is also conserved.	5
04.1	$X=s\bar u$ , with charge $-e$ and strangeness $-1$ ; its antiparticle is $u\bar s$ and has strangeness $+1$ .	Strangeness $-1$ requires a strange quark $s$ , whose charge is $-\frac13e$ . A meson also contains an antiquark. To give total charge $-e$ , that antiquark must have charge $-\frac23e$ , identifying $\bar u$ . Hence $X=s\bar u$ , with $Q=-\frac13e-\frac23e=-e$ and $S=-1+0=-1$ . Replacing both constituents by their antiparticles gives $u\bar s$ ; the antistrange quark makes its strangeness $+1$ .	4
05.1	two up quarks, no down quark and one strange quark; $uus$	Baryon number $+1$ requires three quarks. Strangeness $-1$ requires exactly one strange quark, so let the remaining two constituents include $x$ up quarks and $y$ down quarks, with $x+y=2$ . In units of $e/3$ , the charge condition is $2x-y-1=3$ . Combining $2x-y=4$ with $x+y=2$ gives $x=2$ and $y=0$ . The composition is therefore $uus$ . Its charge is $2(\frac23e)-\frac13e=+e$ and its baryon number is $3(\frac13)=+1$ .	5

3.2.1.7 · Applications of conservation laws

Tier 1 · Easy

Mark scheme for 3.2.1.7 Tier 1 · Easy
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01.1	charge: $+2e=+2e$ ; baryon number: $2=2$ ; lepton number: $0=0$	Initially the charge is $+e+e=+2e$ ; finally it is $+e+0+e=+2e$ . Initially $B=1+1=2$ ; finally $B=1+1+0=2$ . No leptons occur on either side, so lepton number is $0$ before and after.	2
02.1	Any three from charge, baryon number, electron lepton number and muon lepton number.	Charge and baryon number are conserved. Electron and muon lepton numbers are also conserved separately, so either lepton number may supply the third stated quantity.	3

Tier 2 · Standard

Mark scheme for 3.2.1.7 Tier 2 · Standard
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01.1	$X=\bar{\nu}_e$	Charge is conserved because $0=(+1)+(-1)+0$ , so $X$ is neutral. Baryon number is $1=1+0+0$ , so $X$ is not a baryon. Electron lepton number initially is $0$ but the electron contributes $+1$ , so $X$ must contribute $-1$ . The neutral particle with $L_e=-1$ is the electron antineutrino, $X=\bar{\nu}_e$ .	4
02.1	$X=\pi^-$	The initial charge is $+e-e=0$ and the initial baryon number is $+1-1=0$ . The $\pi^+$ has charge $+e$ and baryon number $0$ , so $X$ must have charge $-e$ and baryon number $0$ . The specified pion is therefore $\pi^-$ . Lepton number remains zero throughout.	4
03.1	$u\rightarrow d$ ; charge $+1=0+1+0$ , baryon number $1=1+0+0$ , and electron lepton number $0=0-1+1$	Changing $p=uud$ into $n=udd$ requires one up quark to change into a down quark. Charge is conserved because $+1=0+(+1)+0$ . Baryon number is conserved because $1=1+0+0$ . The positron has electron lepton number $-1$ and the electron neutrino has $+1$ , so the final total is $0-1+1=0$ , equal to the initial value.	4

