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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Critical Thinking

  • question_answer
    Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is                                             [KCET 2004]

    A)            54.4 eV                                   

    B)            13.6 eV

    C)            40.8 eV                                   

    D)            27.2 eV

    Correct Answer: A

    Solution :

               Excitation energy \[\Delta E={{E}_{2}}-{{E}_{1}}\]\[=13.6\ {{Z}^{2}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\]                    \[\Rightarrow 40.8=13.6\times \frac{3}{4}\times {{Z}^{2}}\Rightarrow Z=2.\]                    Now required energy to remove the electron from ground state \[=\frac{+13.6{{Z}^{2}}}{{{(1)}^{2}}}=13.6{{(Z)}^{2}}=54.4\,eV.\]


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