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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Nuclear Physics

NTA Abhyas 2020 · Physics · Nuclear Physics

The binding energy per nucleon of deuterium and helium nuclei are...

The binding energy per nucleon of deuterium and helium nuclei are 1.1MeV1. 1MeV and 7.0MeV7. 0MeV respectively. When two deuterium nuclei fuse together to form a helium nucleus, the energy released in the fusion is

Options

  1. A2.2MeV2. 2MeV
  2. B23.6MeV23. 6MeV
  3. C28.0MeV28. 0MeV
  4. D30.2MeV30. 2MeV

Correct answer

B. 23.6MeV23. 6MeV

Step-by-step solution

The fusion reaction is as given below:

1H2+1H2→2He4+Q{}_{1}H^{2} + {}_{1}H^{2} \to {}_{2}He^{4} + Q

The binding energy of the reacting nuclei

=1.1×2+1.1×2=4.4MeV= 1.1 \times 2 + 1.1 \times 2 = 4.4MeV

The binding energy of the product nuclei

=7.0×4=28.0MeV= 7.0 \times 4 = 28.0MeV

Hence, the energy released in the fusion reaction,

Q=28.0−4.4=23.6MeVQ = 28.0 - 4.4 = 23.6MeV.
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