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Mass
Rest energy

","mcqOptions":[{"text":"

Mass
Rest energy

","correct":true,"explanation":"A particle and its corresponding antiparticle have the same mass, so their rest energies are also identical because $E = mc^{2}$."},{"text":"

Charge
Rest energy

","correct":false,"explanation":"Charge is the tempting one to include, but an antiparticle has the opposite charge to its particle. The identical properties asked for are mass and rest energy."},{"text":"

Mass
Charge

","correct":false,"explanation":"Mass is identical, but charge is not: it is equal in magnitude and opposite in sign. The two identical properties are mass and rest energy."},{"text":"

Charge
Baryon number

","correct":false,"explanation":"Both charge and baryon number reverse sign for the corresponding antiparticle. The shared physical properties in this context are mass and rest energy."}]},{"id":"39ac44ed-d95f-4ee5-891c-64034697879a","cardType":"basic","points":1,"front":"How does the electrical charge of an antiparticle compare to its matter particle?","back":"It is exactly opposite (e.g., -1.60 x 10^-19 C for an electron vs +1.60 x 10^-19 C for a positron).","mcqOptions":[{"text":"It is exactly opposite (e.g., -1.60 x 10^-19 C for an electron vs +1.60 x 10^-19 C for a positron).","correct":true,"explanation":"An antiparticle has the same charge magnitude as its matter particle but the opposite sign, as shown by the electron and positron charges."},{"text":"It is exactly the same (e.g., -1.60 x 10^-19 C for an electron and -1.60 x 10^-19 C for a positron).","correct":false,"explanation":"This confuses identical mass with identical charge. For charged particles, the antiparticle’s charge has the same magnitude but the opposite sign."},{"text":"It is always zero (e.g., 0 C for an electron and 0 C for a positron).","correct":false,"explanation":"Neutral antiparticles can have zero charge, but this is not true in general. An electron and positron have equal-magnitude charges with opposite signs."},{"text":"It is twice as large (e.g., -1.60 x 10^-19 C for an electron vs +3.20 x 10^-19 C for a positron).","correct":false,"explanation":"The magnitude does not double when forming an antiparticle. Only the sign changes, so the electron and positron charges are equal in size and opposite in sign."}]},{"id":"a1cdcc0e-6130-45ea-81cc-5f70ed8724c8","cardType":"cloze","points":null,"front":"Even though it has zero charge, an antineutron is distinct from a neutron because it is made of [ ] and will annihilate if it meets a neutron.","back":"Even though it has zero charge, an antineutron is distinct from a neutron because it is made of antiquarks and will annihilate if it meets a neutron.","mcqOptions":[{"text":"antiquarks","correct":true,"explanation":"A neutron is made from quarks, while an antineutron is made from the corresponding antiquarks. That different internal structure allows neutron-antineutron annihilation."},{"text":"quarks","correct":false,"explanation":"Quarks make up an ordinary neutron, not an antineutron. An antineutron is made of antiquarks, even though its overall electric charge is still zero."},{"text":"leptons","correct":false,"explanation":"Leptons are not the constituents of neutrons or antineutrons. An antineutron is a baryon made from antiquarks, not from leptons."},{"text":"mesons","correct":false,"explanation":"Mesons are quark-antiquark particles, not the constituents inside an antineutron. An antineutron is made from three antiquarks."}]},{"id":"7dbfa3af-fc48-4d1d-b64f-8d04b7251f14","cardType":"cloze","points":null,"front":"Even though it has zero charge, an antineutron is distinct from a neutron because it is made of antiquarks and will [ ] if it meets a neutron.","back":"Even though it has zero charge, an antineutron is distinct from a neutron because it is made of antiquarks and will annihilate if it meets a neutron.","mcqOptions":[{"text":"annihilate","correct":true,"explanation":"A particle and its corresponding antiparticle can annihilate when they meet, converting their mass-energy into photons or other particles."},{"text":"decay","correct":false,"explanation":"Decay is a single unstable particle changing into other particles. A neutron meeting its antineutron is a particle-antiparticle interaction, so they annihilate."},{"text":"ionise","correct":false,"explanation":"Ionisation means removing or adding electrons to an atom, usually by energy transfer. A neutron-antineutron meeting is particle-antiparticle annihilation, not ionisation."},{"text":"scatter","correct":false,"explanation":"Scattering would mean the particles deflect from each other after interacting. A neutron and antineutron are corresponding matter-antimatter partners, so they annihilate."}]},{"id":"f9952a12-823c-4adf-93d9-f36e44165788","cardType":"basic","points":1,"front":"What is the specific name and symbol for the antiparticle of the electron?","back":"Positron ($\\text{e}^{+}$)","mcqOptions":[{"text":"Positron ($\\text{e}^{+}$)","correct":true,"explanation":"The electron’s antiparticle is the positron, which has the same mass as the electron but charge $+e$, so its symbol is $\\text{e}^{+}$."},{"text":"Proton ($\\text{p}^{+}$)","correct":false,"explanation":"A proton is positively charged, which makes it tempting, but it is a baryon and much more massive than an electron. The electron’s antiparticle is the positron, $\\text{e}^{+}$."},{"text":"Electron ($\\text{e}^{-}$)","correct":false,"explanation":"This is the original matter particle, not its antiparticle. The antiparticle must have opposite charge, so the correct answer is the positron, $\\text{e}^{+}$."},{"text":"Muon ($\\mu^{+}$)","correct":false,"explanation":"A positive muon has charge $+e$, but it is the antiparticle of the muon, not of the electron. The electron’s antiparticle is the positron, $\\text{e}^{+}$."}]},{"id":"1e823305-954c-403d-b9f0-2b0ade1d03af","cardType":"cloze","points":null,"front":"To convert energy from mega-electronvolts ($\\text{MeV}$) to Joules ($\\text{J}$), you must multiply by $\\text{[...]}$.","back":"To convert energy from mega-electronvolts ($\\text{MeV}$) to Joules ($\\text{J}$), you must multiply by $\\boxed{1.60 \\times 10^{-13}}$.","mcqOptions":[{"text":"$$1.60 \\times 10^{-13}$","correct":true,"explanation":"One mega-electronvolt is $10^{6}$ electronvolts, and $1\\ \\text{eV} = 1.60 \\times 10^{-19}\\ \\text{J}$, giving $1.60 \\times 10^{-13}\\ \\text{J}$."},{"text":"$$1.60 \\times 10^{-19}$","correct":false,"explanation":"This is the conversion factor for electronvolts to joules, not mega-electronvolts. The extra factor of $10^{6}$ makes the correct multiplier $1.60 \\times 10^{-13}$."},{"text":"$$1.60 \\times 10^{-16}$","correct":false,"explanation":"This uses a factor of $10^{3}$, which corresponds to kilo-electronvolts rather than mega-electronvolts. For MeV, multiply by $10^{6}$ to get $1.60 \\times 10^{-13}$."},{"text":"$$6.25 \\times 10^{12}$","correct":false,"explanation":"This is the reciprocal conversion from joules to mega-electronvolts. To convert MeV into joules, multiply by $1.60 \\times 10^{-13}$ instead."}]},{"id":"2e3867b7-d708-4d15-b59e-29a821965534","cardType":"cloze","points":null,"front":"The energy of a single photon is given by the equation $E = hf$ or $E = \\text{[...]}$.","back":"The energy of a single photon is given by the equation $E = hf$ or $E = \\boxed{\\frac{hc}{\\lambda}}$.","mcqOptions":[{"text":"$$\\frac{hc}{\\lambda}$","correct":true,"explanation":"Using the wave equation $c = f\\lambda$ gives $f = \\frac{c}{\\lambda}$. Substituting into $E = hf$ gives $E = \\frac{hc}{\\lambda}$."},{"text":"$$\\frac{h\\lambda}{c}$","correct":false,"explanation":"This comes from rearranging $c = f\\lambda$ the wrong way round. Since $f = \\frac{c}{\\lambda}$, the correct photon energy is $\\frac{hc}{\\lambda}$."},{"text":"$$\\frac{c\\lambda}{h}$","correct":false,"explanation":"This incorrectly divides by Planck’s constant instead of multiplying by it. Photon energy is $E = hf$, so substituting $f = \\frac{c}{\\lambda}$ gives $\\frac{hc}{\\lambda}$."},{"text":"$$hc\\lambda$","correct":false,"explanation":"This treats frequency as proportional to wavelength, but frequency is inversely proportional to wavelength. The correct substitution is $f = \\frac{c}{\\lambda}$, giving $\\frac{hc}{\\lambda}$."}]},{"id":"1b1c816b-20ba-44ef-ad46-f81ed13a0d37","cardType":"cloze","points":null,"front":"In the photon energy equation, the Planck constant ($h$) has the units $\\text{[...]}$.","back":"In the photon energy equation, the Planck constant ($h$) has the units $\\boxed{ext{J s}}$.","mcqOptions":[{"text":"$$ext{J s}$","correct":true,"explanation":"From $E = hf$, $h = E/f$. Dividing joules by hertz, where $\\text{Hz} = \\text{s}^{-1}$, gives the unit joule-second."},{"text":"$$ext{J s}^{-1}$","correct":false,"explanation":"Inverse seconds are the units of frequency, not Planck’s constant. Rearranging $E = hf$ gives $h = E/f$, so its units are $\\text{J s}$."},{"text":"$$ext{J}$","correct":false,"explanation":"Joules measure energy alone and omit the time factor introduced when dividing by frequency. Planck’s constant has units $\\text{J s}$."},{"text":"$$ext{W}$","correct":false,"explanation":"Watts are joules per second and describe power, which reverses the required time factor. Planck’s constant has units $\\text{J s}$."}]},{"id":"f8fbe399-d856-4f7e-ae33-cf445be19175","cardType":"basic","points":2,"front":"Describe the process of annihilation.","back":"A particle and its antiparticle meet and destroy each other, converting their total mass into two photons of electromagnetic radiation.","mcqOptions":[{"text":"A particle and its antiparticle meet and destroy each other, converting their total mass into two photons of electromagnetic radiation.","correct":true,"explanation":"The particle and antiparticle have opposite quantum properties, so they can annihilate. Their combined mass-energy is released as two photons, allowing both energy and momentum to be conserved."},{"text":"A particle and its antiparticle meet and combine together, converting their total mass into one heavier particle of ordinary matter.","correct":false,"explanation":"Annihilation does not combine the two particles into a heavier particle. They destroy each other, and their total mass-energy is converted into two photons of electromagnetic radiation."},{"text":"A particle and its antiparticle meet and exchange energy, converting their total mass into two lighter particles of ordinary matter.","correct":false,"explanation":"The products are not two lighter particles; annihilation converts the particle-antiparticle pair’s mass-energy into electromagnetic radiation, normally in the form of two photons."