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Conservation laws questions
A proposed interaction is allowed only if the relevant quantum numbers are conserved. Charge, baryon number, the two lepton numbers and strangeness must all balance, and energy and momentum must be conserved as well. If any one of them fails, the interaction does not happen.
18 original questions · 52 marks · the conservation laws notes · Particles
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State four quantities that are always conserved in particle interactions.
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Any four of the following, one mark per two correct (2): charge; baryon number; lepton number; energy; momentum.Describe the change in a quark that occurs in beta-minus decay and state the other particles emitted.
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A down quark changes into an up quark, d → u (1). The decay emits an electron and an electron antineutrino (1).Describe the change in a quark that occurs in beta-plus decay and state the other particles emitted.
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An up quark changes into a down quark, u → d (1); a positron and an electron neutrino are emitted (1).State the electron lepton number of the positron, of the electron neutrino and of the proton.
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Positron: Le = −1 (1). Electron neutrino: Le = +1; proton: Le = 0 (1).State what it means for a quantity to be conserved in a particle interaction.
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Its total after the interaction is the same as its total before (1).Show that the interaction p + p → p + n + π+ conserves both charge and baryon number (π+ is a meson of charge +1).
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Charge: (+1) + (+1) = +2 → (+1) + 0 + (+1) = +2 ✓ (1)
Baryon number: 1 + 1 = 2 → 1 + 1 + 0 = 2 ✓ (1)
Both are conserved (1)Explain what conservation of strangeness tells you about whether an interaction proceeds by the strong or the weak interaction.
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If strangeness is conserved, the interaction can proceed by the strong interaction (1). If strangeness changes (by ±1), it cannot be strong and must proceed by the weak interaction (1), which is the only one that can change strangeness (1).Show that the decay n → p + e− + anti-νe conserves charge, baryon number and lepton number.
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Charge: 0 → (+1) + (−1) + 0 = 0 ✓ (1)
Baryon number: 1 → 1 + 0 + 0 = 1 ✓ (1)
Lepton number: 0 → 0 + (+1) + (−1) = 0 ✓, so all are conserved (1)A student suggests the decay p → n + e+ could occur on its own. Explain why this cannot be a complete interaction.
Explain why the decay μ− → e− + γ is never observed, even though it conserves charge.
A kaon decays by K+ → π+ + π0. The K+ has strangeness +1; pions have strangeness 0. Deduce which interaction is responsible for this decay.
A free proton never decays, even though the proposed decay p → n + e+ + νe conserves charge, baryon number and lepton number. Rest energies: proton 938.3 MeV, neutron 939.6 MeV, positron 0.511 MeV; the neutrino's rest energy is negligible. Explain, with a calculation, why this decay cannot occur for a free proton, and why protons bound in some nuclei can nevertheless decay this way.
A lambda particle (quark content uds) decays into a proton (uud) and a π−. State the quark flavour change in this decay and name the interaction responsible.
In the interaction p + p → p + n + X, the particle X is a meson. Using conservation of charge and baryon number, determine the charge and baryon number of X.
A strong interaction produces two strange particles: π− + p → K0 + X, where K0 has strangeness +1. State the strangeness of X and explain your reasoning.
Describe how conservation laws are used to decide whether a proposed particle interaction can occur.
Three interactions are proposed: (i) p + anti-p → π+ + π−; (ii) p → e+ + γ; (iii) νe + n → p + e−. Deduce which of the proposed interactions are allowed, justifying each answer with the relevant conservation laws.
In the strong interaction K− + p → n + X, the K− has strangeness −1 and baryon number 0. Determine the charge, baryon number and strangeness of X, and hence identify X from the following candidates, all of charge 0: K0 (B = 0, S = +1), anti-K0 (B = 0, S = −1), π0 (B = 0, S = 0), Λ0 (B = +1, S = −1).
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise conservation laws one question at a time
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