| ⇦ |
| ⇨ |
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 potential energy of a system increases if work is done
- The electric current in AC circuit is given by the relation i-3 sinωt+4 cos ωt.
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- In old age arteries carrying blood in the human body become narrow resulting
- If energy(E), velocity(V) and time(T) are chosen as fundamental quantities,
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The potential energy of a system increases if work is done
- The electric current in AC circuit is given by the relation i-3 sinωt+4 cos ωt.
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- In old age arteries carrying blood in the human body become narrow resulting
- If energy(E), velocity(V) and time(T) are chosen as fundamental quantities,
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply