Physics: Nuclear and Particle Physics
Nuclear and particle physics is the A-level Physics topic that extends the core nuclear and particle physics content with three further ideas: estimating the radius of the nucleus from electron diffraction data, using pair production and annihilation to link the particle-antiparticle model to photon energy and frequency, and calculating the energy released in a specific fission or fusion reaction from the mass difference between reactants and products. It draws on ideas of particles, radiation and nuclear structure met earlier in the course, and is examined through multi-step calculations combining E = mc^2, E = hf and the mass-energy data given in the question.
Before you start
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Method
- To estimate nuclear radius from an electron diffraction pattern, use the angle of the first minimum of intensity, theta, in the relationship sin(theta) = 1.22 x lambda / (2 x R), where lambda is the de Broglie wavelength of the electrons (found from lambda = h / (m x v) or lambda = h / p) and R is the nuclear radius; rearrange to make R the subject.
- Use R = r0 x A^(1/3) to estimate the radius of a nucleus of nucleon number A, where r0 is a constant given in the question if needed; to compare the radii of two different nuclides, the ratio of their radii is simply the cube root of the ratio of their nucleon numbers, since r0 cancels.
- For pair production, recognise that a high-energy photon can convert into a particle-antiparticle pair (for example, an electron and a positron) only if its energy is at least equal to the combined rest energy of the pair; find this minimum (threshold) photon energy using E = 2 x m x c^2, where m is the rest mass of one of the particles produced, then use E = h x f to find the corresponding minimum (threshold) frequency.
- For annihilation, recognise that a particle and its antiparticle can annihilate to produce two photons, never just one, since a single photon cannot conserve both energy and momentum; if the particle and antiparticle have negligible kinetic energy before annihilation, the two photons share the total rest energy equally, so each photon has energy E = m x c^2, where m is the rest mass of one of the particles.
- For a fission or fusion reaction energy calculation, use given atomic or nuclear mass data to find the mass difference between the total mass of the reactants and the total mass of the products, then convert this mass difference directly to energy released using E = (delta m) x c^2, converting any mass given in atomic mass units (u) to kilograms first, or working entirely in u and MeV using the conversion 1 u = 931.5 MeV.
- Distinguish antiparticles from their corresponding particles: an antiparticle has identical mass and identical magnitude of any other quantum property (such as baryon number or lepton number) to its particle, but has opposite charge (where the particle is charged) and opposite value of properties such as lepton or baryon number.
- In control of a fission chain reaction, as in a nuclear reactor, link a named component to its function: control rods (made of a neutron-absorbing material such as boron) are raised or lowered to absorb a variable proportion of the neutrons produced, keeping the chain reaction steady (critical) rather than increasing (supercritical) or dying out (subcritical).
Worked example
In a nuclear fusion reaction, a nucleus of deuterium (2/1 H, mass = 2.014102 u) fuses with a nucleus of tritium (3/1 H, mass = 3.016049 u) to form a helium-4 nucleus (4/2 He, mass = 4.002602 u) and a neutron (mass = 1.008665 u): 2/1 H + 3/1 H -> 4/2 He + 1/0 n. Calculate the energy released in this reaction, in MeV, given that 1 u is equivalent to 931.5 MeV.
- Calculate the total mass of the reactants: 2.014102 + 3.016049 = 5.030151 u.
- Calculate the total mass of the products: 4.002602 + 1.008665 = 5.011267 u.
- Calculate the mass difference, the mass converted to energy: delta m = 5.030151 - 5.011267 = 0.018884 u.
- Convert this mass difference to energy using 1 u = 931.5 MeV: E = 0.018884 x 931.5.
- Final answer: E = 17.6 MeV (3 s.f.). This energy is released as the kinetic energy of the helium nucleus and the neutron produced; this is the reaction that powers experimental fusion reactors.
Practice questions
Try each question, then tap to reveal the answer.
Q1State why a single isolated photon cannot undergo pair production in completely empty space, with no nucleus or other particle nearby.Show answer
Answer: A single photon alone cannot simultaneously conserve both energy and momentum when converting into a particle-antiparticle pair; a nearby nucleus (or other particle) is needed to absorb some momentum, allowing both conservation laws to be satisfied.
Q2State one property that is the same for a particle and its antiparticle, and one property that is different.Show answer
Answer: Same: rest mass (and the magnitude of properties such as baryon or lepton number). Different: charge (where the particle is charged, the antiparticle has the opposite charge), and the sign of properties such as lepton or baryon number.
Q3Calculate the minimum (threshold) energy, in MeV, needed for pair production of an electron-positron pair, given the rest mass energy of an electron is 0.511 MeV.Show answer
Answer: Minimum energy = 2 x 0.511 = 1.022 MeV, twice the rest mass energy of one electron, since both an electron and a positron, of equal mass, must be created.
Q4State why pair production cannot occur for a photon with energy less than the combined rest energy of the particle-antiparticle pair it would produce.Show answer
Answer: Below this threshold energy, there would not be enough energy available to account for the total rest mass energy of both particles created, by E = m x c^2, so pair production would violate conservation of energy.
Q5State the equation used to estimate the radius, R, of a nucleus with nucleon number A, in terms of a constant r0.Show answer
Answer: R = r0 x A^(1/3).
Q6Using R = r0 x A^(1/3), calculate the ratio of the radius of a uranium-238 nucleus to the radius of a carbon-12 nucleus.Show answer
Answer: Ratio = (238/12)^(1/3) = (19.8)^(1/3) = 2.71 (3 s.f.). The constant r0 cancels, since it is the same for both nuclei.
Q7State the name and function of control rods in a nuclear fission reactor.Show answer
Answer: Control rods are made of a material that absorbs neutrons, such as boron or cadmium; raising or lowering them into the reactor core adjusts how many neutrons are absorbed, controlling the rate of the fission chain reaction so it remains steady (critical) rather than increasing uncontrollably.
Q8State why a helium-4 nucleus and a neutron produced in deuterium-tritium fusion move apart with significant kinetic energy after the reaction.Show answer
Answer: By conservation of energy, the mass converted into energy during the fusion reaction (the mass defect) appears as kinetic energy of the products, shared between the helium-4 nucleus and the neutron.
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
In an electron diffraction experiment, electrons of wavelength 4.0 x 10^-15 m are directed at a thin sample of a particular nuclide. The first minimum of the resulting diffraction pattern is observed at an angle of 22 degrees from the straight-through direction. Using the relationship sin(theta) = 1.22 x lambda / (2 x R), calculate the radius, R, of the nucleus.
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In a nuclear fission reaction, a slow neutron is absorbed by a nucleus of uranium-235, which then splits into two smaller nuclei and releases further neutrons. The total mass of the reactants (the uranium-235 nucleus plus the incident neutron) is 236.0526 u. The total mass of all the products (the two smaller nuclei plus the released neutrons) is 235.8656 u. Calculate the energy released in this fission reaction, in MeV, given that 1 u is equivalent to 931.5 MeV. Give your answer to 3 significant figures.
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An electron and a positron, both practically at rest, meet and annihilate. Explain why two photons, rather than just one, must be produced, and state one other conservation law, besides energy, that is satisfied by producing two photons of equal energy moving in exactly opposite directions.
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