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NEET Sample Paper NEET Sample Test Paper-81

  • question_answer
    The masses of neutron and proton are 1.0087 and 1.0073 amu respectively. If neutrons and protons combine form helium nucleus of mass 4.0015 amu, the binding energy of the helium nucleus will be

    A)  28.4 MeV                    

    B)  20.8 MeV

    C)  27.3 MeV                    

    D)  14.2 MeV

    Correct Answer: A

    Solution :

    Mass defect \[\Delta m=2({{m}_{n}}+{{m}_{p}})-{{m}_{He}}\] (as helium nucleus has two-protons and two neutrons) \[\Delta m = 2 \left( 1.0087 +1.0073 \right) - 4.0015\] \[=\,\,\,4.032-4.0015=0.0305\] Binding energy \[{{\operatorname{E}}_{B}} = \,\Delta m\times  931 MeV\] \[= \,0.0305 \times  931 = 28.4 MeV\]


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