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BCECE Medical BCECE Medical Solved Papers-2002

  • question_answer
    The energy of an electron in the first Bohr orbit is -13.6 eV, then find the energy of \[H{{e}^{+}}\] in the same Bohr orbit:

    A)  27.2 eV       

    B)  -27.2 eV

    C)  54.4 eV       

    D)  -54.4 eV

    Correct Answer: D

    Solution :

    Key Idea: Energy of electron in H like ions\[=-13.6\times {{Z}^{2}}eV\] for He Z = 2 \[\therefore \] E of electron in \[H{{e}^{+}}=-13.6\times {{(2)}^{2}}\] = - 54.4 eV


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