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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Nucleus, Nuclear Reaction

  • question_answer
    The binding energy of deuteron \[_{1}^{2}H\]is 1.112 MeV per nucleon and an \[\alpha -\]particle \[_{2}^{4}He\]has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction\[_{1}^{2}H+_{1}^{2}H\to _{2}^{4}He+Q\], the energy Q released is [MP PMT 1992; Roorkee 1994; IIT 1996; AIIMS 1997]

    A)            1 MeV                                     

    B)            11.9 MeV

    C)            23.8 MeV                               

    D)            931 MeV

    Correct Answer: C

    Solution :

                       Mass of \[_{1}{{H}^{2}}=2.01478\ a.m.u.\]                    Mass of \[_{2}H{{e}^{4}}=4.00388\ a.m.u.\]                    Mass of two deuterium \[=2\times 2.01478=4.02956\]                    Energy equivalent to \[{{2}_{1}}{{H}^{2}}\]                    \[=4.02956\times 1.112\ MeV=4.48\ MeV\]                    Energy equivalent to \[_{2}{{H}^{4}}\]                    \[=4.00388\times 7.047\ MeV=28.21\ MeV\]                    Energy released\[=28.21-4.48=23.73\ MeV\]= 24 MeV


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