Final answer:
The nuclear reaction involving helium-4 releases 20.58 MeV and is indicative of helium-4's tighter nuclear binding compared to deuterium (²H), which releases 2.22 MeV in its associated reaction.
Step-by-step explanation:
In the context of the nuclear reactions mentioned, the reaction n+ ³He → He + y releases an energy of 20.58 MeV, and the reaction n+ ¹H → ²H+ y releases 2.22 MeV. Comparing the energies released helps to determine which nuclide is more tightly bound. The energy released during a nuclear reaction is indicative of the binding energy of the product nuclide. Since the reaction involving ⁴He releases more energy, it indicates that ⁴He is more tightly bound than ²H. This conclusion is supported by the fact that the nucleus of helium-4 (⁴He) contains two protons and two neutrons and is known to be exceptionally stable, forming a doubly magic nucleus.