| ⇦ | 
| ⇨ |
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 calorimeter contains 0.2 kg of water at 30° C. 0.1kg of water at
- An open knife edge of mass 200g is dropped from height 5m on a cardboard
- A wire having resistance 12Ω is bent in the form of an equilateral triangle.
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A rod lies along the X-axis with one end at the origin and the other at x→∞. It carries
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A calorimeter contains 0.2 kg of water at 30° C. 0.1kg of water at
- An open knife edge of mass 200g is dropped from height 5m on a cardboard
- A wire having resistance 12Ω is bent in the form of an equilateral triangle.
- What will be the ratio of the distance moved by a freely falling body from rest in 4th
- A rod lies along the X-axis with one end at the origin and the other at x→∞. It carries
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