| ⇦ | 
| ⇨ |
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: - A particle falls towards earth from infinity. Its velocity on reaching the earth
- If a small sphere is let fall vertically in a large quantity of still liquid of density
- When one of the slits of Young’s experiment is covered with a transport sheet
- Pure Si at 300 k has equal electrons (nₑ) and hole (nh) concentrations
- A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s,
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle falls towards earth from infinity. Its velocity on reaching the earth
- If a small sphere is let fall vertically in a large quantity of still liquid of density
- When one of the slits of Young’s experiment is covered with a transport sheet
- Pure Si at 300 k has equal electrons (nₑ) and hole (nh) concentrations
- A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s,
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