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A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 a.m.u. The energy liberated per a.m.u. is (Given : 1 a.m.u. = 931 MeV)
Options
(a) 26.7 MeV
(b) 6.675 MeV
(c) 13.35 MeV
(d) 2.67 MeV
Correct Answer:
6.675 MeV
Explanation:
Mass defect m = 0.02866 a.m.u.
Energy = 0.02866 x 931 = 26.7 MeV
As ₁H² + ₁H² → ₂He⁴
Energy liberated per a.m.u. = 13.35/2 MeV = 6.675 MeV
Related Questions: - Cathode rays are produced when the pressure is of the order of
- The number of photo electrons emitted for light of a frequency v
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Cathode rays are produced when the pressure is of the order of
- The number of photo electrons emitted for light of a frequency v
- The work done in which of the following processes is equal to the change in internal energy
- Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively
- According to Bohr model of hydrogen atom, only those orbits are permissible
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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