| ⇦ | 
| ⇨ |
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: - A step-down transformer has 50 turns on secondary and 1000 turns on primary winding.
- A cord is wound round the circumference of wheel of radius
- Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
- The limiting angle of incidence for an optical ray that can be transmitted
- The output of OR gate is 1(one)
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A step-down transformer has 50 turns on secondary and 1000 turns on primary winding.
- A cord is wound round the circumference of wheel of radius
- Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons
- The limiting angle of incidence for an optical ray that can be transmitted
- The output of OR gate is 1(one)
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