},{"text":"A particle and its antiparticle meet and rebound from each other, converting their kinetic energy into two faster particles of ordinary matter.","correct":false,"explanation":"Annihilation is not merely an elastic collision in which the particles survive. The particle and antiparticle disappear, converting their combined mass-energy into two photons."}]},{"id":"bdeea188-8dba-4b15-a522-87beb127f7b9","cardType":"basic","points":2,"front":"Why must annihilation produce two photons instead of just one?","back":"To ensure conservation of momentum (two photons moving in opposite directions maintain a total momentum of zero).","mcqOptions":[{"text":"To ensure conservation of momentum (two photons moving in opposite directions maintain a total momentum of zero).","correct":true,"explanation":"In the centre-of-momentum frame, the initial total momentum is zero. Two equal-energy photons travelling in opposite directions have equal and opposite momenta, preserving this zero total."},{"text":"To ensure conservation of charge (two photons with opposite charges maintain a total charge of zero).","correct":false,"explanation":"Charge conservation does not require two photons because photons are individually uncharged. Two oppositely directed photons are required so their momenta cancel and total momentum remains zero."},{"text":"To ensure conservation of energy (two photons moving in opposite directions maintain a total energy of zero).","correct":false,"explanation":"Energy conservation determines the photons’ total energy, not the need for two opposite directions. Two photons are needed so their equal and opposite momenta conserve momentum."},{"text":"To ensure conservation of lepton number (two photons moving in opposite directions maintain a total lepton number of zero).","correct":false,"explanation":"Lepton number is already zero for a particle-antiparticle pair and photons have zero lepton number. Two photons travelling oppositely are needed specifically to conserve momentum."}]},{"id":"75b93c9a-2d17-4872-b703-827452b1f771","cardType":"basic","points":1,"front":"In electron-positron annihilation, what is the relationship between the rest energy ($E_0$) of one particle and the energy of one produced photon ($hf$)?","back":"$$hf_{\\text{min}} = E_0$","mcqOptions":[{"text":"$$hf_{\\text{min}} = E_0$","correct":true,"explanation":"With negligible initial kinetic energy, the electron and positron contribute equal rest energies. Two equal photons are produced, so each photon has energy equal to one particle’s rest energy."},{"text":"$$hf_{\\text{min}} = 2E_0$","correct":false,"explanation":"Equating one photon’s energy to both particles’ combined rest energy assigns all available energy to one photon. Two photons share the energy, giving $hf_{\\text{min}} = E_0$ for each."},{"text":"$$hf_{\\text{min}} = \\frac{E_0}{2}$","correct":false,"explanation":"This incorrectly divides one particle’s rest energy between the two photons. The total energy $2E_0$ is shared equally, so each photon has energy $E_0$."},{"text":"$$hf_{\\text{min}} = E_0^2$","correct":false,"explanation":"Squaring the rest energy is dimensionally invalid because it would produce units of energy squared. Each photon’s minimum energy equals one particle’s rest energy, so $hf_{\\text{min}} = E_0$."}]},{"id":"20fba73f-9229-48a9-9362-a6a352f22433","cardType":"basic","points":1,"front":"Why is the term 'minimum frequency' used when calculating annihilation or pair production energies?","back":"It assumes the particles have zero kinetic energy.","mcqOptions":[{"text":"It assumes the particles have zero kinetic energy.","correct":true,"explanation":"The threshold calculation includes only the particles’ rest energies. Any initial or final kinetic energy would require additional photon energy and therefore a frequency above the minimum."},{"text":"It assumes the particles have zero rest energy.","correct":false,"explanation":"Zero rest mass applies to photons, not to the massive particles being created or annihilated. The minimum is calculated by assuming the particles have zero kinetic energy."},{"text":"It assumes the particles have zero total momentum.","correct":false,"explanation":"Momentum conservation still applies at the threshold and cannot be ignored. The term minimum instead means that no energy is allocated to the particles’ kinetic energy."},{"text":"It assumes the particles have zero electric charge.","correct":false,"explanation":"Charge does not determine whether the calculated frequency is a minimum. The threshold calculation assumes zero particle kinetic energy, leaving only the required rest energy."}]},{"id":"d1525fa5-e919-4752-ad92-6e6ed9cae6f2","cardType":"basic","points":2,"front":"How does a PET scanner use annihilation to create an image?","back":"It detects pairs of gamma photons emitted in opposite directions from a tracer and traces them back to the annihilation site.","mcqOptions":[{"text":"It detects pairs of gamma photons emitted in opposite directions from a tracer and traces them back to the annihilation site.","correct":true,"explanation":"Detectors register two gamma photons arriving nearly simultaneously on opposite sides of the patient. Their line of response identifies where the positron-electron annihilation occurred within the tracer distribution."