Tier 3 · Hard

Mark scheme for 3.2.1.7 Tier 3 · Hard
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01.1	$\nu_e+n\rightarrow p+e^-$ can occur if energy and momentum permit; the antineutrino reaction violates electron lepton-number conservation	For both proposals, charge is $0+0=+1-1=0$ and baryon number is $0+1=1+0=1$ . For $\nu_e+n\rightarrow p+e^-$ , electron lepton number is $+1+0=0+(+1)$ , so it is conserved. For $\bar{\nu}_e+n\rightarrow p+e^-$ , it would be $-1+0$ initially but $0+(+1)$ finally, so it changes from $-1$ to $+1$ and the reaction is forbidden. The first reaction is allowed by these quantum-number tests, but it can occur only when the initial energy is sufficient and total momentum can also be conserved.	6
02.1	Charge is $0$ on both sides, baryon number is $1$ on both sides and strangeness is $-1$ on both sides. The reaction is allowed by these conservation laws if energy and momentum also permit it.	Initially $Q=-1+1=0$ , $B=0+1=1$ and $S=-1+0=-1$ . Finally $Q=-1+1+0=0$ , $B=1+0+0=1$ and $S=-3+1+1=-1$ . Charge, baryon number and strangeness are therefore all conserved. Because the interaction is proposed to be strong, the conserved strangeness is consistent with it; energy and momentum must still be feasible.	4
03.1	$X=\nu_\mu$ ; weak interaction; total kinetic energy $=104\,\text{MeV}$	Initially the charge is $-1+1=0$ and finally it is $0+0=0$ . Baryon number is $0+1=1$ initially and $1+0=1$ finally. The muon has $L_\mu=+1$ , so $X$ must also have $L_\mu=+1$ and is $\nu_\mu$ , not $\bar\nu_\mu$ . The changes from muon to neutrino and proton to neutron identify the weak interaction. Energy conservation gives total product kinetic energy $106+938-940=104\,\text{MeV}$ , neglecting the neutrino rest energy.	6
04.1	Charge, electron lepton number and muon lepton number are all zero before and after; total product kinetic energy $=24\,\text{MeV}$ .	The initial charge is $-1+1=0$ and the final muon-antimuon charge is also $-1+1=0$ . Initially $L_e=+1-1=0$ and $L_\mu=0$ ; finally $L_e=0$ and $L_\mu=+1-1=0$ . Equal and opposite initial momenta also allow zero total momentum after the collision. The initial energy is $2(118)=236\,\text{MeV}$ . The final rest energy is $2(106)=212\,\text{MeV}$ , so energy conservation leaves $236-212=24\,\text{MeV}$ as total kinetic energy.	5
05.1	The decay is energetically allowed and releases $0.789\,\text{MeV}$ as total kinetic energy; electron-family lepton number is $0$ initially and $0+(+1)+(-1)=0$ finally.	The minimum product energy is the sum of the proton and electron rest energies, since the antineutrino rest energy is negligible: $938.27+0.511=938.781\,\text{MeV}$ . This is less than the neutron rest energy, so the decay is allowed. The energy available as total product kinetic energy is $939.57-938.781=0.789\,\text{MeV}$ . The neutron and proton have electron-family lepton number $0$ , the electron has $+1$ and the electron antineutrino has $-1$ . The total is therefore $0$ before and after the decay.	4

3.2.2.1 · The photoelectric effect

Tier 1 · Easy

Mark scheme for 3.2.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.8\times10^{14}\,\text{Hz}$	Convert the work function: $\phi=2.4\times1.60\times10^{-19}=3.84\times10^{-19}\,\text{J}$ . At threshold $hf_0=\phi$ , so $f_0=(3.84\times10^{-19})/(6.63\times10^{-34})=5.79\times10^{14}\,\text{Hz}$ , or $5.8\times10^{14}\,\text{Hz}$ .	2
02.1	Each photon still has energy below the work function; greater intensity supplies more such photons but does not raise their individual energy.	One electron absorbs energy from one photon. Below threshold, $hf<\phi$ , so an individual photon cannot release an electron. Increasing intensity increases the number of photons arriving per second, not the energy $hf$ of each photon.	2

Tier 2 · Standard

Mark scheme for 3.2.2.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.3\,\text{eV}$ ; $1.3\,\text{V}$	The photon energy in electronvolts is $hf/e=(6.63\times10^{-34})(8.2\times10^{14})/(1.60\times10^{-19})=3.40\,\text{eV}$ . Hence $E_{\text{k,max}}=3.40-2.1=1.30\,\text{eV}$ . Since one electronvolt corresponds to an electron crossing one volt, $eV_s=E_{\text{k,max}}$ gives $V_s=1.30\,\text{V}$ .	3
02.1	$2.30\,\text{eV}$	At threshold, $\phi=hc/\lambda$ . Thus $\phi=(6.63\times10^{-34})(3.00\times10^8)/(540\times10^{-9})=3.683\times10^{-19}\,\text{J}$ . Dividing by $1.60\times10^{-19}\,\text{J eV}^{-1}$ gives $2.302\,\text{eV}$ , or $2.30\,\text{eV}$ to three significant figures.	3
03.1	$6.7\times10^{-34}\,\text{J s}$	For the same metal, subtracting two photoelectric equations removes the work function: $h\Delta f=e\Delta V_s$ . Here $\Delta V_s=0.95-0.45=0.50\,\text{V}$ . Therefore $h=e\Delta V_s/\Delta f=(1.60\times10^{-19})(0.50)/(1.2\times10^{14})=6.67\times10^{-34}\,\text{J s}$ , or $6.7\times10^{-34}\,\text{J s}$ to two significant figures.	3

