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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    Consider the following reaction \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}{{H}^{4}}+Q.\] If \[m\,{{(}_{1}}{{H}^{2}})=2.014\,\,lamu;\] \[m\,{{(}_{2}}{{H}^{4}})=4.0024\,\,lamu.\] The energy Q released (in MeV) in this fusion reaction is

    A)  12  

    B)  6     

    C)  24

    D)  48

    Correct Answer: C

    Solution :

    [c] \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}H{{e}^{4}}+Q\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=m{{(}_{2}}H{{e}^{4}})-2m{{(}_{1}}{{H}^{2}})\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=4.0024-2(2.0141)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=-0.0258\,amu\] Since, \[Q={{c}^{2}}\Delta m\] \[\Rightarrow \,\,\,\,\,\,\,\,\,Q=(0.0258)(931.5)MeV\Rightarrow Q=24\,MeV.\]


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