},{"text":"It detects pairs of positrons emitted in opposite directions from a tracer and traces them back to the radioactive nucleus.","correct":false,"explanation":"PET does not directly detect positrons travelling in the body. It detects oppositely directed gamma-photon pairs and uses their coincident arrival to locate the annihilation site."},{"text":"It detects pairs of visible photons emitted in opposite directions from a tracer and traces them back to the excited atom.","correct":false,"explanation":"A PET scanner detects gamma photons rather than visible photons. Coincident gamma rays moving in opposite directions define a line passing through the annihilation site."},{"text":"It detects pairs of alpha particles emitted in opposite directions from a tracer and traces them back to the decay site.","correct":false,"explanation":"Annihilation produces two gamma photons rather than alpha particles. PET detectors use oppositely directed gamma-photon pairs to reconstruct the location of the annihilation site."}]},{"id":"f3535d50-f5ed-42de-a80a-e3a47a532a64","cardType":"basic","points":2,"front":"What is pair production?","back":"A single high-energy photon vanishes and its energy is converted into a particle-antiparticle pair.","mcqOptions":[{"text":"A single high-energy photon vanishes and its energy is converted into a particle-antiparticle pair.","correct":true,"explanation":"In pair production, a photon’s energy becomes mass and any kinetic energy of two conjugate particles. The products must form a particle-antiparticle pair so conserved quantum numbers remain balanced."},{"text":"A single unstable particle vanishes and its energy is converted into a pair of photons.","correct":false,"explanation":"This describes a particle decay rather than pair production. Pair production begins with one high-energy photon disappearing and creates a particle together with its corresponding antiparticle."},{"text":"A single high-energy photon vanishes and its energy is converted into a pair of lower-energy photons.","correct":false,"explanation":"Pair production does not split a photon into two lower-energy photons. It converts the original photon’s energy into the rest mass and kinetic energy of a particle-antiparticle pair."},{"text":"A single high-energy photon vanishes and its energy is converted into a pair of identical particles.","correct":false,"explanation":"Creating two identical particles would generally violate conserved quantities such as charge. Pair production creates a particle and its antiparticle, whose opposite quantum properties balance."}]},{"id":"738efed8-38a6-4f75-aaa5-b2fc37ef472c","cardType":"basic","points":1,"front":"What is required for pair production to occur besides a high-energy photon?","back":"The photon must pass near a heavy nucleus to help conserve momentum.","mcqOptions":[{"text":"The photon must pass near a heavy nucleus to help conserve momentum.","correct":true,"explanation":"A free photon cannot produce a pair while conserving both energy and momentum. A nearby heavy nucleus recoils, taking momentum with comparatively little kinetic energy."},{"text":"The photon must pass through a strong electric field to help conserve momentum.","correct":false,"explanation":"An electric field is not the required third body in the standard pair-production description. A nearby heavy nucleus recoils and absorbs momentum while taking little energy."},{"text":"The photon must collide with a second photon to help conserve momentum.","correct":false,"explanation":"Another photon can enable a different two-photon pair-production process, but it is not the required condition described here. Standard nuclear pair production occurs near a heavy nucleus."},{"text":"The photon must pass near a free electron to help conserve momentum.","correct":false,"explanation":"An electron is too light to serve as the standard recoil body without substantially affecting the threshold. A heavy nucleus absorbs the required momentum with little kinetic energy."}]},{"id":"5e988058-8a60-4353-9fe1-7611642274ef","cardType":"basic","points":1,"front":"What is the minimum energy required for a photon to undergo pair production?","back":"It must equal the total rest energy of both particles ($hf_{\\text{min}} = 2E_0$).","mcqOptions":[{"text":"It must equal the total rest energy of both particles ($hf_{\\text{min}} = 2E_0$).","correct":true,"explanation":"The photon must supply at least the rest energies of both newly created particles. With no energy left for kinetic energy, the threshold is $hf_{\\text{min}} = 2E_0$."},{"text":"It must equal the rest energy of one particle ($hf_{\\text{min}} = E_0$).","correct":false,"explanation":"Using one particle’s rest energy misses the equal rest energy needed to create its antiparticle. The threshold photon must provide both, giving $hf_{\\text{min}} = 2E_0$."},{"text":"It must equal the total kinetic energy of both particles ($hf_{\\text{min}} = 2E_k$).","correct":false,"explanation":"The threshold is not set by the particles’ total kinetic energy, which is taken as zero at minimum frequency. It equals their combined rest energy, $2E_0$."