Tier 3 · Hard

Mark scheme for 3.2.2.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.66\,\text{V}$ ; emission rate doubles and stopping potential is unchanged	The photon energy is $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(420\times10^{-9})=4.74\times10^{-19}\,\text{J}=2.96\,\text{eV}$ . Thus $E_{\text{k,max}}=2.96-2.30=0.66\,\text{eV}$ and $V_s=0.66\,\text{V}$ . Doubling intensity doubles the photon arrival rate, so the photoelectron emission rate doubles. Each photon still has the same energy, so maximum kinetic energy and stopping potential are unchanged.	5
02.1	$6.64\times10^5\,\text{m s}^{-1}$	The photon energy is $hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(365\times10^{-9})=5.449\times10^{-19}\,\text{J}$ . The work function is $2.15(1.60\times10^{-19})=3.44\times10^{-19}\,\text{J}$ , so $E_{\text{k,max}}=2.009\times10^{-19}\,\text{J}$ . From $E_k=\frac12mv^2$ , $v=\sqrt{2E_k/m}=\sqrt{2(2.009\times10^{-19})/(9.11\times10^{-31})}=6.64\times10^5\,\text{m s}^{-1}$ to three significant figures.	5
03.1	$7.3\times10^{14}\,\text{Hz}$ ; $v_A/v_B=1.8$ (accept $1.79$ )	Metal A has work function $\phi_A=hf_{0A}=(6.63\times10^{-34})(6.0\times10^{14})=3.978\times10^{-19}\,\text{J}$ . For the same photon energy, $\phi_B-\phi_A=e(V_{sA}-V_{sB})=(1.60\times10^{-19})(0.80-0.25)=8.80\times10^{-20}\,\text{J}$ . Thus $\phi_B=4.858\times10^{-19}\,\text{J}$ and $f_{0B}=\phi_B/h=7.33\times10^{14}\,\text{Hz}$ , giving $7.3\times10^{14}\,\text{Hz}$ to two significant figures. Since $\frac12mv^2=eV_s$ , $v_A/v_B=\sqrt{0.80/0.25}=1.79$ , which is $1.8$ to two significant figures.	5
04.1	$h=6.66\times10^{-34}\,\text{J s}$ ; $\phi=2.57\,\text{eV}$ ; $f_0=6.19\times10^{14}\,\text{Hz}$	For each reading, $eV_s=hf-\phi$ . Subtracting gives $h=e\Delta V_s/\Delta f=(1.60\times10^{-19})(1.42-0.38)/[(9.6-7.1)\times10^{14}]=6.656\times10^{-34}\,\text{J s}$ . Using the first point, $\phi=hf-eV_s=(6.656\times10^{-34})(7.1\times10^{14})-(1.60\times10^{-19})(0.38)=4.1178\times10^{-19}\,\text{J}$ . This is $4.1178/1.60=2.5736\,\text{eV}$ . Finally $f_0=\phi/h=(4.1178\times10^{-19})/(6.656\times10^{-34})=6.1865\times10^{14}\,\text{Hz}$ .	4
05.1	$1.36\,\text{V}$ ; $1.26\times10^{-4}\,\text{A}$	One photon has energy $E=hc/\lambda=(1.989\times10^{-25})/(310\times10^{-9})=6.4161\times10^{-19}\,\text{J}=4.010\,\text{eV}$ . Thus $E_{\text{k,max}}=4.010-2.65=1.360\,\text{eV}$ and $V_s=1.36\,\text{V}$ . The incident photon rate is $P/E=(2.8\times10^{-3})/(6.4161\times10^{-19})=4.3640\times10^{15}\,\text{s}^{-1}$ . The emitted-electron rate is $0.18(4.3640\times10^{15})=7.8552\times10^{14}\,\text{s}^{-1}$ , so $I=Ne=(7.8552\times10^{14})(1.60\times10^{-19})=1.2568\times10^{-4}\,\text{A}$ .	6

3.2.2.2 · Collisions of electrons with atoms

Tier 1 · Easy

Mark scheme for 3.2.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.9\times10^{-18}\,\text{J}$	$18\,\text{eV}=18(1.60\times10^{-19})=2.88\times10^{-18}\,\text{J}$ , which is $2.9\times10^{-18}\,\text{J}$ to two significant figures.	1
02.1	$12\,\text{eV}$	An electron accelerated through a potential difference of $V$ volts gains $V$ electronvolts, so it gains $12\,\text{eV}$ .	1

Tier 2 · Standard

Mark scheme for 3.2.2.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10.5\,\text{eV}$ excitation; $1.5\,\text{eV}$ remaining	The electron cannot ionise the atom because $12.0<13.6\,\text{eV}$ . It can supply the larger available discrete excitation energy, $10.5\,\text{eV}$ . Conservation of energy leaves $12.0-10.5=1.5\,\text{eV}$ as the incident electron's kinetic energy.	3
02.1	Excitation moves an electron to a higher bound level using a discrete energy gap; ionisation supplies at least the ionisation energy and removes an electron from the atom.	During excitation an atomic electron remains bound but moves to a higher allowed level, so the transferred energy equals a discrete level difference. During ionisation the transferred energy reaches or exceeds the ionisation energy and an electron becomes free.	3
03.1	total kinetic energy $=4.4\,\text{eV}$ ; it can be shared between the two electrons in different proportions	Ionisation uses $13.6\,\text{eV}$ , so energy conservation leaves $18.0-13.6=4.4\,\text{eV}$ as the combined kinetic energy of the incident and released electrons. The information fixes only their total energy; without their momenta or scattering directions, the way that energy is divided between them is not unique.	3

Tier 3 · Hard

Mark scheme for 3.2.2.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$7.7\,\text{V}$ ; electron impact excites mercury, mercury emits ultraviolet photons on de-excitation, and the coating absorbs these and emits visible photons	An electron accelerated through $V$ volts gains $V\,\text{eV}$ , so the minimum potential for a $7.7\,\text{eV}$ excitation is $7.7\,\text{V}$ . In an inelastic collision the electron transfers this discrete energy to a mercury atom. When the atom returns to a lower level it emits an ultraviolet photon. The fluorescent coating absorbs that ultraviolet photon, becomes excited, and emits lower-energy visible photons as it de-excites.	5
02.1	First excitation $10.5\,\text{eV}$ ; second excitation $6.0\,\text{eV}$ ; final kinetic energy $3.5\,\text{eV}$ .	The first collision transfers the larger allowed energy, $10.5\,\text{eV}$ , leaving $20.0-10.5=9.5\,\text{eV}$ . The electron no longer has enough energy for another $10.5\,\text{eV}$ excitation, but it can transfer $6.0\,\text{eV}$ to the second atom. Its final kinetic energy is $9.5-6.0=3.5\,\text{eV}$ . Only the discrete allowed energies can be transferred.	5
03.1	$13.1\,\text{V}$	The electron's final kinetic energy is $\frac12mv^2=\frac12(9.11\times10^{-31})(1.50\times10^6)^2=1.025\times10^{-18}\,\text{J}$ . This is $(1.025\times10^{-18})/(1.60\times10^{-19})=6.41\,\text{eV}$ . Before the collision it therefore had $6.41+6.7=13.11\,\text{eV}$ . An electron accelerated from rest through $V$ volts gains $V\,\text{eV}$ , so $V=13.1\,\text{V}$ to three significant figures.	4
04.1	$16.3\,\text{eV}$	The incident electron gains $28.0\,\text{eV}$ . The two final kinetic energies are $\frac12mv_1^2/e=\frac12(9.11\times10^{-31})(1.70\times10^6)^2/(1.60\times10^{-19})=8.2275\,\text{eV}$ and $\frac12mv_2^2/e=3.4447\,\text{eV}$ . Energy conservation gives the ionisation energy as $28.0-8.2275-3.4447=16.3278\,\text{eV}$ , which is $16.3\,\text{eV}$ to three significant figures.	5
05.1	$621\,\text{nm}$ (accept $620$ – $621\,\text{nm}$ ); the ultraviolet photon excites an electron, which returns through an intermediate level in two downward transitions that emit the $430\,\text{nm}$ and $621\,\text{nm}$ photons.	Photon-energy conservation gives $hc/(254\,\text{nm})=hc/(430\,\text{nm})+hc/\lambda$ . Cancelling $hc$ gives $1/\lambda=1/254-1/430$ in inverse nanometres. Therefore $\lambda=1/(1/254-1/430)=620.568\,\text{nm}$ , or $621\,\text{nm}$ to three significant figures. Absorption of the ultraviolet photon excites an electron in the coating to a higher allowed level. The electron returns through an intermediate level, emitting the $430\,\text{nm}$ photon in one downward transition and the $621\,\text{nm}$ photon in the other.	5

