Two deuterium nuclei, each having total binding energy 2.2 MeV, fuse to form a helium-4 nucleus having total binding energy 28.4 MeV. Neglect all other products for the purpose of this energy comparison. If the energy released is N MeV, find N rounded to the nearest integer.
Physics · Nuclear physics · radioactivity and law of decay
JEE Mains NEET JEE Advanced

Question

Two deuterium nuclei, each having total binding energy 2.2 MeV, fuse to form a helium-4 nucleus having total binding energy 28.4 MeV. Neglect all other products for the purpose of this energy comparison. If the energy released is N MeV, find N rounded to the nearest integer.

Physics · Nuclear physics · radioactivity and law of decay

G1 Physics · Accepted Answer

Initial total binding energy: BE(initial) = 2×2.2 = 4.4 MeV. Final total binding energy: BE(final) = 28.4 MeV. Energy released: Q = BE(final)−BE(initial) = 28.4−4.4 = 24.0 MeV. Hence N = 24.