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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: - The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- Two point charges +2 coulomb and +10 coulomb repel each other with a force of 12 N.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- Two point charges +2 coulomb and +10 coulomb repel each other with a force of 12 N.
- A particle moves a distance x in time but according to equation x=(t+5)⁻¹
- A wire of length 1 m is placed in a uniform magnetic field of 1.5 tesla at an angle
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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