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Questions on antimatter and photons
Every particle has an antiparticle of equal mass and equal rest energy, with charge and the other quantum numbers reversed. Annihilation turns a pair at rest into two photons; electron-positron pair production needs one photon of at least 1.022 MeV and a nearby nucleus. Photon energy is E = hf = hc/λ, often quoted in MeV.
19 original questions · 57 marks · the antimatter and photons notes · Particles
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Compare a particle and its antiparticle in terms of mass and charge.
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An antiparticle has the same mass (and rest energy) as its corresponding particle (1) but the opposite charge, and opposite values of the other quantum numbers such as baryon number and lepton number (1). For a neutral pair, the neutron and antineutron for example, both members carry zero charge, and it is the reversed baryon number that distinguishes them.A photon has a frequency of 1.0 × 1015 Hz. Calculate its energy (h = 6.63 × 10−34 J s).
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E = hf = 6.63 × 10−34 × 1.0 × 1015 (1)
E = 6.63 × 10−19 J (1)Describe what happens in annihilation.
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A particle meets its antiparticle and the two are destroyed (1), their mass-energy being carried away as photons (1). When the pair meets essentially at rest, the standard case, the products are two equal photons travelling in opposite directions.State what is meant by the rest energy of a particle, and give the rest energy of the electron in MeV.
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Rest energy is the energy equivalent of the particle's mass, the energy locked up in its mass when it is at rest (1). Electron rest energy = 0.511 MeV (1).Name the antiparticle of each of the following: the electron, the proton, the electron neutrino.
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Electron: the positron (1). Proton: the antiproton; electron neutrino: the electron antineutrino (1).Explain why the annihilation of an electron–positron pair produces two photons rather than one.
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The pair's total momentum is (almost) zero and momentum must be conserved (1). A single photon would carry momentum away, whereas two photons emitted in opposite directions can have zero total momentum (1).Calculate the energy of a photon of wavelength 500 nm (c = 3.0 × 108 m s−1).
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E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(500 × 10−9) (1)
E = 3.98 × 10−19 J (1)Calculate the minimum energy, and hence the minimum photon frequency, needed for pair production of an electron and a positron (me = 9.11 × 10−31 kg, c = 3.0 × 108 m s−1).
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Rest energy of one particle = mec2 = 9.11 × 10−31 × (3.0 × 108)2 (1)
= 8.20 × 10−14 J (1)
For a pair, Emin = 2mec2 = 1.64 × 10−13 J (1)
fmin = E/h = 2.47 × 1020 Hz (1)An electron and a positron, effectively at rest, annihilate to produce two identical photons. Calculate the energy and frequency of each photon.
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Each photon carries the rest energy of one particle (1)
E = mec2 = 8.20 × 10−14 J (1)
f = E/h = 1.24 × 1020 Hz (1)A photon has an energy of 6.63 × 10−19 J. Calculate the photon energy in electronvolts (e = 1.60 × 10−19 C).
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E = (6.63 × 10−19)/(1.60 × 10−19) (1)
E = 4.14 eV (1)A dental X-ray set produces photons of energy 60 keV. Calculate the wavelength of these photons (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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E = 60 × 103 × 1.60 × 10−19 = 9.6 × 10−15 J (1). λ = hc/E (1) = (6.63 × 10−34 × 3.0 × 108)/(9.6 × 10−15) = 2.07 × 10−11 m (1).A laser pointer emits light of wavelength 650 nm with a power of 1.5 mW. Calculate the number of photons it emits each second (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1).
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(650 × 10−9) = 3.06 × 10−19 J (1). Number per second = power/photon energy = (1.5 × 10−3)/(3.06 × 10−19) (1) = 4.9 × 1015 photons per second (1).A gamma photon has a wavelength of 2.0 × 10−12 m. Show that the energy of the photon is about 0.6 MeV. Go on to deduce whether this photon could produce an electron–positron pair. The rest energy of the electron is 0.511 MeV (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(2.0 × 10−12) = 9.95 × 10−14 J (1). In MeV: 9.95 × 10−14/(1.60 × 10−13) = 0.62 MeV, which is about 0.6 MeV (1). Pair production needs at least 2 × 0.511 = 1.022 MeV (1). 0.62 MeV < 1.022 MeV, so this photon cannot produce the pair (1).A photon of energy 2.0 MeV passes close to a heavy nucleus and produces an electron–positron pair. The nucleus recoils, taking up momentum but a negligible share of the energy. Given that the rest energy of each particle is 0.511 MeV, calculate the total kinetic energy shared by the two particles, in joules, and state why the nucleus has to be there.
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Rest energy of the pair = 2 × 0.511 = 1.022 MeV (1). Kinetic energy shared = 2.0 − 1.022 = 0.978 MeV (1) = 0.978 × 106 × 1.60 × 10−19 = 1.56 × 10−13 J (1). A lone photon in empty space could not create the pair at all, because energy and momentum cannot both be conserved; the massive nucleus recoils to absorb the surplus momentum while carrying off almost no energy (1).Calculate the maximum wavelength of a photon that can create an electron–positron pair beside a nucleus (rest energy of the pair = 1.022 MeV).
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Emin = 1.022 × 106 × 1.60 × 10−19 (1)
Emin = 1.64 × 10−13 J (1)
λmax = hc/E = (6.63 × 10−34 × 3.0 × 108)/(1.64 × 10−13) (1)
λmax = 1.22 × 10−12 m (1)Explain why pair production normally occurs in the presence of a nearby nucleus.
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A single photon has momentum, and this momentum must be conserved when the pair is created (1). A nearby massive nucleus can recoil and take up momentum (1), allowing both energy and momentum to be conserved (1). Two photons colliding can create a pair with no other body present.A gamma photon of energy 200 MeV passes close to a nucleus. Deduce which, if any, of the following pairs it could create: an electron–positron pair, a muon–antimuon pair, a proton–antiproton pair. Rest energies: electron 0.511 MeV, muon 105.7 MeV, proton 938.3 MeV.
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The minimum photon energy for each pair is twice the rest energy of the particle (1). Thresholds: electron–positron 1.022 MeV, muon–antimuon 211.4 MeV, proton–antiproton 1876.6 MeV (1). 200 MeV exceeds only the 1.022 MeV threshold; it falls short of 211.4 MeV and 1876.6 MeV (1). So only an electron–positron pair can be created, the surplus energy becoming the pair's kinetic energy (1).A positron with 0.35 MeV of kinetic energy annihilates with an electron that is effectively at rest. Calculate the total energy of the photons produced, in joules. The rest energy of the electron is 0.511 MeV (e = 1.60 × 10−19 C).
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The photons carry both rest energies plus the kinetic energy: E = 2 × 0.511 + 0.35 (1) = 1.372 MeV (1). In joules: 1.372 × 106 × 1.60 × 10−19 = 2.2 × 10−13 J (1).A student states that mass is always conserved. Discuss this statement with reference to annihilation and to pair production. In your answer you should describe both processes and give the minimum energies involved for an electron–positron pair (rest energy of the electron = 0.511 MeV).
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In annihilation a particle meets its antiparticle and both are destroyed, so mass is not conserved (1). Their mass-energy is carried away as photons, and for a pair annihilating at rest these are two photons travelling in opposite directions, conserving momentum (1); for an electron–positron pair each photon then carries a minimum of 0.511 MeV (1). In pair production a high-energy photon passing close to a nucleus converts into a particle and its antiparticle, so mass is created (1); this needs a photon of at least the pair's total rest energy, 1.022 MeV for an electron–positron pair (1). So the student is wrong about mass, but total energy, with rest energy included, is conserved in both processes (1).
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