| ⇦ | 
| ⇨ |
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 car wheel is rotated to uniform angular acceleration about its axis
- Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
- For the radioactive nuclei that undergo either α or β decay, which one
- In the Davisson and Germer experiment, the velocity of electrons emitted from the
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A car wheel is rotated to uniform angular acceleration about its axis
- Three unequal resistors in parallel are equivalent to a resistance 1 Ω.
- For the radioactive nuclei that undergo either α or β decay, which one
- In the Davisson and Germer experiment, the velocity of electrons emitted from the
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks
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