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AQA A-level Physics revision notes

Particles and radiation

Section 3.2
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
11 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.2

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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In the exam: Data and formulae booklet provided · calculator allowed in every paper

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3.2.1.1

Constituents of the atom

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The simple atom contains protons and neutrons in the nucleus with electrons outside it. A proton has charge +e+e and relative mass 11, a neutron has zero charge and relative mass 11, and an electron has charge e-e and relative mass about 1/18361/1836; SI masses and charges may also be required.
  • In ZAX{}^{A}_{Z}X, ZZ is proton number, AA is nucleon number and neutron number is AZA-Z.
  • Isotopes share ZZ but have different neutron numbers.
  • Specific charge is net charge divided by total mass in C kg1\text{C kg}^{-1}.
  • Examiners expect ion charge, nuclear composition and sign to be handled separately.
Worked example

An ion contains 1212 protons, 1212 neutrons and 1010 electrons. Find its nuclide notation, charge and approximate specific charge using nucleon mass 1.67×1027kg1.67\times10^{-27}\,\text{kg}.

  1. 1.A=24A=24, Z=12Z=12, so the nuclide is 1224Mg{}^{24}_{12}\text{Mg}.
  2. 2.Net charge is +2e=+3.20×1019C+2e=+3.20\times10^{-19}\,\text{C}.
  3. 3.Mass 24(1.67×1027)\approx24(1.67\times10^{-27}), so q/m=7.98×106C kg1q/m=7.98\times10^6\,\text{C kg}^{-1}.

Answer: 1224Mg2+{}^{24}_{12}\text{Mg}^{2+} with specific charge +8.0×106C kg1+8.0\times10^6\,\text{C kg}^{-1}.

Common mistakes

  • Don't use electron number as proton number ZZ 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 AA, ZZ or net charge.

Tier 1 · Easy

ORIGINAL

State the numbers of protons, neutrons and electrons in a 1225Mg2+{}^{25}_{12}\text{Mg}^{2+} ion.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A helium nucleus has charge +3.20×1019C+3.20\times10^{-19}\,\text{C} and mass 6.64×1027kg6.64\times10^{-27}\,\text{kg}. Calculate its specific charge.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An ion of an isotope with A=40A=40 and Z=20Z=20 has specific charge +7.19×106C kg1+7.19\times10^6\,\text{C kg}^{-1}. Use nucleon mass 1.67×1027kg1.67\times10^{-27}\,\text{kg} and e=1.60×1019Ce=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

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3.2.1.2

Stable and unstable nuclei

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The strong nuclear force stabilises nuclei: it is attractive over separations up to about 3fm3\,\text{fm} but repulsive below about 0.5fm0.5\,\text{fm}, preventing collapse. Unstable nuclei undergo alpha or beta decay.
  • Alpha emission lowers nucleon number by 44 and proton number by 22.
  • In β\beta^- decay a neutron becomes a proton, so AA is unchanged and ZZ rises by 11; 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 AA and ZZ, the correct neutrino and the strong-force range stated with femtometre scale.
Worked example

Complete 614C714N+?{}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+\,?.

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

Answer: 10e+νˉe{}^{0}_{-1}e+\bar{\nu}_e.

Common mistakes

  • Don't change nucleon number during β\beta^- decay.
  • Don't omit the electron antineutrino from a β\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

ORIGINAL

Complete the alpha-decay equation 84210Po82206Pb+?{}^{210}_{84}\text{Po}\rightarrow{}^{206}_{82}\text{Pb}+\,?.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Write the complete equation for the β\beta^- decay of 614C{}^{14}_{6}\text{C} and state how the proton number changes.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

3.2.1.3

Particles, antiparticles and photons

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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/λE=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 0.511MeV0.511\,\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 500nm500\,\text{nm}.

  1. 1.Convert λ=5.00×107m\lambda=5.00\times10^{-7}\,\text{m}.
  2. 2.Use E=hc/λE=hc/\lambda.
  3. 3.E=(6.63×1034)(3.00×108)/(5.00×107)E=(6.63\times10^{-34})(3.00\times10^8)/(5.00\times10^{-7}).

Answer: E=3.98×1019JE=3.98\times10^{-19}\,\text{J}.

Common mistakes

  • Don't assign 1.022MeV1.022\,\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

ORIGINAL

Calculate the energy of a photon of frequency 6.0×1014Hz6.0\times10^{14}\,\text{Hz}. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A slow electron and a slow positron annihilate to produce two identical photons. Each electron has rest energy 0.511MeV0.511\,\text{MeV}. Calculate the wavelength of either photon. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1} and 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A photon of wavelength 3.00×1013m3.00\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 MeV\text{MeV} and explain why the nucleus is required. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1}, 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J} and electron rest energy 0.511MeV0.511\,\text{MeV}.

[5 marks]

Total for this question: 5

3.2.1.4

Particle interactions

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 β\beta^- and β+\beta^+ decay, electron capture and electron-proton collisions, using W+W^+ or WW^- bosons; gluons, Z0Z^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 β\beta^- decay, np+Wn\rightarrow p+W^- followed by We+νˉeW^-\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 β\beta^- decay, then give the two vertices.

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

Answer: Weak interaction via a WW^- 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+W^+ or WW^-.

Exam tip

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

Tier 1 · Easy

ORIGINAL

State the fundamental interaction and exchange particle involved when a neutron undergoes β\beta^- decay.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Complete the weak-interaction equation e+pn+?e^-+p\rightarrow n+\,? and identify the exchange particle involved.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Describe β+\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

3.2.1.5

Classification of particles

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Hadrons experience the strong interaction. Baryons and antibaryons have baryon number +1+1 and 1-1; 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 00, +1+1 or 1-1.
  • Examiners expect classification plus relevant quantum numbers.
Worked example

Classify a proton, kaon and muon.

  1. 1.A proton is a three-quark hadron, so it is a baryon.
  2. 2.A kaon is a quark–antiquark hadron, so it is a meson.
  3. 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

ORIGINAL

Classify the proton, pion and muon.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Complete the decay μe+?+?\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

Tier 3 · Hard

ORIGINAL

In the strong interaction p+pp+Λ0+K+p+p\rightarrow p+\Lambda^0+K^+, the supplied data are B(Λ0)=1B(\Lambda^0)=1, S(Λ0)=1S(\Lambda^0)=-1, B(K+)=0B(K^+)=0 and S(K+)=+1S(K^+)=+1. Explain how the products illustrate the classification and paired production of strange particles.

[5 marks]

Total for this question: 5

3.2.1.6

Quarks and antiquarks

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Only up, down and strange quarks and their antiquarks are required. Their charges are +23e+\tfrac23e, 13e-\tfrac13e and 13e-\tfrac13e; antiquarks have opposite charge and quantum numbers.
  • Each quark has baryon number +13+\tfrac13, each antiquark 13-\tfrac13; strange quark has strangeness 1-1 and antistrange +1+1. Required baryons are p=uudp=uud and n=uddn=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 usˉu\bar{s} has the quantum numbers of a K+K^+ meson.

  1. 1.Charge is +23e+13e=+e+\tfrac23e+\tfrac13e=+e.
  2. 2.Baryon number is +1313=0+\tfrac13-\tfrac13=0.
  3. 3.The antistrange quark gives strangeness +1+1.

Answer: usˉu\bar{s} has charge +e+e, baryon number 00 and strangeness +1+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+1 to the strange quark.

Exam tip

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

Tier 1 · Easy

ORIGINAL

State the quark composition of a proton and show that it has charge +e+e.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A meson has quark composition usˉu\bar{s}. Determine its charge, baryon number and strangeness, and identify the meson.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Describe neutron β\beta^- decay in terms of a quark change and a WW boson. Verify charge conservation at both vertices.

[5 marks]

Total for this question: 5

3.2.1.7

Applications of conservation laws

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 00, +1+1 or 1-1 in weak interactions.
  • In β\beta^- decay a down quark changes to an up quark; in β+\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+pp+n+π+p+p\rightarrow p+n+\pi^+.

  1. 1.Charge: initial +2e+2e; final +e+0+e=+2e+e+0+e=+2e.
  2. 2.Baryon number: initial 22; final 1+1+0=21+1+0=2.
  3. 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

ORIGINAL

Show that charge, baryon number and lepton number are conserved in p+pp+n+π+p+p\rightarrow p+n+\pi^+.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Use conservation of charge, baryon number and electron lepton number to determine XX in np+e+Xn\rightarrow p+e^-+X.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Two proposed reactions are νe+np+e\nu_e+n\rightarrow p+e^- and νˉe+np+e\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

3.2.2.1

The photoelectric effect

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In the photon model, one photon transfers all its energy hfhf to one surface electron.
  • Emission occurs only when hfϕhf\geq\phi, where ϕ\phi is work function, so threshold frequency is f0=ϕ/hf_0=\phi/h.
  • Above threshold, Ek,max=hfϕE_{k,\max}=hf-\phi and stopping potential satisfies eVs=Ek,maxeV_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 2.0eV2.0\,\text{eV} and receives 3.5eV3.5\,\text{eV} photons. Find maximum kinetic energy and stopping potential.

  1. 1.Ek,max=3.52.0=1.5eVE_{k,\max}=3.5-2.0=1.5\,\text{eV}.
  2. 2.Use eVs=Ek,maxeV_s=E_{k,\max}.
  3. 3.An electron energy of 1.5eV1.5\,\text{eV} corresponds to Vs=1.5VV_s=1.5\,\text{V}.

Answer: Ek,max=1.5eVE_{k,\max}=1.5\,\text{eV} and Vs=1.5VV_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 ff exceeds f0f_0; only then calculate kinetic energy.

