| ⇦ | 
| ⇨ |
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 concave lens of focal length f forms an image which is 1/3 times the size
- Dimensional formula for electrical resistance R is given by
- A circular disc of radius R and thickness R/6 has moment of inertia I about an axis
- In a given reaction,
ᴢXᴬ → ᴢ+1Yᴬ → ᴢ-1Kᴬ⁻⁴ → ᴢ-1Kᴬ⁻⁴
- A spherical planet has a mass Mp and diameter Dp. A particle of mass
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A concave lens of focal length f forms an image which is 1/3 times the size
- Dimensional formula for electrical resistance R is given by
- A circular disc of radius R and thickness R/6 has moment of inertia I about an axis
- In a given reaction, ᴢXᴬ → ᴢ+1Yᴬ → ᴢ-1Kᴬ⁻⁴ → ᴢ-1Kᴬ⁻⁴
- A spherical planet has a mass Mp and diameter Dp. A particle of mass
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