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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of \[L{{i}^{2+}}\] is

    A)  30.6 eV                               

    B)  13.6 eV

    C)  3.4 eV                 

    D)         122.4 eV

    Correct Answer: A

    Solution :

    \[E=-{{z}^{2}}\frac{13.6}{{{n}^{2}}}eV\] For first excited state,                 \[{{E}_{2}}=-{{3}^{2}}\times \frac{13.6}{4}\] Ionization energy for first excited state of \[L{{i}^{2+}}\] is\[30.6\,\,eV\].


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