| ⇦ |
| ⇨ |
In the reaction ₁²H+₁³H+₂⁴He+₀¹n, if the binding energies of ₁²H,₁³H and ₂⁴He are respectively a,b and c (in MeV), then the energy (in MeV) released in this reaction is
Options
(a) a+b+c
(b) a+b-c
(c) c-a-b
(d) c+a-b
Correct Answer:
c-a-b
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- In which of the following systems will the radius of the first orbit (n=1) be minimum?
- A particle is executing a simple harmonic motion. Its maximum acceleration
- The drift velocity of the electrons in a copper wire of length 2 m under
- A nucleus ᵐₙ X emits one α- particle and two β- particles. The resulting nucleus is
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- In which of the following systems will the radius of the first orbit (n=1) be minimum?
- A particle is executing a simple harmonic motion. Its maximum acceleration
- The drift velocity of the electrons in a copper wire of length 2 m under
- A nucleus ᵐₙ X emits one α- particle and two β- particles. The resulting nucleus is
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