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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • 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 :

    A)  40.8 eV                               

    B)  27.2 eV

    C)  54.4 eV                               

    D)  13.6 eV

    Correct Answer: C

    Solution :

    The excitation energy in the first excited state is                                 \[E=Rhc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\]                                 \[=(13.6\,eV)\times {{Z}^{2}}\times \frac{3}{4}\]                 \[\therefore \]  \[40.8=13.6\times {{Z}^{2}}\times \frac{3}{4}\]                 \[\Rightarrow \]               \[Z=2\] So, the ion in problem is \[H{{e}^{+}}\]. The energy of the ion in the ground state is                 \[E=-\frac{Rhc\,{{Z}^{2}}}{{{1}^{2}}}\]                 \[=-13.6\times 4\]                 \[=-54.4\,\,eV\] Hence, 54.4 eV is required to remove the electron from the ion,


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