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BVP Medical BVP Medical Solved Paper-2011

  • question_answer
    \[-\frac{1}{2}mg{{R}_{e}}\]= 1.0087 amu, \[{{\alpha }_{1\,}}and\,{{\beta }_{2}}\] = 1.0073 amu\[{{Y}_{1\,}}and\,{{Y}_{2}}\] =4.0015   amu,   (1 amu =931 MeV)I Binding energy of the helium nucleus is

    A)  28.4 MeV                           

    B)  20.8 MeV

    C)  27-3 MeV                          

    D)  14.2 MeV

    Correct Answer: A

    Solution :

                    In helium atom aHe4 (N = 2, P = 2) Mass defect \[1\frac{1}{2}mg{{R}_{e}}\]= (mass of 2 proton + mass of 2 neutrons) =mass of \[-\frac{1}{2}mg{{R}_{e}}\]-particle \[{{\alpha }_{1\,}}and\,{{\beta }_{2}}\] \[{{Y}_{1\,}}and\,{{Y}_{2}}\] Binding energy \[\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{2}{3}\]                                 \[\frac{{{Y}_{1}}}{{{Y}_{2}}}\]


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