},{"text":"It must equal twice the total rest energy of both particles ($hf_{\\text{min}} = 4E_0$).","correct":false,"explanation":"Doubling the combined rest energy imposes an unnecessary extra factor of two. Creating two particles requires one rest energy $E_0$ for each, so the minimum is $2E_0$."}]},{"id":"52adc134-332f-4207-a89b-3242fb1726d7","cardType":"basic","points":2,"front":"What happens in pair production if the photon energy exceeds $2E_0$?","back":"The excess energy is converted into the kinetic energy of the two particles.","mcqOptions":[{"text":"The excess energy is converted into the kinetic energy of the two particles.","correct":true,"explanation":"After supplying the particles’ combined rest energy, any additional photon energy is conserved as their kinetic energy, causing the produced particle and antiparticle to move."},{"text":"The excess energy is converted into the rest energy of the two particles.","correct":false,"explanation":"The rest energy required is fixed at $2E_0$, so energy above this threshold cannot increase it. The excess becomes the kinetic energy of the two particles."},{"text":"The excess energy is converted into the charge energy of the two particles.","correct":false,"explanation":"Electric charge is a conserved property, not an energy store that increases with photon energy. Energy above $2E_0$ becomes the particles’ kinetic energy."},{"text":"The excess energy is converted into the potential energy of the two particles.","correct":false,"explanation":"This confuses the particles’ motion after production with energy stored by their positions. Above the rest-energy threshold, the excess appears as kinetic energy."}]},{"id":"65a5c9bf-226a-463f-be86-5f9d0074b2fb","cardType":"basic","points":1,"front":"Why can a photon not produce two electrons during pair production?","back":"It would violate conservation of charge (the net charge must remain zero).","mcqOptions":[{"text":"It would violate conservation of charge (the net charge must remain zero).","correct":true,"explanation":"A photon has zero charge, whereas two electrons have total charge $-2e$. Producing them alone would therefore change the net charge and violate charge conservation."},{"text":"It would violate conservation of energy (the total energy would become zero).","correct":false,"explanation":"A sufficiently energetic photon can provide the required energy, so energy conservation is not the obstacle. Two electrons are forbidden because their total charge would be $-2e$, not zero."},{"text":"It would violate conservation of momentum (the total momentum must remain zero).","correct":false,"explanation":"This mistakes the general need to conserve momentum for the reason two electrons are impossible. Their combined negative charge violates charge conservation, since the photon is uncharged."},{"text":"It would violate conservation of mass (the total mass must remain zero).","correct":false,"explanation":"Rest mass may be created from photon energy, so mass alone is not conserved relativistically. The problem is that two electrons have non-zero net charge."}]},{"id":"87be99a4-35f4-459e-b451-c6e319a91657","cardType":"cloze","points":null,"front":"In annihilation, $2$ particles produce $2$ photons, so $hf = \\text{[...]}.","back":"In annihilation, $2$ particles produce $2$ photons, so $hf = \\boxed{E_0}.","mcqOptions":[{"text":"$$E_0$","correct":true,"explanation":"The particle and antiparticle provide total rest energy $2E_0$, shared equally between two photons. Each photon therefore has energy $hf=E_0$ when kinetic energy is negligible."},{"text":"$$2E_0$","correct":false,"explanation":"This assigns the pair’s entire rest energy to each photon, doubling the total output. Two photons share $2E_0$, so each has energy $hf=E_0$."},{"text":"$$\\frac{E_0}{2}$","correct":false,"explanation":"This divides one particle’s rest energy between the photons instead of the pair’s combined rest energy. Sharing $2E_0$ equally gives $hf=E_0$."},{"text":"$$4E_0$","correct":false,"explanation":"This incorrectly doubles the total rest energy and then assigns it to each photon. The two photons share the available $2E_0$, giving $hf=E_0$ each."}]},{"id":"58310f9b-fc62-45bf-891c-d55346b3f42a","cardType":"cloze","points":null,"front":"In pair production, $1$ photon produces $2$ particles, so $hf = \\text{[...]}.","back":"In pair production, $1$ photon produces $2$ particles, so $hf = \\boxed{2E_0}.","mcqOptions":[{"text":"$$2E_0$","correct":true,"explanation":"The single photon must supply the rest energy of both the particle and antiparticle. At threshold, their combined rest energy is $2E_0$, so $hf=2E_0$."},{"text":"$$E_0$","correct":false,"explanation":"This accounts for the rest energy of only one produced particle. Pair production creates two particles, so the threshold photon energy is $hf=2E_0$."},{"text":"$$\\frac{E_0}{2}$","correct":false,"explanation":"This wrongly divides one particle’s rest energy despite a single photon supplying both particles. Their combined rest energy requires $hf=2E_0$ at threshold."},{"text":"$$4E_0$","correct":false,"explanation":"This doubles the required combined rest energy without justification. Producing two particles, each with rest energy $E_0$, requires a threshold photon energy of $2E_0$."}]},{"id":"f4f0f9b7-a51e-47aa-85e9-43640a2cda5e","cardType":"basic","points":2,"front":"Identify the symbols for an antiproton and an antineutrino.","back":"

