| ⇦ | 
| ⇨ |
A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 a.m.u. The energy liberated per a.m.u. is (Given : 1 a.m.u. = 931 MeV)
Options
(a) 26.7 MeV
(b) 6.675 MeV
(c) 13.35 MeV
(d) 2.67 MeV
Correct Answer:
6.675 MeV
Explanation:
Mass defect m = 0.02866 a.m.u.
Energy = 0.02866 x 931 = 26.7 MeV
As ₁H² + ₁H² → ₂He⁴
Energy liberated per a.m.u. = 13.35/2 MeV = 6.675 MeV
Related Questions: - Source S₁ is producing, 10¹⁵ photons per second of wavelength 5000Å
- A solid which is transparent to visible light and whose conductivity increases
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
- 300j work is done in sliding a 2kg block up an inclined plane of height 10m
- A particle executes simple harmonic oscillations with an amplitude
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Source S₁ is producing, 10¹⁵ photons per second of wavelength 5000Å
- A solid which is transparent to visible light and whose conductivity increases
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
- 300j work is done in sliding a 2kg block up an inclined plane of height 10m
- A particle executes simple harmonic oscillations with an amplitude
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