| ⇦ |
| ⇨ |
The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV, respectively. In the nuclear reaction ₃⁷Li+₁¹H→ ₂⁴He+ ₂⁴He+Q, the value of energy Q released is
Options
(a) 19.6 MeV
(b) (-24 MeV)
(c) 8.4 MeV
(d) 17.3 MeV
Correct Answer:
17.3 MeV
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- Amout of heat required to raise the temperature of a body through 1 K is called its
- In which of the processes, does the internal energy of the system remain constant?
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- Ice is wrapped in black and white cloth. In which cases more ice will melt
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- Amout of heat required to raise the temperature of a body through 1 K is called its
- In which of the processes, does the internal energy of the system remain constant?
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- Ice is wrapped in black and white cloth. In which cases more ice will melt
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