NEET Physics Nuclei Questions Solved


The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form helium atom, the energy released is

 (a) 19.2 MeV          (b) 23.6 MeV
(c) 26.9 MeV           (d) 13.9 MeV

(b) H12+H12He24+energy 
Binding energy of a deuterium (H12) nuclei = 2×1.1 = 2.2 MeV
Total binding energy of two deuterium nuclei = 2.2×2 = 4.4 MeV

Binding energy of a He24 nuclei = 4×7 = 28 MeV
So, energy released in fusion = 28-4.4 = 23.6 MeV

Difficulty Level:

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