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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: - ₆C¹⁴ absorbs an energitic neutron and emits beta particles. The resulting nucleus is
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- ₆C¹⁴ absorbs an energitic neutron and emits beta particles. The resulting nucleus is
- The mass of a ⁷₃ Li nucleus is 0.042 u less than the sum of the masses of all its nucleons
- 180⁰ phase difference is obtained when light ray is reflected from
- A projectile is fired at an angle of 45° with the horizontal
- In a series resonant LCR circuit, the voltage across R is 100 volts and R =1kΩ
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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