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Atoms and Nuclei question

2014 · Q137
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Atoms and Nuclei question

2014 · Q137

NEETPhysicsAtoms and NucleiMCQ+4 / −1
The binding energy per nucleon of and nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction

37Li+11H→24He+24He+Q{}_3^7Li + {}_1^1H \to {}_2^4He + _2^4He + Q37​Li+11​H→24​He+24​He+Q

the value of energy Q released is
  1. A
    19.6 MeV
  2. B
    −-− 2.4 MeV
  3. C
    8.4 MeV
  4. D
    17.3 MeV
View written solutionFree

Correct answer: D

Given:

Binding energy per nucleon of 3Li7 and 2He4 nuclei are 5.60 MeV and 7.06 MeV

Energy released = 7.06 × 8 – 5.60 × 7

= 17.3 MeV

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