Tier 1 · Easy

ORIGINAL

A metal has work function 2.4eV2.4\,\text{eV}. Calculate its threshold frequency. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s} and 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Radiation of frequency 8.2×1014Hz8.2\times10^{14}\,\text{Hz} illuminates a metal of work function 2.1eV2.1\,\text{eV}. Calculate the maximum kinetic energy in eV\text{eV} and the stopping potential. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s} and e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Light of wavelength 420nm420\,\text{nm} illuminates a metal with work function 2.30eV2.30\,\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=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1} and e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}.

[5 marks]

Total for this question: 5

3.2.2.2

Collisions of electrons with atoms

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 VV gains eVeV, numerically VeVV\,\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 12.0eV12.0\,\text{eV} excites an atom by 8.5eV8.5\,\text{eV}. Find its remaining kinetic energy in eV and J.

  1. 1.Subtract the allowed excitation energy: 12.08.5=3.5eV12.0-8.5=3.5\,\text{eV}.
  2. 2.Convert using 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.
  3. 3.E=3.5(1.60×1019)=5.6×1019JE=3.5(1.60\times10^{-19})=5.6\times10^{-19}\,\text{J}.

Answer: 3.5eV=5.6×1019J3.5\,\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

ORIGINAL

Convert an electron energy of 18eV18\,\text{eV} into joules. Use 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

An electron with kinetic energy 12.0eV12.0\,\text{eV} collides with an atom in its ground state. Excited states are 4.0eV4.0\,\text{eV} and 10.5eV10.5\,\text{eV} above the ground state, and the ionisation energy is 13.6eV13.6\,\text{eV}. Determine the greatest possible excitation and the electron kinetic energy immediately afterwards.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Mercury atoms in a fluorescent tube have an excitation energy of 7.7eV7.7\,\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

3.2.2.3

Energy levels and photon emission

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 E1E_1 to lower E2E_2, hf=E1E2hf=E_1-E_2; the positive energy gap may also be used in λ=hc/ΔE\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 hh, 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 1.5eV-1.5\,\text{eV} to 3.4eV-3.4\,\text{eV}. Find the emitted photon energy and wavelength.

  1. 1.ΔE=(1.5)(3.4)=1.9eV\Delta E=(-1.5)-(-3.4)=1.9\,\text{eV}.
  2. 2.Convert ΔE=1.9(1.60×1019)=3.04×1019J\Delta E=1.9(1.60\times10^{-19})=3.04\times10^{-19}\,\text{J}.
  3. 3.λ=hc/ΔE=6.54×107m\lambda=hc/\Delta E=6.54\times10^{-7}\,\text{m}.

Answer: 1.9eV1.9\,\text{eV} and 6.5×107m6.5\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 hh 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

ORIGINAL

An electron falls from an energy level at 1.5eV-1.5\,\text{eV} to one at 3.4eV-3.4\,\text{eV}. State the photon energy.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

Calculate the wavelength emitted when an electron falls from 1.51eV-1.51\,\text{eV} to 3.40eV-3.40\,\text{eV}. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1} and 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An atom has energy levels 00, 1.2-1.2, 3.0-3.0 and 5.0eV-5.0\,\text{eV}. Electrons are raised to the 0eV0\,\text{eV} level and can return by any sequence. Determine the number of distinct emission lines and calculate the longest wavelength. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1} and 1eV=1.60×1019J1\,\text{eV}=1.60\times10^{-19}\,\text{J}.

[5 marks]

Total for this question: 5

3.2.2.4

Wave-particle duality

Notes
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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 λ=h/p=h/(mv)\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 VV, energy eVeV gives momentum 2meV\sqrt{2meV} and hence λ=h/2meV\lambda=h/\sqrt{2meV}.
  • Examiners expect momentum in kg m s1\text{kg m s}^{-1}, voltage converted through eVeV 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 3.0×1024kg m s13.0\times10^{-24}\,\text{kg m s}^{-1}.

  1. 1.Use λ=h/p\lambda=h/p.
  2. 2.λ=(6.63×1034)/(3.0×1024)\lambda=(6.63\times10^{-34})/(3.0\times10^{-24}).
  3. 3.Express the result in metres and nanometres.

Answer: λ=2.2×1010m=0.22nm\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 λ1/p\lambda\propto1/p before discussing the amount of diffraction.

Tier 1 · Easy

ORIGINAL

Calculate the de Broglie wavelength of a particle with momentum 3.0×1024kg m s13.0\times10^{-24}\,\text{kg m s}^{-1}. Use h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Electrons travel at 2.5×106m s12.5\times10^6\,\text{m s}^{-1}. Calculate their de Broglie wavelength and state how doubling their speed affects the wavelength. Use electron mass 9.11×1031kg9.11\times10^{-31}\,\text{kg} and h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Electrons are accelerated from rest through 150V150\,\text{V} and then diffract from a crystal. Derive an expression for their de Broglie wavelength in terms of VV, calculate it, and explain how increasing VV changes the diffraction. Use e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}, electron mass 9.11×1031kg9.11\times10^{-31}\,\text{kg} and h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}.

[5 marks]

Total for this question: 5

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