| ⇦ | 
| ⇨ |
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: - Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- The workdone in turning a magnet of magnetic moment M by an angle of 90⁰ from the magnetic
- The approximate depth of an ocean is 2700 m. The compressibility of water is
- Acceleration of a particle, executing SHM, at its mean position is
- Two nuclei are said to be mirror nuclei of
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Excitation energy of a hydrogen like ion, in its first excitation state, is 40.8 eV.
- The workdone in turning a magnet of magnetic moment M by an angle of 90⁰ from the magnetic
- The approximate depth of an ocean is 2700 m. The compressibility of water is
- Acceleration of a particle, executing SHM, at its mean position is
- Two nuclei are said to be mirror nuclei of
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