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The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
Options
(a) 30.2 MeV
(b) 23.6 MeV
(c) 2.2 MeV
(d) 28.0 MeV
Correct Answer:
23.6 MeV
Explanation:
Binding energy of two ₁H² nuclei = 2 (1.1 x 2) = 4.4 meV
Binding energy of one ₂He⁴ nucleus = 4 x 7.0 = 28 MeV
Energy released = 28 – 4.4 = 23.6 MeV
Related Questions: - The ratio of specific charge of an α-particle to that of a proton is
- A body undergoes no change in volume. Poisson’s ratio is
- The mass number of nucleus is always
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- The effective length of magnet is 31.4 cm and its pole strength is 0.8Am.
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ratio of specific charge of an α-particle to that of a proton is
- A body undergoes no change in volume. Poisson’s ratio is
- The mass number of nucleus is always
- A radiactive substance emits 100 beta particles in the first 2s and 50 beta
- The effective length of magnet is 31.4 cm and its pole strength is 0.8Am.
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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