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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: - If the kinetic energy of the particle is increased to 16 times its previous value
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the kinetic energy of the particle is increased to 16 times its previous value
- A current of 2A flows through a 2Ω resistor when connected across a battery
- If two vectors are equal in magnitude and their resultant is also equal in magnitude
- Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same
- The string of a pendulum is horizontal.The mass of bob attached to it is m
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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