Questions on the conserved quantities

Question 1

State four quantities that are always conserved in particle interactions.

Accepted Answer

Any four of the following, one mark per two correct (2): charge; baryon number; lepton number; energy; momentum.

Question 2

Describe the change in a quark that occurs in beta-minus decay and state the other particles emitted.

Accepted Answer

A down quark changes into an up quark, d → u (1). The decay emits an electron and an electron antineutrino (1).

Question 3

Describe the change in a quark that occurs in beta-plus decay and state the other particles emitted.

Accepted Answer

An up quark changes into a down quark, u → d (1); a positron and an electron neutrino are emitted (1).

Question 4

State the electron lepton number of the positron, of the electron neutrino and of the proton.

Accepted Answer

Positron: L_e = −1 (1). Electron neutrino: L_e = +1; proton: L_e = 0 (1).

Question 5

State what it means for a quantity to be conserved in a particle interaction.

Accepted Answer

Its total after the interaction is the same as its total before (1).

Question 6

A proposed interaction is found to conserve charge, baryon number and both lepton numbers. State the two further quantities that must also balance before the interaction can occur, and state what follows if either of them does not.

Accepted Answer

Energy and momentum (1). If either one fails to balance the interaction cannot occur, whatever the quantum numbers do (1).

Question 7

A particle at rest decays into two particles. State what conservation of momentum requires of the momenta of the two products.

Accepted Answer

The total momentum before the decay is zero, so the total afterwards must be zero too (1). The two products therefore carry momenta of equal magnitude in opposite directions (1).

Question 8

A bubble-chamber photograph shows a clear gap with no track in it, and then two tracks springing from a point at the end of that gap. State why a neutral particle leaves no track, and state what the gap therefore shows.

Accepted Answer

A track is the trail of ions a charged particle leaves as it ionises the liquid it crosses, and a neutral particle ionises nothing, so it leaves no track (1). The gap shows that a neutral particle crossed it unseen, decaying at the point where the two tracks begin (1).

Question 9

Show that the interaction p + p → p + n + π⁺ conserves both charge and baryon number (π⁺ is a meson of charge +1).

Accepted Answer

Charge: (+1) + (+1) = +2 → (+1) + 0 + (+1) = +2 ✓ (1) Baryon number: 1 + 1 = 2 → 1 + 1 + 0 = 2 ✓ (1) Both are conserved (1)

Question 10

Explain what conservation of strangeness tells you about whether an interaction proceeds by the strong or the weak interaction.

Accepted Answer

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).

Question 11

Show that the decay n → p + e⁻ + anti-ν_e conserves charge, baryon number and lepton number.

Accepted Answer

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)

Question 12

A student suggests the decay p → n + e⁺ could occur on its own. Explain why this cannot be a complete interaction.

Accepted Answer

Lepton number is not conserved (1): the left side has 0, but the positron on the right has L_e = −1, giving −1 (1). An electron neutrino (L_e = +1) must also be emitted to conserve lepton number (1).

Question 13

Explain why the decay μ⁻ → e⁻ + γ is never observed, even though it conserves charge.

Accepted Answer

Lepton number must be conserved separately for each family (1). Muon lepton number falls from 1 to 0, so it is not conserved (1); electron lepton number rises from 0 to 1, also not conserved. Either failure forbids the decay (1).

Question 14

A kaon decays by K⁺ → π⁺ + π⁰. The K⁺ has strangeness +1; pions have strangeness 0. Deduce which interaction is responsible for this decay.

Accepted Answer

Strangeness changes from +1 to 0, so ΔS = −1 (1). The strong interaction conserves strangeness exactly, so it cannot be responsible (1). Only the weak interaction allows strangeness to change by ±1, so the decay proceeds by the weak interaction (1).

Question 15

A free proton never decays, even though the proposed decay p → n + e⁺ + ν_e 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.

Accepted Answer

Total rest energy of the products = 939.6 + 0.511 = 940.1 MeV (1). This exceeds the proton's rest energy of 938.3 MeV (1). A free proton has no source for the extra energy, so energy conservation forbids the decay (1). Inside a nucleus the binding energies of parent and daughter differ, and that difference can supply the shortfall, which is why bound protons can undergo beta-plus decay (1).

Question 16

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.

Accepted Answer

A strange quark changes into an up quark, s → u (1). Strangeness changes (from −1 to 0), so the decay must proceed by the weak interaction (1).

Question 17

A proposed interaction conserves charge, baryon number, every lepton number and strangeness. Explain why these checks are still not enough to show that it occurs, and give one process forbidden by energy and one forbidden by momentum.

