| ⇦ | 
| ⇨ |
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
Options
(a) 30.2 MeV
(b) 23.6 MeV
(c) 2.2 MeV
(d) 28.0 MeV
Correct Answer:
23.6 MeV
Explanation:
Binding energy of two ₁H² nuclei = 2 (1.1 x 2) = 4.4 meV
Binding energy of one ₂He⁴ nucleus = 4 x 7.0 = 28 MeV
Energy released = 28 – 4.4 = 23.6 MeV
Related Questions: - A parallel plate air capacitor is charged to a potential difference of V volts.
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- The surface temperature of the sun which has maximum energy
- A condenser of 250 μF is connected in parallel to a coil of inductance of 0.16 mH
- When the speed of electron beam used in Young’s double slit experiment is increased,
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A parallel plate air capacitor is charged to a potential difference of V volts.
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- The surface temperature of the sun which has maximum energy
- A condenser of 250 μF is connected in parallel to a coil of inductance of 0.16 mH
- When the speed of electron beam used in Young’s double slit experiment is increased,
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