State the quark composition of a proton and of a neutron.
Proton: \( uud \) (two up quarks, one down quark).
Neutron: \( udd \) (one up quark, two down quarks).
Particles & Radiation
10 exam-style questions on the particle zoo, conservation laws, annihilation, pair production and photon interactions. Tap Show answer for full working.
3 Easy 4 Medium 3 Hard
A positron and an electron annihilate at rest, producing two photons of equal energy. Calculate the energy of each photon in MeV. (Rest energy of electron = \(0.511\,\mathrm{MeV}\).)
Total rest energy available: \(2\times 0.511 = 1.022\,\mathrm{MeV}\).
Two identical photons share this energy equally, so each has energy \(0.511\,\mathrm{MeV}\).
Name the four fundamental interactions and state which particles experience the strong nuclear force.
Gravity, electromagnetic, weak nuclear, strong nuclear.
Only hadrons (particles made of quarks) experience the strong nuclear force. Leptons do not.
A \(\pi^{+}\) meson has quark composition \( u\overline{d} \). Write down the quark composition of a \(\pi^{-}\) meson and of a \(\pi^{0}\) meson. Explain why a \(\pi^{0}\) can decay into two photons while a \(\pi^{+}\) cannot.
\(\pi^{-}\): \( \overline{u}d \).
\(\pi^{0}\): a mixture of \( u\overline{u} \) and \( d\overline{d} \) (equal superposition).
The \(\pi^{0}\) is its own antiparticle and has zero charge and zero baryon number, so it can annihilate into two photons. The \(\pi^{+}\) has non-zero charge and cannot decay solely into neutral photons while conserving charge.
A photon of energy \(2.5\,\mathrm{MeV}\) interacts with a stationary nucleus and produces an electron–positron pair. Calculate the total kinetic energy of the electron and positron immediately after production. (Rest energy of electron = \(0.511\,\mathrm{MeV}\).)
Minimum energy required for pair production: \(2\times 0.511 = 1.022\,\mathrm{MeV}\).
Kinetic energy shared by the pair: \(2.5 - 1.022 = 1.478\,\mathrm{MeV}\).
In beta-minus decay a neutron decays into a proton, an electron and an antineutrino. Write the equation in terms of quarks and explain how the weak interaction is involved.
Quark level: \( d \to u + e^{-} + \overline{ u}_{e} \) (one down quark changes to an up quark).
Overall: \( n \to p + e^{-} + \overline{ u}_{e} \).
The change of quark flavour (\(d\to u\)) is mediated by the weak interaction via a virtual \(W^{-}\) boson.
A proton and an antiproton annihilate at rest, producing two photons. Calculate the frequency of each photon. (Rest energy of proton = \(938\,\mathrm{MeV}\); \(h = 4.14\times10^{-15}\,\mathrm{eV\,s}\); \(1\,\mathrm{MeV}=10^{6}\,\mathrm{eV}\).)
Total energy released: \(2\times938 = 1876\,\mathrm{MeV} = 1.876\times10^{9}\,\mathrm{eV}\).
Each photon has energy \(E = 938\times10^{6}\,\mathrm{eV}\).
\(f = E/h = 938\times10^{6} / 4.14\times10^{-15} = 2.27\times10^{23}\,\mathrm{Hz}\).
A \(K^{+}\) meson (\( u\overline{s} \)) decays via the weak interaction into a \(\mu^{+}\) and a neutrino. Explain why this decay cannot proceed via the strong or electromagnetic interaction, and state which conservation laws allow the decay to occur.
The decay changes a strange quark into a non-strange lepton; strangeness is not conserved. The strong interaction conserves strangeness, and the electromagnetic interaction also conserves strangeness (and does not change quark flavour). Only the weak interaction can change quark flavour and violate strangeness conservation.
Allowed by: conservation of charge, lepton number (muon lepton number), energy and momentum. Baryon number is zero throughout.
A high-energy photon of wavelength \(1.2\times10^{-13}\,\mathrm{m}\) undergoes pair production in the field of a nucleus. Calculate the maximum kinetic energy that can be shared by the electron–positron pair. (\(h = 6.63\times10^{-34}\,\mathrm{J\,s}\), \(c = 3.00\times10^{8}\,\mathrm{m\,s}^{-1}\), electron rest mass \(= 9.11\times10^{-31}\,\mathrm{kg}\).)
Photon energy \(E = hc/\lambda = (6.63\times10^{-34}\times3.00\times10^{8})/1.2\times10^{-13} = 1.66\times10^{-12}\,\mathrm{J}\).
Rest energy of pair: \(2mc^{2} = 2\times9.11\times10^{-31}\times(3\times10^{8})^{2} = 1.64\times10^{-13}\,\mathrm{J}\).
Maximum KE total: \(1.66\times10^{-12} - 1.64\times10^{-13} = 1.50\times10^{-12}\,\mathrm{J}\) (or \(9.35\,\mathrm{MeV}\)).
In deep inelastic scattering, electrons are fired at protons and the results provide evidence for quarks. Explain what is meant by deep inelastic scattering and how the results support the existence of quarks inside the proton.
Deep inelastic scattering: high-energy electrons collide with protons; the electrons scatter with large energy and momentum transfer, and the proton is broken up (inelastic).
The scattering pattern shows that the electrons interact with point-like charged constituents inside the proton (Bjorken scaling). The form factors and cross-sections match the predictions for three valence quarks of fractional charge, supporting the quark model.