ΔE = c²Δm applied to the creation and annihilation of matter and antimatter

The units MeV, GeV, MeV/c² and GeV/c², and converting them to joules and kilograms

Particles and antiparticles; annihilation and pair production

The standard model: quarks, baryons (three quarks), mesons (quark–antiquark), leptons and the photon

How the pattern of the standard model predicted the top quark

Conservation of charge, baryon number and lepton number, and writing and interpreting particle equations

Why the longer lifetime of fast-moving unstable particles matters (muons reaching the ground), in words only

1. Mass and energy are equivalent, ΔE = c²Δm, so particles can be created from energy and destroyed into it, but only in reactions that conserve charge, baryon number, lepton number, energy and momentum.

2. An electron's rest energy is about 0.51 MeV; a proton's is about 940 MeV. Particle masses are quoted in MeV/c² or GeV/c².

3. Every particle has an antiparticle with the same mass and opposite charge. A particle and its antiparticle can annihilate into photons; a photon with enough energy can create a particle–antiparticle pair.

4. Hadrons are made of quarks: baryons have three quarks (proton uud, neutron udd), mesons a quark and an antiquark. Leptons (electron, muon, tau and their neutrinos) are fundamental.

5. If any one of charge, baryon number or lepton number is different before and after, the reaction cannot happen.

6. Fast unstable particles survive much longer, as measured by us, than they do at rest: that is why so many muons made high in the atmosphere reach the ground.

Einstein showed that mass is a form of energy. Whenever the energy of a system changes, its mass changes too, and in particle physics the conversion happens in full: mass can disappear and reappear as photons, and photon energy can turn into new particles.

c² = 9.00 × 10¹⁶ m² s⁻² is enormous, so a tiny mass corresponds to a large energy. The total energy of a moving particle is its rest energy plus its kinetic energy. When particles are created, the energy supplied must cover their rest energies, and anything left over becomes kinetic energy.

Energy and momentum are both conserved in every interaction, so mass can be traded for energy but nothing is created from nothing.

WORKED EXAMPLE · QUESTION

(a) Calculate the rest energy of an electron in joules. (b) Express it in MeV (1 MeV = 1.60 × 10⁻¹³ J). (c) A gamma photon has an energy of 1.33 MeV. Calculate the mass equivalent of this energy, and compare it with the mass of an electron.

SOLUTION

(a) ΔE = c²Δm with Δm = the electron's mass: E = 9.11×10⁻³¹ ×(3.00×10⁸)² = 8.20 × 10⁻¹⁴ J

(b) 8.20×10⁻¹⁴ ÷1.60×10⁻¹³ = 0.512 MeV

(c) Convert MeV to joules, then divide by c²: Δm = 1.33 ×1.60×10⁻¹³ ÷(3.00×10⁸)² = 2.13×10⁻¹³ ÷9.00×10¹⁶ = 2.36 × 10⁻³⁰ kg; ÷ 9.11 × 10⁻³¹ = 2.6 electron masses

The accepted rest energy of the electron is 0.511 MeV; data-sheet values rounded to 3 s.f. give 0.512 MeV. Either is fine if your working is shown. Because 1.33 MeV is more than twice the electron's rest energy, this photon has enough energy to create an electron–positron pair.

Joules and kilograms are inconveniently large for single particles. Particle energies are given in electronvolts and their multiples, and masses are given as rest energy divided by c².

Energy in MeV → J: multiply by 1.60 × 10⁻¹³. Mass in MeV/c² → kg: multiply by 1.60 × 10⁻¹³, then divide by c². "MeV/c²" is read literally: an energy in MeV divided by c². Saying "the proton has a mass of 938 MeV/c²" is the same as saying "its rest energy is 938 MeV".

The range of masses is huge, so a logarithmic scale is the only way to show them on one diagram.

WORKED EXAMPLE · QUESTION

(a) Use the data-sheet mass of the proton to show that its mass is about 940 MeV/c². (b) A meson has a mass of 5.28 GeV/c². Calculate its mass in kg and compare it with the mass of the proton.

SOLUTION

(a) Rest energy in joules, then in MeV: 1.67×10⁻²⁷ ×9.00×10¹⁶ = 1.50×10⁻¹⁰ J; ÷ 1.60 × 10⁻¹³ = 939 MeV, so ≈ 940 MeV/c²

(b) GeV → J, then divide by c²: m = 5.28×10⁹ ×1.60×10⁻¹⁹ ÷9.00×10¹⁶ = 9.39 × 10⁻²⁷ kg; ÷ 1.67 × 10⁻²⁷ = 5.62 times the proton's mass

A meson is not "lighter than a baryon" by definition: the names describe the quark content, not the mass.

A neutral particle can still have a distinct antiparticle: the antineutron and the antineutrino are told apart by their baryon or lepton number. A photon is its own antiparticle.

Antimatter is made in collisions in accelerators, in cosmic-ray showers and in some radioactive decays (a positron is emitted in β⁺ decay).

Annihilation

When a particle meets its antiparticle they annihilate: their total mass is converted into energy, usually as two gamma photons. If the pair is almost at rest, the total momentum is zero. A single photon always carries momentum, so momentum could not be conserved; two photons moving in opposite directions with equal energies keep the total at zero. For energy, 2mc² (plus any kinetic energy) = total photon energy, so each photon from an electron–positron pair at rest has E = mₑ c² ≈0.51 MeV.

WORKED EXAMPLE · QUESTION

A positron and an electron, both almost at rest, annihilate. (a) Explain why two photons are produced rather than one. (b) Calculate the energy, the frequency and the wavelength of each photon.

SOLUTION

(a) Total momentum before ≈ 0. One photon would carry momentum, so two photons are emitted in opposite directions with equal momenta: total momentum stays zero

(b) Each photon carries the rest energy of one of the particles: E = mₑ c² = 8.20 × 10⁻¹⁴ J (0.512 MeV). From E = hf: f = 8.20×10⁻¹⁴ ÷6.63×10⁻³⁴ = 1.24 × 10²⁰ Hz. From c = fλ: λ= 3.00×10⁸ ÷1.237×10²⁰ = 2.43 × 10⁻¹² m

This wavelength is far shorter than that of X-rays used in medicine: these are gamma photons.

Pair production

A high-energy photon can turn into a particle–antiparticle pair, for example γ→e⁻ + e⁺. The photon must supply at least the rest energy of both particles: E_(min) = 2mc². For an electron–positron pair this is about 1.02 MeV, and any extra energy becomes the kinetic energy of the pair.

Pair production happens close to a nucleus. The nucleus takes up some momentum, which lets both energy and momentum be conserved; its share of the energy is tiny and is usually ignored. Charge is conserved (0 → −1 + 1), and lepton number too (0 → +1 − 1). That is why the products must be a particle and its antiparticle.

WORKED EXAMPLE · QUESTION

(a) Calculate the minimum energy, in GeV, a photon needs to create a proton–antiproton pair. (b) Calculate the minimum frequency of the photon. (c) A photon of energy 2.40 GeV creates a proton–antiproton pair. Calculate the total kinetic energy of the pair.

SOLUTION

(a) Energy for two proton rest masses: E_(min) = 2mₚ c² = 2 ×1.67×10⁻²⁷ ×9.00×10¹⁶ = 3.01×10⁻¹⁰ J; ÷ 1.60 × 10⁻¹⁰ = 1.88 GeV

(b) f_(min) = 3.01×10⁻¹⁰ ÷6.63×10⁻³⁴ = 4.53 × 10²³ Hz

(c) Energy left after paying for the rest energies: Eₖ = 2.40 - 1.88 = 0.52 GeV

The same photon could not make just one proton: charge and baryon number would not be conserved.

The standard model describes all known matter with a small set of fundamental particles: six quarks, six leptons and the force-carrying particles, of which you need the photon. Each has an antiparticle.

Hadrons are particles made of quarks (baryons and mesons), and quarks are never seen on their own. Quark charges are u = +2/3, d = −1/3 and s = −1/3 (in units of e), and each quark has baryon number +1/3. Antiquarks have the opposite charge and baryon number −1/3. The charge of a hadron is the sum of its quark charges, and is always a whole number. The neutrino is a lepton with no charge and a very small mass; it hardly interacts with matter, which is why it was first proposed to explain the "missing" energy in beta decay.

How the pattern predicted the top quark

Leptons come in three pairs: (e, νe), (μ, νμ), (τ, ντ). The quarks were expected to follow the same symmetry: three pairs (u, d), (c, s), (t, b). When the bottom quark was discovered in 1977, the model required a partner with charge +2/3 to complete the third pair. Physicists searched for it at ever higher energies, and the top quark was found in 1995 with a mass of about 173 GeV/c², which is why it took so long.

This is a good example of a scientific model making a successful prediction that was later confirmed by experiment.

Write the equation with each particle as a symbol, then add up three numbers on each side. If any one of them is different before and after, the interaction cannot happen.

Energy (including rest energy) and momentum must also be conserved. A particle can only decay into particles whose total rest mass is less than its own; collisions can make heavier particles if there is enough kinetic energy.