Accepted Answer

The quantum-number checks are necessary but not sufficient: energy and momentum must be conserved as well, and either one can forbid a process on its own (1). Energy forbids the decay of a free proton, p → n + e⁺ + ν_e, because the rest energies of the products come to more than the proton's (1). Momentum forbids a lone photon in empty space becoming an electron and a positron, which is why pair production needs a nearby nucleus to take up the recoil (1).

Question 18

The interaction ν_μ + n → μ⁻ + p conserves charge, baryon number and both lepton numbers. Rest energies: neutron 939.6 MeV, proton 938.3 MeV, muon 105.7 MeV. Calculate the least energy the incident neutrino must bring for the interaction to be possible.

Accepted Answer

Rest energy of the products = 938.3 + 105.7 = 1044.0 MeV (1)
The neutron supplies 939.6 MeV of that, leaving a shortfall of 1044.0 − 939.6 = 104.4 MeV (1)
The neutrino must bring at least that much, and in practice more, because the products must also carry momentum away and so carry kinetic energy (1)

Question 19

A neutral particle decays in flight into two charged particles. One leaves with momentum 3.0 × 10⁻²⁰ kg m s⁻¹ and the other with momentum 4.0 × 10⁻²⁰ kg m s⁻¹, and their directions are at right angles to one another. Calculate the momentum of the neutral particle before it decayed.

Accepted Answer

Momentum is a vector, and the total after the decay equals the total before (1)
The two momenta are perpendicular, so p = √((3.0 × 10⁻²⁰)² + (4.0 × 10⁻²⁰)²) (1)
p = 5.0 × 10⁻²⁰ kg m s⁻¹, directed along the line the two momenta add to (1)

Question 20

In a bubble-chamber photograph two tracks spring from a single point, curling in opposite senses, and no track leads into that point. Deduce what can be said about the particle that decayed and about the charges of the two particles produced.

Accepted Answer

Nothing leads into the vertex, so the particle that decayed left no track and was therefore neutral: only a charged particle ionises the liquid, and only that ionisation makes a track (1). The two tracks curl opposite ways in one field, so the two products carry charges of opposite sign (1). The parent was neutral, so those two charges must cancel exactly rather than merely differ in sign (1).

Question 21

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.

Accepted Answer

Charge: 2 = 1 + 0 + q_X (1) so q_X = +1 (1) Baryon number: 2 = 1 + 1 + B_X, so B_X = 0 (1) X is therefore a positive meson (for example a π⁺) (1)

Question 22

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.

Accepted Answer

The initial particles (π⁻ and p) have strangeness 0 (1), so the total must stay 0 in a strong interaction (1). Since K⁰ has strangeness +1 (1), X must have strangeness −1 so that (+1) + (−1) = 0 (1).

Question 23

Describe how conservation laws are used to decide whether a proposed particle interaction can occur.

Accepted Answer

Check each conserved quantity in turn: charge, baryon number and each lepton number must total the same before and after (1); strangeness must be unchanged if the interaction is strong (1); energy and momentum must also be conserved. If any one check fails, the interaction cannot occur (1).

Question 24

Three interactions are proposed: (i) p + anti-p → π⁺ + π⁻; (ii) p → e⁺ + γ; (iii) ν_e + n → p + e⁻. Deduce which of the proposed interactions are allowed, justifying each answer with the relevant conservation laws.

Accepted Answer

(i) Charge: +1 − 1 = 0 → +1 − 1 = 0 ✓; baryon number: +1 − 1 = 0 → 0 + 0 = 0 ✓; all lepton numbers zero throughout, so (i) is allowed (1). (ii) Baryon number: 1 → 0, not conserved (and electron lepton number 0 → −1), so (ii) is forbidden (1). (iii) Charge: 0 + 0 = 0 → +1 − 1 = 0 ✓; baryon number: 0 + 1 = 1 → 1 + 0 = 1 ✓ (1); electron lepton number: +1 + 0 = +1 → 0 + 1 = +1 ✓, so (iii) is allowed (1).

Question 25

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).

Accepted Answer

Charge: −1 + 1 = 0 → 0 + q_X, so q_X = 0 (1). Baryon number: 0 + 1 = 1 → 1 + B_X, so B_X = 0 (1). The interaction is strong, so strangeness is conserved: −1 + 0 → 0 + S_X, giving S_X = −1 (1). The candidate with charge 0, B = 0 and S = −1 is the anti-K⁰ (1).

Question 26

A charged particle crosses a bubble chamber in a uniform magnetic field of flux density 0.80 T directed into the page. Its track curls clockwise, starting on a wide arc of radius 0.15 m and tightening steadily, turn after turn, to a radius of 0.050 m before it disappears. In this field, a positively charged particle moving into the chamber curls anticlockwise. The particle carries a charge of magnitude 1.60 × 10⁻¹⁹ C. State the sign of the particle's charge, calculate its momentum on the wide arc, and explain what the tightening of the track shows about the particle, including which end of the track, the wide end or the tight end, it reached first.

