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JIPMER Jipmer Medical Solved Paper-2000

  • question_answer
    12.      If the binding energy per nucleon in \[L{{i}^{7}}\] and \[H{{e}^{4}}\] nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction             \[L{{i}^{7}}+p\xrightarrow{{}}2\,{{\,}_{2}}H{{e}^{4}}\] is:

    A)  17.3 MeV                           

    B)  8.4 MeV             

    C)         2.4 MeV             

    D)         19.6 MeV

    Correct Answer: A

    Solution :

    Binding energy of \[L{{i}^{7}}=39.20\,MeV\] Binding energy of \[H{{e}^{4}}=28.24\,MeV\] Therefore binding energy of \[2H{{e}^{4}}=56.48\,MeV\] Now energy of reaction is \[=56.48-39.20=17.28\text{ }MeV\] \[=17.3\text{ }MeV\]


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