3.2.2.3 · Energy levels and photon emission

Tier 1 · Easy

Mark scheme for 3.2.2.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.9\,\text{eV}$	The emitted energy is the positive level difference: $\Delta E=(-1.5)-(-3.4)=1.9\,\text{eV}$ .	1
02.1	Electrons occupy discrete energy levels, so only photons with particular energy differences are emitted.	Each line is produced by a transition between two allowed energy levels. Because the levels are discrete, only particular energy gaps and photon frequencies occur.	2

Tier 2 · Standard

Mark scheme for 3.2.2.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.58\times10^{-7}\,\text{m}$	The energy gap is $(-1.51)-(-3.40)=1.89\,\text{eV}=1.89(1.60\times10^{-19})=3.024\times10^{-19}\,\text{J}$ . Therefore $\lambda=hc/\Delta E=(6.63\times10^{-34})(3.00\times10^8)/(3.024\times10^{-19})=6.58\times10^{-7}\,\text{m}$ .	3
02.1	$4.98\times10^{14}\,\text{Hz}$ ; $2.01\times10^{-15}\,\text{s}$	Use $\Delta E=hf$ . Therefore $f=(3.30\times10^{-19})/(6.63\times10^{-34})=4.977\times10^{14}\,\text{Hz}$ , which is $4.98\times10^{14}\,\text{Hz}$ to three significant figures. The period is $T=1/f=2.009\times10^{-15}\,\text{s}$ , or $2.01\times10^{-15}\,\text{s}$ to three significant figures.	3
03.1	initial level $2.1\,\text{eV}$ ; final level $4.9\,\text{eV}$	A photon is absorbed only if its energy equals an available upward energy gap. From the ground state there is no level at $2.8\,\text{eV}$ , so a ground-state atom cannot absorb the photon. From the $2.1\,\text{eV}$ level, the gap is $4.9-2.1=2.8\,\text{eV}$ , so absorption raises the atom to $4.9\,\text{eV}$ .	3

