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Question
- Write one nuclear fusion reaction.
- State the approximate value of energy released in the reaction mentioned in part (a).
- Give reason for the release of energy stated in part (b).
Short Answer
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Solution
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- \[\ce{\underset{\text{(deuterium)}}{^2_1H} + \underset{\text{(deuterium)}}{^2_1H} -> \underset{\text{(helium isotope)}}{^3_2He}} + \underset{\text{(neutron)}}{^1_0n} + 3.3 \text{MeV}\]
- \[\ce{\underset{\text{(helium isotope)}}{^3_2He} + \underset{\text{(deuterium)}}{^2_1H} -> \underset{\text{(helium) }}{^4_2He}} + \underset{\text{(proton)}}{^1_1H} + 18.3 \text{MeV}\]
- Approximately 21.6 MeV of energy is emitted in the reaction.
- When two deuterium nuclei \[\ce {^2_1H}\] fuse, nucleus of helium isotope \[\ce {^3_2He}\] is formed and 3.3 MeV energy is released. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus \[\ce {^4_2He}\] and 18.3 MeV of energy is released in this process.
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