JEE Main & Advanced Physics Nuclear Physics And Radioactivity JEE PYQ-Nuclear Physics

  • 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 :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner