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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    Two deuterons udnergo nuclear fusion to form a Helium nucleus. Energy released in this process is :      [JEE Online 08-04-2017] (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)

    A)  23.6 MeV                           

    B)  25.8 MeV

    C)  30.2 MeV                           

    D)  32.4 MeV

    Correct Answer: A

    Solution :

    \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{1}}\to 2{{H}_{c}}^{4}\] initiate \[\Rightarrow \] 1.1 × 4 = 4.4 final \[\Rightarrow \] 4 × 7 = 28 release \[\Rightarrow \] 28 - 4.4 = 23.6


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