Beta-minus decay as a particle equation is n →p + e⁻ + ν̄. Check: Q 0 → 1 − 1 + 0 = 0; B 1 → 1; L 0 → 0 + 1 − 1 = 0. The antineutrino is what makes the lepton number balance.

At quark level the same decay is d →u + e⁻ + ν̄: one down quark (−1/3) becomes an up quark (+2/3), and the charge difference of +1 is balanced by the electron.

WORKED EXAMPLE · QUESTION

(a) A baryon has a charge of +1 and contains one s quark. Deduce its other two quarks. (b) Use conservation laws to decide which of these interactions are possible.

SOLUTION

(a) Charges must add to +1. s = −1/3, so the other two must total +4/3: u u (+2/3 + 2/3). The baryon is uus.

(b) Tabulate Q, B, L for each side:

One failure is enough to rule an interaction out. The pion's real decay is π⁺ →μ⁺ + ν, with a neutrino (L = +1), not an antineutrino. The second line shows why the proton is stable: there is no lighter baryon for it to decay into.

Cosmic rays (mostly protons) hit nuclei high in the atmosphere, about 10–15 km up, and produce pions, which quickly decay into muons. A muon at rest has a mean lifetime of about 2.2 µs. Even at almost the speed of light, it would travel only about 660 m in that time, so almost none should reach sea level.

In fact, large numbers are detected at the ground. For particles moving at speeds close to c, the lifetime measured by an observer they are moving past is much longer than the lifetime at rest; at 0.995c it is about ten times longer. This relativistic increase in lifetime is only significant at speeds that are a large fraction of c. It also matters in accelerators: fast unstable particles travel much further before decaying, which is how their tracks can be recorded.

WORKED EXAMPLE · QUESTION

Muons are produced about 12 km above the ground and travel downwards at 0.995c. Their mean lifetime at rest is 2.2 µs. Explain why muons detected at sea level are evidence that the lifetime of a fast-moving particle is increased. Include a calculation in your answer.

SOLUTION

Speed = 0.995 × 3.00 × 10⁸ = 2.985 × 10⁸ m s⁻¹; in one lifetime at rest a muon travels 2.985 × 10⁸ × 2.2 × 10⁻⁶ ≈ 660 m.

The journey of 12 km takes 12 × 10³ ÷ 2.985 × 10⁸ ≈ 40 µs, about 18 lifetimes. If the lifetime were 2.2 µs, only a minute fraction of the muons would still exist at sea level, far too few to detect in large numbers.

Many muons are in fact detected at sea level, so they must survive much longer than 2.2 µs as measured in the Earth's frame.

This is the relativistic increase in lifetime: it is significant because the muons travel at a speed very close to c. At everyday speeds the effect is far too small to notice.

7 Atomic structure and particle accelerators

Particles are given the energy for pair creation in linacs and cyclotrons, and the curved tracks of pair production are interpreted with r = p/BQ.

Photons (Unit 2) and momentum (4.1)

E = hf, c = fλ and the electronvolt are used for every annihilation and pair-production photon. Conservation of momentum explains why annihilation gives two photons and why pair production needs a nucleus nearby.

Nuclear decay and binding energy (Unit 5)

ΔE = c²Δm gives the energy released in fission and fusion, and β± decay equations use the same conservation checks as this page.

1. State the quark composition of a proton and of a neutron, and show that their charges are correct.

2. A muon has a mass of 106 MeV/c². What is its mass in kg?

3. What is the minimum photon energy needed to make a muon–antimuon pair?

4. Is n → p + e⁻ possible? Give a reason.

5. Why is the π⁺ classed as a meson and not a baryon?

6. An electron and a positron, each with negligible kinetic energy, annihilate. State the total energy released in MeV and explain why it is shared equally between two photons.

1. Proton uud: +2/3 + 2/3 − 1/3 = +1. Neutron udd: +2/3 − 1/3 − 1/3 = 0.

2. 106 × 1.60 × 10⁻¹³ = 1.70 × 10⁻¹¹ J; ÷ 9.00 × 10¹⁶ = 1.88 × 10⁻²⁸ kg.

3. Twice the rest energy: 2 × 106 MeV = 212 MeV (0.212 GeV).

4. No. Charge (0 → 0) and baryon number (1 → 1) balance, but lepton number does not: 0 before, +1 after. An antineutrino must also be emitted.

5. It is made of a quark and an antiquark (ud̄), so its baryon number is 1/3 − 1/3 = 0. Baryons have three quarks and baryon number +1.

6. Total energy = 2mₑc² ≈ 2 × 0.51 = 1.02 MeV. The total momentum before is zero, so the two photons must have equal and opposite momenta; since p = E/c for a photon, equal momenta mean equal energies, 0.51 MeV each.