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Atoms and Nuclei question

2005 · Q129
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Atoms and Nuclei question

2005 · Q129

NEETPhysicsAtoms and NucleiMCQ+4 / −1
In the reaction 12{}_1^212​H + 13{}_1^313​H →\to→ 24{}_2^424​He + 01{}_0^101​n, if the binding
energies of 12{}_1^212​ H, 13{}_1^313​H and 24{}_2^424​He are respectively a, b and c (in MeV),
then the energy (in MeV) released in this reaction is
  1. A
    a + b + c
  2. B
    a + b −-− c
  3. C
    c −-− a −-− b
  4. D
    c + a −-− b
View written solutionFree

Correct answer: C

Energy released in given reaction

= BE of products – BE of reactants = c – (a+ b) = c – a – b

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