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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    Energy required to remove both the electrons of helium is 79 eV. The energy required to remove one electron from the two electrons of neutral helium atom is

    A)  54.4 eV           

    B)  24.6 eV

    C)  32.2 eV                        

    D)  42.2 eV

    Correct Answer: B

    Solution :

    After removal of one electron, remaining atom is hydrogen like. Its energy in first orbit is: \[E=-{{2}^{2}}(13.6)=-54.4\,eV\] So, binding energy = 79 - 54.4 = 24.6 eV


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