Tier 3 · Hard

Mark scheme for 3.2.2.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6$ lines; $1.04\times10^{-6}\,\text{m}$	Four levels give one line for each pair of levels, so the distinct gaps are $1.2$ , $3.0$ , $5.0$ , $1.8$ , $3.8$ and $2.0\,\text{eV}$ : six lines. The longest wavelength comes from the smallest gap, $1.2\,\text{eV}=1.92\times10^{-19}\,\text{J}$ . Thus $\lambda_{\max}=hc/\Delta E=(6.63\times10^{-34})(3.00\times10^8)/(1.92\times10^{-19})=1.04\times10^{-6}\,\text{m}$ .	5
02.1	$266\,\text{nm}$	The direct transition energy equals the sum of the two cascade photon energies. Since $E=hc/\lambda$ , $1/\lambda=1/(450\,\text{nm})+1/(650\,\text{nm})$ . Thus $\lambda=(450\times650)/(450+650)=265.909\,\text{nm}$ , which is $266\,\text{nm}$ to three significant figures.	5
03.1	$6.61\times10^{-34}\,\text{J s}$ and $6.62\times10^{-34}\,\text{J s}$ ; percentage difference $=0.16\%$	From $\Delta E=hc/\lambda$ , $h=\Delta E\lambda/c$ . The first transition gives $h_1=(2.10)(1.60\times10^{-19})(590\times10^{-9})/(3.00\times10^8)=6.608\times10^{-34}\,\text{J s}$ . The second gives $h_2=(3.40)(1.60\times10^{-19})(365\times10^{-9})/(3.00\times10^8)=6.619\times10^{-34}\,\text{J s}$ . Using the unrounded values, the percentage difference relative to their mean is $0.161\%$ , or $0.16\%$ to two significant figures.	5
04.1	Six photon energies: $1.65$ , $2.25$ , $2.30$ , $3.90$ , $4.55$ and $6.20\,\text{eV}$ ; the longest wavelength is $752\,\text{nm}$ from the $1.65\,\text{eV}$ transition. Each photon energy is one allowed level difference, so no other energies are emitted.	With four levels, the six possible downward gaps are $1.65-0=1.65\,\text{eV}$ , $3.90-1.65=2.25\,\text{eV}$ , $6.20-3.90=2.30\,\text{eV}$ , $3.90-0=3.90\,\text{eV}$ , $6.20-1.65=4.55\,\text{eV}$ and $6.20-0=6.20\,\text{eV}$ . They are all distinct, so six photon energies occur. The longest wavelength corresponds to the smallest gap: $\lambda=hc/E=1240/1.65=751.5\,\text{nm}$ , or $752\,\text{nm}$ . Photon energies equal differences between allowed levels, so the set of six gaps exhausts the possible emissions.	5
05.1	intermediate level $-3.70\,\text{eV}$ ; upper level $-1.80\,\text{eV}$ ; wavelength $2.49\times10^{-7}\,\text{m}$	The intermediate energy is $-6.80+3.10=-3.70\,\text{eV}$ . The upper energy is $-3.70+1.90=-1.80\,\text{eV}$ . A direct return to the ground state releases $(-1.80)-(-6.80)=5.00\,\text{eV}=8.00\times10^{-19}\,\text{J}$ . Hence $\lambda=hc/\Delta E=(1.989\times10^{-25})/(8.00\times10^{-19})=2.486\times10^{-7}\,\text{m}$ , or $2.49\times10^{-7}\,\text{m}$ .	5

3.2.2.4 · Wave-particle duality

Tier 1 · Easy

Mark scheme for 3.2.2.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.2\times10^{-10}\,\text{m}$	Use $\lambda=h/p$ : $\lambda=(6.63\times10^{-34})/(3.0\times10^{-24})=2.21\times10^{-10}\,\text{m}$ , or $2.2\times10^{-10}\,\text{m}$ .	2
02.1	Electrons produce a diffraction pattern, showing that electrons have wave properties.	A beam of electrons forms a diffraction pattern after passing through or reflecting from a suitable crystal. Diffraction is characteristic wave behaviour.	2

Tier 2 · Standard

Mark scheme for 3.2.2.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.9\times10^{-10}\,\text{m}$ ; doubling speed halves the wavelength	The momentum is $p=mv=(9.11\times10^{-31})(2.5\times10^6)=2.28\times10^{-24}\,\text{kg m s}^{-1}$ . Hence $\lambda=h/p=(6.63\times10^{-34})/(2.28\times10^{-24})=2.91\times10^{-10}\,\text{m}$ . Since $\lambda=h/(mv)$ , doubling $v$ doubles momentum and halves the wavelength.	3
02.1	$3.68\times10^{-24}\,\text{kg m s}^{-1}$ ; $2.21\times10^3\,\text{m s}^{-1}$	Rearrange $\lambda=h/p$ to $p=h/\lambda$ . With $\lambda=0.180\,\text{nm}=1.80\times10^{-10}\,\text{m}$ , $p=(6.63\times10^{-34})/(1.80\times10^{-10})=3.683\times10^{-24}\,\text{kg m s}^{-1}$ . Using $p=mv$ , $v=(3.683\times10^{-24})/(1.67\times10^{-27})=2.205\times10^3\,\text{m s}^{-1}$ . To three significant figures, $p=3.68\times10^{-24}\,\text{kg m s}^{-1}$ and $v=2.21\times10^3\,\text{m s}^{-1}$ .	3
03.1	$p_{\text{final}}/p_{\text{initial}}=1.40$	For the unchanged apparatus, ring diameter $D$ is proportional to $\lambda$ , while $\lambda=h/p$ , so $D\propto1/p$ . Therefore $p_{\text{final}}/p_{\text{initial}}=D_{\text{initial}}/D_{\text{final}}=42/30=1.40$ .	3

