| ⇦ | 
| ⇨ |
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 circular disc X of radius R is made from an iron plate of thickness t and another disc
- In Young’s double slit interference experiment, using two coherent waves
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- The alternating current in a circuit is given by I=50 sin314t. The peak value
- When current I is flowing through a conductor, the drift velocity is v.
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A circular disc X of radius R is made from an iron plate of thickness t and another disc
- In Young’s double slit interference experiment, using two coherent waves
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- The alternating current in a circuit is given by I=50 sin314t. The peak value
- When current I is flowing through a conductor, the drift velocity is v.
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