Question bank — Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF

False. It releases energy only while climbing toward the peak of the Binding Energy per Nucleon Curve near $A\approx 56$; fusing nuclei past iron absorbs energy.

False. The product is lighter; the "lost" rest mass $\Delta m$ is exactly what reappears as the released kinetic energy $\Delta m{c}^2$.

False. The strong force only acts once the nuclei nearly touch; before that the long-range Coulomb repulsion forms a barrier of hundreds of keV that must be overcome first.

False. The Sun burns ordinary protons through the p–p chain; D–T is our Earth choice because it fuses easiest at achievable temperatures, but the Sun has no free tritium.

False. The 14.1 MeV neutron is neutral, so magnetic fields cannot confine it; only the charged 3.5 MeV $\alpha$ is held to heat the plasma.

False. Temperature supplies energy to the Maxwell tail, but real fusion at solar/reactor temperatures also relies on Quantum Tunnelling to cross a barrier most particles cannot climb over.

False. The strong force per nucleon saturates (stays roughly constant); the fall is due to long-range Coulomb repulsion growing like $Z^2$ and eventually winning.

Roughly true but stated carelessly. They carry about 2% of the ~26.7 MeV per helium, small but real — the "26.7 MeV of sunlight" figure already excludes what the neutrinos remove.

True, all else equal, since the Lawson triple product $nT{\tau }_E$ rises with $n$ — but pushing $n$ too high shortens confinement in practice, so it is a trade-off, not a free win.

The subtraction is backwards. Mass defect is (sum of free nucleons) $-$ (nucleus); the nucleus is the lighter one, so $\Delta m>0$ and $B=\Delta m{c}^2>0$.

Momentum conservation forces equal and opposite momenta, and $E=p^2/2m$ makes kinetic energy inversely proportional to mass — so the light neutron takes ~14.1 MeV, the heavier $\alpha$ ~3.5 MeV.

The core is hot enough; the bottleneck is step 1 of the p–p chain, which needs a proton to convert to a neutron via the rare weak interaction mid-collision.

Both can release energy. Fission of heavy nuclei and fusion of light nuclei both move toward the $B/A$ peak — they approach the same summit from opposite sides.

Opposite strategies. Tokamaks use magnetic fields for low density and long ${\tau }_E$; ICF uses lasers to crush a pellet to huge density for a nanosecond ${\tau }_E$ — no magnetic bottle involved.

The barrier heights are similar (both are charge-1 nuclei), but D–T has a far larger cross-section at reachable temperatures, which is why it is the preferred lab fuel.

Iron is the least useful for energy release — sitting at the peak, it can be neither fused nor fissioned to release energy; it is the "ash," not the fuel.

The short-range strong force gives each added nucleon new bonds early on (steep rise), but once a nucleon only borders neighbours it can reach, adding more stops helping — the force saturates and the curve flattens.

Any material wall would instantly cool and be eroded by the plasma; charged particles must instead be steered along magnetic field lines into helical paths that never touch the walls.

The per-proton fusion rate is tiny because the trigger is a rare weak-force decay, so an enormous fuel reservoir converts a minuscule individual rate into a huge, steady total power over billions of years.

Net energy gain balances fusion output against losses, and the physics makes that condition depend on all three together — a shortfall in one (e.g. density) can be compensated by another (e.g. confinement time), which is why tokamaks and ICF succeed by different routes.

Being charged, it stays confined and deposits its energy back into the plasma, sustaining the temperature — this "$\alpha$-heating" is what lets a burn become self-sustaining (ignition).

Quantum particles have a nonzero probability of appearing on the far side of an energy barrier they could not climb classically, so nuclei in the energetic tail can fuse even without the full barrier energy.

No — reaching the barrier top only removes the classical block; whether the strong force actually captures them still depends on cross-section and a probability, so fusion is likely but not certain.

A lone proton has no binding energy (nothing to bind to), so energy release begins only once a bound product like deuterium forms — the p–p first step is uphill in difficulty precisely because it must also flip a proton to a neutron.

Neither. The peak is the balance point where both moving up in $A$ (fusion) and down in $A$ (fission) lead away from maximum binding, so both cost energy.

Reactant and product rest masses are equal, so no kinetic energy is released or absorbed — the reaction is energetically neutral and useless as a power source.

Purely from thermal energy, no — but because tunnelling probability is nonzero even for slow particles, an astronomically tiny rate persists; it is just far too small to matter without heat.