Tier 3 · Hard

Mark scheme for 3.2.2.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\lambda=\dfrac{h}{\sqrt{2meV}}=1.00\times10^{-10}\,\text{m}$ ; increasing $V$ reduces the wavelength and the diffraction angle	Energy conservation gives $eV=\frac{1}{2}mv^2$ . Since $p=mv$ , squaring gives $p^2=m^2v^2=2meV$ , so $p=\sqrt{2meV}$ and $\lambda=h/\sqrt{2meV}$ . Substitution gives $\lambda=(6.63\times10^{-34})/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(150)}=1.00\times10^{-10}\,\text{m}$ . Increasing $V$ increases momentum as $\sqrt{V}$ , so $\lambda$ decreases as $1/\sqrt{V}$ and the diffraction angle or ring diameter decreases.	5
02.1	$\lambda_e/\lambda_p=42.8$ ; the electron produces the larger diffraction angle	For equal kinetic energy $E_k$ , $p=\sqrt{2mE_k}$ and $\lambda=h/p$ , so $\lambda\propto1/\sqrt{m}$ . Therefore $\lambda_e/\lambda_p=\sqrt{m_p/m_e}=\sqrt{(1.67\times10^{-27})/(9.11\times10^{-31})}=42.8$ to three significant figures. The electron has the longer wavelength, so it produces the larger diffraction angle for the same crystal spacing.	4
03.1	Diffraction shows wave properties but does not show a particle changing into a wave; the de Broglie model assigns wave properties to moving particles. Repeated quantitative tests over different momenta, uncertainty analysis, peer review and independent replication are needed.	An electron diffraction pattern is evidence that electrons display wave behaviour, but it does not show that an electron switches from being a particle into a wave inside the crystal. The de Broglie model instead predicts an associated wavelength for a moving particle and must make quantitative predictions such as $\lambda\propto1/p$ . One observation cannot exclude apparatus effects or establish a general law. Measurements should be repeated across a range of momenta with uncertainties assessed, then methods and conclusions should undergo peer review and independent replication before the scientific community accepts a revised model.	4
04.1	$p=7.4\times10^{-24}\,\text{kg m s}^{-1}$ ; $v_p=4.4\times10^3\,\text{m s}^{-1}$ and $v_\alpha=1.1\times10^3\,\text{m s}^{-1}$ ; $E_{k,p}=0.10\,\text{eV}$ and $E_{k,\alpha}=0.025\,\text{eV}$	The common momentum is $p=h/\lambda=(6.63\times10^{-34})/(0.090\times10^{-9})=7.3667\times10^{-24}\,\text{kg m s}^{-1}$ . Using $v=p/m$ gives $v_p=(7.3667\times10^{-24})/(1.67\times10^{-27})=4.411\times10^3\,\text{m s}^{-1}$ and $v_\alpha=(7.3667\times10^{-24})/(6.68\times10^{-27})=1.103\times10^3\,\text{m s}^{-1}$ . With $E_k=p^2/(2m)$ , the energies are $0.10155\,\text{eV}$ for the proton and $0.025387\,\text{eV}$ for the alpha particle. To two significant figures, the results are $7.4\times10^{-24}\,\text{kg m s}^{-1}$ , $4.4\times10^3\,\text{m s}^{-1}$ , $1.1\times10^3\,\text{m s}^{-1}$ , $0.10\,\text{eV}$ and $0.025\,\text{eV}$ .	6
05.1	$h=(6.72\pm0.17)\times10^{-34}\,\text{J s}$ with percentage uncertainty $2.5\%$ ; it is consistent with $6.63\times10^{-34}\,\text{J s}$	The experimental value is $h=(7.50\times10^{-11})\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(275)}=6.7152\times10^{-34}\,\text{J s}$ . Since $h\propto\lambda V^{1/2}$ , the percentage uncertainty is $100[0.15/7.50+\frac12(3/275)]=2.545\%$ , or $2.5\%$ . The absolute uncertainty is $0.02545(6.7152\times10^{-34})=0.1709\times10^{-34}\,\text{J s}$ , giving $(6.72\pm0.17)\times10^{-34}\,\text{J s}$ . Its interval, approximately $6.55$ to $6.89\times10^{-34}\,\text{J s}$ , includes $6.63\times10^{-34}\,\text{J s}$ , so the values are consistent.	6