| ⇦ | 
| ⇨ |
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:
- Weber ampere per metre is equal to
- In which of the following systems will the radius of the first orbit (n=1) be minimum?
- Which is incorrect about thermal process
- A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three
- If the ionisation energy for the hydrogen atom is 13.6 eV, the energy required
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