NEET Physics Nuclei Questions Solved

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The binding energy of deuteron H12 is 1.112 MeV per nucleon and an α-particle He24 has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction H12+H12He24+Q, the energy Q released is

(a) 1 MeV             (b) 11.9 MeV
(c) 23.8 MeV         (d) 931 MeV

(c) Mass of H21 = 2.01478 a.m.u
Mass of He42 = 4.00388 a.m.u
Mass of two deuterium = 2×0.1478=4.02956
Energy equivalent to 2H21

= 4.02956×1.112 MeV=4.48 MeV

Energy equivalent to He42
= 4.00388×7.047 MeV = 28.21 MeV
Energy released = 28.21-4.48 = 23.73 MeV = 24 MeV

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