Accepted Answer

The track curls clockwise, the opposite way to the anticlockwise curl stated for a positive particle, so the particle is negatively charged (1)
p = BQr = 0.80 × 1.60 × 10⁻¹⁹ × 0.15 (1)
p = 1.9 × 10⁻²⁰ kg m s⁻¹ (1)
Since r = p/(BQ) with B and Q fixed, the shrinking radius means the particle's momentum is falling as it crosses the chamber, which happens because it is losing energy by ionising the liquid it passes through (1)
A particle only slows as it travels, so the wide end of the track was reached first and the tight end last (1)

Question 27

In a fixed-target experiment a proton is fired at a stationary proton and the interaction p + p → p + p + π⁰ occurs. The rest energy of the π⁰ is 135.0 MeV. Explain why the incident proton's kinetic energy must be greater than 135.0 MeV, even though that much energy would account for the pion's rest energy on its own.

Accepted Answer

Energy conservation requires the incident proton to supply the pion's rest energy of 135.0 MeV (1)
Momentum must be conserved as well, and the momentum before the collision is not zero, since one proton is moving and the other is at rest (1)
The products cannot therefore all be left at rest: they have to carry that momentum away, and so they carry kinetic energy too (1)
The incident kinetic energy must cover both the rest energy created and that kinetic energy, so it is greater than 135.0 MeV (1)

Question 28

A Λ⁰ baryon at rest decays into a proton and a π⁻. Rest energies: Λ⁰ 1115.7 MeV, proton 938.3 MeV, π⁻ 139.6 MeV. Calculate the total kinetic energy shared by the products, and explain why energy conservation permits this decay while it forbids the decay of a free proton into a neutron (939.6 MeV) and a positron (0.511 MeV).

Accepted Answer

Rest energy of the products = 938.3 + 139.6 = 1077.9 MeV (1)
Energy released = 1115.7 − 1077.9 = 37.8 MeV, which appears as the kinetic energy of the proton and the pion (1)
The parent's rest energy exceeds the products', so there is energy to spare and the decay can occur (1)
For the free proton it is the other way round: the neutron and the positron carry 940.1 MeV of rest energy against the proton's 938.3 MeV, a shortfall a free proton has nothing to supply (1)

Question 29

Two tracks spring from a point in a bubble chamber, curling in opposite senses, with no track leading into that point. The chamber sits in a uniform magnetic field of flux density 0.60 T. One track has radius 0.24 m and the other 0.080 m, and each particle carries a charge of magnitude 1.60 × 10⁻¹⁹ C. Determine the momentum of each particle, and state how the two momenta are related to the momentum of the particle that decayed.

Accepted Answer

p = BQr for the wider track = 0.60 × 1.60 × 10⁻¹⁹ × 0.24 (1)
p = 2.3 × 10⁻²⁰ kg m s⁻¹ (1)
For the tighter track, p = 7.7 × 10⁻²¹ kg m s⁻¹, smaller by a factor of 3 because the radius is smaller by that factor (1)
Momentum is a vector, so the two must add to the momentum of the particle that decayed, along the invisible line running back from the vertex (1)

Question 30

A photon of energy 2.00 MeV passes close to a lead nucleus of mass 3.4 × 10⁻²⁵ kg and converts into an electron and a positron. A photon of energy E carries momentum E/c. Calculate the momentum of the photon, and, taking the nucleus to recoil with the whole of that momentum, calculate the kinetic energy of the recoiling nucleus. Explain what your two answers show about why this conversion can happen beside a nucleus but not in empty space. 1 MeV = 1.60 × 10⁻¹³ J; c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

Photon energy = 2.00 × 1.60 × 10⁻¹³ = 3.20 × 10⁻¹³ J, so p = E/c = (3.20 × 10⁻¹³)/(3.00 × 10⁸) (1)
p = 1.07 × 10⁻²¹ kg m s⁻¹ (1)
The nucleus carries that momentum, and it is far too massive to be moving at any appreciable fraction of the speed of light, so E_k = p²/(2M) = (1.07 × 10⁻²¹)²/(2 × 3.4 × 10⁻²⁵) (1)
E_k = 1.7 × 10⁻¹⁸ J, which is 1.0 × 10⁻⁵ MeV (1)
That is about 5 × 10⁻⁶ of the photon's 2.00 MeV, so the nucleus takes the whole of the momentum while leaving practically all of the energy to the pair (1)
Momentum must be conserved as well as energy, and the photon's momentum cannot simply vanish. In empty space there is no third body to take it, and the pair on its own cannot balance both quantities at once, so the conversion is forbidden there and permitted beside a nucleus (1)