| ⇦ | 
| ⇨ |
If the binding energy of the nucleon in ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction: p+₃⁷Li→2 ₂⁴He, energy of proton must be
Options
(a) 39.2 MeV
(b) 28.24 MeV
(c) 17.28 MeV
(d) 1.46 MeV
Correct Answer:
17.28 MeV
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - In Young’s double slit experiment, the slit seperation is 1 mm and the screen
- A wire mesh consisting of very small squares is viewed at a distance
- A proton of mass 1.6×10⁻²⁷ kg goes round in a circular orbit of radius 0.10 m under
- Which scientist experimentally proved the existence of electromagnetic waves?
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment, the slit seperation is 1 mm and the screen
- A wire mesh consisting of very small squares is viewed at a distance
- A proton of mass 1.6×10⁻²⁷ kg goes round in a circular orbit of radius 0.10 m under
- Which scientist experimentally proved the existence of electromagnetic waves?
- Three capacitors each of capacity 4 μF are to be connected in such a way that the effective
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