Antiproton: $\\overline{\\text{p}}$
Antineutrino: $\\overline{\\nu}$

","mcqOptions":[{"text":"

Antiproton: $\\overline{\\text{p}}$
Antineutrino: $\\overline{\\nu}$

","correct":true,"explanation":"A bar over a particle symbol denotes its antiparticle. Therefore the antiproton is written $\\overline{\\text{p}}$ and the antineutrino is written $\\overline{\\nu}$."},{"text":"

Antiproton: $\\text{p}$
Antineutrino: $\\nu$

","correct":false,"explanation":"These are the symbols for the proton and neutrino themselves, not their antiparticles. Antiparticles are indicated by bars: $\\overline{\\text{p}}$ and $\\overline{\\nu}$."},{"text":"

Antiproton: $\\text{p}^{-}$
Antineutrino: $\\nu^{-}$

","correct":false,"explanation":"Antiparticles are not generally identified by attaching minus signs to particle symbols. The standard antiparticle notation uses bars, giving $\\overline{\\text{p}}$ and $\\overline{\\nu}$."},{"text":"

Antiproton: $\\overline{\\text{n}}$
Antineutrino: $\\overline{\\text{e}}$

","correct":false,"explanation":"The bar indicates antiparticles, but the underlying symbols name the wrong particles: neutron and electron. The required symbols are $\\overline{\\text{p}}$ and $\\overline{\\nu}$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Particles, antiparticles and photons","title":"Particles, antiparticles and photons Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"99b594b3-645d-5fc9-816e-b3fd089efc1a","walled":"$L39"}],"$L3a","$L3b"]